
We propose and analyze a mathematical model for wave propagation in infinite trees with self-similar structure at infinity. This emphasis is put on the construction and approximation of transparent boundary conditions. The performance of the constructed boundary conditions is then illustrated by numerical experiments.
Citation: Patrick Joly, Maryna Kachanovska, Adrien Semin. Wave propagation in fractal trees. Mathematical and numerical issues[J]. Networks and Heterogeneous Media, 2019, 14(2): 205-264. doi: 10.3934/nhm.2019010
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We propose and analyze a mathematical model for wave propagation in infinite trees with self-similar structure at infinity. This emphasis is put on the construction and approximation of transparent boundary conditions. The performance of the constructed boundary conditions is then illustrated by numerical experiments.
In the recent years, there has been a surge of interest in the investigation of problems defined by partial differential equations along the edges of a network (or graph), with particular transmission conditions at the nodes (or vertexes) of the graph (please see [12,14] for the usual terminology in graph theory). To cite only a few representative examples, see [26] for the case of an elliptic operator in a ramified domain, [30,31] for the Helmholtz equation in a network seen as a limit of a two-dimensional thin domain, or [1] for the resolution of the Hamilton-Jacobi equation. Respective problems are typically referred to as problems posed on quantum graphs, see [11].
The works that inspired the present article concern the modeling of the respiratory system [23]. To a first approximation, the human lung can be seen as a network of many small tubes (the bronchioli) inside which the air flows. In some models [23], the tubes are assumed to be thin enough so that the air pressure is constant in each cross-section: as a consequence, each tube can be represented by a (1D) edge of a graph. In addition, to take into account that the number of bronchioli is very large, the bronchioli network is modeled as an infinite tree with some fractal and self-similarity properties (as defined in the reference monograph [22]). Finally, one models the air flow by solving the Laplace equation in such a network (which includes implicitly nodal transmission conditions). The infinite nature of the tree is indeed the main source of difficulty from both mathematical and numerical points of view. In particular, to complete the model, one needs to impose some "boundary condition at infinity" whose precise meaning requires to work with the weak formulation of the problem and an adequate functional framework, as explained in more detail later in the paper.
Our motivation was to study the propagation of sound in such a structure. This is important in applications since sound propagation in the human lung can be used for the detection of some pathologies of the respiratory system [27,32]. Therefore, we have to study the wave equation (in short
In particular, as far as numerical computations are concerned, the main source of difficulty is the infinite number of edges in the tree. Thus, we need to address the problem of truncating the computational domain to a finite subtree, which raises the question of the identification of the boundary condition [15] to be put at the artificial extremities of the truncated tree. This is the main motivation of this work. As we are going to see, we are able to give an answer to this question under the assumption that, after a certain generation, the subtrees are self-similar. It is worthwhile mentioning that the same type of questions was considered in a series of papers by Y. Achdou, N. Tchou and their collaborators: in these works, they do not study fractal trees but (particular) domains with a fractal boundary. In many papers they treat the Laplace operator [5,6,7], but also the time harmonic wave (Helmholtz) equation [3,2,4], for the solution of which they propose a particular iterative algorithm.
As a matter of fact, applying the Fourier-Laplace transform equation to the wave equation leads to study a family of Helmholtz equations parametrized by the frequency. This is the point of view that we shall adhere to for the construction of transparent boundary conditions. This approach emphasizes the close link between the properties of the solution of the wave equation and the spectral theory of the underlying elliptic operator
The paper is organized as follows. In section 1, we provide a geometrical and functional framework for studying wave propagation problems defined on infinite trees. In particular, we define weighted Sobolev spaces on such trees, which allows to formulate rigorously the Dirichlet and Neumann problem on a tree, and to obtain the corresponding well-posedness result for the time-domain wave equation. Next, we discuss the question of the construction of transparent boundary conditions for truncating the computational domain to a finite tree. Such construction is based on a use of the Dirichlet-to-Neumann (DtN) operator. Finally, we recall some classical results about the well-posedness of the Helmholtz equation for complex frequencies, as well as the representation of the solution to the Helmholtz equation in the case when the resolvent of the Laplace operator is compact.
In section 2, we define a notion of a self-similar
Section 4 is dedicated to the analysis of the properties of the solutions of a family of Helmholtz equations parametrized by frequency. In particular, in section 4.1 we introduce the notion of quasi-self-similarity for functions depending on frequency, and show that on self-similar trees the solutions of the Helmholtz equation are quasi-self-similar. In section 4.2 we consider a particular case of the Lapace equation, for which the solutions are self-similar and can be obtained in explicit form.
Section 5 is devoted to various characterizations of transparent boundary conditions (the DtN operator) for the Helmholtz equation on self-similar trees. First of all, based on the results of the previous sections, we show the meromorphicity of the symbol of the DtN operator in section 5.1. In section 5.2 we demonstrate that it satisfies a certain non-linear equation. We prove in particular the uniqueness of the solution to this equation under appropriate conditions. Next, section 5.3 is dedicated to certain positivity properties of the symbol of the DtN, related to the stability of the transparent boundary conditions in the time domain. In section 5.4, we provide an algorithm for the evaluation of the symbol of the DtN for complex frequencies. The numerical results obtained with the help of this algorithm are given in section 5.5.
In section 6 we propose an approximation of the DtN operators that stems from the truncation of the Taylor series for their symbol. We thus obtain first- and second-order transparent boundary conditions, for which we prove the stability. Their efficiency is validated with the help of the numerical experiments.
Finally, section 7 is dedicated to the open questions and possible extensions of this work.
In this work we will conciliate the view of the graph as an algebraic structure with its vision as a geometric object, see [11]. A graph
$ Σe={Mv+(be−ae)−1(Mv′−Mv)(se−ae),se∈[ae,be]}. $ |
By definition,
$ \mathbb{G} = \bigcup\limits_{e\in\mathcal{E}}\Sigma_e. $ |
For any
$ Ev={e∈E/Mv∈Σe}. $ | (1) |
We assume in the following that
Let us finally remark that the dimension
Definition 1.1. (Weight). A weight is a function
Remark 1.2. Choosing the weight function
Remark 1.3. We will denote a graph
Definition 1.4. [Wave equation] A solution of the weighted wave equation on
$ ∂2tue−∂2sue=0,on Σe×R+,∀e∈E, $ | (2) |
and at each node
$ {ue(Mv,t)=ue′(Mv,t),∀(e,e′)∈E2v(i)∑e∈Evεv,eμe∂sue(Mv,t)=0,(ii) $ | (3) |
where
$ εv,e={1, if v is the origin of e,−1, if v is the end of e. $ | (4) |
Note that (3)
Equations (2) and (3) can be collected in a single equation, using a (very intuitive) notion of distributional derivative along
$ μ∂2tu−∂s(μ∂su)=0,on G×R+, $ | (5) |
where
One of the particularities of the model (3), at least with respect to more standard cases, is the presence of the weight function
$ \mathbb{G}^\delta = \bigcup\limits_{e\in \mathcal{E}} \Sigma_e^\delta, \quad \Sigma_e^\delta = \bigcup\limits_{x \in \Sigma_e} \lbrace x + {\mu_e^{\frac{1}{d-1}}} \; B(0, \delta) \rbrace, $ |
where
$ ∂2tuδ−Δuδ=0 in Gδ,∂nuδ=0 on ∂Gδ. $ |
In [18], it is explained how the conditions (3) can be improved to get a more accurate model with respect to
In the following, we consider a particular case where the graph
First of all, let us provide an auxiliary definition.
Definition 1.5. [Child/parent of an edge and of a vertex] Given two oriented edges
Definition 1.6. [Rooted graph and a root] We shall say that a graph
Definition 1.7. [Tree] We will call a tree a rooted connected graph
The above definition is equivalent to saying that
● each edge (apart from the root) has only one parent;
● except from the origin of the root, each vertex has one parent.
Definition 1.8. [Generations of a tree] In a tree, we define a generation
● the generation
● the generation
Definition 1.9. [Infinite trees] If for all
$ T=⋃k∈N0Gk, $ | (6) |
and in such trees each vertex
With the above definition, we can introduce the following notation:
$ Gn(T)=J(n)⋃j=0Σn,j. In particular, with this notation, Σ0,0=Σ. $ | (7) |
Let us define the set of children indices of a given vertex
$ Cn,j:={k∈[0,J(n)]/ Σn+1,k is a child of Mn,j}. $ | (8) |
According to (8) and to the orientation of the tree, the Kirchhoff condition (3)
$ μn,j∂sun,j(Mn,j)=∑k∈Cn,jμn+1,k∂sun+1,k(Mn,j), $ | (9) |
where
For convenience, we denote the value of a continuous function
$ un,j=u(Mn,j). $ | (10) |
Definition 1.10. [Truncated tree] We denote by
Definition 1.11. [Subtree] For any
For an illustration of the notion of the subtree see figure 2 for
One of the goals of this work is to study the wave equation on such infinite trees. Of course, in this case the wave equation needs to be completed by a boundary condition at the root
$ u(M⋆,t)=f(t), $ | (11) |
by the boundary conditions on the leaves of the tree (which we will discuss in a moment), by initial conditions, for instance homogeneous initial conditions
$ u(⋅,0)=∂tu(⋅,0)=0, in T, $ | (12) |
but also, in general, by a "boundary condition at infinity", which is trickier to define and will be made precise in Section 1.5.
In the following we shall very soon restrict our discussion to compact trees.
Definition 1.12. [Compact tree] Let
In what follows, we will study only the following subclass of infinite trees.
Assumption 1.13. A tree
Let us first consider a very degenerate case of an infinite tree. Given
$ xn+1:=L(1−α)n∑ℓ=0αℓ=L(1−αn+1),n∈N, $ |
that form a strictly increasing sequence of real numbers
$ J(n)=0,Mn=xn+1,Σn,0=[xn,xn+1],∀n⩾0. $ |
In this case, each generation
$ T=+∞⋃n=0[xn,xn+1]=[0,L), $ | (13) |
It is then easy to check that the wave equation (2, 3) on such a tree is nothing but the 1D wave equation along
Definition 1.14. [Functional spaces] Let
1. Lebesgue space of square-integrable functions. We denote by
$ ‖ $ | (14) |
2. Sobolev space. We denote by
$ \begin{equation} \left\lvert {u} \right\rvert_{{ {\rm H}_\mu^1( \mathcal{T})}}^2 = \int_{ \mathcal{T}} \mu \, \lvert \partial_s u \rvert^2 : = \sum\limits_{n \geqslant 0} \sum\limits_{j = 0}^{J(n)} \mu_{n, j} \left\lVert {\partial_s u} \right\rVert_{{ {\rm L}^2(\Sigma_{n, j})}}^2 < + \infty, \end{equation} $ | (15) |
and the
$ \begin{equation} \left\lVert {u} \right\rVert_{{ {\rm H}_\mu^1( \mathcal{T})}}^2 : = \left\lVert {u} \right\rVert_{{ {\rm L}_\mu^2( \mathcal{T})}}^2 + \left\lvert {u} \right\rvert_{{ {\rm H}_\mu^1( \mathcal{T})}}^2. \end{equation} $ | (16) |
All the above spaces are then obviously equipped with a Hilbert space structure and will provide an adequate framework for studying the wave equation on
Notation. In what follows, for any
$ \begin{equation} \int_{ \mathcal{T}} \mu \, u \, v : = \sum\limits_{n \geqslant 0} \; \sum\limits_{j = 0}^{J(n)} \; \mu_{n, j} \; \int_{\Sigma_{n, j}} u_{n, j} \, {\overline{v_{n, j}}} \, ds, \quad {u_{n, j}: = u|_{\Sigma_{n, j}}, v_{n, j}: = v|_{\Sigma_{n, j}}}. \end{equation} $ | (17) |
We are now in position to provide a rigorous definition of the problems that we are interested in. Let us first explain our approach for the case when
$ \partial_s u(L, t) = 0 \quad (\mbox{Neumann condition}), \quad u(L, t) = 0 \quad (\mbox{Dirichlet condition}). $ |
These conditions are perfectly reflecting: in particular, they are energy preserving in the absence of the source term. We consider below the generalization of these boundary conditions for a general infinite tree
The Neumann initial boundary value problem
$ \begin{equation} V_ \mathfrak{n} = \lbrace v \in {\rm H}_\mu^1( \mathcal{T}) \; / \; v(M_\star) = 0 \; \rbrace, \ \end{equation} $ | (18) |
which is a closed subspace of
$ \begin{equation} \tag{{\mathcal{P}_ \mathfrak{n}}} \left\lbrace \ \begin{aligned} & \\ & \text{Find } u \in C^2\big(0, T; {\rm L}_\mu^2( \mathcal{T})\big) \cap C^1\big(0, T; {\rm H}_\mu^1( \mathcal{T})\big) \quad \text{s. t. } \; u(M_\star, t) = f(t), \\[6pt] & {u(., 0) = \partial_t u(., 0) = 0}\, \mbox{ and}\\[6pt] & \frac{d^2}{dt^2} \int_{ \mathcal{T}} \mu \, u(\cdot, t) \, v + \int_{ \mathcal{T}} \mu \, \partial_s \, u(\cdot, t) \partial_s v = 0, \ \forall \; v \in V_ \mathfrak{n}. \end{aligned} \right. ~~(\mathcal{P}_\mathfrak{n})\end{equation} $ |
In the case of the degenerate tree (13),
Definition 1.15. Let
1.
$ \begin{equation} {\rm H}_{\mu, \text{c}}^1( \mathcal{T}) = \left\lbrace v \in {\rm H}_\mu^1( \mathcal{T}) \quad \text{such that} \quad \exists \; N { \in \mathbb{N}}\ / \ v = 0 \text{ in } \mathcal{T} \setminus \mathcal{T}^N \right\rbrace. \end{equation} $ | (19) |
2.
$ \begin{equation} {\rm H}_{\mu, \text{0}}^1( \mathcal{T}) = \overline{ {\rm H}_{\mu, \text{c}}^1( \mathcal{T})}^{\, {\rm H}_\mu^1( \mathcal{T})}. \end{equation} $ | (20) |
Remark 1.16. As one can expect, in certain cases, the space
The Dirichlet initial boundary value problem
$ \begin{equation} V_ \mathfrak{d} = \lbrace v \in {\rm H}_{\mu, \text{0}}^1( \mathcal{T}) \; / v(M_\star) = 0 \; \rbrace, \ \end{equation} $ | (21) |
which is a closed subspace of
$ \begin{equation} \tag{{\mathcal{P}_ \mathfrak{d}}} \left\lbrace \ \begin{aligned} & \text{Find } u \in C^2\big(0, T; {\rm L}_\mu^2( \mathcal{T})\big) \cap C^1\big(0, T; {\rm H}_{\mu, \text{0}}^1( \mathcal{T})\big) \quad \text{s. t. } \; u(M_\star, t) = f(t), \\[6pt] &{u(., 0) = \partial_t u(., 0) = 0} \, \mbox{ and}\\[6pt] & \frac{d^2}{dt^2} \int_{ \mathcal{T}} \mu \, u(\cdot, t) \, v + \int_{ \mathcal{T}} \mu \, \partial_s u(\cdot, t) \, \partial_s v = 0, \ \forall v \in V_ \mathfrak{d}. \end{aligned} \right. ~~~(\mathcal{P}_\mathfrak{d})\end{equation} $ |
Let us state, without proof, a classical result about problems (
Proposition 1.17. Let
When one considers the problem of the numerical approximation of (
Numerically, a natural objective would be to restrict the computation to the solution
$ \begin{align} \Lambda_{n+1, {k}} \; \varphi(t) : = -\partial_s \widetilde{u}^\varphi_{n+1, k}(M_{n, j}, t), \end{align} $ | (22) |
where
The transparent condition at the end point
$ \begin{equation} \quad \mu_{n, j} \, \partial_s u_{n, j} (M_{n, j}, \cdot)+\mathcal{B}_{n, j}\, u_{n, j}(M_{n, j}, \cdot) = 0, \end{equation} $ | (23) |
where
$ \begin{align} \mathcal{B}_{n, j}\, u_{n, j}(M_{n, j}, \cdot) = \sum\limits_{k \in \mathcal{C}_{n, j}} \mu_{n+1, k} \; \Lambda_{n+1, k} \, u_{n, j}(M_{n, j}, \cdot). \end{align} $ | (24) |
Since the wave equation has constant coefficients in time, it is clear that
$ \begin{equation} \widehat{g}(\omega) = ( \mathcal{F} g)(\omega) = \int_{0}^{\infty} g(t) \exp(\imath \omega t) \, dt, \quad \ \omega \in \mathbb{C}, \quad {\mathcal{I}}m \, \omega > 0, \end{equation} $ | (25) |
we get a relation of the form
$ \begin{equation} ( \mathcal{F} \Lambda_{n+1, {k}}\varphi)(\omega) = \mathbf{\Lambda}_{n+1, {k}}(\omega) \; \mathcal{F} \varphi(\omega). \end{equation} $ | (26) |
The symbol
$ \begin{equation} \mathbf{\Lambda}_{n+1, {k}}(\omega) : = -\partial_s \widehat{u}_{n+1, {k}} (M_{n, j}, \omega), \end{equation} $ | (27) |
where
$\left\{ {\begin{array}{*{20}{l}} { - \mu {\mkern 1mu} {\omega ^2}\;{{\hat u}_{n + 1,k}}( \cdot ,\omega ) - {\partial _s}(\mu {\mkern 1mu} {\partial _s}{{\hat u}_{n + 1,k}})( \cdot ,\omega ) = 0,\quad s \in {{\cal T}_{n + 1,k}},{\mkern 1mu} k \in {{\cal C}_{n,j}},}\\ {{{\hat u}_{n + 1,k}}({M_{n,j}},\omega ) = 1.} \end{array}~~~~({{\cal P}_{\rm{H}}}}) \right.$ |
At this point, we have not advanced much, since the computation of the symbol
We finish this section by some results on the Helmholtz equation in general trees, in particular, the well-posedness and the meromorphicity of the solution with respect to the frequency for some particular classes of trees.
We consider the following problem on a weighted tree
$ \begin{equation} \left\lbrace \begin{array}{ll} - \mu \, \omega^2 u - \partial_s ( \mu \, \partial_s u) = 0 & \quad \text{in } \mathcal{T}, \\ u({M_\star}) = 1, & \end{array} \right. \end{equation} $ | (28) |
completed, like the wave equation in section 1.5, by a homogeneous (Dirichlet or Neumann) condition at infinity. More rigorously, using the functional framework of section 1.4 and the Hilbert spaces
$ \begin{equation} \tag{{\mathcal{P}_{ \mathfrak{n}, \omega}}} \left\lbrace \qquad \begin{aligned} & \text{Find } u \in {\rm H}_\mu^1( \mathcal{T}) \ / \ u({M_\star}) = 1, \text{such that}\\ & \int_{ \mathcal{T}} \mu \, \partial_s u \, \partial_s v - \omega^2 \int_{ \mathcal{T}} \mu \, u \, v = 0, \quad \forall v \in V_ \mathfrak{n}, \end{aligned} \right.~~~(\mathcal{P}_{\mathfrak{n}, \omega }) \end{equation} $ |
$ \begin{equation} \tag{{\mathcal{P}_{ \mathfrak{d}, \omega}}} \left\lbrace \qquad \begin{aligned} & \text{Find } u \in {\rm H}_{\mu, \text{0}}^1( \mathcal{T}) \ / \ u({M_\star}) = 1, \text{such that}\\ & \int_{ \mathcal{T}} \mu \, \partial_s u \, \partial_s v - \omega^2 \int_{ \mathcal{T}} \mu \, u \, v = 0, \quad \forall v \in V_ \mathfrak{d}. \end{aligned} \right.~~~(\mathcal{P}_{\mathfrak{d}, \omega }) \end{equation} $ |
For these two problems, we can immediately state the well-posedness result for non-real frequencies.
Proposition 1.18. For each
$ \begin{equation} u_{ \mathfrak{n}}(\cdot, \omega) \qquad (\mathit{\text{resp.}}u_{ \mathfrak{d}}(\cdot, \omega)). \end{equation} $ | (29) |
Proof. It is a simple application of the Lax-Milgram theorem left to the reader.
Since
$ \begin{equation} \mbox{for } \mathfrak{a} = \mathfrak{n}, \mathfrak{d}, \quad \forall \; \omega \not\in \mathbb{R}, \quad u_{ \mathfrak{a}}(\cdot, -\omega) = u_{ \mathfrak{a}}(\cdot, \omega), \quad u_{ \mathfrak{a}}(\cdot, \overline{\omega}) = \overline{u_{ \mathfrak{a}}(\cdot, \omega)}. \end{equation} $ | (30) |
A complementary point of view consists in introducing the two unbounded positive self-adjoint operators in
$ a(u, v) : = \int_{ \mathcal{T}} \mu \, \partial_s u \, \overline{\partial_s v}, $ |
we define these operators as follows:
$ \begin{equation} \left\{ \begin{array}{l} D( \mathcal{A}_ \mathfrak{d}) = \big\{ u \in V_ \mathfrak{d}\; / \; \exists \; C > 0 \mbox{ such that } |a(u, v)| \leqslant C \; \|v\|_{ {\rm L}_\mu^2( \mathcal{T})}, \forall \; v \in V_ \mathfrak{d} \big\}, \\ \forall \; u \in D( \mathcal{A}_ \mathfrak{d}), \quad ( \mathcal{A}_ \mathfrak{d} u, v)_{ {\rm L}_\mu^2( \mathcal{T})} = a(u, v), \quad \forall \; v \in V_ \mathfrak{d}; \end{array} \right. \end{equation} $ | (31) |
$ \begin{equation} \left\{ \begin{array}{l} D( \mathcal{A}_ \mathfrak{n}) = \big\{ u \in V_ \mathfrak{n} \; / \; \exists \; C > 0 \mbox{ such that } |a(u, v)| \leqslant C \; \|v\|_{ {\rm L}_\mu^2( \mathcal{T})}, \forall \; v \in V_ \mathfrak{n} \big\}, \\ \forall \; u \in D( \mathcal{A}_ \mathfrak{n}), \quad ( \mathcal{A}_ \mathfrak{n} u, v)_{ {\rm L}_\mu^2( \mathcal{T})} = a(u, v), \quad \forall \; v \in V_ \mathfrak{n}. \end{array} \right. \end{equation} $ | (32) |
It is easy to check that, defining
$ \begin{align} \widetilde{ {\rm H}}_\mu^2( \mathcal{T})& = \Big\{v\in {\rm H}_\mu^1( \mathcal{T}): \, v_{n, j}\in {\rm H}^2(\Sigma_{n, j}), \, 0 \leqslant j \leqslant p^n-1, \; n \geqslant 0, \text{ and satisfies (33)}\Big\}, \end{align} $ |
$ \begin{align} &\sum\limits_{n = 0}^{\infty}\sum\limits_{j = 0}^{p^n-1}\int\limits_{\Sigma_{n, j}}\mu|\partial_s^2 v_{n, j}|^2 < \infty. \end{align} $ | (33) |
the domains of the operators
$ \begin{equation} \left\{ \begin{array}{l} D( \mathcal{A}_ \mathfrak{d}) = \big\{ u \in V_ \mathfrak{d}\; / u\in \widetilde{ {\rm H}}_\mu^2( \mathcal{T}), \mbox{ and } {(9)} \mbox{ holds} \big\}, \\ D( \mathcal{A}_ \mathfrak{n}) = \big\{ u \in V_ \mathfrak{n} \; / u\in \widetilde{ {\rm H}}_\mu^2( \mathcal{T}), \mbox{ and } {(9)} \mbox{ holds} \big\}, \end{array} \right. \end{equation} $ | (34) |
and thus
$ \begin{equation} \forall \; u \in V_ \mathfrak{n}, \quad a(u, u) = \int_ \mathcal{T} \mu \, |\partial_s u|^2 \quad \mbox{and} \quad a(u, u) = 0 \Rightarrow u = 0 \quad (u({M_\star}) = 0). \end{equation} $ | (35) |
The solutions to the problems (
$ \begin{equation} f_{r}(\omega) : = {\mu^{-1} \; \big[ \, \partial_s \big(\mu \, \partial_s u_{r}\big) + \omega^2 \, \mu \; u_{r} \, \big]} \in {\rm L}_\mu^2( \mathcal{T}), \end{equation} $ | (36) |
the functions
$ \begin{equation} u_{ \mathfrak{n}}(\cdot, \omega) = u_{r} + \big( \mathcal{A}_ \mathfrak{n} - \omega^2)^{-1} \; f_{r}(\omega), \quad u_{ \mathfrak{d}}(\cdot, \omega) = u_{r} + \big( \mathcal{A}_ \mathfrak{d} - \omega^2)^{-1} \; f_{r}(\omega). \end{equation} $ | (37) |
From standard properties of the resolvent of self-adjoint operators [19,29], we deduce the
Proposition 1.19. The functions
Let us consider the case when one of the two following assumptions holds true:
$ \begin{equation} (Compactness) \quad \left\{ \begin{array}{ll} ( \mathfrak{d}) \quad \mbox{The injection } V_ \mathfrak{d} \subset {\rm L}_\mu^2( \mathcal{T}) \mbox{ is compact}.\\ ( \mathfrak{n}) \quad \mbox{The injection } V_ \mathfrak{n} \subset {\rm L}_\mu^2( \mathcal{T}) \mbox{ is compact}. \end{array} \right. \end{equation} $ | (38) |
Of course, (38)-
The properties (38)-
$ \begin{equation} \left\{ \begin{array}{ll} \mbox{(38)-$ \mathfrak{n}$ } \; \Rightarrow \; \sigma( \mathcal{A}_ \mathfrak{n}) = \lbrace (\omega_ \mathfrak{n}^n)^2, n \geqslant 1 \rbrace, \quad \omega_ \mathfrak{n}^{n+1} \geqslant \omega_ \mathfrak{n}^{n} > 0, \quad \lim\limits_{n\rightarrow + \infty} \omega_ \mathfrak{n}^{n} = + \infty, \\ \mbox{(38)-$ \mathfrak{d}$ } \; \Rightarrow \;\sigma( \mathcal{A}_ \mathfrak{d}) = \lbrace (\omega_ \mathfrak{d}^n)^2, n \geqslant 1 \rbrace, \quad \omega_ \mathfrak{d}^{n+1} \geqslant \omega_ \mathfrak{d}^{n} > 0, \quad \lim\limits_{n\rightarrow + \infty} \omega_ \mathfrak{d}^{n} = + \infty. \end{array} \right. \end{equation} $ | (39) |
Remark 1.20. Let us remark that
The corresponding eigenfunctions, which form a Hilbert basis in
$ \begin{equation} \left\{ \begin{array}{ll} \lbrace \varphi_ \mathfrak{n}^n, n \geqslant 1 \rbrace, \quad \varphi_ \mathfrak{n}^n \in D( \mathcal{A}_ \mathfrak{n}), \quad \mathcal{A}_ \mathfrak{n} \, \varphi_ \mathfrak{n}^n = (\omega_ \mathfrak{n}^n)^2 \; \varphi_ \mathfrak{n}^n, \\ \lbrace \varphi_ \mathfrak{d}^n, n \geqslant 1 \rbrace, \quad \varphi_ \mathfrak{d}^n \in D( \mathcal{A}_ \mathfrak{d}), \quad \mathcal{A}_ \mathfrak{d} \, \varphi_ \mathfrak{d}^n = (\omega_ \mathfrak{d}^n)^2 \; \varphi_ \mathfrak{d}^n . \end{array} \right. \end{equation} $ | (40) |
In particular, under the assumption (38)-
Lemma 1.21 (Poincaré inequality). If (38)-
$ \begin{align} \|u\|_{ {\rm L}_\mu^2( \mathcal{T})} \leqslant C\|\partial_s u\|_{ {\rm L}_\mu^2( \mathcal{T})}, \qquad \mathit{\text{for all}}~~u\in V_{ \mathfrak{n}} \mathit{\text{(resp.}}~u\in V_{ \mathfrak{d}} \mathit{\text{)}}. \end{align} $ | (41) |
The well-posedness result then reads.
Lemma 1.22. If (38)-
We will use the above lemma and (37) to express the solution to the (Dirichlet or Neumann) Helmholtz problems in the basis of the corresponding eigenfunctions.
Proposition 1.23. If (38)-
$ \begin{equation} u_{ \mathfrak{n}}(\cdot, \omega) = u_{ \mathfrak{n}}(., 0) + \sum\limits_{n = 0}^{+\infty} \; \frac{ \omega^2 c_{ \mathfrak{n}}^n}{(\omega_ \mathfrak{n}^n)^2- \omega^2}\; \varphi_ \mathfrak{n}^n, \qquad c_{ \mathfrak{n}}^n = \partial_s \varphi_{ \mathfrak{n}}^n({M_\star})\left(\omega_{ \mathfrak{n}}^n\right)^{-2}. \end{equation} $ | (42) |
Similarly, if (38)-
$ \begin{equation} u_{ \mathfrak{d}}(\cdot, \omega) = u_{ \mathfrak{d}}(., 0) + \sum\limits_{n = 0}^{+\infty} \; \frac{ \omega^2 c_{ \mathfrak{d}}^n}{(\omega_ \mathfrak{d}^n)^2- \omega^2}\; \varphi_ \mathfrak{d}^n, \qquad c_{ \mathfrak{d}}^n = \partial_s \varphi_{ \mathfrak{d}}^n({M_\star})\left(\omega_{ \mathfrak{d}}^n\right)^{-2}. \end{equation} $ | (43) |
Proof. We will show the proof for
$ \begin{align*} u = u_{ \mathfrak{n}}(\cdot, \omega)-u_{ \mathfrak{n}}(\cdot, 0), \qquad u\in V_{ \mathfrak{n}}. \end{align*} $ |
Defining
$ \begin{align*} f_{ \mathfrak{n}}: = \mu^{-1}\left(\partial_s(\mu\partial_s u_{ \mathfrak{n}}(., 0))+\omega^2\mu u_{ \mathfrak{n}}(., 0)\right) = \omega^2 u_{ \mathfrak{n}}(., 0)\in {\rm L}_\mu^2( \mathcal{T}), \end{align*} $ |
we deduce that
$ \begin{align} u = \left(\mathcal{A}_{ \mathfrak{n}}-\omega^2\right)^{-1}f_{ \mathfrak{n}}, \text{ hence } u_{ \mathfrak{n}}(\cdot, \omega) = u_{ \mathfrak{n}}(\cdot, 0)+\omega^2 \left(\mathcal{A}_{ \mathfrak{n}}-\omega^2\right)^{-1}u_{ \mathfrak{n}}(., 0). \end{align} $ | (44) |
Next, we expand
$ \begin{align*} (u_{ \mathfrak{n}}(., 0), \varphi_{ \mathfrak{n}}^n)_{ {\rm L}_\mu^2( \mathcal{T})}& = -\left(\omega_{ \mathfrak{n}}^n\right)^{-2} (u_{ \mathfrak{n}}(., 0), \mu^{-1}\partial_s(\mu \partial_s \varphi_{ \mathfrak{n}}^n))_{ {\rm L}_\mu^2( \mathcal{T})}, \end{align*} $ |
where we use that
$ \begin{align*} (u_{ \mathfrak{n}}(., 0), \varphi_{ \mathfrak{n}}^n)_{ {\rm L}_\mu^2( \mathcal{T})}& = \left(\omega_{ \mathfrak{n}}^n\right)^{-2}\partial_s \varphi_{ \mathfrak{n}}^n({M_\star})u_{ \mathfrak{n}}({M_\star}, 0) = \left(\omega_{ \mathfrak{n}}^n\right)^{-2}\partial_s \varphi_{ \mathfrak{n}}^n({M_\star}). \end{align*} $ |
Thus, the desired result follows by inserting the expansion of
Remark 1.24. The series (42) (resp. (43)) converges uniformly in
In this section, we introduce a notion of a self-similar weighted
Definition 2.1. [
Remark 2.2. Any
Later on, we will need to provide a numbering for edges and vertices of a
$ \begin{equation} {(i_1 i_2 \cdots i_n)_p : = } \sum\limits_{j = 1}^n \; i_j \, p^{n-j}. \end{equation} $ | (45) |
The map
Definition 2.3. [Self-similar
$ \begin{align*} \sigma_i = \tau_i \circ h_i \circ {\Theta_i}, \qquad 0 \leqslant i \leqslant p-1, \end{align*} $ |
where
●
●
●
Let additionally these similitudes satisfy the following assumption: for all
$\begin{array}{*{20}{l}} {{i_\ell },{j_1}, \ldots ,{j_k} \in \{ 0, \ldots ,p - 1\} ,}\\ {{\sigma _{{i_1}}}{\sigma _{{i_2}}} \ldots {\sigma _{{i_\ell }}}({M_{0,0}}) = {\sigma _{{j_1}}}{\sigma _{{j_2}}} \ldots {\sigma _{{j_k}}}({M_{0,0}}){\rm{ if }}\;\;{\rm{and }}\;\;{\rm{only}}\;\;{\rm{ if }}}\\ {k = \ell {\rm{ }}\;\;{\rm{and}}\;\;{\rm{ }}({i_1}, \ldots ,{i_\ell }) = ({j_1}, \ldots ,{j_k}).} \end{array}$ | (46) |
Then a tree
$ \begin{align} \mathcal{G}^0 = \{\Sigma_{0, 0}\}, \qquad \mathcal{G}^k = \bigcup\limits_{i = 0}^{p-1}\sigma_i(\mathcal{G}^{k-1}), \qquad k \geqslant 1, \end{align} $ | (47) |
is called a self-similar
The assumption (46) ensures that the object constructed in the definition 2.3 defines a
By construction, for a self-similar tree as in definition 2.3, any
$ \begin{equation} \Sigma_{n, j} = \sigma_{n, j} (\Sigma_{0, 0}), \qquad j = 0, \ldots, p^n-1, \end{equation} $ | (48) |
where
$ \begin{equation} \sigma_{n, j} : = \sigma_{j_1} \, \sigma_{j_2} \cdots \sigma_{j_n} \quad \mbox{for} \quad j = {(j_1 j_2 \cdots j_n)_p}. \end{equation} $ | (49) |
Obviously
$ \begin{equation} \sigma_{n, j} = \tau_{n, j} \circ h_{n, j} \circ {\Theta_{n, j}} \end{equation} $ | (50) |
where
$ \begin{equation} \alpha_{n, j} = \alpha_{j_1} \, \alpha_{j_2} \cdots \alpha_{j_n}, \quad \mbox{for} \quad j = {(j_1 j_2 \cdots j_n)_p}. \end{equation} $ | (51) |
By construction,
The fact that definition 2.3 does construct a connected
Lemma 2.4. Let
Moreover,
$ \lbrace \Sigma_{n+1, pj + i}, 0 \leqslant i \leqslant p-1 \rbrace $ |
are numbered in a consecutive way, from
Proof. This proof is left to the reader. In particular, one can use the assumption (46) to show that the constructed graph has no loops. In order to show that it is connected, one can employ the numbering (48).
For the clarity of some proofs, we will need following notation:
● we shall distinguish an integer
●
● with
$ \begin{align*} \Sigma_{\boldsymbol{j}k_1\cdots k_m} = \Sigma_{n+m, \ell}, \qquad u_{\boldsymbol{j}k_1\cdots k_m} = u_{n+m, \ell}, \qquad \mathbf{u}_{\boldsymbol{j}k_1\cdots k_m} = \mathbf{u}_{n+m, \ell}. \end{align*} $ |
From the definition 1.11 of a subtree, it is clear that, a subtree
$ \begin{equation*} \mathcal{T}_{1, i} = \sigma_i( \mathcal{T}), \quad \forall \; 0 \leqslant i \leqslant p-1, \end{equation*} $ |
and that, as a consequence,
$ \begin{equation} \mathcal{T} = {\Sigma_{0, 0}} \cup \bigcup\limits_{i = 0}^{p-1} \sigma_i( \mathcal{T}). \end{equation} $ | (52) |
In fact, the above property can be seen as an alternative to definition 2.3.
Definition 2.5. [Reference self-similar
For the reference tree, the length of
$ \begin{equation} {\boldsymbol{\alpha} : = (\alpha_{0}, \alpha_{2}, \ldots, \alpha_{p-1}) \in \big( \mathbb{R}^+_*\big)^p.} \end{equation} $ | (53) |
When convenient, we shall denote
Compact self-similar trees. The reader will easily remark that
$ \begin{equation} \mathcal{T} \equiv \mathcal{T}_{\boldsymbol{\alpha}} \mbox{ is compact (cf. definition 1.12) if and only if } \quad |\boldsymbol{\alpha}|_\infty : = \max\limits_{0 \leqslant i \leqslant p-1} \alpha_i < 1 . \end{equation} $ | (54) |
Definition 2.6. [Self-similar weighted
$ \begin{equation} \mu_{0, 0} = 1 \;\text{ and }\; \mu(\sigma_i(s)) = \mu_i \, \mu(s), \quad s \in \mathcal{T}, \quad 0 \leqslant i < p. \end{equation} $ | (55) |
In particular, we have
$ \begin{equation} \mu(s) = \mu_{n, j} : = \mu_{j_1} \mu_{j_2} \cdots \mu_{j_n}\; \mbox{ along } \; \Sigma_{n, j} \; \mbox{ if } \; j = {(j_1 j_2 \cdots j_n)_p}. \end{equation} $ | (56) |
A weighted self-similar tree is thus characterized by two
We shall often use in the sequel the following computational trick.
Proposition 2.7. Let
$ \begin{equation} \sum\limits_{j = 0}^{p^n-1} \mu_{n, j}, \alpha_{n, j}^{{\zeta}} = \left( \sum\limits_{i = 0}^{p-1} \mu_i \, \alpha_i^{{\zeta}} \right)^n. \end{equation} $ | (57) |
Proof. Taking all possible
$ \begin{equation} \sum\limits_{j = 0}^{p^n-1} \mu_{n, j} \alpha_{n, j}^{{\zeta}} = \sum\limits_{j_1 = 0}^{p-1} \sum\limits_{j_2 = 0}^{p-1} \dots \sum\limits_{j_n = 0}^{p-1} \; (\mu_{j_1}\alpha_{j_1}^{{\zeta}}) \, (\mu_{j_2} \alpha_{j_2}^{{\zeta}}) \cdots (\mu_{j_n} \alpha_{j_n}^{{\zeta}}). \end{equation} $ | (58) |
The formula (57) follows then from the discrete version of Fubini's theorem.
Example : Regular and geometric trees. By definition, a self-similar
$ \begin{equation*} \alpha_i = \alpha , \quad \forall \; 0 \leqslant i \leqslant p-1. \end{equation*} $ |
We illustrate in figure 4 all the notions and notations introduced above in the case of a symmetric regular dyadic tree, for which
$ \begin{equation*} d = 2, \quad p = 2, \quad \alpha_1 = \alpha_2 = 1/2 \end{equation*} $ |
and where
As discussed before, the principal goal of this work is to provide a theoretical and numerical basis for approximating the DtN operator, cf. section 1.6. For this we need to understand the following:
● whether the solutions to the problems (
● whether (38)-
The goal of this section is to answer these questions in the case of self-similar trees. In all the derivations of this section, we will use the following simplifying assumption.
Assumption 3.1. A tree
All the results of this section are valid for compact self-similar trees, and some of them hold for arbitrary, not necessarily compact, self-similar trees. This will be stated explicitly.
In this section, we will introduce a notion of the trace at infinity for functions in the Sobolev space
Let us first define the "boundary at infinity"
$ \begin{equation} \Gamma_\infty : = [0, 1]. \end{equation} $ | (59) |
Next, we wish to define the trace at infinity of a function
$ \begin{equation} \Gamma_n = \bigcup\limits_{j = 0}^{p^n-1} [a_{n, j} , a_{n, j+1}] \quad (\equiv \Gamma_\infty), \quad a_{n, 0} = 0, \quad a_{n, p^n} = 1. \end{equation} $ | (60) |
To define intermediate values
$ \begin{align} &\Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle : = \sum\limits_{i = 0}^{p-1}\frac{\mu_i}{\alpha_i}. \end{align} $ | (61) |
Then (the reason for the choice of this particular partition will be explained later),
$ \begin{align} a_{n, 0} = 0, \qquad a_{n, j+1} = a_{n, j}+\frac{\mu_{n, j}}{\alpha_{n, j}}\Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle^{-n}, \qquad j = 0, \ldots, p^n-1. \end{align} $ | (62) |
First of all, notice that using (57) with
$ \begin{align} \gamma_i : = \frac{\mu_i}{\alpha_i} \, \left \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \right\rangle^{-1}, \quad &0 \leqslant i \leqslant p-1, \; \mbox{ so that } \sum\limits_{i = 0}^{p-1} \gamma_i = 1. \end{align} $ | (63) |
Then the passage from step
$ \begin{equation} \left\{ \begin{array}{l} [a_{n, j} , a_{n, j+1}] = \bigcup\limits_{i = 0}^{p-1} \; [ \, a_{n+1, p j+i} , a_{n+1, p j+i+1} \, ], \\[8pt] a_{n+1, p j} = a_{n, j}, \quad a_{n+1, p j+i+1} - a_{n+1, p j+i} = \gamma_i \; ( a_{n, j+1} - a_{n, j} ), \end{array} \right. \\[4pt] \end{equation} $ | (64) |
which leads, after some calculations, to (62).
Next, for any
$ \begin{equation} ~~ \tau _n(x) : = {\mathbf u}_{n, j} \equiv u(M_{n, j}), \quad \mbox{for } x \in [a_{n, j} , a_{n, j+1}], \quad 0 \leqslant j \leqslant p^n-1 \; . \end{equation} $ | (65) |
Theorem 3.2. Assume that
$ \begin{equation} \Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle > 1. \end{equation} $ | (66) |
Then, for any
$ \begin{equation} ~~ \tau _\infty u: = \lim\limits_{n \rightarrow + \infty} \; \tau_n u \quad \mathit{\mbox{exists in}}~ L^2(\Gamma_\infty), \end{equation} $ | (67) |
and the application
$ \begin{equation} \forall \; u \in {\rm H}_\mu^1( \mathcal{T}), \quad \|~~ \tau _\infty u\|_{L^2(\Gamma_\infty)} \leqslant C_{\infty} \; \|u\|_{ {\rm H}_\mu^1( \mathcal{T})}. \end{equation} $ | (68) |
Moreover,
$ \begin{equation} {\rm H}_{\mu, \mathit{\text{0}}}^1( \mathcal{T}) \subseteq \mathit{\mbox{Ker}} ~~ \tau _\infty = \{ u \in {\rm H}_\mu^1( \mathcal{T}) \; / \; ~~ \tau _\infty u = 0 \}. \end{equation} $ | (69) |
Proof. Let
The difference
$ [\, a_{n+1, pj+\ell}, a_{n+1, pj+\ell+1}], \; 0 \leqslant j \leqslant p^{n}-1, \; 0 \leqslant \ell \leqslant p-1, $ |
where it takes the value
$ \begin{equation} \| \tau_{n+1}u - \tau_n u \|_{L^2(\Gamma_\infty)}^2 = \left \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \right \rangle^{-n-1} \sum\limits_{j = 0}^{p^n-1} \sum\limits_{\ell = 0}^{p-1} \; \frac{\mu_{n, j}}{\alpha_{n, j}} \, \frac{\mu_{\ell}}{\alpha_{\ell}} \; |\mathbf{u}_{n+1, pj+\ell} - \mathbf{u}_{n, j}|^2. \end{equation} $ | (70) |
Recall that
$ \begin{equation} {\mathbf u}_{n+1, pj+\ell} - {\mathbf u}_{n, j} = \int_{\Sigma_{n+1, pj+\ell}} \partial_s u. \end{equation} $ | (71) |
Using the Cauchy-Schwarz inequality, we have
$ \begin{align} |{\mathbf u}_{n+1, pj+\ell} - {\mathbf u}_{n, j}|^2 \leqslant \left( \int_{\Sigma_{n+1, pj+\ell}} \mu^{-1} \right) \; \left(\int_{\Sigma_{n+1, pj+\ell}} \mu \, |\partial_s u|^2 \right), \end{align} $ | (72) |
that is to say
$ \left|\begin{array}{lll} \nonumber |{\mathbf u}_{n+1, pj+\ell} - {\mathbf u}_{n, j}|^2 & \leqslant & \frac{\alpha_{n+1, pj+\ell}}{\mu_{n+1, pj+\ell}} \; \int_{\Sigma_{n+1, pj+\ell}} \mu \, |\partial_s u|^2 \\ & { = } & \frac{\alpha_\ell}{\mu_\ell} \, \frac{\alpha_{n, j}} {\mu_{n, j}} \; \int_{\Sigma_{n+1, pj+\ell}} \mu \, |\partial_s u|^2. \end{array} \right. $ |
After multiplication by
$ \left|\begin{array}{lll} \nonumber \sum\limits_{j = 0}^{p^n-1} \sum\limits_{\ell = 0}^{p-1} \; \frac{\mu_{n, j}}{\alpha_{n, j}} \, \frac{\mu_{\ell}}{\alpha_{\ell}} \; |{\mathbf u}_{n+1, pj+\ell} - {\mathbf u}_{n, j}|^2 \\ \qquad \qquad \leqslant \;\sum\limits_{j = 0}^{p^n-1} \sum\limits_{\ell = 0}^{p-1} \int_{\Sigma_{n+1, pj+\ell}} \mu \, |\partial_s u|^2 = \int_{ \mathcal{G}^{n+1}} \mu \, |\partial_s u|^2 . \end{array} \right. $ |
Thus, we deduce from (70) that
$ \| \tau_{n+1}u - \tau_n u \|_{L^2(\Gamma_\infty)}^2 \leqslant \Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle^{-n-1} \int_{ \mathcal{G}^{n+1}} \mu \, |\partial_s u|^2 \leqslant \Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle^{-n-1} \; \int_{ \mathcal{T}} \mu \, |\partial_s u|^2 . $ |
This proves, since
$ \left| \begin{array}{lll} \nonumber \| \tau_\infty u \|_{L^2(\Gamma_\infty)} & \leqslant & \| \tau_0 u\| _{L^2(\Gamma_\infty)} + \sum\limits_{n = 0}^{+\infty} \Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle^{-\frac{n+1}{2}} \; \|\partial_s u\|_{ {\rm L}_\mu^2( \mathcal{T})} \\ & = & \| \tau_0 u\| _{L^2(\Gamma_\infty)} + \Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle^{-\frac{1}{2}}\Big( \Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle^{\frac{1}{2}}-1 \Big)^{-1} \; \|\partial_s u\|_{ {\rm L}_\mu^2( \mathcal{T})}. \end{array} \right. $ |
To estimate
$ \| \tau_0 u\| _{L^2(\Gamma_\infty)} = |u(M_\star)| \leqslant C_0 \; \|u\|_{H^1(\Sigma_{0, 0})} \leqslant C_0 \; \|u\|_{ {\rm H}_\mu^1( \mathcal{T})}. $ |
For the embedding (69), note that if
Remark 3.3. It is not difficult to see that (66) is a necessary condition for the existence of the trace, at least for compact regular trees with regular weights, i.e. when
$ \begin{equation*} \alpha_i = \alpha , \quad \mu_i = \mu , \quad \forall \; 0 \leqslant i \leqslant p-1, \mbox{ in which case } \Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle \equiv \frac{p \mu}{\alpha}. \end{equation*} $ |
Let the length of the root edge be
$ \big \{u_{n, j}(x) : I_n \longrightarrow \mathbb{C}, \quad 0 \leqslant j \leqslant p^n-1 \big \} \mbox{ with } I_n = [x_n, x_{n+1}] \mbox{ and } x_n = 1- \alpha^n. $ |
A function
$ u_{n, j}(x) = u_n(x), \quad \text{ for all }\;0 \leqslant j \leqslant p^n-1. $ |
Any symmetric function can be identified to a 1D function defined on an interval:
$ \hat u(x) : I \longrightarrow \mathbb{C} \mbox{ with } I = \bigcup\limits_{j = 0}^{p^n-1} I_n \equiv [0, 1], \, \text{ s.t. } \hat u|_{I_n} = u_n, \quad \forall \; n \in \mathbb{N}. $ |
Let us introduce the space
$ {\rm H}_{\mu, \text{s}}^1( \mathcal{T}) = \{ u \in {\rm H}_\mu^1( \mathcal{T}) \; / \; u \mbox{ is symmetric} \}. $ |
According to the identification process
$ \begin{equation} \|u\|^2_{ {\rm H}_{\mu, \text{s}}^1( \mathcal{T})} = \int_0^1 \big( |\hat u'(x)|^2 + |\hat u(x)|^2 \big) \, w_d(x) \; dx, \end{equation} $ | (73) |
where the piecewise constant weight function
$ w_d(x) = w_n : = ( p \mu)^n \; \mbox{ for } x \in I_n = [x_n, x_{n+1}]. $ |
Noticing that
$ \begin{equation} \|\hat u\|^2_{1, w} = \int_0^1 \big( |\hat u'(x)|^2 + |\hat u(x)|^2 \big) \, w(x) \; dx, \quad w(x) = (1-x)^\beta, \quad \beta = \frac{\log (p\mu)}{\log \alpha}. \end{equation} $ | (74) |
Since
When
$ \int_0^1 \hat u'(x)^2 \, w(x) \; dx = \int_0^1 (1-x)^{\beta - 2} \; dx < +\infty \quad \mbox{since } \beta > 1. $ |
For the limit case
$ \begin{align*} \hat u(x) = \left\{ \begin{array}{ll} \log | \log (1-x)| & x > 1/2, \\ \log\log 2, & x \leqslant 1/2, \end{array}\right. \end{align*} $ |
has a finite norm
In what follows, we will use the notation
$ \begin{align} \|\tau_n u\|^2 = \Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle^{-n}\sum\limits_{j = 0}^{p^n-1}\frac{\mu_{n, j}}{\alpha_{n, j}}|\mathbf{u}_{n, j}|^2. \end{align} $ | (75) |
Remark 3.4. Theorem 3.2 holds both for compact and non-compact trees.
We are now going to prove that, just like for the usual Sobolev spaces on the interval, the inclusion (69) is in fact an equality. This provides a useful characterization of
Theorem 3.5. Assume that (66) holds, so that the trace operator
$ \begin{equation} {\rm H}_{\mu, \mathit{\text{0}}}^1( \mathcal{T}) = \mathit{\mbox{Ker}} ~~\tau_\infty. \end{equation} $ | (76) |
The proof of this theorem is quite long. It will use the following lemma that provides a sufficient condition for a function in
Lemma 3.6. Let
$ \begin{align} n^{-1} \Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle^n\|\tau_n u\|^2\rightarrow 0, \qquad n\rightarrow +\infty, \end{align} $ | (77) |
then
Proof. The proof relies on an approximation process adapted from [33]. Let
$ \left\{ \begin{array}{l} \left. \varphi_{n}\right|_{\mathcal{T}^n}\equiv 1, \qquad \left. \varphi_{n}\right|_{\mathcal{T}\setminus\mathcal{T}^{2n}}\equiv 0, \\ \varphi_{n}(M_{n+\ell, j}) = 1-\frac{\ell}{n}, \qquad j = 0, \ldots , p^{n+\ell}-1, \qquad 0 \leqslant \ell \leqslant n.\\[2pt] \end{array} \right. $ |
Notice that the support of
$ \begin{equation} \left. \varphi'_{n}\right|_{\Sigma_{n+\ell, j}} = (n \, \alpha_{n+\ell, j})^{-1}, \quad j = 0, \ldots, p^{n+\ell}-1. \end{equation} $ | (78) |
Our goal is to prove that
$ \begin{align} u_n\rightarrow u \;\text{ and }\; u_n' = \varphi_{n}'u+ \varphi_{n}u'\rightarrow u'\qquad \text{ in } {\rm L}_\mu^2( \mathcal{T}). \end{align} $ | (79) |
By Lebesgue's dominated convergence theorem,
$ \begin{align} u_n\rightarrow u, \quad \varphi_{n}u'\rightarrow u', \; n\rightarrow +\infty, \qquad \text{ in } {\rm L}_\mu^2( \mathcal{T}). \end{align} $ | (80) |
Therefore, it remains to show that
$ \begin{equation} \| \varphi_{n}'u\|_{L^2_{\mu}(\mathcal{T})}^2 = \! \! \sum\limits_{m = n+1}^{2n} \!\! \sum\limits_{j = 0}^{p^{m}-1}\int\limits_{\Sigma_{m, j}}\mu \, ( \varphi'_{n})^2 \, |u|^2 = \! \! \sum\limits_{m = n+1}^{2n} \sum\limits_{j = 0}^{p^{m}-1}(n \, \alpha_{m, j})^{-2}\int\limits_{\Sigma_{m, j}}\mu \, |u|^2. \end{equation} $ | (81) |
Since we want to bound the above using the traces
For this we will use the following 1D Poincaré inequality
$ \begin{equation} \forall\; v \in {\rm H}^1(0, L) \; \mbox{ with } \; v(L) = 0, \quad \int\limits_0^L |v|^2ds \leqslant \frac{4}{\pi^2} \, L^2\int\limits_0^L |v'|^2 \, ds. \end{equation} $ | (82) |
Let us introduce a piecewise-constant interpolant
$ \begin{align} \left.\Pi u\right|_{\Sigma_{n, j}} = \tilde{\mathbf{ u}}_{n, j} : = \mbox{ the constant function ${\mathbf u}_{n, j}$}, \quad 0 \leqslant j \leqslant p^n-1, \quad n \geqslant 0. \end{align} $ | (83) |
Then, thanks the the Poincaré inequality, applied to the function
$ \begin{align} \|u - \Pi u\|_{ {\rm L}_\mu^2(\Sigma_{m, j})}^2 \leqslant \frac{4}{\pi^2} \, \alpha_{m, j}^2 \|\partial_s u\|_{ {\rm L}_\mu^2(\Sigma_{m, j})}^2. \end{align} $ | (84) |
With the above and
$ \|u\|_{ {\rm L}_\mu^2(\Sigma_{m, j})}^2 \leqslant 2 \, \|\Pi u\|_{ {\rm L}_\mu^2(\Sigma_{m, j})}^2+2 \, \|u - \Pi u\|_{ {\rm L}_\mu^2(\Sigma_{m, j})}^2, $ |
we deduce the following upper bound on
$ \begin{align} \|u\|_{ {\rm L}_\mu^2(\Sigma_{m, j})}^2 \leqslant \frac{8}{\pi^2} \, \alpha_{m, j}^2 \, \|\partial_s u\|^2_{ {\rm L}_\mu^2(\Sigma_{m, j})}+2 \, \mu_{m, j}\alpha_{m, j} \; |{\mathbf u}_{m, j}|^2. \end{align} $ | (85) |
Plugging in the above bound into (81), we end up with the following expression:
$ \begin{align*} \| \varphi_{n}'u\|_{L^2_{\mu}(\mathcal{T})}^2& \leqslant \frac{8}{n^2 \pi^2} \sum\limits_{m = n+1}^{2n}\sum\limits_{j = 0}^{p^m-1}\|\partial_s u\|^2_{ {\rm L}_\mu^2(\Sigma_{m, j})}+\frac{2}{n^2}\sum\limits_{m = n+1}^{2n}\sum\limits_{j = 0}^{p^m-1} \frac{\mu_{m, j}}{\alpha_{m, j}} \, |{\mathbf u}_{m, j}|^2\\ & = \frac{8}{n^2 \pi^2} \, \|\partial_s u\|^2_{ {\rm L}_\mu^2(\mathcal{T}^{2n}\setminus\mathcal{T}^ n)}+\frac{2}{n^2} \, \sum\limits_{m = n+1}^{2n}\Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle^{m}\|\tau_{m}u\|^2, \qquad \text{ cf. }(75). \end{align*} $ |
Obviously, the first term in the above tends to
$ \Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle^{m}\|\tau_{m}u\|^2 = m \, \varepsilon_{m}, \quad \varepsilon_{m}\rightarrow 0 \quad (m \rightarrow + \infty). $ |
Then the second term in the above bound can be estimated as follows:
$ \begin{align*} \frac{2}{n^2} \, \sum\limits_{m = n+1}^{2n}\Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle^{m}\|\tau_{m}u\|^2& = \frac{2}{n^2} \, \sum\limits_{m = n+1}^{2n} m \, \varepsilon_{m} \leqslant 4\max\limits_{m = n, \ldots, 2n}\varepsilon_{m}\rightarrow 0, \quad (n\rightarrow +\infty). \end{align*} $ |
Thus,
To prove theorem 3.5, it remains to show that (77) holds for any
$ \begin{align} \big\{{\mathbf u}_{\boldsymbol{j}\ell_1....\ell_N}, \, 0 \leqslant \ell_k \leqslant p-1, \, 1 \leqslant k \leqslant N \big \}. \end{align} $ | (86) |
Lemma 3.7. Let
$ \begin{align*} q_{\ell} \geqslant 0, \quad 0 \leqslant \ell \leqslant p-1, \qquad \sum\limits_{\ell = 0}^{p-1}q_{\ell} = 1. \end{align*} $ |
Then the nodal value
$ \begin{align} {\mathbf u}_{n, j} & = P_{n, j}^N - \sum\limits_{k = 1}^N D_{n, j}^{k}, \end{align} $ | (87) |
$ \begin{align} P_{n, j}^N & = \sum\limits_{\ell_1 = 0}^{p-1} \cdots \sum\limits_{\ell_N = 0}^{p-1}q_{\ell_1}\cdots q_{\ell_N} \, {\mathbf u}_{ \boldsymbol{j} \ell_1 \cdots \ell_N}, \end{align} $ | (88) |
$ \begin{align} D_{n, j}^k & = \sum\limits_{\ell_1 = 0}^{p-1} \cdots \sum\limits_{\ell_k = 0}^{p-1} \; q_{\ell_1} \cdots q_{\ell_k} \int_{\Sigma_{\boldsymbol{j}\ell_1 \cdots \ell_k}} \! \! \partial_s u. \end{align} $ | (89) |
In the above
Proof. The proof is done by induction. It consists essentially in playing with the basic identity (71) and in exploiting, in order to get an optimal result, all the paths that connect
Let us first consider the case
$ \forall \; 0 \leqslant \ell_1 \leqslant p, \quad {\mathbf u}_{ \boldsymbol{j}} = {\mathbf u}_{\boldsymbol{j}\ell_1} - \int_{\Sigma_{ \boldsymbol{j}\ell_1}} \partial_s u. $ |
To exploit all the possible paths between
$ \begin{equation} {\mathbf u}_{ \boldsymbol{j}} = \sum\limits_{\ell_1 = 0}^{p-1} q_{\ell_1} {\mathbf u}_{\boldsymbol{j}\ell_1} - \sum\limits_{\ell_1 = 0}^{p-1} \; q_{\ell_1} \int_{\Sigma_{\boldsymbol{j}\ell_1}} \partial_s u \end{equation} $ | (90) |
which is (87) for
$ {\mathbf u}_{\boldsymbol{j}\ell_1 \cdots \ell_N} = \sum\limits_{\ell_{N+1} = 0}^{p-1} q_{\ell_{N+1}} {\mathbf u}_{\boldsymbol{j}\ell_1 \cdots \ell_N \ell_{N+1}} - \sum\limits_{\ell_{N+1} = 0}^{p-1} \; q_{\ell_{N+1}} \int_{\Sigma_{\boldsymbol{j}\ell_1\cdots \ell_N \ell_{N+1}}} \partial_s u, $ |
which we substitute into (87), using (88) and (89),
$ {\mathbf u}_{ \boldsymbol{j}} = P_{n, j}^{N+1} - \sum\limits_{\ell_1 = 0}^{p-1} \cdots \sum\limits_{\ell_N = 0}^{p-1} \sum\limits_{\ell_{N+1} = 0}^{p-1} \; \; q_{\ell_1} \cdots q_{\ell_N} q_{\ell_{N+1}} \int_{\Sigma_{\boldsymbol{j}\ell_1\cdots \ell_N \ell_{N+1}}} \partial_s u - \sum\limits_{k = 1}^N D_{n, j}^{k}, $ |
which is the desired result since the central term above is nothing but
Remark 3.8. For the functions of the class
In order to prove theorem 3.5, we first need to relate the traces
$ \begin{equation} C_{{\boldsymbol \alpha}{\boldsymbol \mu}}^N = \sum\limits_{k = 1}^N \Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle^{-k} . \end{equation} $ | (91) |
Lemma 3.9. For all
$ \begin{equation} \Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle^n \|\tau_n u\|^2 \leqslant 2C_{{\boldsymbol \alpha}{\boldsymbol \mu}}^N \, \|\partial_s u\|^2_{ {\rm L}_\mu^2(\mathcal{T}\setminus\mathcal{T}^n)} + 2 \, \Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle^n \|\tau_{n+N} u\|^2. \end{equation} $ | (92) |
Proof. By definition of
$ \begin{equation} \Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle^n \, \|\tau_n u\|^2 = \sum\limits_{j = 0}^{p^n-1} \frac{\mu_{n, j}}{\alpha_{n, j}} \, \big|{\mathbf u}_{n, j}\big|^2. \end{equation} $ | (93) |
Thus, using (87) with
$ \begin{equation} \Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle^n \; \|\tau_n u\|^2 \leqslant 2 \sum\limits_{j = 0}^{p^n-1} \frac{\mu_{n, j}}{\alpha_{n, j}} \, \big|P_{n, j}^N \big|^2 + 2 \sum\limits_{j = 0}^{p^n-1} \frac{\mu_{n, j}}{\alpha_{n, j}} \, \Big| \sum\limits_{k = 1}^N D_{n, j}^k \Big|^2. \end{equation} $ | (94) |
By convexity of
$ \begin{align} \begin{split} \left| \begin{array}{lll} |P_{n, j}^N|^2 & \leqslant & \sum\limits_{\ell_1 = 0}^{p-1} \cdots \sum\limits_{\ell_N = 0}^{p-1}\gamma_{\ell_1}\cdots \gamma_{\ell_N} \, |{\mathbf u}_{\boldsymbol{j}\ell_1 \cdots \ell_N}|^2 \\ & = & \Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle^{-N} \sum\limits_{\ell_1 = 0}^{p-1} \cdots \sum\limits_{\ell_N = 0}^{p-1}\frac{\mu_{\ell_1}}{\alpha_{\ell_1}}\cdots \frac{\mu_{\ell_N}}{\alpha_{\ell_N}} \, |{\mathbf u}_{\boldsymbol{j}\ell_1 \cdots \ell_N}|^2. \end{array} \right. \end{split} \end{align} $ | (95) |
After multiplication by
$ \sum\limits_{j = 0}^{p^n-1} \frac{\mu_{n, j}}{\alpha_{n, j}} \, \big|P_{n, j}^N \big|^2 \leqslant \Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle^{-N} \; {{\sum\limits_{j = 0}^{p^n-1}}} \sum\limits_{\ell_1 = 0}^{p-1} \cdots\sum\limits_{\ell_N = 0}^{p-1}\frac{\mu_{n, j}}{\alpha_{n, j}} \, \frac{\mu_{\ell_1}}{\alpha_{\ell_1}}\cdots \frac{\mu_{\ell_N}}{\alpha_{\ell_N}} \, |{\mathbf u}_{ \boldsymbol{j}\ell_1 \cdots \ell_N}|^2, $ |
or, alternatively, thanks to (93) for
$ \begin{equation} \sum\limits_{j = 1}^{p^n-1} \frac{\mu_{n, j}}{\alpha_{n, j}} \, \big|P_{n, j}^N \big|^2 \leqslant \Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle^{n} \; \|\tau_{n+N} u\|^2. \end{equation} $ | (96) |
In the same way, by convexity again, we deduce from (89) that
$ \begin{align} |D_{n, j}^k|^2 \leqslant\Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle^{-k}\sum\limits_{\ell_1 = 0}^{p-1} \cdots \sum\limits_{\ell_k = 0}^{p-1}\frac{\mu_{\ell_1}}{\alpha_{\ell_1}}\cdots \frac{\mu_{\ell_k}}{\alpha_{\ell_k}} \, \Big|\int_{\Sigma_{ \boldsymbol{j}\ell_1 \cdots \ell_k}} \! \! \partial_s u \, \Big|^2. \end{align} $ | (97) |
Using the Cauchy-Schwarz inequality (like in (72))
$ \begin{equation} |D_{n, j}^k|^2 \leqslant \Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle^{-k} \Big(\frac{\mu_{n, j}}{\alpha_{n, j}}\Big)^{-1} \sum\limits_{\ell_1 = 0}^{p-1} \cdots \sum\limits_{\ell_k = 0}^{p-1} \int_{\Sigma_{\boldsymbol{j}\ell_1 \cdots \ell_k}} \! \! \mu \, |\partial_s u|^2. \end{equation} $ | (98) |
Next, using the discrete Cauchy-Schwarz inequality and the definition of
$ \begin{align} \Big| \sum\limits_{k = 1}^N D_{n, j}^k \Big|^2 = \Big| \sum\limits_{k = 1}^N \Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle^{-\frac{k}{2}} \Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle^{\frac{k}{2}} D_{n, j}^k \Big|^2 \leqslant C_{{\boldsymbol \alpha}{\boldsymbol \mu}}^N \sum\limits_{k = 1}^N \Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle^{k} |D_{n, j}^k|^2. \end{align} $ | (99) |
Multiplying the above by
$ \begin{equation} \frac{\mu_{n, j}}{\alpha_{n, j}} \, \Big|\sum\limits_{k = 1}^N D_{n, j}^k \Big|^2 \leqslant C_{{\boldsymbol \alpha}{\boldsymbol \mu}}^N \sum\limits_{k = 1}^N \sum\limits_{\ell_1 = 0}^{p-1} \cdots \sum\limits_{\ell_k = 0}^{p-1} \int_{\Sigma_{\boldsymbol{j}\ell_1 \cdots \ell_k}} \! \! \mu \, |\partial_s u|^2. \end{equation} $ | (100) |
Since the sets
$ \begin{align} \sum\limits_{j = 0}^{p^n-1}\sum\limits_{\ell_1 = 0}^{p-1} \cdots \sum\limits_{\ell_k = 0}^{p-1}\int_{\Sigma_{\boldsymbol{j}\ell_1 \cdots \ell_k}} \! \! \mu \, |\partial_s u|^2 = \int_{ \mathcal{G}^{n+k}} \mu \, |\partial_s u|^2. \end{align} $ | (101) |
Thus, after summation of (100) over
$ \begin{equation} \sum\limits_{j = 0}^{p^{n}-1} \frac{\mu_{n, j}}{\alpha_{n, j}} \, \Big|\sum\limits_{k = 1}^N D_{n, j}^k \Big|^2 \leqslant C_{{\boldsymbol \alpha}{\boldsymbol \mu}}^N \; \sum\limits_{k = 1}^N \int_{ \mathcal{G}^{n+k}} \mu \, |\partial_s u|^2 = C_{{\boldsymbol \alpha}{\boldsymbol \mu}}^N \int_{ \mathcal{T}^{n+N}\setminus \mathcal{T}^{n}} \mu \, |\partial_s u|^2. \end{equation} $ | (102) |
Finally, the inequality (92) is obtained by gathering (94), (96) and (102).
Now we have all the ingredients necessary to prove theorem 3.5.
Proof of theorem 3.5. By theorem 3.2, see (69), it suffices to prove that
$ \begin{equation} \forall \; u \in \mbox{Ker } \tau_\infty, \quad \Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle^n\|\tau_n u\|^2\rightarrow 0, \qquad n\rightarrow \infty. \end{equation} $ | (103) |
By lemma 3.6 this will imply that
First, since (66) holds, we can define
$ \begin{equation} { C_{{\boldsymbol \alpha}{\boldsymbol \mu}} : = \sum\limits_{k = 1}^{+\infty} \Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle^{-k} = \left(\Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle-1\right)^{-1}.} \end{equation} $ | (104) |
From lemma 3.9 it follows that for all
$ \begin{equation*} \Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle^n \|\tau_n u\|^2 \leqslant 2C_{{\boldsymbol \alpha}{\boldsymbol \mu}} \, \|\partial_s u\|^2_{ {\rm L}_\mu^2(\mathcal{T}\setminus\mathcal{T}^n)} + 2 \, \Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle^n \|\tau_{n+N} u\|^2. \end{equation*} $ |
Let us assume
$ \begin{equation} \Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle^n \|\tau_n u\|^2 \leqslant 2 C_{{\boldsymbol \alpha}{\boldsymbol \mu}}\|\partial_s u\|^2_{ {\rm L}_\mu^2(\mathcal{T}\setminus\mathcal{T}^n)}. \end{equation} $ | (105) |
Since
It is natural to ask how big the image
$ \begin{equation} \big \langle \boldsymbol{\mu}\boldsymbol{\alpha} \big \rangle : = \sum\limits_{i = 0}^{p-1}{\mu_i}\, {\alpha_i}, \quad \Big( \; \big \langle {\boldsymbol{\mu}}{\boldsymbol{\alpha}} \big \rangle < \Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle \mbox{ since } |{\boldsymbol{\alpha}}|_\infty < 1 \Big). \end{equation} $ | (106) |
Theorem 3.10. Assume that
$ \begin{equation} \big \langle {\boldsymbol{\mu}}{\boldsymbol{\alpha}} \big \rangle < 1 < \Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle. \end{equation} $ | (107) |
Then, for any
Proof. Let
$ \begin{equation} \big \langle {\boldsymbol{\mu}}{\boldsymbol{\alpha}} \big \rangle < 1 \quad \Longleftrightarrow \quad { \mathbb{1}_{ \mathcal{T}}}\in {\rm L}_\mu^2( \mathcal{T}). \end{equation} $ | (108) |
Let
$ \left\{ \begin{array}{ll} u \equiv 0, & \mbox{ in the truncated tree } \mathcal{T}^{n-1}, \\ u = \boldsymbol{\varphi}_{n, j} \; { \mathbb{1}_{ \mathcal{T}}}, & \mbox{ in each subtree } \mathcal{T}_{n, j}, 0 \leqslant j \leqslant p^n-1, \\ u \mbox{ is affine}, & \mbox{ along each edge of the generation } \mathcal{G}^n.\\ \end{array} \right. $ |
Remark that
Remark 3.11. Theorem 3.5, lemmas 3.6, 3.7, 3.9 and theorem 3.10 hold both for compact and non-compact trees.
Remark 3.12. The following result proves that
$ \begin{equation*} \tau_\infty \in {\mathcal{L}} \big( {\rm H}_\mu^1( \mathcal{T}), {\rm{H}}^\nu(\Gamma_\infty)\big) \quad \text{for any } \nu < \nu^*, \end{equation*} $ |
where the critical Sobolev regularity exponent
$ \begin{equation*} \nu^* = \min\left(\frac{1}{2}, \min\limits_{0 \leqslant i \leqslant p-1}\left(\frac{1}{2}-\frac{\log\frac{\mu_i}{\alpha_i}}{2\log\gamma_i}\right)\right). \end{equation*} $ |
When
Theorem 3.13. Assume that
$ \begin{equation} \forall \; u \in {\rm H}_\mu^1( \mathcal{T}), \quad ~~ \tau _\infty u = 0 \quad \mathit{\mbox{(i.e. Im $~~ \tau _\infty = \{0\}$ or $\mathit{\mbox{Ker}} ~~ \tau _\infty = {\rm H}_\mu^1( \mathcal{T}). \, )$}} \end{equation} $ | (109) |
Proof. By theorem 3.18, see section 3.3,
$ {\rm H}_{\mu, \text{0}}^1( \mathcal{T}) = \mbox{Ker } ~~ \tau _\infty. $ |
In this section we present the conditions on
Theorem 3.14. If the condition (107) holds, then
Proof. The result is an immediate consequence of the trace theorem 3.10, since the equality
As we are going to see, when
We shall state this result as two theorems, whose proofs are quite different:
● theorem 3.15 for
● theorem 3.18 for
Theorem 3.15. If
To prove this result, we shall use the following technical lemma.
Lemma 3.16. Let
(i)
(ii)
Then the sequence
Proof. See appendix A.
Proof of theorem 3.15. By lemma 3.6, it suffices to prove that (77) holds for any
$ \begin{equation} \left\{ \begin{array}{l} \| \tau_{n}u\|^2 = \Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle^{-n} \sum\limits_{j = 0}^{p^n-1} \; \frac{\mu_{n, j}}{\alpha_{n, j}} \; | \mathbf{u}_{\boldsymbol{j}}|^2 , \\ \| \tau_{n+1}u\|^2 = \Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle^{-n-1} \sum\limits_{j = 0}^{p^n-1} \sum\limits_{\ell = 0}^{p-1} \; \frac{\mu_{n, j}}{\alpha_{n, j}} \, \frac{\mu_{\ell}}{\alpha_{\ell}} \; | \mathbf{u}_{\boldsymbol{j}\ell}|^2 . \end{array} \right. \end{equation} $ | (110) |
Because
$ \begin{align*} \mathbf{u}_{\boldsymbol{j}\ell} = \mathbf{u}_{\boldsymbol{j}}+\int\limits_{\Sigma_{\boldsymbol{j}\ell}} \partial_s u, \end{align*} $ |
we deduce, using the Young's inequality (with a parameter
$ \begin{equation*} | \mathbf{u}_{\boldsymbol{j}\ell}|^2 \leqslant (1+\eta_n) \, | \mathbf{u}_{\boldsymbol{j}}|^2+(1+\eta_n^{-1}) \, \frac{\alpha_{n, j}}{\mu_{n, j}} \, \frac{\alpha_{\ell}}{\mu_{\ell}} \; \int\limits_{\Sigma_{\boldsymbol{j}\ell}} \mu \, |\partial_s u|^2. \end{equation*} $ |
Multiplying the above by
$ \begin{equation} t_{n+1} \leqslant (1+\eta_n) \, \Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle \, t_{n}+(1+\eta_n^{-1}) \, \|\partial_s u\|^2_{\mathcal{G}^{n+1}}, \quad \mbox{where} \quad t_n : = \Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle^n \|\tau_n u\|^2. \end{equation} $ | (111) |
In the above the term
Case 1.
$ \gamma : = (1+\eta) \, \Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle < 1. $ |
Then, by Lemma 3.16, case (ⅰ), we prove that
Case 2. the limit case
$ \begin{equation} \widehat t_{n+1} \leqslant \big(1+\eta_n \big) \, \frac{n}{n+1} \, \widehat t_{n} +(1+\eta_n^{-1}) \, (n+1)^{-1} \|\partial_s u\|^2_{\mathcal{G}^{n+1}}, \end{equation} $ | (112) |
We choose
$ \big(1+\eta_n \big) \, \frac{n}{n+1} = 1 - \frac{1}{2(n+1)} \quad \Longrightarrow \quad \eta_n = \frac{1}{2n} > 0. $ |
Then, as
Remark 3.17. We have shown that when
This does not hold when
$ { \mathbb{1}_{ \mathcal{T}}\in {\rm H}_\mu^1( \mathcal{T})}, \mbox{ cf. (108)}, \; \; \mbox{ and } \; \; \Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle^n\|\tau_n { \mathbb{1}}_{ \mathcal{T}}\|^2 = \|\tau_n{ \mathbb{1}}_{ \mathcal{T}}\|^2 = 1. $ |
Nonetheless, in this case
Finally, we consider the case when
Theorem 3.18. If
The proof of this result is, in its structure, quite similar to the proof of
Lemma 3.19. For all
$ \begin{equation} \left\{ \begin{array}{ll} \frac{1}{2} \; \|\Pi u\|_{ {\rm L}_\mu^2( \mathcal{T}\setminus \mathcal{T}^{n})}^2-\frac{4}{\pi^2} \, |\boldsymbol{\alpha}|^{2n+2}_{\infty} \, \|\partial_s u\|_{ {\rm L}_\mu^2( \mathcal{T}\setminus \mathcal{T}^{n})}^2 \leqslant \|u\|_{ {\rm L}_\mu^2( \mathcal{T}\setminus \mathcal{T}^{n})}^2, & (i) \\ \|u\|_{ {\rm L}_\mu^2( \mathcal{T}\setminus \mathcal{T}^{n})}^2 \leqslant 2 \; \|\Pi u\|^2_{ {\rm L}_\mu^2(\mathcal{T}\setminus\mathcal{T}^n)}+\frac{8}{\pi^2} \, |\boldsymbol{\alpha}|^{2n+2}_{\infty} \, \|\partial_s u\|_{L^2_{\mu}(\mathcal{T}\setminus\mathcal{T}^{n})}^2. & (ii) \end{array} \right. \end{equation} $ | (113) |
In particular, for all
Proof. The bounds (113) follow from
$ \begin{align} \|u\|_{ {\rm L}_\mu^2( \mathcal{T}\setminus \mathcal{T}^{n})}^2 \leqslant 2 \; \|\Pi u\|^2_{ {\rm L}_\mu^2(\mathcal{T}\setminus\mathcal{T}^n)} + 2 \; \|u -\Pi u\|^2_{ {\rm L}_\mu^2(\mathcal{T}\setminus\mathcal{T}^n)}, \\ \|\Pi u\|_{ {\rm L}_\mu^2( \mathcal{T}\setminus \mathcal{T}^{n})}^2 \leqslant 2 \; \|u\|^2_{ {\rm L}_\mu^2(\mathcal{T}\setminus\mathcal{T}^n)} + 2 \; \|u -\Pi u\|^2_{ {\rm L}_\mu^2(\mathcal{T}\setminus\mathcal{T}^n)}, \end{align} $ | (114) |
and the bound (84) for
$ \begin{align*} \|u-\Pi u\|^2_{ {\rm L}_\mu^2(\Sigma_{n, j})} \leqslant \frac{4}{\pi^2}|\boldsymbol{\alpha}|_{\infty}^{2n}\|\partial_s u\|^2_{ {\rm L}_\mu^2(\Sigma_{n, j})}. \end{align*} $ |
The fact that
The interest of working with
$ \begin{align} \widehat{C}_{{\boldsymbol \alpha}{\boldsymbol \mu}}^N = \sum\limits_{\ell = 1}^N \big \langle \boldsymbol{\mu}\boldsymbol{\alpha} \big \rangle^{-\ell}|\boldsymbol{\alpha}|_{\infty}^{2\ell}. \end{align} $ | (115) |
Lemma 3.20. For any
$ \begin{equation} \Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle^n \|\tau_n u\|^2 \leqslant 2\, \widehat{C}_{{\boldsymbol \alpha}{\boldsymbol \mu}}^N \, \|\partial_s u\|^2_{ {\rm L}_\mu^2(\mathcal{T}\setminus\mathcal{T}^n)} + 2 \, \big \langle \boldsymbol{\mu}\boldsymbol{\alpha} \big \rangle^{-N} |\boldsymbol{\alpha}^{-1}|_{\infty}^{2n}\, \|\Pi u\|_{ {\rm L}_\mu^2( \mathcal{G}^{n+N})}^2. \end{equation} $ | (116) |
Proof. Let us first remark that
$ \begin{align} \| \Pi u\|_{ {\rm L}_\mu^2( \mathcal{G}^n)}^2 = \sum\limits_{j = 0}^{p^n-1} \; {\mu_{n, j}}{\alpha_{n, j}} \; | \boldsymbol{u}_{n, j}|^2 \end{align} $ | (117) |
resembles
Thus, we will use the same idea as in the proof of lemma 3.9; we start with the inequality (94), however, we define
As a consequence, proceeding as in the proof of lemma 3.9, we see that (94) is still valid but with different
$ \begin{equation} \begin{array}{lll} |P_{n, j}^N|^2 & \leqslant & \big \langle \boldsymbol{\mu}\boldsymbol{\alpha} \big \rangle^{-N} \sum\limits_{\ell_1 = 0}^{p-1} \cdots \sum\limits_{\ell_N = 0}^{p-1} {\mu_{\ell_1}}{\alpha_{\ell_1}}\cdots {\mu_{\ell_N}}{\alpha_{\ell_N}} \, |{\mathbf u}_{\boldsymbol{j}\ell_1 \cdots \ell_N}|^2. \end{array} \end{equation} $ | (118) |
After multiplication by
$ \begin{align} \sum\limits_{j = 0}^{p^n-1} \frac{\mu_{n, j}}{\alpha_{n, j}} \, \big|P_{n, j}^N \big|^2 & \leqslant \big \langle \boldsymbol{\mu}\boldsymbol{\alpha} \big \rangle^{-N} \; \sum\limits_{j = 0}^{p^n-1} \frac{\mu_{n, j}}{\alpha_{n, j}}\sum\limits_{\ell_1 = 0}^{p-1} \cdots\sum\limits_{\ell_N = 0}^{p-1}\, \mu_{\ell_1}\alpha_{\ell_1}\cdots \mu_{\ell_N}\alpha_{\ell_N} \, |{\mathbf u}_{ \boldsymbol{j} \ell_1 \cdots \ell_N}|^2. \end{align} $ |
Since
$ \begin{align} \sum\limits_{j = 0}^{p^n-1} \frac{\mu_{n, j}}{\alpha_{n, j}} \, \big|P_{n, j}^N \big|^2 & \leqslant \big \langle \boldsymbol{\mu}\boldsymbol{\alpha} \big \rangle^{-N}|\boldsymbol{\alpha}^{-1}|_{\infty}^{2n}\sum\limits_{k = 0}^{p^{n+N}-1}\mu_{n+N, k}\, \alpha_{n+N, k}\, |\mathbf{u}_{n+N}, k|^2\\ & = \big \langle \boldsymbol{\mu}\boldsymbol{\alpha} \big \rangle^{-N}|\boldsymbol{\alpha}^{-1}|_{\infty}^{2n}\, \|\Pi u\|_{ {\rm L}_\mu^2( \mathcal{G}^{n+N})}^2, \end{align} $ | (119) |
where the last equality follows from (117).
To obtain an upper bound for
$ \begin{align*} |D_{n, j}^k|^2 \leqslant \big \langle \boldsymbol{\mu}\boldsymbol{\alpha} \big \rangle^{-k}\sum\limits_{\ell_1 = 0}^{p-1} \cdots \sum\limits_{\ell_k = 0}^{p-1} {\mu_{\ell_1}}{\alpha_{\ell_1}}\cdots {\mu_{\ell_k}}{\alpha_{\ell_k}} \, \Big|\int_{\Sigma_{ \boldsymbol{j}\ell_1 \cdots \ell_k}} \! \! \partial_s u \, \Big|^2. \end{align*} $ |
Using the Cauchy-Schwarz inequality, like in (72),
$ \begin{align} |D_{n, j}^k|^2 & \leqslant \big \langle \boldsymbol{\mu}\boldsymbol{\alpha} \big \rangle^{-k} \; \frac{\alpha_{n, j}}{\mu_{n, j}} \sum\limits_{\ell_1 = 0}^{p-1} \cdots \sum\limits_{\ell_k = 0}^{p-1} \alpha_{\ell_1}^2\cdots \alpha_{\ell_k}^2 \, \int_{\Sigma_{ \boldsymbol{j}\ell_1 \cdots \ell_k}} \mu \, |\partial_s u|^2 \\ & \leqslant \big \langle \boldsymbol{\mu}\boldsymbol{\alpha} \big \rangle^{-k} \; | \boldsymbol{\alpha}|_\infty^{2k} \; \frac{\alpha_{n, j}}{\mu_{n, j}} \sum\limits_{\ell_1 = 0}^{p-1} \cdots \sum\limits_{\ell_k = 0}^{p-1} \int_{\Sigma_{ \boldsymbol{j}\ell_1 \cdots \ell_k}} \mu \, |\partial_s u|^2 \end{align} $ | (120) |
Using the discrete Cauchy-Schwartz inequality, introducing
$ \begin{align} \sum\limits_{j = 0}^{p^{n}-1} \frac{\mu_{n, j}}{\alpha_{n, j}} \, \Big|\sum\limits_{k = 1}^N D_{n, j}^k \Big|^2 & = \sum\limits_{j = 0}^{p^{n}-1} \frac{\mu_{n, j}}{\alpha_{n, j}} \, \Big|\sum\limits_{k = 1}^N\left(\nu^{k}\cdot \nu^{-k}D_{n, j}^k \right)\Big|^2 \\ & \leqslant \widehat{C}_{{\boldsymbol \alpha}{\boldsymbol \mu}}^N {\sum\limits_{k = 1}^N}\sum\limits_{j = 0}^{p^{n}-1} \, \sum\limits_{\ell_1 = 0}^{p-1} \cdots \sum\limits_{\ell_k = 0}^{p-1} \int_{\Sigma_{ \boldsymbol{j}\ell_1 \cdots \ell_k}} \mu \, |\partial_s u|^2, \end{align} $ |
where we used the bound (120) and the definition (115) of
$ \begin{align} \sum\limits_{j = 0}^{p^{n}-1} \frac{\mu_{n, j}}{\alpha_{n, j}} \, \Big|\sum\limits_{k = 1}^N D_{n, j}^k \Big|^2 & \leqslant \widehat{C}_{{\boldsymbol \alpha}{\boldsymbol \mu}}^N \; \sum\limits_{k = 1}^N \int_{ \mathcal{G}^{n+k}} \mu \, |\partial_s u|^2 = \widehat{C}_{{\boldsymbol \alpha}{\boldsymbol \mu}}^N \|\partial_s u\|^2_{ {\rm L}_\mu^2( \mathcal{T}^{n+N}\setminus \mathcal{T}^n)}. \end{align} $ | (121) |
The desired result is obtained by substituting (129) and (121) into (94).
Now we have all the components necessary to prove theorem 3.18.
Proof of theorem 3.18. We will use the characterization of the space
Since the tree is compact, i.e.
$ \begin{align*} \widehat{C}_{{\boldsymbol \alpha}{\boldsymbol \mu}}: = \big \langle \boldsymbol{\mu}\boldsymbol{\alpha} \big \rangle^{-1}|\boldsymbol{\alpha}|_{\infty}^2\left(1-\big \langle \boldsymbol{\mu}\boldsymbol{\alpha} \big \rangle^{-1}|\boldsymbol{\alpha}|_{\infty}^2\right)^{-1}. \end{align*} $ |
This allows us to take a limit
$ \begin{align*} \Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle^n \|\tau_n u\|^2 \leqslant 2 \, \widehat{C}_{{\boldsymbol \alpha}{\boldsymbol \mu}} \; \|\partial_s u\|^2_{ {\rm L}_\mu^2(\mathcal{T}\setminus\mathcal{T}^n)}. \end{align*} $ |
This shows that
Remark 3.21. Theorem 3.15 holds both for compact and non-compact trees. The proof of Theorem 3.18, however, uses the compactness of the tree.
In this section we will summarize the results of the previous sections about the trace operator and relationship between the spaces
This difference between different values of
$ \begin{align*} \mathbb{P}: = \{(\boldsymbol{\mu}, \, \boldsymbol{\alpha})\in \left(\mathbb{R}^*_{+}\right)^p\times \left(\mathbb{R}^*_{+}\right)^p: \, |\boldsymbol{\alpha}|_{\infty} < 1\}. \end{align*} $ |
According to figure 6, we can partition it into the three regions:
$ \begin{align} \begin{split} &\mathbb{P}_N: = \{(\boldsymbol{\mu}, \, \boldsymbol{\alpha})\in\mathbb{P}: \; \Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle \leqslant 1\}, \\ &\mathbb{P}_{ND}: = \{(\boldsymbol{\mu}, \, \boldsymbol{\alpha})\in\mathbb{P}: \; \Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle > 1, \, \big \langle \boldsymbol{\mu}\boldsymbol{\alpha} \big \rangle < 1\}, \\[4pt] &\mathbb{P}_{D}: = \{(\boldsymbol{\mu}, \, \boldsymbol{\alpha})\in\mathbb{P}: \; \big \langle \boldsymbol{\mu}\boldsymbol{\alpha} \big \rangle \geqslant 1\}. \end{split} \end{align} $ | (122) |
Note that (this will be used later)
$ \begin{equation} \Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle > 1 \; \Longleftrightarrow \; (\boldsymbol{\mu}, \, \boldsymbol{\alpha}) \in \mathbb{P}_{ND} \cup \mathbb{P}_{D}, \quad \big \langle \boldsymbol{\mu}\boldsymbol{\alpha} \big \rangle < 1 \; \Longleftrightarrow \; (\boldsymbol{\mu}, \, \boldsymbol{\alpha}) \in \mathbb{P}_{ND} \cup \mathbb{P}_{N}. \end{equation} $ | (123) |
The choice of the notation with indices
It appears that independently of
Lemma 3.22. Let
$ \begin{equation} \forall \; u \in V, \quad \left\lVert {u} \right\rVert_{{ {\rm L}_\mu^2( \mathcal{T} \setminus \mathcal{T}^n)}} \leqslant \gamma_n \left\lVert {u} \right\rVert_{{ {\rm H}_\mu^1( \mathcal{T})}}. \end{equation} $ | (124) |
Proof. See appendix B.
Depending on the approach taken to prove the inequality (124), we split the compactness result into two theorems:
● the compactness of the embedding
● the compactness of the embedding
We start with the case
Theorem 3.23. If
Proof. We use the criterion of lemma 3.22. Because of theorem 3.18, it suffices to demonstrate that (124) holds for any function from
Let us first assume that
$ \begin{align} \|\Pi u\|_{ {\rm L}_\mu^2( \mathcal{T}\setminus \mathcal{T}^n)}^2 = \sum\limits_{m = n+1}^{N}\|\Pi u\|_{ {\rm L}_\mu^2( \mathcal{G}^{m})}^2 = \sum\limits_{m = n+1}^{N}\sum\limits_{j = 0}^{p^m-1}\alpha_{m, j}\mu_{m, j}|\mathbf{u}_{m, j}|^2. \end{align} $ | (125) |
First, we apply lemma 3.7 with
$ \begin{align} |\mathbf{u}_{m, j}|^2 = \left|\sum\limits_{k = 1}^{N-m} D_{m, j}^k\right|^2. \end{align} $ | (126) |
Together with (100), the above results in
$ \begin{align*} \sum\limits_{j = 0}^{p^m-1}\alpha_{m, j}\mu_{m, j}|\mathbf{u}_{m, j}|^2 \leqslant C_{\boldsymbol{\alpha}\boldsymbol{\mu}}^{N-m}\sum\limits_{k = 1}^{N-m}\; \sum\limits_{j = 0}^{p^m-1}\alpha_{m, j}^2\sum\limits_{\ell_1 = 0}^{p-1}\cdots \sum\limits_{\ell_k = 0}^{p-1}\|\partial_s u\|^2_{ {\rm L}_\mu^2(\Sigma_{ \boldsymbol{j}\ell_1\cdots\ell_k})}. \end{align*} $ |
With the bound
$ \begin{align*} \sum\limits_{j = 0}^{p^m-1}\alpha_{m, j}\mu_{m, j}|\mathbf{u}_{m, j}|^2 \leqslant C_{\boldsymbol{\alpha}\boldsymbol{\mu}}|\boldsymbol{\alpha}|_{\infty}^{2m}\|\partial_s u\|^2_{ {\rm L}_\mu^2( \mathcal{T}^{N}\setminus \mathcal{T}^m)} = C_{\boldsymbol{\alpha}\boldsymbol{\mu}}|\boldsymbol{\alpha}|_{\infty}^{2m}\|\partial_s u\|^2_{ {\rm L}_\mu^2( \mathcal{T}\setminus \mathcal{T}^m)}. \end{align*} $ |
Thus, for any
$ \begin{align} \|\Pi u\|_{ {\rm L}_\mu^2( \mathcal{T}\setminus \mathcal{T}^n)}^2 \leqslant C_{\boldsymbol{\alpha}\boldsymbol{\mu}}\, \frac{|\boldsymbol{\alpha}|_{\infty}^{{2n+2}}}{1-|\boldsymbol{\alpha}|_{\infty}^2}\|\partial_s u\|^2_{ {\rm L}_\mu^2( \mathcal{T}\setminus \mathcal{T}^n)}. \end{align} $ | (127) |
With (113)-(ⅱ) and the density argument, a similar inequality holds for
The case
$ \begin{align} \|\Pi u\|_{ {\rm L}_\mu^2( \mathcal{T}\setminus \mathcal{T}^n)}^2 \leqslant \gamma_n \; \|\partial_s u\|^2_{ {\rm L}_\mu^2( \mathcal{T}^n)}, \quad \mbox{with } \lim\limits_{n \rightarrow + \infty} \gamma_n = 0. \end{align} $ | (128) |
Compared with (127),
The reason is that when
Theorem 3.24. If
Proof. Like in the proof of theorem 3.23, it suffices to show that for all
$ \mbox{(124) holds with $\|u\|_{ {\rm L}_\mu^2( \mathcal{T} \setminus \mathcal{T}^n)}$ replaced by $\|\Pi u\|_{ {\rm L}_\mu^2( \mathcal{T} \setminus \mathcal{T}^n)}.$} $ |
Without loss of generality, we may assume that
$ \begin{align*} \mathbf{u}_{n, j} = \mathbf{u}_{j_1\cdots j_n} = \sum\limits_{\ell = 1}^n\; \int_{\Sigma_{j_1\cdots j_{\ell}}}\partial_s u. \end{align*} $ |
Using discrete and continuous (cf. (72)) Cauchy-Schwarz inequalities, we have
$ \begin{align*} |\mathbf{u}_{j_1\cdots j_n}|^2 \leqslant n \sum\limits_{\ell = 1}^n\; \Big|\;\int_{\Sigma_{j_1\cdots j_\ell}} \partial_s u\, \Big|^2 = n\, \sum\limits_{\ell = 1}^n\; \frac{\alpha_{j_1} \cdots \alpha_{j_\ell}}{\mu_{j_1} \cdots \mu_{j_\ell}} \, \|\partial_s u\|^2_{ {\rm L}_\mu^2(\Sigma_{j_1\cdots j_{\ell}})}. \end{align*} $ |
After the multiplication by
$ \begin{align*} \alpha_{n, j} \mu_{n, j}|\mathbf{u}_{j_1\cdots j_n}|^2& \leqslant n \sum\limits_{\ell = 1}^n \alpha_{j_1}^2\cdots \alpha_{j_{\ell}}^2 \alpha_{j_{\ell+1}}\mu_{j_{\ell+1}}\cdots \alpha_{j_n}\mu_{j_n}\|\partial_s u\|^2_{ {\rm L}_\mu^2(\Sigma_{j_1\cdots j_{\ell}})}\\ & \leqslant n\sum\limits_{\ell = 1}^n|\boldsymbol{\alpha}|_{\infty}^{2\ell} \, \alpha_{j_{\ell+1}}\mu_{j_{\ell+1}}\cdots \alpha_{j_n}\mu_{j_n}\|\partial_s u\|^2_{ {\rm L}_\mu^2(\Sigma_{j_1\cdots j_{\ell}})}. \end{align*} $ |
According to (117), by summation of the above over
$ \|\Pi u\|^2_{ {\rm L}_\mu^2(\mathcal{G}^n)} \leqslant n\sum\limits_{\ell = 1}^n \sum\limits_{j_{1} = 0}^{p-1}\cdots\sum\limits_{j_n = 0}^{p-1} \, |\boldsymbol{\alpha}|_{\infty}^{2\ell} \, \alpha_{j_{\ell+1}}\mu_{j_{\ell+1}}\cdots \alpha_{j_n}\mu_{j_n} \, \|\partial_s u\|^2_{ {\rm L}_\mu^2(\Sigma_{j_1\cdots j_{\ell}})}. $ |
Setting
$ \sum\limits_{j_{1} = 0}^{p-1}\cdots\sum\limits_{j_n = 0}^{p-1} = \sum\limits_{j_{1} = 0}^{p-1}\cdots\sum\limits_{j_\ell = 0}^{p-1} \Big( \sum\limits_{j_{\ell+1} = 0}^{p-1} \cdots\sum\limits_{j_n = 0}^{p-1} \Big), $ |
we get
$ \left| \begin{array}{lll} \|\Pi u\|^2_{ {\rm L}_\mu^2(\mathcal{G}^n)} & \leqslant & n\, \sum\limits_{\ell = 1}^n \, A_{n, \ell} \, |\boldsymbol{\alpha}|_{\infty}^{2\ell} \sum\limits_{j_{1} = 0}^{p-1}\cdots\sum\limits_{j_\ell = 0}^{p-1} \, \|\partial_s u\|^2_{ {\rm L}_\mu^2(\Sigma_{j_1\cdots j_{\ell}})}\\ & = & n\, \sum\limits_{\ell = 1}^n|\boldsymbol{\alpha}|_{\infty}^{2\ell}\big \langle \boldsymbol{\mu}\boldsymbol{\alpha} \big \rangle^{n-\ell}\|\partial_s u\|^2_{ {\rm L}_\mu^2( \mathcal{G}^{\ell})} \\ & \leqslant & n \; \Big( \sum\limits_{\ell = 1}^n|\boldsymbol{\alpha}|_{\infty}^{2\ell}\big \langle \boldsymbol{\mu}\boldsymbol{\alpha} \big \rangle^{n-\ell} \Big) \; \|\partial_s u\|^2_{ {\rm L}_\mu^2( \mathcal{T}^n)}. \end{array} \right. $ |
A direct computation yields
$ \begin{align*} \left\{ \begin{array}{l} \text{if } \big \langle \boldsymbol{\mu}\boldsymbol{\alpha} \big \rangle = |\boldsymbol{\alpha}|_{\infty}^{2}, \qquad \sum\limits_{\ell = 1}^n|\boldsymbol{\alpha}|_{\infty}^{2\ell}\, \big \langle \boldsymbol{\mu}\boldsymbol{\alpha} \big \rangle^{n-\ell} = n\big \langle \boldsymbol{\mu}\boldsymbol{\alpha} \big \rangle^{n}, \\ \text{if } \big \langle \boldsymbol{\mu}\boldsymbol{\alpha} \big \rangle\neq |\boldsymbol{\alpha}|_{\infty}^{2}, \qquad \sum\limits_{\ell = 1}^n|\boldsymbol{\alpha}|_{\infty}^{2\ell}\, \big \langle \boldsymbol{\mu}\boldsymbol{\alpha} \big \rangle^{n-\ell} = |\boldsymbol{\alpha}|_{\infty}^{2}\frac{\big \langle \boldsymbol{\mu}\boldsymbol{\alpha} \big \rangle^n-|\boldsymbol{\alpha}|_{\infty}^{2n}}{\big \langle \boldsymbol{\mu}\boldsymbol{\alpha} \big \rangle-|\boldsymbol{\alpha}|_{\infty}^{2}}. \end{array}\right. \end{align*} $ |
Combining the above two expressions, we thus obtained (128) with
$ \begin{align*} \gamma_n = C \; \sum\limits_{k = n}^{\infty}k^2\max \Big(\big \langle \boldsymbol{\mu}\boldsymbol{\alpha} \big \rangle, |\boldsymbol{\alpha}|_{\infty}^2 \Big)^k \quad ( \; \rightarrow 0 \; \mbox{ when } n \rightarrow + \infty), \end{align*} $ |
where
Remark 3.25. The statement of lemma 3.22 is valid both for compact and non-compact trees. The results of theorems 3.24, 3.23 do not hold for non-compact trees. It appears that the necessary [33,Theorem 6.1.7] (and, as we have shown, sufficient) condition for the compactness of the embedding
As we will see later, in order to construct transparent boundary conditions, it will be necessary to understand the structure of the solutions to the Neumann (
● self-similarity (in a certain sense) of the solutions to (
● the continuity of the solutions to (
● the difference between the solutions to (
Let us introduce a notion of a quasi-self-similar function, which will play an important role in understanding of the structure of the solutions to the Helmholtz equation on self-similar trees.
Definition 4.1. [Quasi-self-similarity] A function
$ \begin{equation} u\big(\sigma_i(s), \omega\big) = r(\omega)\, u(s, \alpha_i \omega), \qquad s\in\mathcal{T}. \end{equation} $ | (129) |
The above notion reduces to a classical notion of self-similarity in
$ \begin{align} u\big(\sigma_i(s), 0\big) = r(0)\, u(s, 0), \qquad s\in\mathcal{T}. \end{align} $ | (130) |
For any quasi-self-similar function
$ \begin{equation} u(\sigma_{n, j}(s), \omega) = r(\omega)\left[\prod\limits_{k = 1}^{n-1}r(\alpha_{j_1}\cdots\alpha_{j_k}\omega)\right] \, u(s, \alpha_{n, j}\omega), \end{equation} $ | (131) |
where
Theorem 4.2. The function
Proof. We provide the proof for
$ \begin{equation} r(\omega): = u(M_{0, 0}, \omega), \qquad \omega\in\mathbb{C}\setminus\mathbb{R}. \end{equation} $ | (132) |
Our goal is to show that
It is not difficult to notice that
Let us define the following quantity:
$ \begin{equation} u_{i}(s, \omega): = r(\omega)^{-1} \; u\big(\sigma_i(s), \omega\big). \end{equation} $ | (133) |
Notice that
$ \begin{equation} \partial_s u_{ i}(s, \omega) = r (\omega)^{-1}\, \alpha_i\, \partial_s u\circ \sigma_i. \end{equation} $ | (134) |
Choosing
$ \begin{equation} \int_{ \mathcal{T}_{1, i}} \mu(s_i) \, \partial_{s_i} u \, \partial_{s_i} v_i - \omega^2 \int_{ \mathcal{T}_{1, i}} \mu(s_i) \ u(s_i) \ v_i (s_i) = 0, \quad \forall v_i \in V_{ \mathfrak{d}, i} \; . \end{equation} $ | (135) |
Performing in the above integrals the change of variables
$ \begin{equation} \int_{ \mathcal{T}} \mu(\sigma_i(s)) \, \partial_{s_i} u(\sigma_i(s)) \, \partial_{s_i} v_i (\sigma_i(s)) - \omega^2 \int_{ \mathcal{T}} \mu(\sigma_i(s)) \ u(\sigma_i(s))\ v_i (\sigma_i(s)) = 0, \end{equation} $ | (136) |
for all
$ \begin{equation} { \alpha_i^{-2} \int_{ \mathcal{T}} \mu \, \partial_s u_{i} \, \partial_s v - \omega^2 \int_{ \mathcal{T}} \mu \ u_{i} v = 0, \quad \forall v \in V_ \mathfrak{d}.} \end{equation} $ | (137) |
Since
$ \begin{equation*} u (\sigma_i(s), \omega) = r (\omega)\, u (s, \alpha_i \, \omega), \end{equation*} $ |
which concludes the proof.
In this section we address the question of the well-posedness of the problems (
Let us now find explicitly the solutions to the Dirichlet/Neumann problem for the Laplace equation. This problem is well-posed, according to Lemma 1.22, in particular, because the compactness assumptions (38)-
Theorem 4.3. The function
$ \begin{align} u(\sigma_i(s)) = r_0 \, u(s), \qquad s\in\mathcal{T}. \end{align} $ | (138) |
Proof. The statement follows by extending the proof of theorem 4.2 to
In the following theorem we calculate explicitly the solutions to (
Theorem 4.4. The solutions to (
● if
● if
$ \begin{align} r_0 = \Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle^{-1}. \end{align} $ | (139) |
● if
Proof. Let us find a general form of the solution to (
$ \begin{align*} \partial_s u\circ \sigma_i = r_0 \, \alpha_i^{-1}\partial_s u. \end{align*} $ |
Then the Kirchhoff condition (9) at the node
$ \begin{align} \partial_s u_{0, 0}(M_{0, 0}) = \sum\limits_{i = 0}^{p-1}r_0 \, \frac{\mu_i}{\alpha_i} \, \partial_s u_{0, 0}({M_\star}). \end{align} $ | (140) |
Since
● either
● or
$ \begin{equation} u_{0, 0}(s) = 1+(r_0-1)s. \end{equation} $ | (141) |
To summarize,
Case 1.
Case 2.
The function
Case 3.
Since in this case
Thus, we see that in the region
$ \begin{align} u_{ \mathfrak{d}}(., \omega_k)\rightarrow u_{ \mathfrak{d}}(., 0), \qquad u_{ \mathfrak{n}}(., \omega_k)\rightarrow u_{ \mathfrak{n}}(., 0), \qquad \text{ strongly in } {\rm H}_\mu^1( \mathcal{T}). \end{align} $ | (142) |
This is an immediate corollary of proposition 1.23 (notice that (38)-
Corollary 4.5. The following holds true for the solutions of (
● if
● if
Proof. The result in
When
$ \begin{align*} u_{ \mathfrak{d}}(., \omega_{\ell}) = u_{ \mathfrak{n}}(., \omega_{\ell}), \qquad \lim\limits_{\ell\rightarrow \infty}\omega_{\ell} = \omega\in\mathbb{C}\setminus\mathbb{R}. \end{align*} $ |
Then, by the uniqueness continuation theorem for holomorphic vector-valued functions, see [9,Proposition A.2,p.462],
The above property shows that the transparent boundary conditions, which we aim to construct, should take into account the fact that the solutions of the Dirichlet and of the Neumann problem differ when
In this section, we investigate some properties of the DtN operator, as it was introduced in section 1.6, and more precisely the computation of its symbol as a function of the frequency
Before entering into the details, let us remind that depending on the value
● if
● otherwise, if
Let us define the two Dirichlet and Neumann symbols as
$ \begin{equation} \mathbf{\Lambda}_{ \mathfrak{d}}(\omega) : = - \, \partial_s u_{ \mathfrak{d}}({M_\star}, \omega), \quad \mathbf{\Lambda}_{ \mathfrak{n}}(\omega) : = - \, \partial_s u_{ \mathfrak{n}}({M_\star}, \omega). \end{equation} $ | (143) |
According to proposition 1.23, the functions
$ \begin{align} \left\{ \begin{array}{l} \mathbf{\Lambda}_{ \mathfrak{d}}(\omega) = \, \mathbf{\Lambda}_{ \mathfrak{d}}(0) - \sum\limits_{n = 0}^{+\infty} \; \frac{\alpha_{ \mathfrak{d}}^n\omega^2}{(\omega_ \mathfrak{d}^n)^2- \omega^2}, \quad \alpha_{ \mathfrak{d}}^n = \left( \frac{\partial_s \varphi_{ \mathfrak{d}}^n({M_\star})}{\omega_{ \mathfrak{d}}^n}\right)^2, \\ \mathbf{\Lambda}_{ \mathfrak{n}}(\omega) = \, \mathbf{\Lambda}_{ \mathfrak{n}}(0) - \sum\limits_{n = 0}^{+\infty} \; \frac{\alpha_{ \mathfrak{n}}^n\omega^2}{(\omega_ \mathfrak{n}^n)^2- \omega^2}, \quad \alpha_{ \mathfrak{n}}^n = \left( \frac{\partial_s \varphi_{ \mathfrak{n}}^n({M_\star})}{\omega_{ \mathfrak{n}}^n}\right)^2. \end{array} \right. \end{align} $ | (144) |
The convergence of the above series is uniform on the compact subsets of
Remark 5.1. The formulas (144) show that the set of poles of
Another property of the symbol of the DtN operator follows naturally from the explicit form of the zero-frequency solutions for the Helmholtz equation. It is formulated below.
Lemma 5.2. The symbols
● in
$ \mathbf{\Lambda}_{ \mathfrak{d}}(\omega)\equiv\mathbf{\Lambda}_{ \mathfrak{n}}(\omega), \mathit{\text{and}} ~~\mathbf{\Lambda}_{ \mathfrak{d}}(0) = \mathbf{\Lambda}_{ \mathfrak{n}}(0) = 0. $ |
● in
$ \mathbf{\Lambda}_{ \mathfrak{d}}(\omega)\neq\mathbf{\Lambda}_{ \mathfrak{n}}(\omega), \quad \mathbf{\Lambda}_{ \mathfrak{d}}(0) = 1-\Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle^{-1} \mathit{\text{and}}~ \; \mathbf{\Lambda}_{ \mathfrak{n}}(0) = 0. $ |
● in
$ \mathbf{\Lambda}_{ \mathfrak{d}}(\omega)\equiv \mathbf{\Lambda}_{ \mathfrak{n}}(\omega)~~ \mathit{\text{and}}~~ \mathbf{\Lambda}_{ \mathfrak{d}}(0) = \mathbf{\Lambda}_{ \mathfrak{n}}(0) = 1-\Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle^{-1}. $ |
Proof. Let us consider
By theorem 4.4, and more precisely (141),
$ u_{0, 0}(s) = 1+(r_0-1)s, \quad r_0 = \Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle^{-1}. $ |
Thus,
In this section we will demonstrate how one can compute the symbols
Lemma 5.3. Each function
$ \begin{equation} \left\{ \begin{array}{l} \mathit{\mbox{Find}}~~ \mathbf{\Lambda}(\omega) : \mathbb{C} \setminus \mathbb{R} \mapsto \mathbb{C}~~ \mathit{\mbox{such that}} \\[4pt] \omega \sin\omega + \mathbf{\Lambda}(\omega) \cos \omega = \Big( \cos\omega - \mathbf{\Lambda}(\omega) \frac{\sin\omega}{\omega} \Big) \sum\limits_{i = 0}^{p-1} \frac{\mu_i}{\alpha_i} \mathbf{\Lambda}(\alpha_i \omega). \end{array} \right. \end{equation} $ | (145) |
Proof. Let
$ \partial^2_s u_{0, 0} + \omega^2 u_{0, 0} = 0, \quad u_{0, 0}(0, \omega) = 1, \quad \partial_s u_{0, 0}(0, \omega) = -\mathbf{\Lambda}(\omega), $ |
which leads to
$ \begin{equation} \partial_s u_{0, 0} (M_{0, 0}, \omega) = - \, \omega \sin \omega - \mathbf{\Lambda}(\omega) \cos \omega. \end{equation} $ | (146) |
Along
$ \begin{equation} \partial_s u_{1, i} (M_{0, 0}, \omega) = \alpha_i^{-1} \, r(\omega) \, \partial_s u_{0, 0} ({M_\star}, \alpha_i\omega) \equiv\, - \, \alpha_i^{-1}\, r(\omega) \; \mathbf{\Lambda}(\alpha_i \omega), \end{equation} $ | (147) |
where the self-similarity ratio
$ \begin{equation} r(\omega) = u_{0, 0}(M_{0, 0}, \omega) \equiv \cos \omega - \frac{\mathbf{\Lambda}(\omega)}{\omega} \, \sin \omega. \end{equation} $ | (148) |
One gets (145) by substituting (146), (147), (148) into the Kirchhoff equation (9) at
$ \qquad \qquad \partial_s u_{0, 0}(M_{0, 0}, \omega) = \sum\limits_{i = 0}^{p-1} \mu_i \, \partial_s u_{1, i}(M_{0, 0}, \omega). $ |
Remark 5.4. Depending on the values of
Since this equation (145) is quadratic, one expects that it admits several (naively at least two) solutions. However, as we know, when
● when
● when
First of all, notice that because the symbol of the DtN operator satisfies (145) and is an even function, analytic in the origin, a priori one of the solutions of the equation (145) is even and analytic in the origin. Moreover, in the origin this solution satisfies lemma 5.2. A priori it is not obvious that such a solution is unique. However, the uniqueness can be shown, and the corresponding result is formulated in the following lemma.
Let us draw the attention of the reader to the fact that in the lemma below, we use
Lemma 5.5. Any solution of (145), continuous in the origin, satisfies
$ \begin{align*} \mathbf{\Lambda}(0) = 0\qquad \mathit{\text{or}}\qquad \mathbf{\Lambda}(0) = 1- \Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle^{-1}. \end{align*} $ |
If
$ \begin{align*} \mathbf{\Lambda}_N(\omega) = -\left(1-\big \langle \boldsymbol{\mu}\boldsymbol{\alpha} \big \rangle\right)^{-1}\omega^2+O(\omega^4). \end{align*} $ |
If
$ \mathbf{\Lambda}_D(0) = 1- \Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle^{-1}. $ |
Moreover, its Taylor expansion at the origin is given, as
$ \begin{align*} \mathbf{\Lambda}_D(\omega)& = \mathbf{\Lambda}_D(0)-\frac{1}{3}\left(\Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle^2+\Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle+1\right)\left(\Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle^2-\big \langle \boldsymbol{\mu}\boldsymbol{\alpha} \big \rangle\right)^{-1}\omega^2+O(\omega^4) \end{align*} $ |
Proof. Please see the appendix C. Let us remark that it is possible to obtain a higher-order expansion of
Combining the above with lemma 5.2, we arrive at the following conclusion.
Corollary 5.6. Let
● if
● if
● if
Remark 5.7. The corollary clarifies the notation by explaining why, when
Remark 5.8. The uniqueness results of lemma 5.5 fail to be true if one looks for not necessarily smooth solutions
$ \begin{array}{l} \mathbf{\Lambda}(\omega) = \mathbf{\Lambda}_D(\omega), \quad \mbox{if } |\omega| \in \mathbb{Q}, \\ \mathbf{\Lambda}(\omega) = \mathbf{\Lambda}_N(\omega), \quad \mbox{if } |\omega| \notin \mathbb{Q}, \end{array} $ |
is a solution of (145), different from
$ \begin{array}{l} \mathbf{\Lambda}(\omega) = \mathbf{\Lambda}_N(\omega), \quad \mbox{if } |\omega| \in \mathbb{Q}, \\ \mathbf{\Lambda}(\omega) = \mathbf{\Lambda}_D(\omega), \quad \mbox{if } |\omega| \notin \mathbb{Q}, \end{array} $ |
is a solution of (145), different from
An important property of the DtN operator is its positivity, which is related to the energy conservation. This property will be crucial for obtaining an approximation of symbols
$ \begin{align*} \mathbb{C}^+ = \{z\in\mathbb{C}: \, \operatorname{Im}{z} > 0\}. \end{align*} $ |
Theorem 5.9. The symbol
$ \begin{align} { \operatorname{Im}\left(\omega^{-1}\mathbf{\Lambda}(\omega)\right) < 0, \qquad \mathit{\text{for all}}~~\, \omega\in\mathbb{C}^+.} \end{align} $ | (149) |
In other words,
Proof. This property can be shown directly, by examining the expressions (144); however, we provide a more general proof, which relies only on the properties of the underlying sesquilinear form. We show the result for
$ \begin{align*} -\int\limits_{\mathcal{T}}\mu\omega^2 \left|u_{ \mathfrak{d}}(., \omega) \right|^2+\int\limits_{\mathcal{T}}\mu \left|\partial_s u_{ \mathfrak{d}}(., \omega)\right|^2+\partial_s u_{ \mathfrak{d}}({M_\star}, \omega) = 0. \end{align*} $ |
Dividing the above by
$ \begin{align*} \omega^{-1}\boldsymbol{\Lambda}_{ \mathfrak{d}}(\omega) = \omega^{-1}\int\limits_{\mathcal{T}}\mu \, |\partial_s u_{ \mathfrak{d}}|^2-|\omega|^2 \, \omega\int\limits_{\mathcal{T}}\mu \, |u_{ \mathfrak{d}}|^2. \end{align*} $ |
It remains to notice that
Remark 5.10. The above property will be employed to prove that the algorithm for the evaluation of the symbol of the DtN operator, which we present in the next section, is well-defined (i.e. no division by zero occurs in the course of this algorithm). But the meaning of this positivity property is much more important than this: it is fundamental for the stability of the boundary-value problems. This is implicitly used, in particular, in theorem 6.2.
Let us state the following two trivial properties of the Herglotz functions, useful further.
Lemma 5.11. Let
$ \begin{align} \operatorname{Im} (\omega^{-1}f(\omega)) < 0. \end{align} $ | (150) |
Then if
Proof. For
$ \begin{align*} \operatorname{sign} \operatorname{Im} \left(\omega^{-1}f(\omega)\right) = \operatorname{sign} \operatorname{Im}\left(f_0 \, \omega^{-1}+f_2 \, \omega\right). \end{align*} $ |
It is easy to conclude with (150).
Remark 5.12. The solutions
$ \mathbf{\Lambda}_N(\omega) \; \mbox{ for } \; \big \langle \boldsymbol{\mu}\boldsymbol{\alpha} \big \rangle \geqslant 1 \quad \mbox{ and } \quad \mathbf{\Lambda}_D(\omega) \; \mbox{ for } \; \Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle \leqslant 1 $ |
as solutions of the equation (145). However, as already seen by verifying the conditions of lemma 5.11 for their expansions in the origin (cf. lemma 5.5), these functions no longer satisfy (150) in these regions (with an exception of a special case
Let us now provide an algorithm for the numerical approximation of
$ \begin{equation} \mathbf{\Lambda}(\omega) = - \, \omega \; \frac{\omega\sin\omega-\cos\omega f_{\alpha, \mu}(\omega)}{\omega\cos\omega+\sin\omega f_{\alpha, \mu}(\omega)}, \qquad f_{\alpha, \mu}(\omega) = \sum\limits_{i = 0}^{p-1}\frac{\mu_i}{\alpha_i}\mathbf{\Lambda}(\alpha_i\omega). \end{equation} $ | (151) |
This expression defines the values of the function
First of all, let us consider
$ \begin{align} \mathcal{B}_n = \{ \omega \in \mathbb{C} \setminus \mathbb{R} \; \text{ s.t. }\; |\omega| < r_n : = |\boldsymbol{\alpha}|_{\infty}^{-n} \, r_0 \} \end{align} $ | (152) |
where
$ 0 < r_0 < \omega_ \mathfrak{a}^0. $ |
Then, at the continuous level, the iterative algorithm proceeds as follows.
● Initialization: given a truncation parameter
$ \begin{align} \mathbf{\Lambda}^a_ \mathfrak{a}(\omega) : = \sum\limits_{n = 0}^N \lambda_{ \mathfrak{a}, {2n}} \; \omega^{2n}, \qquad \omega \in \mathcal{B}_0, \end{align} $ | (153) |
where the (real) coefficients
● Induction: Supposing that
$ \begin{equation} \mathbf{\Lambda}_{ \mathfrak{a}}^a(\omega) = -\omega \; \frac{\omega\sin\omega-\cos\omega f_{\alpha, \mu}^a(\omega)}{\omega\cos\omega+\sin\omega f_{\alpha, \mu}^{a}(\omega)}, \qquad f_{\alpha, \mu}^a(\omega) = \sum\limits_{i = 0}^{p-1}\frac{\mu_i}{\alpha_i}\mathbf{\Lambda}_{ \mathfrak{a}}^a(\alpha_i\omega). \end{equation} $ | (154) |
Note that the above expression completely defines
$ \text{for }\omega \in \mathcal{B}_{n+1} \setminus \mathcal{B}_n, \quad \alpha_i \, \omega \in \mathcal{B}_n, \quad \forall \; 0 \leqslant i \leqslant p-1.\\[4pt] $ |
One could ask whether the above algorithm is well-defined, in the sense that a division by zero never occurs in (154). This is the case, provided a certain condition on (153); this fact is a corollary of the following two lemmas.
Lemma 5.13. Let
1.
2. the function
$ \operatorname{Im} \left(\omega^{-1}F(\omega)\right) < 0, \qquad \omega\in \Omega. $ |
Proof. Let us prove the first assertion. Notice that
$ \begin{align*} h(\omega) = \omega\cos\omega+\sin\omega f(\omega) = \omega\sin\omega \, g(\omega), \qquad g(\omega) = \cot \omega+\omega^{-1}f(\omega), \end{align*} $ |
vanishes for some
$ \begin{align} \operatorname{Im}{g}(\omega) < 0 \text{ in } \mathbb{C}^+. \end{align} $ | (155) |
To prove 2., we compute
$ \begin{array}{ll} \operatorname{sign} \operatorname{Im}(\omega^{-1}F(\omega))& = -\operatorname{sign} \operatorname{Im} \big((1-\omega^{-1}f(\omega)\cot \omega )\overline{(\cot \omega+\omega^{-1}f(\omega))} \, \big)\\[4pt] & = -\operatorname{sign} \operatorname{Im}\big( \; \overline{g(\omega)}-\omega^{-1}f(\omega)|\cot\omega|^2-|\omega^{-1}f(\omega)|^2\cot\omega \, \big). \end{array} $ |
The above is negative, for the same reasons as (155), and because
The above result shows that if
Lemma 5.14. Let
If
$ \begin{align*} f_0 \geqslant 0, \qquad f_2 \leqslant 0, \qquad f_{2\ell} = 0, \qquad \mathit{\text{for all}}~~ \ell > 1. \end{align*} $ |
Otherwise, if either
Proof. The first part of the statement follows by contradiction, by taking
To formulate the principal result about the feasibility of the algorithm (153, 154), let us introduce an auxiliary quantity
$ \begin{align} \begin{split} &\text{if }N > 1, \qquad \operatorname{Im}\left(\omega^{-1}\boldsymbol{\Lambda}_{ \mathfrak{a}}^a(\omega)\right) < 0 \text{ for all } \omega\in \mathcal{B}(0, r_*(N))\cap\mathbb{C}^+, \\ &\text{ if }N = 1, \qquad r_*(N) = +\infty. \end{split} \end{align} $ | (156) |
Then we can formulate the following lemma about the properties of the algorithm.
Lemma 5.15. Let
$ \begin{equation} \operatorname{Im}\left(\omega^{-1}\mathbf{\Lambda}_{ \mathfrak{a}}^a(\omega)\right) < 0, \qquad \mathit{\text{for all}}\, ~~\omega\in\mathbb{C}^+. \end{equation} $ | (157) |
Proof. It is easy to see that it is sufficient to check that the stated result holds for the first step of the algorithm (i.e. the construction of
First of all, thanks to lemma 5.14 (in the case
$ \begin{align*} \operatorname{Im}\left(\omega^{-1}\boldsymbol{\Lambda}_{ \mathfrak{a}}^a(\omega)\right) < 0, \qquad \forall \; \omega \in B\big(0, r_*(N)\big). \end{align*} $ |
Next, using the definition of the function
$ \forall \; \omega \in \mathcal{B}_1 \cap \mathbb{C}^+, \quad \operatorname{Im} \, \left( \omega^{-1} \, f_{\alpha, \mu}(\omega) \right) < 0. $ |
Then applying lemma 5.13, item 1, with
$ \Omega : = \mathcal{B}_1 \cap \mathbb{C}^+ \quad \mbox{ and }\quad f(\omega) = f_{\alpha, \mu}(\omega), $ |
we deduce that
$ \forall \; \omega \in \mathcal{B}_1 \cap \mathbb{C}^+, \quad \omega \, \cos \omega + \sin \omega \, f_{\alpha, \mu}(\omega) \neq 0. $ |
Thus, there is no problem to extend the function
$ \operatorname{Im} \, \left(\omega^{-1} \, \mathbf{\Lambda}^a_ \mathfrak{a}(\omega) \right) < 0 \quad \mbox{in } \mathcal{B}_{1} \cap \mathbb{C}^+. $ |
Remark 5.16. We did not investigate the (expected) convergence of our algorithm when
In practice, we compute a discrete approximation of
Next, we consider the following polar mesh of the quarter plane
$ \begin{align*} \omega_j^n = n \; \Delta r \; e^{ij \Delta \theta}, \quad n \geqslant 1, \quad 1 \leqslant j \leqslant N_\theta, \end{align*} $ |
where
$ \begin{align} { N_0 > (|\boldsymbol{\alpha}|_{\infty}^{-1}-1)^{-1}.} \end{align} $ | (158) |
Then, for each fixed
● As long as
● As soon as
$ { \alpha_i \, \omega_j^n = \eta \, \omega_{j}^{\ell} + (1 - \eta) \, \omega_{j}^{\ell-1} \quad \mbox{with } \ell < n.} $ |
Remark that the condition (158) ensures that
$ \mathbf{\Lambda}^a_ \mathfrak{d}(\alpha_i \omega_j^n) \longrightarrow \eta \, \mathbf{\Lambda}^a_ \mathfrak{d}(\omega^{\ell}_j) + (1 - \eta) \, \mathbf{\Lambda}^a_ \mathfrak{d}( \omega^{\ell-1}_j). $ |
We consider a dyadic symmetric tree (cf. figure 4) characterized by
The approximation of
The property (149) of
Meromorphic solutions of (145)
In this section, we present a stable low-order local approximation of the DtN operator, based on the expansions provided in lemma 5.5, see corollary 5.6.
First of all, in practice we are interested in arbitrary
$ \begin{equation} \mathbf{\Lambda}_{ \mathfrak{a}, \ell}(\omega) = \ell^{-1} \, \mathbf{\Lambda}_ \mathfrak{a}(\omega \, \ell). \end{equation} $ | (159) |
Let us come back to the transparent conditions (23) of section 1.6. For simplicity, but without any loss of generality, let us consider the case where the tree is the self-similar tree
$ \begin{equation} \quad \mu_{n, j} \, \partial_s u = -\sum\limits_{k \in \mathcal{C}_{n, j}} \mu_{n+1, k} \; \Lambda_{n+1, k} u \qquad \text{ at the point }M_{n, j}, \end{equation} $ | (160) |
where we recall that
$ \mathcal{C}_{n, j} = \{ pj + i, \, 0 \leqslant i \leqslant p-1 \}, \quad \mu_{n+1, pj+i} = \mu_i \, \mu_{n, j}. $ |
Moreover, as the length of the root edge of the tree of
$ \Lambda_{n+1, pj+i} = \Lambda_{ \mathfrak{a}, \alpha_i \ell_{n, j}}, $ |
where we have used (159) for notation, with
$ \begin{equation} \partial_s u = -\sum\limits_{i = 0}^{p-1} \, \mu_{i} \; \Lambda_{ \mathfrak{a}, \alpha_i \ell_{n, j}} \, u \quad \mbox{at the point }M_{n, j}. \end{equation} $ | (161) |
From the compactness property of the tree, one expects that, since
$ \begin{equation} \mathbf{\Lambda}_ \mathfrak{a}(\omega) \simeq \mathbf{\Lambda}_ \mathfrak{a}^{(2)}(\omega) : = \lambda_{ \mathfrak{a}, 0} + \lambda_{ \mathfrak{a}, 0} \, \omega^2, \end{equation} $ | (162) |
where according to lemma 5.5, we have for the Dirichlet problem (and
$ \begin{equation} \lambda_{ \mathfrak{d}, 0} = 1-\Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle^{-1} , \quad \lambda_{ \mathfrak{d}, 2} = -\frac{1}{3} \; \left( 1 + \Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle+\Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle^2\right)\left( \Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle^2-\big \langle \boldsymbol{\mu}\boldsymbol{\alpha} \big \rangle\right)^{-1} , \end{equation} $ | (163) |
while for the Neumann case (which means that
$ \begin{equation} \lambda_{ \mathfrak{n}, 0} = 0, \qquad \lambda_{ \mathfrak{n}, 2} = -\left(1-\big \langle \boldsymbol{\mu}\boldsymbol{\alpha} \big \rangle\right)^{-1}. \end{equation} $ | (164) |
Remark 6.1. We will refer to these approximations as to the second-order conditions, because they are constructed using the first three terms of the expansion of the symbol of the DtN in
Note that one has the following sign properties:
$ \begin{equation} \lambda_{ \mathfrak{a}, 0} \geqslant 0, \quad \lambda_{ \mathfrak{a}, 2} \leqslant 0, \quad \mathfrak{a} = \mathfrak{d}, \mathfrak{n}. \end{equation} $ | (165) |
Then, (162) combined with (159), suggests the following approximation of the operators
$ \begin{equation} \Lambda_{ \mathfrak{a}, \alpha_i \ell_{n, j}} \sim \Lambda_{ \mathfrak{a}, \alpha_i \ell_{n, j}}^{(2)} : = \lambda_{ \mathfrak{a}, 0} \; \alpha_i ^{-1} \; \ell_{n, j}^{-1} - \lambda_{ \mathfrak{a}, 2} \; \alpha_i \; \ell_{n, j}\, \partial_t^2. \end{equation} $ | (166) |
This leads to the following boundary value problem on the truncated tree
$ \begin{equation} \left\{ \begin{array}{l} \mu \, \partial^2_t u_{ \mathfrak{a}, 2}^n - \partial_s ( \mu \, \partial_s u_{ \mathfrak{a}, 2}^n) = 0, \; \; \mbox{on } \mathcal{T}^n \times \mathbb{R}^+, \\[6pt] \partial_s u_{ \mathfrak{a}, 2}^n + \sum\limits_{i = 0}^{p-1} \mu_i \, \Big( \lambda_{ \mathfrak{a}, 0} \; \alpha_i ^{-1} \; \ell_{n, j}^{-1} \, u_{ \mathfrak{a}, 2}^n - \lambda_{ \mathfrak{a}, 2} \; \alpha_i \; \ell_{n, j} \, \partial_t^2 u_{ \mathfrak{a}, 2}^n \Big) = 0, \\[15pt] \mbox{at $M_{n, j}$, }~ 0 \leqslant j \leqslant J(n), \\ {u^{n}_{ \mathfrak{a}, 2}(., 0) = \partial_t u^n_{ \mathfrak{a}, 2}(., 0) = 0, } \end{array} \right. \end{equation} $ | (167) |
completed by initial conditions and the Dirichlet condition at the entrance of the tree. The weak formulation of the above problem reads
$ \begin{equation} \tag{{\mathcal{P}_{ \mathfrak{a}, n}}} \left\lbrace \quad \begin{aligned} & \text{Find } u^n_{ \mathfrak{a}, 2}(\cdot, t) : [0, T] \rightarrow {\rm H}_\mu^1( \mathcal{T}^n) \; / \; u^n_{ \mathfrak{a}, 2}(M_\star, t) = f(t) \; \mbox{and}\\[6pt] & \frac{d^2}{dt^2} \int_{ \mathcal{T}^n} \mu \, u^n_{ \mathfrak{a}, 2}(\cdot, t) \, v + \int_{ \mathcal{T}^n} \mu \, \partial_s \, u_{ \mathfrak{a}, 2}(\cdot, t) \, \partial_s v \\ & - \big \langle \boldsymbol{\mu}\boldsymbol{\alpha} \big \rangle\, \lambda_{ \mathfrak{a}, 2} \frac{d^2}{dt^2} \sum\limits_{j = 0}^{J(n)}\; {\mu_{n, j}}\, \ell_{n, j} \; u_{ \mathfrak{a}, 2}^n(M_{n, j}, t) \, v(M_{n, j}) , \\[6pt] & +\Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle \; \lambda_{ \mathfrak{a}, 0} \; \sum\limits_{j = 0}^{J(n)} {\mu_{n, j}}\, \ell_{n, j}^{-1} \, u_{ \mathfrak{a}, 2}^n(M_{n, j}, t) \, v(M_{n, j}) = 0, \\[6pt] & \forall \; v \in V( \mathcal{T}^n) = \{ v \in {\rm H}_\mu^1( \mathcal{T}^n) \, / \, v(M_\star) = 0\}, \\ &{u^{n}_{ \mathfrak{a}, 2}(., 0) = \partial_t u^n_{ \mathfrak{a}, 2}(., 0) = 0.} \end{aligned} \right.~~~~(\mathcal{P}_ {\mathfrak{a}, n} ) \end{equation} $ |
Remark that, contrary to the case of the exact problems, where the distinction between Neumann and Dirichlet problems occurred in the variational spaces, for the approximate problems, the difference appears in the bilinear forms, via the coefficients
Our main theoretical result is the following stability result.
Theorem 6.2. Let
$ \begin{array}{lll} \mathcal{E}_{ \mathfrak{a}, 2}^n(t) & : = & \frac{1}{2} \int_{ \mathcal{T}^n} \mu \Big\lbrace |\partial_t u_{ \mathfrak{a}, 2}^n(\cdot, t)|^2 + |\partial_t u_ \mathfrak{a}(\cdot, t)|^2 \Big\rbrace\\[10pt] & - & \frac{1}{2}\big \langle \boldsymbol{\mu}\boldsymbol{\alpha} \big \rangle \lambda_{ \mathfrak{a}, 2} \; \sum\limits_{j = 0}^{J(n)}\; \ell_{n, j}\, \mu_{n, j}\, |\partial_t u_{ \mathfrak{a}, 2}^n(M_{n, j}, t)|^2\\[10pt] & + & \frac{1}{2} \Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle \, \lambda_{ \mathfrak{a}, 0} \;\sum\limits_{j = 0}^{J(n)} \; \ell_{n, j}^{-1} \, \mu_{n, j}\, |u_{ \mathfrak{a}, 2}^n(M_{n, j}, t) |^2 \end{array} $ |
is constant for
Proof. Again the existence and uniqueness result is a classical exercise on the theory of second order linear hyperbolic problems. The result about the energy is obtained in the usual way after multiplying the first equation of (167) by
In this section we validate the performance of our approximate conditions numerically on the example of the dyadic symmetric tree (cf. figure 4). Our goal is to check the influence of the order (1 or 2) of the absorbing boundary condition together with the influence of the truncation order
We shall not discuss in detail the method that we used for the discretization of the truncated problem since it is quite classical.
The spatial discretization is done on a uniform spatial mesh with step size
For the time discretization, we use an explicit scheme coupled with an implicit discretization of the boundary terms. In particular, given a time step
$ {\partial_t^2 u_{ \mathfrak{a}, 2}(\cdot, t^k) \sim \frac{ u_{ \mathfrak{a}, 2}^{n, k} - 2 \, u_{ \mathfrak{a}, 2}^{n, k} + u_{ \mathfrak{a}, 2}^{n, k}}{\Delta t^2}, } $ |
the volumic term in the stiffness bilinear form of (
$ \int_{ \mathcal{T}^n} \mu \, \partial_s \, u_{ \mathfrak{a}, 2}(\cdot, t^k) \, \partial_s v \; \sim \; \int_{ \mathcal{T}^n} \mu \, \partial_s \, u_{ \mathfrak{a}, 2}^{n, k} \, \partial_s v, $ |
and, finally, the boundary
$ \begin{array}{l} \Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle\; \lambda_{ \mathfrak{a}, 0} \;\sum\limits_{j = 0}^{J(n)}\, \mu_{n, j} \, \ell_{n, j}^{-1}\; u_{ \mathfrak{a}, 2}^ n(M_{n, j}, t^k) \, v(M_{n, j}) \\ \quad \sim \Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle\; \lambda_{ \mathfrak{a}, 0} \; \sum\limits_{j = 0}^{J(n)} \Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle \;\mu_{n, j} \ell_{n, j}^{-1} \; \Big(\frac{ u_{ \mathfrak{a}, 2}^{n, k} + 2 \, u_{ \mathfrak{a}, 2}^{n, k} + u_{ \mathfrak{a}, 2}^{n, k}}{4}\Big) (M_{n, j}) \, v(M_{n, j}). \end{array} $ |
As a consequence, the resulting numerical scheme is stable under the classical CFL condition
We present the numerical simulations for the scattering problem with the Dirichlet condition at infinity
$ \begin{equation} \left\lbrace \begin{array}{ll} \mu \partial_t^2 u(s, t) - \partial_s \big( \mu \partial_s u \big)(s, t) = 0, & (s, t) \in \mathcal{T}\times \mathbb{R}^+, \\[8pt] u(s, 0) = f(s), & s \in \mathcal{T}, \\[8pt] \partial_t u(s, 0) = g(s), & s \in \mathcal{T}, \\[8pt] \partial_t u(M_\star, t) + \partial_s u(M_\star, t) = 0, & t \in \mathbb{R}^+ , \end{array} \right. \end{equation} $ | (168) |
on the dyadic tree with the parameters
$ \alpha_0 = \alpha_1 = \alpha = \mu_0 = \mu_1 = 0.6, $ |
with the length of the first branch
$ \begin{equation*} f(s) = \exp\big(-30(s-1)^2\big), \quad g(s) = - \partial_s f(s). \end{equation*} $ |
According to figure 8, the smallest positive pole of
$ \omega_ \mathfrak{d}^0 \approx 1.37, $ |
Thus we cannot expect a good approximation of
Due to the choice of the source term, for time
$ \widehat{f}(\omega) = \exp \left(-\frac{\omega^2}{120}\right). $ |
We observe in particular that for
$ |\omega| \geqslant \omega_{\text{cut}} = 30, \quad |\widehat{f}(\omega)| \leqslant e^{-7.5} \lesssim 6.10^{-4}. $ |
We perform the simulations with this source. To truncate the tree (find
$ \begin{equation} \ell\alpha^{n+1} \omega_{\text{cut}} < \omega_ \mathfrak{d}^0 \quad \Longrightarrow \quad n > \frac{\ln \omega_ \mathfrak{d}^0 - \ln \left(\omega_{\text{cut}}\ell\right)}{\ln \alpha}-1, \quad \mbox{which gives } n \geqslant 7. \end{equation} $ | (169) |
In what follows, we compare the reference solution, computed on the tree
● the solution
● the solution
● the solution
In figure 11 we plot these computed solutions at the middle of the root branch of the tree
These qualitative results are quantified by table 1 and figure 12 where we computed the
Number of generations | Dirichlet condition | First order condition | Second order condition | Gain with first order | Gain with second order |
| | | | 1.34 | 3.05 |
| | | | 1.80 | 7.35 |
| | | | 2.89 | 15.83 |
| | | | 4.53 | 30.5 |
| | | | 7.47 | 59.9 |
In figure 12 we demonstrate that the convergence of the absorbing boundary conditions with respect to
The contributions of this work are of both theoretical and numerical nature. First of all, from the theoretical point of view, we have presented an extensive analysis of the properties of a weighted wave equation in an infinite compact tree with self-similar endings. One particularly tricky question is the treatment of the boundary conditions (Neumann or Dirichlet) at 'infinity', the understanding of which requires a deep analysis of particular weighted Sobolev spaces on compact fractal
The authors are grateful to Konstantin Pankrashkin (University of Paris-Sud) for fruitful discussions.
The first case of the lemma is obvious. First of all, it is not difficult to verify by induction that:
$ \begin{equation} x_{n} \leqslant \gamma^n \; x_0+\sum\limits_{\ell = 0}^{n-1}\varepsilon_{\ell} \; \gamma^{n-1-\ell}. \end{equation} $ | (170) |
The first term in the rhs of (170) converges to zero because
$ { \left|\begin{array}{ll} \sum\limits_{\ell = 0}^{\lceil \frac{n}{2}\rceil } \; \varepsilon_{\ell} \; \gamma^{n-1-\ell} \leqslant \left(\lceil \frac{n}{2}\rceil+1\right) \; |\varepsilon|_\infty \; \gamma^{n-\lceil \frac{n}{2}\rceil-1} & \quad \longrightarrow 0 \quad (\mbox{since } \gamma < 1) \\ \sum\limits_{\ell = \lceil \frac{n}{2}\rceil+1}^{n-1}\varepsilon_{\ell} \; \gamma^{n-1-\ell} \leqslant \frac{1}{1-\gamma} \; \sup\limits_{\ell \geqslant \lceil n/2\rceil } \varepsilon_{\ell} & \quad \longrightarrow 0 \quad (\mbox{since } \varepsilon_{n} \rightarrow 0) \end{array} \right.} $ |
The second case is slightly trickier. First of all, we obtain by induction the equivalent of the inequality (170), that is:
$ \begin{equation} { x_{n} \leqslant x_0\prod\limits_{\ell = 0}^{n-1}\gamma_{\ell}+\sum\limits_{k = 0}^{n-1}\varepsilon_{k} \left(\prod\limits_{\ell = k+1}^{n-1}\gamma_{\ell}\right). } \end{equation} $ | (171) |
or equivalently, setting
$ \begin{equation} x_{n} \leqslant \Gamma_{n-1} \; x_0 +\sum\limits_{k = 0}^{n-1} \big( \, \Gamma_{n-1} / \, \Gamma_{k} \, \big) \, \varepsilon_{k}. \end{equation} $ | (172) |
The first term in the rhs of (171) converges to
$ \begin{align*} \gamma_\ell = 1-\frac{a}{\ell +1 } \quad \Longrightarrow \quad \Gamma_n : = \prod\limits_{\ell = 0}^n \gamma_{\ell} \; \leqslant \; \mathit{\boldsymbol{e}}^{-\sum\limits_{\ell = 0}^n\frac{a}{\ell+1}}\rightarrow 0, \qquad n \rightarrow \infty. \end{align*} $ |
On the other hand, for
$ \sum\limits_{k = 0}^{n-1} \big( \, \Gamma_{n-1} / \, \Gamma_{k} \, \big) \, \varepsilon_{k} \leqslant \sum\limits_{k = 0}^{N} \big( \, \Gamma_{n-1} / \, \Gamma_{k} \, \big) \, \varepsilon_{k} + \sum\limits_{k = N+1}^{+\infty} \, \varepsilon_{k}. $ |
Since
$ \ n \geqslant N_\varepsilon+1 \quad \Longrightarrow \quad \sum\limits_{k = 0}^{n-1} \big( \, \Gamma_{n-1} / \, \Gamma_{k} \, \big) \, \varepsilon_{k} \; \leqslant \; \Gamma_{n-1} \, S_\varepsilon + \frac{\varepsilon}{2}, \quad S_\varepsilon : = \sum\limits_{k = 0}^{N_\varepsilon} \frac{\varepsilon_{k}}{\Gamma_{k}}\; . $ |
Since
Step 1: proof that (124) implies the compactness. Let
Let us consider the sequence
$ \left. u_n^0\right|_{ \mathcal{T}^0} \to u^0 \equiv u_{\vert \mathcal{T}^0} \quad \text{in } L _\mu^2(\mathcal{T}^0). $ |
Similarly,
$ \left.u_n^1\right|_{ \mathcal{T}^1} \to u^1 \equiv u_{\vert \mathcal{T}^1} \quad \text{in } L_\mu^2(\mathcal{T}^1). $ |
By induction, we can thus build a double-indexed sequence
$ \begin{equation} n \rightarrow u_n^{k+1} \text{ is a subsequence of } n \rightarrow u_n^k \quad \mbox{and} \quad u_n^k \to u^k \equiv u_{\vert \mathcal{T}^k} \quad \text{in } {\rm L}_\mu^2( \mathcal{T}^k). \end{equation} $ | (173) |
Let us now define the diagonal subsequence of
$ \begin{equation} \widetilde{u}_n = u_n^n \in V, \end{equation} $ | (174) |
for which
$ \begin{equation} \widetilde{u}_n \to u \quad \text{in } {\rm L}_\mu^2( \mathcal{T}^k), \quad \forall k \geqslant 1. \end{equation} $ | (175) |
Indeed, for all
Next, let us demonstrate that
$ \begin{equation} \left\lVert {\widetilde{u}_n-u} \right\rVert_{{ {\rm L}_\mu^2( \mathcal{T})}}^2 = \left\lVert {\widetilde{u}_n-u} \right\rVert_{{ {\rm L}_\mu^2( \mathcal{T}^k)}}^2 + \left\lVert {\widetilde{u}_n-u} \right\rVert_{{ {\rm L}_\mu^2( \mathcal{T} \setminus \mathcal{T}^k)}}^2, \qquad k\in\mathbb{N}. \end{equation} $ | (176) |
It remains to apply (124) to
$ \begin{equation} \left\lVert {\widetilde{u}_n-u} \right\rVert_{{ {\rm L}_\mu^2( \mathcal{T})}}^2 \leqslant \left\lVert {\widetilde{u}_n-u} \right\rVert_{{ {\rm L}_\mu^2( \mathcal{T}^k)}}^2 + C \, \gamma_k^2, \qquad k\in\mathbb{N}. \end{equation} $ | (177) |
Let
$ n \geqslant N_k( \varepsilon) \quad \Longrightarrow \quad \left\lVert {\tilde{u}_n-u} \right\rVert_{{ {\rm L}_\mu^2( \mathcal{T})}}^2 < \varepsilon. $ |
Thus,
Step 2: proof that the compactness implies (124). Assume that the embedding of
$ \begin{equation} \gamma_n = \sup\limits_{u \in \mathcal{B}} \left\lVert {u} \right\rVert_{{ {\rm L}_\mu^2( \mathcal{T} \setminus \mathcal{T}^n)}}, \quad \mathcal{B} = \{ u \in V \; / \; \|u\|_{ {\rm H}_\mu^1( \mathcal{T})} = 1 \}. \end{equation} $ | (178) |
Note that
$ \begin{equation} \gamma_n = \left\lVert {u_n} \right\rVert_{{ {\rm L}_\mu^2( \mathcal{T} \setminus \mathcal{T}^n)}} \; . \end{equation} $ | (179) |
Again, by compactness of
$ \begin{equation} u_{ \varphi(n)} \longrightarrow u\in V, \quad \text{strongly in } {\rm L}_\mu^2( \mathcal{T}), \quad \text{weakly in } {\rm H}_\mu^1( \mathcal{T}). \end{equation} $ | (180) |
Writing
$ \begin{equation} \gamma_{ \varphi(n)} \leqslant \left\lVert {u} \right\rVert_{{ {\rm L}_\mu^2( \mathcal{T} \setminus \mathcal{T}^{ \varphi(n)})}} + \left\lVert {u_{ \varphi(n)} - u} \right\rVert_{{ {\rm L}_\mu^2( \mathcal{T})}}. \end{equation} $ | (181) |
This shows that
Let
$ \begin{align*} \mathbf{\Lambda}(0) = \Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle \big(1-\mathbf{\Lambda}(0)\big)\mathbf{\Lambda}(0), \;\text{ or, alternatively, }\; \mathbf{\Lambda}(0)\left(1-\Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle+\Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle\mathbf{\Lambda}(0)\right) = 0. \end{align*} $ |
The above equation has the following solutions:
● when
$ \begin{align} \mathbf{\Lambda}(0) = 0\, \text{ or }\, \mathbf{\Lambda}(0) = 1-\Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle^{-1}. \end{align} $ | (182) |
● when
Hence the first part of the lemma. To prove the rest of the lemma we will show that under the appropriate assumptions on
$ \begin{align} \mathbf{\Lambda}(\omega) = \sum\limits_{\ell = 0}^{\infty}\lambda_{2\ell} \, \omega^{2\ell}. \end{align} $ | (183) |
Inserting this expansion into (145), we obtain the following equation:
$ \begin{equation} \begin{array}{l} \omega \; \sum\limits_{n = 0}^\infty \frac{(-1)^n \omega^{2n+1}}{(2n+1){!}} + \Big( \sum\limits_{n = 0}^\infty \lambda_{2n} \, \omega^{2n} \Big) \Big( \sum\limits_{n = 0}^\infty \frac{(-1)^n \omega^{2n}}{(2n){!}} \Big) \\[15pt] \qquad = \sum\limits_{i = 0}^{p-1} \; \frac{\mu_i}{\alpha_i} \Big(\; \sum\limits_{n = 0}^\infty \frac{(-1)^n \omega^{2n}}{(2n){!}} \Big)\Big( \; \sum\limits_{n = 0}^\infty \lambda_{2n} \, \alpha_i^{2n}\, \omega^{2n} \Big) \\[15pt] \qquad - \sum\limits_{i = 0}^{p-1} \; \frac{\mu_i}{\alpha_i} \Big(\sum\limits_{n = 0}^\infty \lambda_{2n} \, \omega^{2n}\Big) \Big(\sum\limits_{n = 0}^\infty \frac{(-1)^n \omega^{2n}}{(2n+1){!}} \Big) \Big(\sum\limits_{n = 0}^\infty \lambda_{2n} \, \alpha_i^{2n}\, \omega^{2n} \Big). \end{array} \end{equation} $ | (184) |
Next, we wish to identify the terms in
For
$ \begin{align} L_{2m}&: = \frac{(-1)^{m-1}}{(2m-1){!}} +\sum\limits_{k = 0}^m \frac{(-1)^{m -k} \, \lambda_{2k}}{(2(m-k)){!}}\\ & = \frac{(-1)^{m-1}}{(2m-1){!}}+\lambda_{2m} + \sum\limits_{k = 0}^{m-1} \frac{(-1)^{m -k} \, \lambda_{2k}}{(2(m-k)){!}}. \end{align} $ | (185) |
As for the right hand side, we next observe that
$ \sum\limits_{n = 0}^\infty \frac{(-1)^n \omega^{2n}}{(2n){!}} - \Big(\sum\limits_{n = 0}^\infty \lambda_{2n} \omega^{2n}\Big) \Big(\sum\limits_{n = 0}^\infty \frac{(-1)^n \omega^{2n}}{(2n+1){!}} \Big) = \sum\limits_{n = 0}^\infty c_{2n} \; \omega^{2n} , $ |
where we have set
$ \begin{equation} c_{2n} : = \frac{(-1)^n}{(2n){!}} - \sum\limits_{q = 0}^n\frac{(-1)^{n-q} \lambda_{2q} }{(2n-2q+1){!}}. \end{equation} $ | (186) |
The coefficient in
$ R_{2m}: = \sum\limits_{k = 0}^m \Big(\sum\limits_{i = 0}^{p-1} \; \mu_i \, \alpha_i^{2k-1} \Big) \, \lambda_{2k} \, c_{2m-2k} \equiv \sum\limits_{k = 0}^m \eta_{2k} \, \lambda_{2k} \, c_{2m-2k}, $ |
where we defined
$ \begin{align*} \eta_{n}: = \sum\limits_{i = 0}^{p-1}\mu_{i}\alpha_{i}^{n-1}, \qquad n \geqslant 0. \end{align*} $ |
In other words, with (186),
$ \begin{align*} R_{2m} = \sum\limits_{k = 0}^m \; \eta_{2k} \, \lambda_{2k} \, \frac{(-1)^{m-k}}{(2m-2k){!}} - \sum\limits_{k = 0}^m \sum\limits_{q = 0}^{m-k} \eta_{2k} \, \lambda_{2k} \, \frac{(-1)^{m-k-q} \lambda_{2q} }{(2m-2k-2q+1){!}}. \end{align*} $ |
Isolating the terms involving
$ \begin{equation} \begin{array}{l} R_{2m} = \big(\eta_{2m} \, - (\eta_0 + \eta_{2m}) \lambda_0 \big) \, \lambda_{2m} + \sum\limits_{k = 0}^{m-1} \, \frac{(-1)^{m-k}}{(2m-2k){!}} \, \eta_{2k} \, \lambda_{2k} \\ \quad - \; \eta_0 \, \lambda_{0} \, \sum\limits_{q = 0}^{m-1} \frac{(-1)^{m-q} \, \lambda_{2q} }{(2m-2q+1){!}} - \sum\limits_{k = 1}^{m-1} \sum\limits_{q = 0}^{m-k} \eta_{2k} \, \lambda_{2k} \frac{(-1)^{m-k-q} \, \lambda_{2q} }{(2m-2k-2q+1){!}}. \end{array} \end{equation} $ | (187) |
With
$ \begin{align} \begin{split} A_{m}\lambda_{2m}& = \frac{(-1)^{m}}{(2m-1){!}}- \sum\limits_{k = 0}^{m-1} \frac{(-1)^{m -k} \, \lambda_{2k}}{(2(m-k)){!}} \\ &+\sum\limits_{k = 0}^{m-1} \, \frac{(-1)^{m-k}}{(2m-2k){!}} \, \eta_{2k} \, \lambda_{2k}- \; \eta_0 \, \lambda_{0} \, \sum\limits_{q = 0}^{m-1} \frac{(-1)^{m-q} \lambda_{2q} }{(2m-2q+1){!}} \\ & -\, \sum\limits_{k = 1}^{m-1} \sum\limits_{q = 0}^{m-k} \eta_{2k} \, \lambda_{2k} \frac{(-1)^{m-k-q} \lambda_{2q} }{(2m-2k-2q+1){!}}, \\[6pt] \end{split} \end{align} $ | (188) |
$ \begin{equation} \mbox{where } \quad A_m = 1\, - \eta_{2m} \, + (\eta_0 + \eta_{2m}) \lambda_0. \end{equation} $ | (199) |
Thus, to prove the uniqueness of an even solution analytic in the origin, it is sufficient to fix
Now it remains to consider the two cases of the statement of the lemma, see also (123):
● If
$ \lambda_0 \equiv \boldsymbol{\Lambda}(0) = 0. $ |
In this case, we get the following expression for (189):
$ \begin{align} A_m = 1\, -\eta_{2m}, \qquad m \geqslant 1. \end{align} $ | (190) |
Notice that
$ \begin{align} A_1 = 1-\big \langle \boldsymbol{\mu}\boldsymbol{\alpha} \big \rangle > 0 \quad \Longrightarrow \quad \mbox{$A_m > 0$ for all $m \geqslant 1$}, \end{align} $ | (191) |
hence the uniqueness of the solution with
● If
$ \lambda_0 \equiv \boldsymbol{\Lambda}(0) = 1-\Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle^{-1}. $ |
In this case (189) becomes
$ \begin{align} A_m = 1\, - \eta_{2m} \, + (\eta_0 + \eta_{2m})\, (1-\eta_0^{-1}) = \eta_0 - \eta_{2m}\, \eta_0^{-1}. \end{align} $ | (192) |
Like before, it is easy to see that
$ \begin{align} A_1 = \Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle-\big \langle \boldsymbol{\mu}\boldsymbol{\alpha} \big \rangle \, \Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle^{-1} = \Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle^{-1}\left(\Big \langle \frac{\boldsymbol{\mu}}{\boldsymbol{\alpha}} \Big \rangle^2-\big \langle \boldsymbol{\mu}\boldsymbol{\alpha} \big \rangle\right) > 0, \end{align} $ | (193) |
because
Hence
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Number of generations | Dirichlet condition | First order condition | Second order condition | Gain with first order | Gain with second order |
| | | | 1.34 | 3.05 |
| | | | 1.80 | 7.35 |
| | | | 2.89 | 15.83 |
| | | | 4.53 | 30.5 |
| | | | 7.47 | 59.9 |
Number of generations | Dirichlet condition | First order condition | Second order condition | Gain with first order | Gain with second order |
| | | | 1.34 | 3.05 |
| | | | 1.80 | 7.35 |
| | | | 2.89 | 15.83 |
| | | | 4.53 | 30.5 |
| | | | 7.47 | 59.9 |