
In this paper we consider a non-standard dynamical inverse problem for the wave equation on a metric tree graph. We assume that positive masses may be attached to the internal vertices of the graph. Another specific feature of our investigation is that we use only one boundary actuator and one boundary sensor, all other observations being internal. Using the Dirichlet-to-Neumann map (acting from one boundary vertex to one boundary and all internal vertices) we recover the topology and geometry of the graph, the coefficients of the equations and the masses at the vertices.
Citation: Sergei Avdonin, Julian Edward. An inverse problem for quantum trees with observations at interior vertices[J]. Networks and Heterogeneous Media, 2021, 16(2): 317-339. doi: 10.3934/nhm.2021008
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In this paper we consider a non-standard dynamical inverse problem for the wave equation on a metric tree graph. We assume that positive masses may be attached to the internal vertices of the graph. Another specific feature of our investigation is that we use only one boundary actuator and one boundary sensor, all other observations being internal. Using the Dirichlet-to-Neumann map (acting from one boundary vertex to one boundary and all internal vertices) we recover the topology and geometry of the graph, the coefficients of the equations and the masses at the vertices.
This paper concerns inverse problems for differential equations on quantum graphs. Under quantum graphs or differential equation networks (DENs) we understand differential operators on geometric graphs coupled by certain vertex matching conditions. Network-like structures play a fundamental role in many problems of science and engineering. The range for the applications of DENs is enormous. Here is a list of a few.
–Structural Health Monitoring. {DENs, classically, arise in the study of stability, health, and oscillations of flexible structures that are made of strings, beams, cables}, and struts. Analysis of these networks involve DENs associated with heat, wave, or beam equations whose parameters inform the state of the structure, see, e.g., [44].
–Water, Electricity, Gas, and Traffic Networks. An important example of DENs is the Saint-Venant system of equations, which model hydraulic networks for water supply and irrigation, see, e.g., [33]. Other important examples of DENs include the telegrapher equation for modeling electric networks, see, e.g., [3], the isothermal Euler equations for describing the gas flow through pipelines, see, e.g., [21], and the Aw-Rascle equations for describing road traffic dynamics, see e.g., [29].
–Nanoelectronics and Quantum Computing. Mesoscopic quasi-one-dimensional structures such as quantum, atomic, and molecular wires are the subject of extensive experimental and theoretical studies, see, e.g., [37], the collection of papers in [38,39,40]. The simplest model describing conduction in quantum wires is the Schrödinger operator on a planar graph. For similar models appear in nanoelectronics, high-temperature superconductors, quantum computing, and studies of quantum chaos, see, e.g., [42,41,45].
–Material Science. DENs arise in analyzing hierarchical materials like ceramic and metallic foams, percolation networks, carbon and graphene nano-tubes, and gra-phene ribbons, see, e.g., [1,46,47].
–Biology. Challenging problems involving ordinary and partial differential equations on graphs arise in signal propagation in dendritic trees, particle dispersal in respiratory systems, species persistence, and biochemical diffusion in delta river systems, see, e.g., [7,24,48].
Quantum graph theory gives rise to numerous challenging problems related to many areas of mathematics from combinatoric graph theory to PDE and spectral theories. A number of surveys and collections of papers on quantum graphs appeared in previous years; we refer to the monograph by Berkolaiko and Kuchment, [25], for a complete reference list. The inverse theory of network-like structures is an important part of a rapidly developing area of applied mathematics---analysis on graphs. Being tremendously important for all aforementioned applications these theories have not been, however, sufficiently developed. To date, there are relatively few results related to inverse problems on graphs, and almost exclusively they concern trees, i.e. graphs without cycles.
The first question to be asked when studying inverse problems is how to establish the uniqueness result, i.e. to characterize spectral, or scattering, or dynamical data ensuring uniques solution of the inverse problem. It was shown that inverse boundary spectral and scattering problems for differential equations on graphs with cycles do not have in general a unique solution [41,34,43]. The results on stable identification are known only for trees, and only for the case of boundary inputs (controls) and observations. It was proved that a DEN is identifiable if the actuators and sensors are placed at all or all but one boundary vertices.
There are two groups of uniqueness results in this direction: for trees with a priori known topology and lengths of the edges [28,49,32] and for trees with unknown topology [22,23,11,13]. The most significant result of the last two cited papers is developing a constructive and robust procedure for the recovery tree's parameters, which became known as the leaf peeling method. This method was extended to boundary inverse problems for various types of PDEs on trees in a series of our subsequent papers [7,15,20].
The boundary control method in inverse theory demonstrates [11,23] that inverse (identification) problems for DENs are closely related to control and observation problems for PDEs on graphs. The latter problems were studied in numerous papers, see, e.g. [5,16,30,35,44,50] and references therein.
In this paper, we solve a non-standard dynamical inverse problem for the wave equation on a metric tree graph. Let
The graph
We will assume that for each internal vertex
utt−uxx+qu=0, (x,t)∈(Ω∖V)×[0,T], | (1) |
u|t=0=ut|t=0=0, x∈Ω | (2) |
ui(vk,t)−uj(vk,t)=0, i,j∈J(vk), vk∈V∖Γ, t∈[0,T], | (3) |
∑j∈J(vk)∂uj(vk,t)=Mkutt(vk,t), vk∈V∖Γ, t∈[0,T], | (4) |
u(γ0,t)=f(t), t∈[0,T], | (5) |
u(γk,t)=0, k=1,…,m, t∈[0,T]. | (6) |
Here
Inverse Problem 1. Assume an observer knows the topology of the tree, i.e. the number of boundary vertices and interior vertices, the adjacency relations for this tree, i.e. for each pair of vertices, whether or not there is an edge joining them. Assume the observer also knows the boundary condition (5), and that (6) holds at the other boundary vertices. The unknowns are the lengths
Let
(R01f)(t):=∂uf1(γ0,t). | (7) |
Physically, this corresponds to applying a Dirichlet control and placing a tension sensor, both at
Theorem 1.1. From operator
The proof of this, appearing in Section 2, is an adaptation of an argument well known for the massless case,
We now define the other measurements required for the inverse problem. For interior vertex
(Rkjf)(t):=∂ufj(vk,t), j=2,...,Υk−1. | (8) |
We remark in passing that because the control and sensors are at different places, Theorem 1.1 does not apply. We will show that it is not required to measure
Let
Theorem 1.2. Assume
Placement of internal sensors has been considered in the engineering and computer science literature, see, e.g. [26,27]. We are unaware of any mathematical works treating the inverse problem on general tree graphs with measurements at the internal vertices, except for [8] where the interior vertices are assumed to satisfy delta-prime matching conditions instead of (3), (4). Internal measurements might have advantages in situations where some boundary vertices are inaccessible. In future work, we will study inverse problems of graphs with cycles, in which case both boundary and internal observations appear to be necessary. For a discussion on inverse problems for graphs with cycles see [4] and references therein.
We briefly mention some of the ideas used in the proof of Theorem 1.2. Denote by
The iterative nature of our solution actually allows us to solve what at first glance seems to be a much harder inverse problem.
Inverse Problem 2. Assume an observer knows
To solve this inverse problem, for
Theorem 1.3. Assume
We remark that in the theorem,
In an engineering setting, Inverse Problem 2 could be solved using the following process. One begins with only the control and sensor at
We now compare our paper with the literature. All papers referred to in this paragraph assume all controls and measurements take place at boundary vertices. In [11], the authors consider trees with no masses, and assume that controls and measurements are placed at
In the present paper we develop a new version of the dynamical leaf peeling method. A special feature of our paper is that we use only one control together internal observations. This may be useful in some physical settings where some or most boundary points are inaccessible, or where use of more than one control might be difficult. The extension of dynamical leafing peeling to systems with attached masses, for which the underlying analysis is more complicated than in the mass-free setting, should also be of interest. Another potential advantage of the method presented here is that we recover all parameters of the graphs, including its topology, from the
In this section, we prove well-posedness of our IBVP for a star shape graph. We also derive representations of both the solution and the Schwartz kernel of the components of the response operator. The representations will be used in Section 3 to solve the inverse problem. We then indicate how these results can be extended from star graphs to arbitrary trees.
In what follows, it will convenient to denote
Fn={f∈Hn(R): f(t)=0 if t≤0}, |
where
dndtnHn=H; |
at times we will use
Consider a star shaped graph with edges
Recall the notation
∂2u∂t2−∂2u∂x2+qu=0, x∈ej, j=1,...,N, t∈×[0,T], | (9) |
u|t=0=ut|t=0=0, | (10) |
u(0,t)=ui(0,t)=uj(0,t), i≠j, t∈[0,T], | (11) |
N∑j=1∂uj(0,t)=M∂2u∂t2(0,t), t∈[0,T], | (12) |
u1(ℓ1,t)=f(t), t∈[0,T], | (13) |
uj(ℓj,t)=0, j=2,...,N, t∈[0,T]. | (14) |
Let
g(t)=uf(0,t). | (15) |
For (10), it is standard that the waves have unit speed of propagation on the interval, so
We will use a representation of
Define
{∂w2∂t2(x,s)−∂w2∂x2(x,s)+qj(x)w(x,s)=0, 0<x<s<∞,w(0,s)=0, w(x,x)=−12∫x0qj(η)dη, x>0. |
A proof of solvability of the Goursat problem can be found in [14].
Consider the IBVP on the interval
˜utt−˜uxx+qj(x)˜u=0, 0<x<ℓj, t∈(0,T), | (16) |
˜u(x,0)=˜ut(x,0)=0, 0<x<ℓj,˜u(0,t)=h(t),˜u(ℓj,t)=0, t>0. | (17) |
Then the solution to (16)-(17) on
˜uh(x,t)=∑n≥0: 0≤2nℓj+x≤t(h(t−2nℓj−x)+∫t2nℓj+xwj(2nℓj+x,s)h(t−s)ds) |
−∑n≥1: 0≤2nℓj−x≤t(h(t−2nℓj+x)+∫t2nℓj−xwj(2nℓj−x,s)h(t−s)ds). | (18) |
Setting
Define the "reduced response operator" on
(˜R0jh)(t)=∂˜uj(0,t), t∈[0,T], |
associated to the IBVP (16)-(17). From (18) we immediately obtain:
Lemma 2.1. For
(˜R0jh)(t)=∫t0˜R0j(s)h(t−s)ds, |
with
˜R0j(s)=−δ′(s)−2∑n≥1δ′(s−2nℓj)−2∑n≥1wj(2nℓj,2nℓj)δ(s−2nℓj)+˜r0j(s). |
and
In what follows we will refer to
It will be useful also to represent the solution of a wave equation on an interval when the control is on the right end. Thus consider the IBVP:
vtt−vxx+q1(x)v=0, 0<x<ℓ1, t>0,v(x,0)=vt(x,0)=0, 0<x<ℓ1,v(0,t)=0,v(ℓ1,t)=f(t), t>0. | (19) |
Set
{∂ω2∂t2(x,s)−∂ω2∂x2(x,s)+˜qj(x)ω(x,s)=0, 0<x<s,ω(0,s)=0, ω(x,x)=−12∫x0˜qj(η)dη, x<ℓj. |
By changing coordinates in (18), we get
vf(x,t)= f(t−ℓ1+x)+∫tℓ1−xω1(ℓ1−x,s)f(t−s)ds−f(t−ℓ1−x)−∫tℓ1+xω1(ℓ1+x,s)f(t−s)ds+f(t−3ℓ1+x)+∫t3ℓ1−xω1(3ℓ1−x,s)f(t−s)ds−f(t−3ℓ1−x)−∫t3ℓ1+xω1(3ℓ1+x,s)f(t−s)ds… | (20) |
We begin section by proving an analog of Lemma 2.1 for
Lemma 2.2. The response function for
R01(s)=r01(s)+∑n≥0(anδ′(s−2nℓ1)+bnδ(s−2nℓ1)). |
Here
a1=−2and,b1=−2ω1(2ℓ1,2ℓ1)+4/M. | (21) |
If
Proof. We see that on
uf(x,t)=vf(x,t)+˜ug(x,t). | (22) |
Thus by (18) with
(R01)f(t)=−ufx(ℓ1,t)=−vfx(ℓ1,t)−˜ugx(ℓ1,t) |
=−f′(t)−2∑n≥1f′(t−2nℓ1)−2∑n≥1ω1(2nℓ1,2nℓ1)f(t−2nℓ1)+∫t0∂ω1(0,s)f(t−s)ds |
+2∑n≥1∫t2nℓ1∂ω1(2nℓ1,s)f(t−s)ds+2∑n≥0g′(t−(2n+1)ℓ1) |
+2∑n≥0w1((2n+1)ℓ1,(2n+1)ℓ1)g(t−(2n+1)ℓ1) |
−2∑n≥0∫t(2n+1)ℓ1∂w1((2n+1)ℓ1,s)g(t−s)ds. | (23) |
Next, we study the structure of
Mg″(t)+Ng′(t)=2∑n≥0f′(t−(2n+1)ℓ1)+2∑n≥0ω1((2n+1)ℓl,(2n+1)ℓ1)f(t−(2n+1)ℓ1)−2∑n≥0∫t(2n+1)ℓ1∂ω1((2n+1)ℓ1,s)f(t−s)ds+N∑j=1∫t0∂wj(0,s)g(t−s)ds+2N∑j=1∑n≥1[−g′(t−2nℓj)−wj(2nℓj,2nℓj)g(t−2nℓj)+2∫t2nℓj∂wj(2nℓj,s)g(t−s)ds]. | (24) |
In what follows, we will assume
up(x,t)=(p∗uδ)(x,t)=∫ts=0p(t−s)uδ(x,s)ds. |
As a consequence,
For
g(t)=∑m≥0cmH(t−(2m+1)ℓ1)+tmH1(t−βm)+˜a(t−ℓ1). | (25) |
Here
We will now justify the claim. Substituting (25) into (24), and matching the
M∑m≥0cmδ′(t−(2m+1)l1)=2∑n≥0δ′(t−(2n+1)ℓ1), t∈[0,T]. | (26) |
We conclude that
M(∑m≥0tmδ(t−βm))=−Nc(∑m≥0δ(t−(2m+1)l1))−2cN∑j=1∑n≥1∑m≥0δ(t−2nℓj−(2m+1)ℓ1)+2∑n≥0ω1((2n+1)ℓ1,(2n+1)ℓ1)δ(t−(2n+1)ℓ1). | (27) |
We can solve for
{2nℓj+(2m+1)ℓ1: m≥0; n≥0; j=1,2,...,N}. |
Next, for any given
β0=ℓ1, t0=2(ω1(ℓ1,ℓ1)/M−N/M2). | (28) |
For larger
Case 1.
Case 2. For some positive integer
Case 3. For some positive integer
tm=−NcM−2LcM+2ω1((2nl+1)ℓ1,(2nl+1)ℓ1)M. |
Accounting for these cases, we thus solve for
Next, we solve for
M˜a″(t−ℓ1)+N˜a′(t−ℓ1)+2N∑j=1∑n≥1˜a′(t−ℓ1−2nℓj) |
=˜b0(t)+N∑j=1∫t0∂wj(0,s)˜a(t−s−ℓ1)ds |
+2N∑j=1∑n≥1[−wj(2nℓj,2nℓj)˜a(t−2nℓj−ℓ1)+2∫t2nℓj∂wj(2nℓj,s)˜a(t−s−ℓ1)ds], |
where
M˜a′(t−ℓ1)+N˜a(t−ℓ1)+2N∑j=1∑n≥1˜a(t−2nℓj−ℓ1) |
=˜b1(t−ℓ1)+∫ts=ℓ1(N∑j=1∫s0∂wj(0,r)˜a(s−r−ℓ1)dr |
+2N∑j=1∑n≥1−wj(2nℓj,2nℓj)˜a(s−2nℓj−ℓ1) |
+2∫s2nℓj∂wj(2nℓj,r)˜a(s−r−ℓ1)dr) ds, | (29) |
with
˜a(w−ℓ1) |
=1M∫wt=0(eN(w−t)/M∫ts=ℓ1∫sr=0(N∑j=1∂wj(0,r))˜a(s−r−ℓ1)) dr ds dt+˜b2(w−ℓ1), |
with
˜a(w−ℓ1) |
=1M∫wt=0(eN(w−t)/M∫ts=ℓ1∫sr=0(N∑j=1∂wj(0,r))˜a(s−r−ℓ1)) dr ds dt+˜b2(w−ℓ1). |
Again we solve this Volterra equation to determine
Lemma 2.3. Let
g(t)=∫t0A(s)f(t−s)ds, | (30) |
where
A(t)=∑m≥0amH(t−(2m+1)ℓ1)+tmH1(t−νm)+˜a(t−ℓ1). | (31) |
Furthermore,
Proof. Since
Proposition 1. Let
Proof. One can determine
F(t)=c1δ′(t−2ℓ1)+c2δ(t−2ℓ1)+2H(t−2ℓ1)(t0+w1(ℓ1,ℓ1)ψ)+G(t), |
where
Lemma 2.4. Label the central vertex
R1j(s)=r1j(s)+∑n≥1(bnδ(s−βn)+rnH0(s−βn)). |
Here
Proof. the lemma follows immediately from Lemmas 2.2 and 2.3; the details are left to the reader.
We can adapt the methods of the previous subsection to the case the internal vertex is massless (also see [11] for a proof of the results below). Here we will only mention the modifications necessary. In Subsection 2.3, the argument carries through word for word until (24), which becomes a first order integral-differential equation, since
Lemma 2.5. Let
g(t)=∫t0A(s)f(t−s)ds, | (32) |
where
A(s)=∑k≥0akδ(t−ξk)+bkH(t−ξk)+˜a(t−ℓ1). | (33) |
Here
Inserting
Lemma 2.6. The response function for
R01(s)=r01(s)+∑n≥0(znδ′(s−ζn)+ynδ(s−ζn)). |
Here
Proposition 2. Let
The reader is referred to [11] for a proof of this.
We conclude this section with the following lemma, whose proof is similar to that of Lemma 2.4 and is left to the reader.
Lemma 2.7. Label the central vertex
R1j(s)=r1j(s)+∑n≥1(anδ′(s−βn)+bnδ(s−βn)). |
Here
This lemma should be compared with Lemma 2.4. For this lemma, the lead singularity is of the form
In this subsection, we extend some of the previous results to trees. The extensions will be used in Section 3 in solving the inverse problem on trees.
We begin by discussing the wellposed of the system (1)-(6). Let
Theorem 2.8. If
The proof of the theorem is based on the analysis of the waves incoming to, transmitted through and reflected from an interior vertex, and the waves reflected from the boundary vertices. The details are left to the reader; also see [6].
Theorem 2.9. Let
a) The response function for
R01(s)=r01(s)+∑n≥1(anδ′(s−ζn)+bnδ(s−ζn)). |
Here
b) from
Proof. We sketch this proof, leaving the details to the reader. The key point is the waves propagate at unit speed. Hence for
For an internal vertex
Lemma 2.10. Let
Rkj(s)=rkj(s)+∑n≥1(bnHK−2(s−βn)+rnHK−1(s−βn)). |
Here
Proof. The proof follows from the proof of Lemma 2.2, together with the transmission and reflection properties of waves at interior vertices, and reflection properties at boundary vertices. The details are left to the reader; see also [11]- where, however, the formula analogous to Lemma 2.2 should have the terms of the form
In this section we prove Theorem 1.2. In the first subsection, we establish some notation, and give an outline of the solution method, Steps 1-3. Then in Subsection 3.2, we present the technical heart of our argument, using the equation
˜R12∗A=R12 |
and the expansions for
We begin by establishing some notation. Let
We will define an associated response operator as follows. Suppose
∂2w∂t2−∂2w∂x2+qw=0, x∈Ωkj∖Vkj, | (34) |
wi(vl,t)=wj(vl,t), i,j∈J(vl), vl∈(Vkj∖Γ)∖{vk}, | (35) |
∑i∈J(vl)∂wi(vl,t)=Ml∂2w∂t2(vl,t), vl∈(Vkj∖Γ)∖{vk}, | (36) |
w(vk,t)=p(t), | (37) |
w(γl,t)=0, γl∈Γ∩Vkj, | (38) |
and initial conditions
w|t=0=wt|t=0=0. | (39) |
Then we define an associated reduced response operator
(˜Rkjp)(t)=∂wpj(vk,t), wpj:=wp|ekj, |
with associated response function
We have the following important result is essentially a restatement of Theorem 2.9.
Theorem 3.1. For vertex
In this section we will present an iterative method to determine the operator
Step 1. Let
Step 2. Consider
Step 3. In Subsection 3.3, we will solve for determine
Proposition 3. The function
The rest of this subsection will be devoted to proving this proposition.
Let
utt−uxx+q1u=0, x∈(0,ℓ1), t>0,u(ℓ1,t)=f(t), t>0,ux(ℓ1,t)=(R01f)(t), t>0,u(x,0)=0, x∈(0,ℓ1). |
Since the operator
∫t0˜R12(s)(f∗A)(t−s)ds=∂uf2(v1,t)=A(t)=∫t0R12(s)f(t−s)ds, ∀f∈L2(0,T). |
This follows from the definition of the response operators for any
˜R12∗A=R12. | (40) |
Below, we will use (40) to determine
Since
˜R12(s)=˜r12(s)+∑p≥0zpδ′(s−ζp)+∑l≥0ylδ(s−ηl). | (41) |
Here
We now must separately consider the cases
Case A.
In what follows, it will be convenient to extend
R12(s)=r12(s)+∑n≥1[bnδ(s−βn)+rnH(s−βn)], r12|s∈(0,β1)=0, β1=ℓ1; | (42) |
A(s)=˜a(s−ℓ1)+∑k≥1[akH(s−αk)+tkH1(s−νk)],α1=ν1=ℓ1,a1=2M1. | (43) |
Here
Lemma 3.2. The sets
Proof. By (40), (41), (42), and (43), we get by matching delta functions:
∑n≥1bnδ(t−βn)=∑p≥1∑k≥1akzpδ(t−ζp−αk). | (44) |
Step 1. We solve for
{b1,β1,a1,α1}⟹{z1,ζ1}. |
Step 2. We solve for
Case 1.
In this case, we must have
ζ2=α1−β2, z2=b2/a1. |
Case 2a.
ζ2=α1−β2, z2=(b2−a2z1)/a1. |
Case 2b.
Again there are three cases:
Case 2bⅰ.
Case 2bⅱ.
Case 2bⅲ.
Repeating this procedure as necessary, say for a total of
{bk,βk,ak,αk}N2k=1⟹{zk,ζk}2k=1. |
We must have
Step
Iterating the procedure above, suppose for
{bk,βk,ak,αk}Npk=1⟹{zk,ζk}pk=1. |
Here
We can again distinguish three cases:
Case 1.
Case 2. There exists an integer
β(Np+1)=ζj1+αj1=...=ζjQ+αjQ. |
Note that all the numbers
Case 2ⅰ.
b(Np+1)=zp+1a1+zj1aj1+...+zjQajQ. |
We thus solve for
Case 2ⅱ.
Repeating the reasoning in Case 2ii as often as necessary, we will eventually solve for
{bk,βk,ak,αk}Np+1k=1⟹{zk,ζk}p+1k=1. |
Hence we can solve for
Lemma 3.3. The sets
Proof. We identify the Heavyside functions in (40). By (41), (42), and (43), we get
∑n≥1rnH(t−βn)ds−∑k≥1∑p≥1zptkH(t−ζp−νk)=∑k≥1∑l≥1akylH(t−ηl−αk). |
Since the left hand side is known, we can argue as in Lemma 3.2 to solve for
Lemma 3.4. The function
Proof. We solve for
ddt(˜R12∗A)=ddtR12. |
Hence by (41), (42), and (43), we calculate
C(t)=∫t0˜r(s)˜a′(t−s−ℓ1)ds+∑k≥1ak˜r(t−αk)+∑k≥1tk∫t−νk0˜r(s)ds. |
We set
C(z)=∫z0˜r(s)(˜a′(z−s)+t1)ds+a1˜r(z)+∑k≥2ak˜r(z+α1−αk) |
+∑k≥2tk∫z+ν1−νk0˜r(s)ds. | (45) |
Setting
α:=min(mink≥0(αk+1−αk),mink≥0(νk+1−νk)). |
Since we will be choosing finite
1) For
2) Suppose we have solved for
z∈((n−1)α,nα), |
and identify terms in (45) that we already know. We have for
∫z+ν1−νk0˜r(s)ds=C(z)+∫z+ν1−νk(n−1)α˜r(s)ds. |
For
z+ν1−νk−s≤nα+ν1−νk−s≤nα−(k−1)α−s≤(n−1)α−s≤0, |
so
∑k≥2tk∫z+ν1−νk0˜r(s)ds=C(z). | (46) |
Similarly, for
∑k≥2ak˜r(z+α1−αk)=C(z). | (47) |
Combining (46) and (47) with (45), we get
C(z)=a1˜r(z)+∫z0˜r(s)(˜a′(z−s)+t1) ds. |
This is a Volterra equation of the second kind, and thus we solve for
[(n−1)α,nα). |
Iterating this argument finitely many times, we will have solved for
Case B:
In this case, we must replace (42), (43) by
R12(s)=r12(s)+∑n≥1bnδ′(t−βn)+rnδ(s−βn), r12|s∈(0,β1)=0, β1=ℓ1,A(s)=˜a(s−ℓ2)+∑k≥1akδ(s−αk)+tkH(s−νk), α1=ν1=ℓ1. |
with piecewise continuous
Careful reading of Steps 2, 3 shows that we can choose any
The purpose of this subsection is to determine
∫t0˜R22(s) (B∗f)(t−s)ds=∫t0R22(s)f(t−s)ds. |
Of course
ytt−yxx+q2y=0, x∈(0,ℓ2), t>0y(ℓ2,t)=a(t), t>0yx(ℓ2,t)=(R12∗δ)(t), t>0y(x,0)=0, x∈(0,ℓ2). |
Since
The rest of the argument here is a straightforward adaptation of the previous subsection. The details are left to the reader.
We would like to thank the referee for his many suggestions that improved the exposition in this paper.
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