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Research article Special Issues

Shifted-Legendre orthonormal method for high-dimensional heat conduction equations

  • Received: 01 October 2021 Revised: 02 March 2022 Accepted: 07 March 2022 Published: 14 March 2022
  • MSC : 65M12, 65N12

  • In this paper, a numerical alogorthm for solving high-dimensional heat conduction equations is proposed. Based on Shifted-Legendre orthonormal polynomial and εbest approximate solution, we extend the algorithm from low-dimensional space to high-dimensional space, and prove the convergence of the algorithm. Compared with other numerical methods, the proposed algorithm has the advantages of easy expansion and high convergence order, and we prove that the algorithm has α-Order convergence. The validity and accuracy of this method are verified by some numerical experiments.

    Citation: Liangcai Mei, Boying Wu, Yingzhen Lin. Shifted-Legendre orthonormal method for high-dimensional heat conduction equations[J]. AIMS Mathematics, 2022, 7(5): 9463-9478. doi: 10.3934/math.2022525

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  • In this paper, a numerical alogorthm for solving high-dimensional heat conduction equations is proposed. Based on Shifted-Legendre orthonormal polynomial and εbest approximate solution, we extend the algorithm from low-dimensional space to high-dimensional space, and prove the convergence of the algorithm. Compared with other numerical methods, the proposed algorithm has the advantages of easy expansion and high convergence order, and we prove that the algorithm has α-Order convergence. The validity and accuracy of this method are verified by some numerical experiments.



    Let ΩRn (n2, possibly n=1971) be a bounded smooth domain and consider the problem

    {Δu=1inΩ,u=0,uν=conΩ, (1.1)

    where cR is a constant and ν denotes the outward unit normal to Ω. Imposing both Dirichlet and Neumann conditions on Ω makes the problem overdetermined so that, in general, (1.1) has no solution. In a celebrated paper published in 1971, Serrin [15] proved that if (1.1) admits a smooth solution, then Ω must necessarily be a ball. This paper has raised a great interest and, nowadays (half a century later), it has reached almost 600 citations in the Mathscinet. Serrin's original proof combines analytic arguments, such as the Maximum Principle and a refinement of Hopf's boundary Lemma, with geometric techniques such as the moving plane method inspired to the Alexandrov characterization of spheres [1,2]. Starting from [18], several different approaches have been devised as an alternative to Serrin's original proof, see [14] and the references therein for a fairly complete survey.

    The moving plane method has been fruitfully used in symmetry results for semilinear elliptic equations, see [4,10] for second order equations and [3] for higher order problems. Let BRn be the unit ball and consider the semilinear polyharmonic problem under Dirichlet boundary conditions:

    {(Δ)mu=f(u)in B,u=ur==m1urm1=0on B. (1.2)

    Here r=|x| denotes the radial variable and, hence, the outward normal direction to B. The following result, valid for second order equations with m=1, is a restatement of [10,Theorem 1], combined with deep remarks by Spruck [10,Remark 1].

    Theorem 1.1. ([4,10]) Let n2 and m=1. Assume that

    eitherfLiploc(R+;R)orfC(R+;R)is nondecreasing. (1.3)

    Then, every positive strong solution uW2,nloc(B)C(¯B) (u>0 in B) of (1.2) is radially symmetric (u=u(r)) and u(r)<0 for all r(0,1). Moreover, if

    fLiploc(R+;R)andf(0)0, (1.4)

    then every nonnegative and non-trivial strong solution uW2,nloc(B)C(¯B) (u0, u0 in B) of (1.2) is radially symmetric (u=u(r)) and u(r)<0 for all r(0,1).

    This result was extended in [3,Theorem 1] to the case m2, in the following form:

    Theorem 1.2. ([3]) Let n2 and m1. Assume that

    fC(R+;R)is nondecreasing withf(0)0. (1.5)

    Then, every nonnegative and non-trivial strong solution uHm0(B)L(B) (u0, u0 in B) of (1.2) is radially symmetric (u=u(r)) and u(r)<0 for all r(0,1).

    In its original version, Theorem 1.2 was stated by requiring that the solution u was strictly positive, u>0 in B. But this is not necessary since [9,Theorem 5.1] ensures that if u0 is nontrivial, then u>0 (recall that f(u)0 in view of (1.5)). Assumption (1.5) is stronger than (1.3) and one may wonder whether Theorem 1.2 also holds under weaker assumptions. A nice example by Sweers [16] shows that, for a smooth and decreasing f, positive radial solutions may not be radially decreasing. Therefore, (1.3) is not enough to obtain the full statement of Theorem 1.1 (radial symmetry and monotonicity) in the higher order case m2. This discrepancy of assumptions between the cases m=1 and m2 is due to the lack of a maximum principle for polyharmonic operators; see [9,Section 1.2].

    To find conditions, different from (1.5), ensuring both radial symmetry and radial monotonicity for (1.2) in the case m2 is a challenging problem. The first purpose of this paper is precisely to discuss some conditions on the source f which ensure the radial symmetry of the solutions of (1.2) and their radial monotonicity. In Section 2, particular attention is devoted to the regularity and monotonicity of f: for the biharmonic equation in BRn, in Theorem 2.3 we exhibit a Hölder-continuous function f for which the radial symmetry of solutions fails, while Proposition 2.4 displays a strictly decreasing and sign-changing f that still ensures the existence of a radially symmetric and strictly decreasing solution.

    Contrary to the case of Navier boundary conditions, considered in [17], (1.2) cannot be reduced to a second order system when m2. This is why the moving plane procedure used in the proof of Theorem 1.2 was carried on by using fine estimates of the Green function G associated to the polyharmonic operator (Δ)m, see [3,Section 2]. The second purpose of the present paper is to obtain new properties of the Green function in the so-called conformal dimensions n=2m. In Section 3 we prove that some estimates become explicit identities in conformal dimensions, see formula (3.6). Combined with the inversion in the sphere and the Kelvin transform, this enables us to obtain an elegant symmetry property of the Green function, see Theorem 3.2 and the sketchy representation in Figures 3 and 4. In turn, this result is used in Corollary 3.5 where we present an alternative formula for the computation of the partial derivatives of solutions of (1.2): since the sign of these partial derivatives is the fundamental feature to implement the moving plane procedure, this formula can become useful under suitable assumption on the source f. Indeed, based on this formula, Proposition 3.7 suggests that the radial symmetry and monotonicity of the solutions of (1.2) is ensured if fL(R+) is small.

    Finally, as an application of Proposition 3.7, in Section 4 we revisit the counterexample by Sweers [16] where radial monotonicity of the solution fails: by appropriately modifying the source f (enforcing the conditions given in Proposition 3.7) we numerically obtain radially symmetric and strictly decreasing solutions of the biharmonic equation in the unit ball of R4 for a decreasing function f with sufficiently small derivative. Throughout this work, some open problems and questions are posed.

    In this section we discuss radial properties (symmetry and monotonicity) of solutions of (1.2). Let us recall the weak formulation of (1.2) within the Sobolev space Hm0(B), which is a Hilbert space if endowed with the scalar product

    (u,v)m={B(Δm/2u)(Δm/2v)if m is evenB(Δ(m1)/2u)(Δ(m1)/2v)if m is oddu,vHm0(B).

    We denote the induced norm by m; in particular, 0 is the L2(B)-norm. A function uHm0(B) is called a weak solution of (1.2) if f(u)Hm(B) (the dual space of Hm0(B)) and (1.2) is satisfied in a weak sense, that is

    (u,v)m=f(u),vfor all vHm0(B),

    where , stands for the duality product between Hm(B) and Hm0(B). In the present article, however, we will mostly deal with slightly more regular solutions: when f is continuous and uHm0(B)L(B), we say that u is a strong solution of (1.2) if

    (u,v)m=Bf(u)vfor all vHm0(B); (2.1)

    the integral exists since uL(B) and f is continuous. By elliptic regularity, any such strong solution u belongs to C2m1,α(¯B), for some α(0,1), and all partial derivatives of order less than m vanish on B. Moreover, if f is Hölder continuous, then uC2m,α(¯B) is a classical solution, see [9]. In the sequel, we always take (at least) fC(R;R).

    Our first (elementary) result is a restatement of [3,Remark (iv)] and gives a different condition for radial symmetry (but not for radial monotonicity).

    Theorem 2.1. Let n2 and m1. Let λ1 be the first Dirichlet eigenvalue for (Δ)m in B. If

    fC(R;R)satisfiesf(t)f(s)ts<λ1t>s, (2.2)

    then there exists at most one strong solution uHm0(B)L(B) of problem (1.2) which is, moreover, radially symmetric.

    Proof. By contradiction, assume that (1.2) admits two strong solutions u1,u2Hm0(B)L(B). Then by (2.1) we have both that

    (u1,v)m=Bf(u1)vand(u2,v)m=Bf(u2)vfor all vHm0(B).

    Set wu1u2, subtract these two equations and choose v=w as a test function to obtain

    (w,w)m=B(f(u1)f(u2))w. (2.3)

    By (2.2) we know that

    (f(u1(x))f(u2(x)))w(x)λ1w(x)2xB,

    with strict inequality in a set of positive measure since w0 (recall u2u1). Therefore, (2.3) yields

    w2m<λ1w20,

    which contradicts the Poincaré inequality and proves uniqueness.

    Once uniqueness is established, it suffices to remark that if u is a nonradial solution of (1.2), then for any given nontrivial rotation ASO(n), also the function uAuA is a (different) strong solution of (1.2), which contradicts the just proved uniqueness statement.

    We have so shown that, if (2.2) holds, then there exists at most a unique solution of (1.2) which is necessarily radially symmetric. This proves the theorem.

    Theorem 2.1 states that one cannot expect radial monotonicity under the sole assumption (2.2). Note also that Theorem 2.1 does not require the positivity of the solution and its elementary proof is based on the invariance properties of polyharmonic operators under rotations, a fact that ensures uniqueness. Let us recall that symmetry is also ensured if f(s)<λ2, for all sR (with λ2 being the second Dirichlet eigenvalue for (Δ)m in B, see [6]), and uniqueness is guaranteed if fC(R+;R) is sublinear at 0 and + (see [7]). All this raises a natural question:

    Problem 2.2. Is radial symmetry related to uniqueness? Are there examples of multiple radial positive solutions? Except, of course, when f(u)=λ1u!

    Some attention also deserves the regularity of f, as this feature is related to the existence of non-radial solutions. The counterexample in [10,Remark 1] shows that Theorem 1.1 (m=1) does not hold if the Lipschitz continuity assumption on f is relaxed to Hölder continuity. We extend this example to the biharmonic equation (m=2), showing that the monotonicity requirement in (1.5) cannot be dropped even in an arbitrarily small interval. We only deal with the biharmonic operator since the amount of computations grows very quickly as m increases.

    Theorem 2.3. Let n2. For every ε>0 there exists a Hölder-continuous function gε:R+R such that

    gε(0)=0,gε(w)>εwR+,gε(ε)>0,gεis strictly increasing over[ε,),

    and such that the problem

    {Δ2u=gε(u)inBu=ur=0onB,

    admits both

    a radial solution (u=u(r)) in C4(B), nonnegative (but not strictly positive) and radially decreasing (but not strictly radially decreasing): namely, u(r0)=0 and u(r0)=0 for some r0(0,1);

    infinitely many nonnegative nontrivial solutions in C4(B) which are not radially symmetric.

    Proof. For any p>4 define the function g:R+R by

    g(w)=4p(p1)w14p[4(p3)(p2)4(n(p2)+2(p2)2)w1/p+(n2+4(p2)(p1)+2n(2p3))w2/p]w0. (2.4)

    Then gC0,14p([0,);R)C((0,);R) and

    g(w)=4(p2)(p1)w4/p[4(p4)(p3)+4(3(n4)2p2p(n3)+7p)w1/p+(n26n+8+4p2+4p(n3))w2/p]w>0.

    The quantities between brackets in both the expressions of g(w) and g(w) can be seen as second order polynomials in the variable w1/p. In particular, there must exist finite numbers M=M(n,p)>0, Γ1=Γ1(n,p)>0, Γ2=Γ2(n,p)(0,Γ1) such that

    M(n,p)minw0g(w),g(w)>0w1/p>Γ1(n,p),g(w)>0w1/p>Γ2(n,p). (2.5)

    Two plots of g, for given values of p>4 and n2, are displayed in Figure 1.

    Figure 1.  Plot of g on the interval [0,1] for (p,n)=(4.5,3) (left) and (p,n)=(4.2,6) (right).

    Since m=2, the Eq (1.2) in radial coordinates becomes

    {u(4)(r)+2(n1)ru(r)+(n1)(n3)r2u(r)(n1)(n3)r3u(r)=g(u(r))r(0,1)u(0)=u(0)=u(1)=u(1)=0, (2.6)

    where the conditions u(0)=u(0)=0 are needed to ensure the smoothness of u in B, see [8]. By putting u(r)(1r2)p, for r[0,1], one sees that such u satisfies u(1)=u(1)=0 (fulfilling the Dirichlet boundary conditions) and also the equation in (2.6), where g(u(r)) is as in (2.4). Hence,

    u(x)=(1|x|2)pfor xB

    is a C4(¯B)-positive, radial and strictly radially decreasing solution (u(r)<0 in (0,1)) of the problem

    {Δ2u=g(u)inBu=ur=0onB.

    In order to obtain the statement, we need some rescaling. First we extend u to all Rn by setting

    u(x)={(1|x|2)pif xB0if xB, (2.7)

    so that uC4(Rn) is nonnegative and nontrivial (but not strictly positive), radial and radially decreasing (but not strictly decreasing). We now argue as in [10,p. 220]: take any point x0Rn such that |x0|=3 and consider the function ˆuC4(Rn) defined as

    ˆu(x)=u(x)+u(xx0)={(1|x|2)pif |x|<1(1|xx0|2)pif |xx0|<10otherwise, (2.8)

    which is a nonnegative solution of the problem

    {Δ2ˆu=g(ˆu)inB5ˆu=ˆur=0onB5,

    where B5Rn is the (open) ball of radius 5 centered at the origin of Rn, and g is given by (2.4). Nevertheless, ˆu is not radially symmetric!

    Then we rescale the problem as follows. For u=u(r) as in (2.7) and for some γ>0, define v(r)u(5r)/γ for r[0,1], so that

    v(r)={1γ(125r2)pif 0r<150if r15,

    and v solves

    {Δ2v=gγ(v)inBv=vr=0onBwheregγ(w)625γg(γw)w0.

    Finally, fix ε>0 and let us construct the function gε in the statement. Choose γ0>0 such that

    γ0>1εmax{Γ1,625M}Γ2γ0<Γ1γ0<ε   and   625Mγ0<ε.

    Then gεgγ0 satisfies all the assumptions of the statement and the function v is a radial solution, nonnegative but not strictly positive (since it vanishes in BB1/5), and radially decreasing but not strictly radially decreasing (since it is constant in BB1/5). Moreover, after rescaling the function ˆu in (2.8), we obtain infinitely many nonnegative nontrivial solutions which are not radially symmetric, one for each x0 and, possibly with multiple bumps.

    The next result shows that the monotonicity assumption in (1.5) is not necessary to obtain the statement of Theorem 1.2. For n2, consider the function

    fn(w)1632+576(n1)+32(n1)(n3)[1680+672(n1)+48(n1)(n3)]ww0, (2.9)

    which is strictly decreasing and sign-changing, see Figure 2 below.

    Figure 2.  Plot of fn on the interval [0,2] for n=5 (right).

    However, we can prove

    Proposition 2.4. Let fn be as in (2.9). The unique strong solution of the problem

    {Δ2u=fn(u)inBu=ur=0onB, (2.10)

    is given by

    u(x)=(1|x|4)2xB,

    and, hence, it is positive, radially symmetric and strictly decreasing in the radial variable.

    Proof. The function fn defined in (2.9) satisfies the assumption (2.2) so that, by arguing as in the proof of Theorem 2.1, we infer that (2.10) admits at most one strong solution, which is necessarily radially symmetric.

    In radial coordinates, when m=2, the Eq (1.2) becomes (2.6). The fact that u(r)=(1r4)2, for r[0,1], is a (radial) solution follows by noticing that u(1)=u(1)=0 (fulfilling the Dirichlet boundary conditions), and then, by inserting the expressions of u(r) and fn(u(r)) into (2.6).

    We introduce here the Green function of the polyharmonic operator (Δ)m under Dirichlet boundary conditions in BRn and we determine some of its properties. Let us firstly define

    θ(x,y)(1|x|2)(1|y|2)x,yB.

    Then for x,yB, with xy, Boggio [5,p.126] (see also [9,Section 2.6]) gave the following explicit representation of the Green function:

    G(x,y)=kmn|xy|2mn(θ(x,y)|xy|2+1)1/21(z21)m1zn1dz=kmn2|xy|2mn θ(x,y)|xy|20zm1(z+1)n/2dz=kmn2H(|xy|2,θ(x,y)), (3.1)

    where kmn is a positive constant defined by

    kmn=Γ(1+n2)nπn/24m1((m1)!)2, (3.2)

    see [9,Lemma 2.27], and H:(0,)×[0,)R is defined as

    H(s,t)=smn2ts0zm1(z+1)n/2dzs>0,t0. (3.3)

    The following statement is a direct consequence of Boggio's work [5], elliptic regularity (see [9,Section 2.5]) and the estimates in [12].

    Proposition 3.1. Let hL(B), and let uHm0(B) be a weak solution of (Δ)mu=h in B under Dirichlet boundary conditions, that is

    u,vm=Bhvfor allvHm0(B). (3.4)

    Then uC2m1,α(¯Ω) and it satisfies

    Dku(x)=BDkxG(x,y)h(y)dyx¯B, (3.5)

    where Dk stands for any partial derivative of order |k|<2m. In particular, u is a strong solution and Dku0 on B for |k|m1.

    In conformal dimensions n=2m, the Green function admits a simpler representation. Indeed, the change of variables ξ=1z+1 yields

    \begin{aligned} H(s,t) & = \int_0^{\frac{t}{s}}\frac{z^{m-1}}{(z+1)^{m}}\,dz = \int_{0}^{\frac{t}{s}} \left( 1 - \frac{1}{z+1} \right)^{m-1} \frac{1}{z+1}\,dz = \int_{\frac{s}{s+t}}^{1} \dfrac{1}{\xi} (1-\xi)^{m-1} \, d\xi \\[6pt] & = \int_{\frac{s}{s+t}}^{1} \left[ \dfrac{1}{\xi} + \sum\limits_{k = 1}^{m-1} (-1)^{k} \binom{m-1}{k} \xi^{k-1} \right] d\xi \\[6pt] & = \log \left(1 + \dfrac{t}{s} \right) + \sum\limits_{k = 1}^{m-1} \dfrac{(-1)^{k}}{k} \binom{m-1}{k} \left( 1 - \left( \dfrac{s}{s+t} \right)^{k} \right) \qquad \forall s > 0, \ t \geq 0. \end{aligned}

    Therefore, by setting \kappa_{m} \doteq k_{2m}^{m} , for every x, y \in {\bf B} we have

    \begin{equation} \label{{Greenconf}} G(x,y) = \dfrac{\kappa_{m}}{2} \left[ \log \left(1 + \dfrac{\theta(x,y)}{|x-y|^2} \right) + \sum\limits_{k = 1}^{m-1} \dfrac{(-1)^{k}}{k} \binom{m-1}{k} \left( 1 - \left( \dfrac{|x-y|^2}{|x-y|^2+\theta(x,y)} \right)^{k} \right) \right]. \end{equation} (3.6)

    The main result of this section (Theorem 3.2) shows that, in conformal dimensions, the Green function satisfies an elegant reflection property. For this, given the point x = (x_{1}, x')\in {\bf B} with x_{1} > 0 , we define

    \begin{equation} R_x\doteq\frac{1-\vert x\vert^2}{2x_{1}}\, ,\qquad y_{x}\doteq(x_{1}+R_x,x')\, ,\qquad \mathcal C_{x} \doteq\left\{y\in {\bf B}\; ; \, \vert y - y_{x} \vert > R_x\; \right\}, \end{equation} (3.7)

    and the set

    \begin{equation} \label{{esferaS}} \mathcal{S}_{x} \doteq \left\{y\in \mathbb{R}^{n}\; ; \; (1-\vert x\vert^2)(x_{1}-y_{1})+x_{1}\vert x-y\vert^2 = 0\right\} , \end{equation} (3.8)

    which is a sphere with center at y_x and radius R_x (see the proof of Theorem 3.2 below and also Figure 4). We then denote by \mathcal{P}_{x} : \mathbb{R}^{n} \setminus \{y_{x}\} \longrightarrow \mathbb{R}^{n} the inversion in \mathcal{S}_{x} , given by the expression

    \mathcal{P}_{x}(y) = y_{x} + R_x^2\frac{y-y_{x}}{\vert y-y_{x}\vert^2} \qquad \forall y \in \mathbb{R}^{n} \setminus \{y_{x}\} \, .

    Finally, the Kelvin transform of a function h : \mathcal C_{x} \longrightarrow \mathbb{R} with respect to \mathcal{S}_{x} is defined as

    \mathcal{K}_{x}(h)(y) = h(\mathcal{P}_{x}(y)) \qquad \forall y \in \mathcal C_{x} \, .

    Then we prove

    Theorem 3.2. For any x = (x_{1}, x')\in {\bf B}\subset\mathbb{R}^{2m} with x_1 > 0 and any u \in H^{2m}({\bf B}) \cap H_0^m({\bf B}) we have

    \begin{equation} \begin{aligned} \dfrac{1 - \vert x \vert^{2}}{\kappa_{m} \, x_{1}} \frac{\partial u}{\partial x_{1}}(x) = \int_{\mathcal C_{x}} & \dfrac{\vert y-y_{x}\vert^2 - R_{x}^2}{|x-y|^2}\left(\frac{\theta(x,y)}{|x-y|^2+\theta(x,y)}\right)^m (-\Delta_{y})^{m} \left[ \mathcal{K}_{x}(u)(y) - u(y) \right] dy \, . \end{aligned} \end{equation} (3.9)

    Proof. The following identities hold for the function H defined in (3.3): for all s, t > 0 we have

    \begin{equation} \dfrac{\partial H}{\partial t}(s,t) = \frac{t^{m-1}}{(t+s)^m}\, ,\qquad \dfrac{\partial H}{\partial s} (s,t) = -\frac{t^{m}}{s(t+s)^m}\, , \qquad \dfrac{\partial^{2} H}{\partial s \partial t} (s,t) = -\frac{mt^{m-1}}{(t+s)^{m+1}}\, , \end{equation} (3.10)

    see also [3]. Moreover, for every x, y \in {\bf B} we have

    \begin{equation} \dfrac{\partial G}{\partial x_{1}} (x,y) = \kappa_{m} \left[ (x_1-y_1) \dfrac{\partial H}{\partial s} \left(|x-y|^2,\theta(x,y) \right)- x_{1}(1-|y|^2) \dfrac{\partial H}{\partial t} \left( |x-y|^2,\theta(x,y) \right) \right]. \end{equation} (3.11)

    We then prove the announced symmetry property, namely that the zero level sets of \partial_{x_{1}} G(x, \cdot) are spherical caps. Let us denote the zero level set by

    \mathcal{L}_{x} \doteq \left\{y\in {\bf B}\; ; \; \frac{\partial G}{\partial x_{1}}(x,y) = 0\right\}\, .

    By (3.10) and by (3.11) we have

    \begin{equation} \begin{aligned} -\frac{1}{\kappa_{m}} \frac{\partial G}{\partial x_{1}}(x,y) & = x_{1}(1-|y|^2) \dfrac{\partial H}{\partial t}(|x-y|^2,\theta(x,y))-(x_1-y_1)\dfrac{\partial H}{\partial s}(|x-y|^2,\theta(x,y)) \\[6pt] & = x_{1}(1-|y|^2) \frac{\theta(x,y)^{m-1}}{(|x-y|^2+\theta(x,y))^m} + (x_{1}-y_{1})\frac{\theta(x,y)^m}{|x-y|^2(|x-y|^2 + \theta(x,y))^m}\\[6pt] & = \frac{\theta(x,y)^{m-1}}{|x-y|^2(|x-y|^2+\theta(x,y))^m}(1-\vert y\vert^2)\left[ x_{1}\vert x-y\vert^2 + (1-\vert x\vert^2)(x_{1}-y_{1}) \right]. \end{aligned} \end{equation} (3.12)

    Therefore, \partial_{x_{1}} G(x, y) vanishes if and only if y\in {\bf B} belongs to the set \mathcal{S}_{x} defined in (3.8). Then we notice that y\in {\bf B}\cap\mathcal{S}_{x} if and only if

    \begin{align} 0 = \frac1{x_{1}}\left[(1-\vert x\vert^2)(x_{1}-y_{1})+x_{1}\vert x-y\vert^2 \right]& = \frac{1-\vert x\vert^2}{x_{1}}(x_{1}-y_{1}) + (x_{1}-y_{1})^2 + \vert x'-y'\vert^2\\[6pt] & = 2R_x(x_{1}-y_{1}) + (x_{1}-y_{1})^2 + \vert x'-y'\vert^2\\[6pt] & = (R_x+x_{1}-y_{1})^2 + \vert x'-y'\vert^2 -R_x^2\\[6pt] & = \vert y_{x}-y\vert^2 - R_x^2\, . \end{align} (3.13)

    Hence, \mathcal{S}_{x} is a sphere with center at y_x and radius R_x and the set of negativity of \partial_{x_{1}} G(x, y) is the cap \mathcal C_{x} defined in (3.7).

    Next, we notice that, by definition of y_{x} and R_x , we have

    \vert y_{x}\vert^2 = \left(x_{1}+\frac{1-\vert x\vert^2}{2x_{1}}\right)^2+\vert x'\vert^2 = x_{1}^2+(1-\vert x\vert^2)+R_x^2+\vert x'\vert^2 = 1+R_x^2\, .

    Then we compute

    \begin{array}{l}\vert \mathcal{P}_{x}(y) \vert^2 = \vert \mathcal{P}_{x}(y) -y_{x}+y_{x}\vert^2 = \vert \mathcal{P}_{x}(y) -y_{x}\vert^2 + 2(\mathcal{P}_{x}(y)-y_{x})\cdot y_{x} + \vert y_{x}\vert^2\\[3pt] = \frac{R_x^4}{\vert y-y_{x}\vert^2} + 2R_x^2\frac{y-y_{x}}{\vert y-y_{x}\vert^2}\cdot y_{x} + \vert y_{x}\vert^2 \\= \frac{R_x^2}{\vert y-y_{x}\vert^2}\left(R_x^2 + 2(y-y_{x})\cdot y_{x} + \vert y_{x}\vert^2\frac{\vert y-y_{x}\vert^2}{R_x^2}\right)\\[3pt] = \frac{R_x^2}{\vert y-y_{x}\vert^2}((R_x^2 -\vert y-y_{x}\vert^2)+(\vert y-y_{x}\vert^2 + 2(y-y_{x})\cdot y_{x}+\vert y_{x}\vert^2)\\-\vert y_{x}\vert^2 + \vert y_{x}\vert^2\frac{\vert y-y_{x}\vert^2}{R_x^2})\\[3pt] = \frac{R_x^2}{\vert y-y_{x}\vert^2}\left((R_x^2 -\vert y-y_{x}\vert^2)+|y|^{2}-\vert y_{x}\vert^2 + \vert y_{x}\vert^2\frac{\vert y-y_{x}\vert^2}{R_x^2}\right)\\[3pt] = \frac{R_x^2}{\vert y-y_{x}\vert^2}\left((R_x^2 +|y|^{2}-\vert y_{x}\vert^2) + \frac{(\vert y_{x}\vert^2-R_x^2) \vert y-y_{x}\vert^2}{R_x^2} \right)\\[3pt] = \frac{R_x^2}{\vert y-y_{x}\vert^2}\left( (|y|^{2} -1) +\frac{\vert y-y_{x}\vert^2}{R_x^2}\right) = 1 - \frac{R_x^2}{\vert y-y_{x}\vert^2} (1 - |y|^{2}) \, . \end{array}

    This computation has two main consequences. First, that the unit sphere \partial {\bf B} is invariant under the inversion in \mathcal{S}_{x} , that is,

    y\in\partial {\bf B}\ \Longleftrightarrow\ \mathcal{P}_x(y)\in\partial {\bf B}\, .

    Second, that

    \begin{equation} \theta(x,\mathcal{P}_{x}(y)) = \frac{R_{x}^2}{\vert y_{x}-y\vert^2} \, \theta(x,y)\qquad\forall y\in {\bf B}\, . \end{equation} (3.14)

    We next claim that if \mathcal C_{x} is the cap defined in (3.7), then {\bf B} = \mathcal C_{x} \cup \mathcal{P}_{x}(\mathcal C_{x}) \cup \mathcal{L}_{x} . To this end, it suffices to show that

    \mathcal{P}_{x}( \mathcal C_{x}) = \left\{y\in {\bf B}\; ; \, \vert y - y_{x} \vert < R_x\right\}.

    Indeed, let y \in \mathcal C_{x} , so that \mathcal{P}_{x}(y) \in \mathcal{P}_{x}(\mathcal C_{x}) and \vert y - y_{x} \vert > R_x . By definition of \mathcal{P}_{x} we have

    | \mathcal{P}_{x}(y) - y_{x} | = \dfrac{R_x^{2}}{| y - y_{x} |} < R_x,

    thus proving that \mathcal{P}_{x}(\mathcal C_{x}) \subset \left\{y\in {\bf B}\; ; \, \vert y - y_{x} \vert < R_x\right\} . On the other hand, let y \in {\bf B} be such that \vert y - y_{x} \vert < R_x . Since the function \mathcal{P}_{x} is bijective, there exists a point y_{0} \in \mathbb{R}^{n} such that \mathcal{P}_{x}(y_{0}) = y . By definition, we then have

    | y_{0} - y_{x} | = \dfrac{R_x^{2}}{| y - y_{x} |} > R_x,

    implying y_{0} \in \mathcal C_{x} , and subsequently, that y \in \mathcal{P}_{x}(\mathcal C_{x}) and \left\{y\in {\bf B}\; ; \, \vert y-y_{x} \vert < R_x \right\} \subset \mathcal{P}_{x}(\mathcal C_{x}) .

    Since {\bf B} = \mathcal C_{x} \cup \mathcal{P}_{x}(\mathcal C_{x}) \cup \mathcal{L}_{x} , by Proposition 3.1, we have

    \begin{aligned} \frac{\partial u}{\partial x_{1}}(x) & = \int_{ \mathcal C_{x}} \frac{\partial G}{\partial x_{1}}(x,y) (-\Delta_{y})^{m}u(y) \;dy + \int_{\mathcal{P}_{x}( \mathcal C_{x})} \frac{\partial G}{\partial x_{1}}(x,y)(-\Delta_{y})^{m}u(y)\;dy\\[6pt] & = \int_{ \mathcal C_{x}} \frac{\partial G}{\partial x_{1}}(x,y)(-\Delta_{y})^{m}u(y)\;dy + \int_{ \mathcal C_{x}} \left(\frac{R_{x}}{\vert y-y_{x}\vert}\right)^{2n} \, \frac{\partial G}{\partial x_{1}}(x,\mathcal{P}_{x}(y)) ((-\Delta_{y})^{m} u)(\mathcal{P}_{x}(y)) \;dy\\[6pt] & = \int_{ \mathcal C_{x}} \frac{\partial G}{\partial x_{1}}(x,y)(-\Delta_{y})^{m}u(y)\;dy + \int_{ \mathcal C_{x}} \left(\frac{R_{x}}{\vert y-y_{x}\vert}\right)^{4m} \, \frac{\partial G}{\partial x_{1}}(x,\mathcal{P}_{x}(y)) ((-\Delta_{y})^{m} u)(\mathcal{P}_{x}(y))\;dy,\\[6pt] \end{aligned}

    where we used that the Jacobian matrix \mathcal{P}'_{x} satisfies

    \left| \text{det}(\mathcal{P}'_{x}(y)) \right| = \left(\frac{R_{x}}{\vert y-y_{x}\vert}\right)^{2n}\qquad\forall y\in {\bf B}\, .

    In view of (3.12) and (3.13) we can write

    \begin{equation} -\frac{1}{\kappa_{m}} \dfrac{1 - \vert x \vert^{2}}{x_{1}} \frac{\partial G}{\partial x_{1}}(x,y) = \dfrac{\vert y_{x}-y\vert^2 - R_x^2}{|x-y|^2} \left(\frac{\theta(x,y)}{\theta(x,y)+|x-y|^2}\right)^m \qquad \forall y \in {\bf B} \, , \end{equation} (3.15)

    so that

    \begin{equation} -\frac{1}{\kappa_{m}} \dfrac{1 - \vert x \vert^{2}}{x_{1}} \frac{\partial u}{\partial x_{1}}(x) = I_{1}(x) + I_{2}(x) \, , \end{equation} (3.16)

    where

    To compute I_{2}(x) , we first notice that

    \begin{equation} \vert y_{x}-\mathcal{P}_{x}(y)\vert^2-R_x^2 = \frac{R_x^4}{\vert y_{x}-y\vert^2}-R_x^2 = -\frac{R_x^2}{\vert y_{x}-y\vert^2}(\vert y_{x}-y\vert^2-R_x^2) < 0 \qquad \forall y \in \mathcal C_{x} \, , \end{equation} (3.17)

    whereas, for every y \in {\bf B} we have

    \begin{equation} \begin{aligned} |x-\mathcal{P}_{x}(y)|^{2} & = \vert x-y_{x}\vert^2 + 2(x-y_{x})\cdot(y_{x}-\mathcal{P}_{x}(y)) +\vert y_{x}-\mathcal{P}_{x}(y)\vert^2\\[3pt] & = \vert x-y_{x}\vert^2 + \frac{2R_x^2}{\vert y_{x}-y\vert^2}(x-y_{x})\cdot(y_{x}-y) +\frac{R_x^4}{\vert y-y_{x}\vert^2}\\[3pt] & = \frac{R_x^2}{\vert y_{x}-y\vert^2}\left( \frac{\vert y_{x}-y\vert^2}{R_x^2}\vert x-y_{x}\vert^2 + 2(x-y_{x})\cdot(y_{x}-y) + {R_x^2}\right)\\[3pt] & = \frac{R_x^2}{\vert y_{x}-y\vert^2}\left( {\vert y_{x}-y\vert^2} + 2(x-y_{x})\cdot(y_{x}-y) + \vert x-y_{x}\vert^2\right) = \frac{R_x^2}{\vert y_{x}-y\vert^2} \, |x-y|^2 \, . \end{aligned} \end{equation} (3.18)

    After plugging (3.14), (3.17) and (3.18) into I_{2}(x) we get

    I_{2}(x) = - \int_{\mathcal C_{x}} \dfrac{\vert y-y_{x}\vert^2 - R_{x}^2}{|x-y|^2}\left(\frac{\theta(x,y)}{|x-y|^2+\theta(x,y)}\right)^m \left(\frac{R_{x}}{| y - y_{x} |}\right)^{4m} ((-\Delta_{y})^{m} u)(\mathcal{P}_{x}(y)) dy

    and, after recalling (3.16), we deduce

    \begin{array}{l} -\frac{1}{\kappa_{m}} \dfrac{1 - \vert x \vert^{2}}{x_{1}} \frac{\partial u}{\partial x_{1}}(x) \\[6pt] = \int_{\mathcal C_{x}} \dfrac{\vert y-y_{x}\vert^2 - R_{x}^2}{|x-y|^2}\left(\frac{\theta(x,y)}{|x-y|^2+\theta(x,y)}\right)^m \\ \left[ (-\Delta_{y})^{m} u(y) - \left(\frac{R_{x}}{| y - y_{x} |}\right)^{4m} ((-\Delta_{y})^{m} u)(\mathcal{P}_{x}(y)) \right] dy \, . \end{array}

    A direct computation (see [13,Proposition 7.3]) shows that

    \begin{equation} (-\Delta_{y})^{m} \mathcal{K}_{x}(u)(y) = \left(\frac{R_{x}}{| y - y_{x} |}\right)^{4m} ((-\Delta_{y})^{m} u)(\mathcal{P}_{x}(y)) \qquad \forall y \in \mathcal{C}_{x} \, , \end{equation} (3.19)

    which concludes the proof after changing the sign.

    Problem 3.3. It is not clear how to derive an expression similar to (3.9) in non-conformal dimensions n \geq 2 , n \neq 2m . This can be seen after computing

    \dfrac{\partial H}{\partial s}(s,t) = \left( n - \dfrac{m}{2} \right) s^{m-\frac{n}{2}-1}\int_0^{\frac{t}{s}}\frac{z^{m-1}}{(z+1)^{n/2}}\,dz - \dfrac{t^{m}}{s(t+s)^{n/2}} \qquad \forall s > 0, \ t \geq 0 \,,

    so that the determination of the level sets of \partial_{x_{1}} G(x, \cdot) as in (3.12) becomes a delicate task. Is it possible to obtain a representation formula similar to (3.9) in any dimension n \geq 2 ?

    Clearly, a representation formula similar to (3.9) holds for any other directional derivative of u . Note that (3.7) may be rewritten as y_x = x+(R_x, 0) , which implies that y_x is an "horizontal translation" of x \in {\bf B} . Moreover, R_x\to0 if x\to\partial {\bf B} while R_x\to+\infty if x_1\to0 : the limit of the map x\mapsto R_x does not exist when x approaches the equator \partial {\bf B}\cap \{x_1 = 0\} . The level surfaces of x\mapsto R_x are spheres:

    R_x = k\in(0,\infty)\ \Longrightarrow\ (x_1+k)^2+|x'|^2 = 1+k^2\, .

    The center of the level-surface sphere is then (x_1, x') = (-k, 0) and its radius is \sqrt{1+k^2} . For a fixed k > 0 , when x runs over the level surface R_x = k , the corresponding y_x , defined by (3.7), runs over a portion of the sphere centered in the origin with the same radius \sqrt{1+k^2} , see Figure 3.

    Figure 3.  Value of R_x , positions of x and y_x , see (3.7). In blue: the level surface R_x = k > 0 .

    Note also that \mathcal C_{x} , the domain of negativity of \partial_{x_{1}} G(x, y) , is the intersection between two balls. In Figure 4 we sketch the mutual position of the sphere \mathcal{S}_x and the ball {\bf B} . In particular, we emphasize that

    |y_x|^2 = \left(x_1+\frac{1-|x|^2}{2x_1}\right)^2+|x'|^2 = 1+\frac{(1-|x|^2)^2}{4x_1^2} > 1\qquad\forall x\in {\bf B}\quad(x_1 > 0)

    which, again, tells us that y_x is exterior to {\bf B} and it approaches \partial {\bf B} as x approaches \partial {\bf B} , while it goes to infinity if x approaches the plane x_1 = 0 .

    Figure 4.  Representation of the sphere \mathcal{S}_{x} for different positions of x \in {\bf B} .

    Theorem 3.2 has several relevant consequences. Firstly, we deduce

    Corollary 3.4. Let m \geq 1 . For every x \in {\bf B} \subset \mathbb{R}^{2m} such that x_{1} > 0 , the following identities hold

    \begin{equation} \begin{aligned} (1 - \vert x \vert^{2})^{m} = \dfrac{1}{\pi^{m}} \dfrac{(2m-1)!}{(m-1)!} \int_{\mathcal C_{x}} \dfrac{\vert y-y_{x}\vert^2 - R_{x}^2}{|x-y|^2} \\ \left(\frac{\theta(x,y)}{|x-y|^2+\theta(x,y)}\right)^m \left[ 1 - \left(\frac{R_{x}}{| y - y_{x} |}\right)^{4m} \right] dy \, , \end{aligned} \end{equation} (3.20)
    \begin{equation} \begin{aligned} \mathbb{K}_m\doteq \pi^{m} \dfrac{(m-1)!}{(2m-1)!} = \int_{\mathcal C_{x}} \dfrac{\vert y-y_{x}\vert^2 - R_{x}^2}{|x-y|^2} \left(\frac{1 - \vert y \vert^{2}}{|x-y|^2+\theta(x,y)}\right)^m \left[ 1 - \left(\frac{R_{x}}{| y - y_{x} |}\right)^{4m} \right] dy \, . \end{aligned} \end{equation} (3.21)

    Proof. We start by noticing that

    \begin{aligned} & \Delta r^{2m} = 2m(2m + n -2) r^{2m-2} \, , \qquad \Delta^{2} r^{2m} = 2m(2m-2)(2m + n -2)(2m + n -4) r^{2m-4} \, , \\[6pt] & \quad \Delta^{3} r^{2m} = 2m(2m-2)(2m-4)(2m + n -2)(2m + n -4)(2m + n -6) r^{2m-6} \, , \end{aligned}

    where n = 2m . By induction we then obtain

    \Delta^{m} r^{2m} = (2m)!! \, \dfrac{(2m + n -2)!!}{(n-2)!!} = (2m)!! \, \dfrac{(4m -2)!!}{(2m-2)!!} = 2^{2m-1} (2m)! \, .

    Consider the polyharmonic version of the original problem (1.1) proposed by Serrin:

    \begin{equation} \left\{ \begin{array}{ll} (-\Delta)^{m}u = 1\qquad & \mbox{in } {\bf B}\\[3pt] u = \dfrac{\partial u}{\partial r} = \dots = \dfrac{\partial^{m-1}u}{\partial r^{m-1}} = 0\qquad & \mbox{on }\partial {\bf B}. \end{array}\right. \end{equation} (3.22)

    A direct application of the binomial expansion yields

    (-\Delta)^{m} (1 - r^{2})^{m} = (-\Delta)^{m} \left[ \sum\limits_{k = 0}^{m} \binom{m}{k} (-1)^{k} r^{2k} \right] = \Delta^{m}r^{2m} = 2^{2m-1}(2m)! \, ,

    that is

    \begin{equation} (-\Delta)^{m} U_{m}(r) = 1 \qquad \text{with} \qquad U_{m}(r) \doteq \dfrac{(1 - r^{2})^{m}}{2^{2m-1} (2m)!} \qquad \forall r \in [0,1]. \end{equation} (3.23)

    Since U_{m} satisfies the Dirichlet boundary conditions in (3.22) _{2} , it is a strong \mathcal{C}^{\infty}(\overline{ {\bf B}}) -solution of (3.22). Applying identity (3.9) to the function U_{m} and recalling (3.2) (with n = 2m ), we derive equality (3.20). The identity (3.21) is a straightforward consequence of (3.20), after replacing \theta(x, y) .

    It is remarkable that the right hand side of (3.21) does not depend on x\in {\bf B} . As a further consequence of Theorem 3.2, we give an alternative formula to compute the partial derivatives of the solutions of (1.2), which should be compared with Proposition 3.1.

    Corollary 3.5. Let {\bf B}\subset \mathbb{R}^{2m} be the unit ball. Suppose that f \in \mathcal{C}(\mathbb{R}; \mathbb{R}) and that u\in H_0^m({\bf B}) \cap L^{\infty}({\bf B}) is a strong solution of (1.2). Then, for every x \in {\bf B} such that x_{1} > 0 , the following formula holds:

    \begin{equation} \dfrac{\vert x \vert^{2}-1}{\kappa_{m} \, x_{1}} \frac{\partial u}{\partial x_{1}}(x) = \int_{\mathcal C_{x}}\dfrac{\vert y-y_{x}\vert^2 - R_{x}^2}{|x-y|^2}\left[\frac{\theta(x,y)}{|x-y|^2+\theta(x,y)}\right]^m \left[ f(u(y)) - \left(\frac{R_{x}}{| y - y_{x} |}\right)^{4m} f(u(\mathcal{P}_{x}(y))) \right]\!dy. \end{equation} (3.24)

    Formula (3.24) follows directly from (3.9)–(3.19) and suggests the problem:

    Problem 3.6. Since the moving plane procedure developed in [3,Section 3] requires that

    \frac{\partial u}{\partial x_{1}}(x) < 0 \qquad \mathit{\text{for all}} \ x \in {\bf B} \ \ \mathit{\text{with}} \ \ x_{1} > 0 \ ,

    is it possible to use (3.24) to prove this inequality and to relax assumption (1.5), but still ensuring the statement of Theorem 1.2?

    In connection with Problem 3.6 we notice that, by looking at (3.24), it is clear that the overall sign of the integral over \mathcal{C}_{x} is determined by the behavior of the quantity

    \Phi_{x}(y) \doteq f(u(y)) - \left(\frac{R_{x}}{| y - y_{x} |}\right)^{4m} f(u(\mathcal{P}_{x}(y))) \qquad \forall y \in \mathcal{C}_{x} \, .

    Fixing x \in {\bf B} such that x_{1} > 0 and u\in H_0^m({\bf B}) , notice that \Phi_{x}(y) = 0 for all y \in \partial \mathcal{C}_{x} \cap {\bf B} , while for y \in \partial \mathcal{C}_{x} \cap \partial {\bf B} (so that \mathcal{P}_{x}(y) \in \partial {\bf B} ) we have that \Phi_{x}(y) = 0 if f(0) = 0 , and \Phi_{x}(y) > 0 if f(0) > 0 .

    Under suitable assumptions on the source f , in the next result we give an upper bound for the partial derivative of u (recall that |x| < 1 !), which shows that \partial_{x_{1}}u(x) < 0 for all x \in {\bf B} located far away from \partial {\bf B} , with x_1\geq\varepsilon for some \varepsilon > 0 , see Figure 5 for a schematic representation. Therefore, in order to solve Problem 3.6, one should mainly focus the attention on the complement of this region.

    Figure 5.  In gray: subset of {\bf B} in which the strict negativity of \partial_{x_{1}}u(x) holds.

    To this end, we define

    \frac1\Gamma \doteq \inf\limits_{v\in\mathcal{C}^{2m}(\overline{ {\bf B}})\cap H^m_0( {\bf B})\setminus\{0\}}\frac{\|\Delta^mv\|_{L^{\infty}( {\bf B})}}{\|\nabla v\|_{L^{\infty}( {\bf B})}}\, ,

    and we state:

    Proposition 3.7. Assume that f \in W^{1, \infty}(\mathbb{R}_+; \mathbb{R}_+) satisfies

    \begin{equation} f(s) \geq M \qquad \forall s \geq 0 \, , \ \ \mathit{\text{for some}} \ \ M > 0, \end{equation} (3.25)

    and let u\in H_0^m({\bf B}) \cap L^{\infty}({\bf B}) be a strong solution of (1.2). For any point x = (x_{1}, x')\in {\bf B} with x_1 > 0 we have

    \begin{aligned} \dfrac{\vert x \vert^{2}-1}{\kappa_{m} \, x_{1}} \frac{\partial u}{\partial x_{1}}(x) & \geq \left( M - 2\Gamma\|f\|_{L^{\infty}( \mathbb{R}_+)}\| f' \|_{L^{\infty}( \mathbb{R}_+)} \right) \mathbb{K}_m (1-|x|^2)^{m} \\[6pt] & \quad - 2\Gamma\|f\|_{L^{\infty}( \mathbb{R}_+)}\| f' \|_{L^{\infty}( \mathbb{R}_+)} \dfrac{1 - \vert x \vert^{2}}{\kappa_{m} \, x_{1}} \int_{\mathcal C_{x}} \left(\frac{R_{x}}{| y - y_{x} |}\right)^{4m} \frac{\partial G}{\partial x_{1}}(x,\mathcal{P}_x(y)) \, dy \, , \end{aligned}

    where \mathbb{K}_m > 0 is defined in (3.21).

    Proof. In view of Corollary 3.4, by the Maximum Principle and assumption (3.25), any strong solution u\in H_0^m({\bf B}) \cap L^{\infty}({\bf B}) of (1.2) satisfies

    u(x) \geq \dfrac{M}{2^{2m-1} (2m)!} (1 - |x|^{2})^{m} \qquad \forall x \in {\bf B}\, .

    By embedding theorems and elliptic regularity we also have that

    \| \nabla u \|_{L^{\infty}( {\bf B})}\leq\Gamma\|f\|_{L^{\infty}( \mathbb{R}_+)} \, .

    Then, by the Mean Value Theorem, we obtain the inequality

    |u(\mathcal{P}_x(y))-u(y)|\leq \Gamma\|f\|_{L^{\infty}( \mathbb{R}_+)}|\mathcal{P}_{x}(y) - y | \qquad \forall y \in \mathcal{C}_{x} \, .

    From this we deduce that

    |f(u(\mathcal{P}_x(y)))-f(u(y))|\leq\Gamma\|f\|_{L^{\infty}( \mathbb{R}_+)}\|f'\|_{L^{\infty}( \mathbb{R}_+)}\,| \mathcal{P}_{x}(y) - y | \qquad \forall y \in \mathcal{C}_{x} \, ,

    thus yielding

    f(u(\mathcal{P}_{x}(y))) \leq f(u(y)) + \Gamma\|f\|_{L^{\infty}( \mathbb{R}_+)}\| f' \|_{L^{\infty}( \mathbb{R}_+)} \, | \mathcal{P}_{x}(y) - y | \qquad \forall y \in \mathcal{C}_{x}

    and, by using (3.25), we finally obtain

    \begin{aligned} f(u(y))-\left(\frac{R_{x}}{|y-y_x|}\right)^{4m} f(u(\mathcal{P}_{x}(y))) & \geq \left[1-\left(\frac{R_{x}}{| y - y_{x} |}\right)^{4m}\right]M \\[6pt] & \quad -\left(\frac{R_{x}}{| y - y_{x} |}\right)^{4m}\Gamma\|f\|_{L^{\infty}( \mathbb{R}_+)}\| f' \|_{L^{\infty}( \mathbb{R}_+)} \, | \mathcal{P}_{x}(y)-y|\, . \end{aligned}

    By inserting this bound within (3.24), we get

    \begin{equation} \dfrac{\vert x \vert^{2}-1}{\kappa_{m} \, x_{1}} \frac{\partial u}{\partial x_{1}}(x) \geq I_1(x)-I_2(x) \, , \end{equation} (3.26)

    where

    From Corollary 3.4 we infer that

    \begin{equation} I_1(x) = M \mathbb{K}_m (1-|x|^2)^m\, . \end{equation} (3.27)

    In order to bound I_2(x) , we notice that

    |\mathcal{P}_{x}(y)-y| < 2 \quad \text{and} \quad \frac{R_{x}}{| y - y_{x} |} < 1 \qquad \forall y\in\mathcal C_{x}\, .

    Hence,

    \begin{equation} I_2(x) \leq 2\Gamma\|f\|_{L^{\infty}( \mathbb{R}_+)}\| f' \|_{L^{\infty}( \mathbb{R}_+)} \int_{\mathcal C_{x}}\dfrac{\vert y-y_{x}\vert^2 - R_{x}^2}{|x-y|^2}\left[\frac{\theta(x,y)}{|x-y|^2+\theta(x,y)}\right]^m dy \, . \end{equation} (3.28)

    From (3.15) and Proposition 3.1, and by proceeding as in the proof of Corollary 3.4, we obtain

    \begin{aligned} & \int_{\mathcal C_{x}}\dfrac{\vert y-y_{x}\vert^2 - R_{x}^2}{|x-y|^2}\left[\frac{\theta(x,y)}{|x-y|^2+\theta(x,y)}\right]^m dy = -\frac{1}{\kappa_{m}} \dfrac{1 - \vert x \vert^{2}}{x_{1}} \int_{\mathcal C_{x}} \frac{\partial G}{\partial x_{1}}(x,y) \, dy \\[6pt] & = -\frac{1}{\kappa_{m}} \dfrac{1 - \vert x \vert^{2}}{x_{1}} \left( \int_{ {\bf B}} \frac{\partial G}{\partial x_{1}}(x,y) \, dy - \int_{\mathcal{P}_{x}(\mathcal{C}_{x})} \frac{\partial G}{\partial x_{1}}(x,y) \, dy \right) \\[6pt] & = -\frac{1}{\kappa_{m}} \dfrac{1 - \vert x \vert^{2}}{x_{1}} \left[ \dfrac{1}{2^{2m-1} (2m)!} \dfrac{\partial}{\partial x_{1}} \left( (1 - |x|^{2})^{m}\right) - \int_{\mathcal C_{x}} \left(\frac{R_{x}}{| y - y_{x} |}\right)^{4m} \frac{\partial G}{\partial x_{1}}(x,\mathcal{P}_x(y)) \, dy \right] \\[6pt] & = \mathbb{K}_m (1-|x|^2)^m + \dfrac{1 - \vert x \vert^{2}}{\kappa_{m} \, x_{1}} \int_{\mathcal C_{x}} \left(\frac{R_{x}}{| y - y_{x} |}\right)^{4m} \frac{\partial G}{\partial x_{1}}(x,\mathcal{P}_x(y)) \, dy \, . \end{aligned}

    After plugging this into (3.28) we deduce that

    \begin{equation} I_2(x) \leq 2\Gamma\|f\|_{L^{\infty}( \mathbb{R}_+)}\| f' \|_{L^{\infty}( \mathbb{R}_+)} \left[ \mathbb{K}_m (1-|x|^2)^m + \dfrac{1 - \vert x \vert^{2}}{\kappa_{m} \, x_{1}} \int_{\mathcal C_{x}} \left(\frac{R_{x}}{| y - y_{x} |}\right)^{4m} \frac{\partial G}{\partial x_{1}}(x,\mathcal{P}_x(y)) \, dy \right]. \end{equation} (3.29)

    The conclusion is reached after inserting (3.27) and (3.29) into (3.26).

    Proposition 3.7 suggests that the radial symmetry and monotonicity of the solutions of (1.2) can be ensured if \|f'\|_{L^{\infty}(\mathbb{R}_+)} is small. This assumption is not needed if m = 1 (see Theorem 1.1), while for m \geq 2 , the required condition is that f' be nonnegative (see Theorem 1.2). Therefore, one is led to analyze the cases where the negative part of f' is small, that is, \|(f')_{-}\|_{L^{\infty}(\mathbb{R}_+)} small. This issue is tackled numerically in the next section, for the biharmonic equation in {\bf B}\subset\mathbb{R}^{4} .

    As a direct consequence of the Hopf-type lemma by Grunau-Sweers [11,Theorem 3.2] we obtain

    Proposition 4.1. Assume that f \in W^{1, \infty}(\mathbb{R}_+; \mathbb{R}_+) satisfies (3.25), and let u\in H_0^m({\bf B}) \cap L^{\infty}({\bf B}) be a strong solution of (1.2). Then, there exists \gamma_{f} > 0 such that

    x \cdot \nabla u(x) < 0 \qquad \mathit{\text{for all}} \ x \in {\bf B} \ \mathit{\text{such that}} \ \gamma_{f} < |x| < 1 \, .

    Guido Sweers [16] provided an explicit example of a non-decreasing radial solution of a linear biharmonic problem in {\bf B}\subset\mathbb{R}^{2} (equation (1.2) with m = 2 ). In this section we take advantage of his example and give numerical evidence to claim that Proposition 4.1 might be complemented with the statement that \gamma_{f} can be made arbitrarily small (possibly zero) provided that \|(f')_{-}\|_{L^{\infty}(\mathbb{R}_+)} is sufficiently small.

    Consider the following radial function defined in \mathbb{R}^{4} (the conformal dimension for the biharmonic operator):

    v(x) = \sum\limits_{k = 0}^{\infty} a_{k} \, r^{4k+2} \doteq \sum\limits_{k = 0}^{\infty} \dfrac{(-1)^{k}}{2^{4k+2}} \dfrac{r^{4k+2}}{(2k+2)! (2k+1)!} \qquad \forall x \in \mathbb{R}^{4} \, ,

    whose plot in the interval [-10, 10] is displayed in Figure 6 below.

    Figure 6.  Plot of v in the interval [-10, 10] .

    Since m = 2 and n = 4 , the biharmonic operator in radial coordinates becomes

    \Delta^2 \psi(r) = \psi^{(4)}(r) + \dfrac{6}{r}\psi'''(r) + \dfrac{3}{r^{2}}\psi''(r) - \dfrac{3}{r^{3}}\psi'(r) \qquad \forall r \in (0,1).

    After noticing that

    a_{k+1} = - \dfrac{1}{64 (k+2)(k+1)(2k+3)^{2}} \, a_{k} \qquad \forall k \in \mathbb{N} \, ,

    one readily sees that

    \Delta^{2} v = - v \quad \text{in} \quad \mathbb{R}^{4} \, .

    Let r_0\approx 9.2218 be the first nonzero local minimum of v ; numerically, we find that v(r_0)\approx-14.8388 . Then define the function

    \begin{equation} u(r) \doteq v(r_{0}r) - v(r_{0}) \qquad \forall r \in [0,1] \, , \end{equation} (4.1)

    which is radially symmetric and strictly positive in [0, 1) , but not decreasing, in the interval [0, 1] ; see the left picture in Figure 7 below.

    Figure 7.  Left: graph of u in (4.1). Right: graph of the numerical solution of (4.2), for \gamma = 1 .

    By defining

    f(w) = r_{0}^{4} (|v(r_{0})|-w) \qquad \forall w \geq 0 \, ,

    we observe that f'(w)\equiv-r_{0}^{4} < 0 , so that f is decreasing and (2.2) is satisfied. By Theorem 2.1 we then know that there exists at most one strong solution of the following problem:

    \left\{ \begin{aligned} & \Delta^2 u = f(u)\quad\text{in} \; {\bf B} \\[6pt] & u = \frac{\partial u}{\partial r} = 0\quad\text{on} \; \partial {\bf B}. \end{aligned} \right.

    The solution is radially symmetric and a simple computation shows that it coincides with u defined in (4.1). Note that u(0) = |v(r_{0})| and u(0.5) > 16.5 > |v(r_{0})| , so that f(u) is sign-changing in [0, 1] .

    We now modify f in order to reduce \|(f')_{-}\|_{L^{\infty}(\mathbb{R}_+)} . We take

    f_{\gamma}(w) \doteq r_{0}^{4}(|v(r_{0})|-\gamma w) \qquad \forall w \geq 0 \, ,

    being \gamma > 0 a variable parameter. Then we consider the problem

    \begin{equation} \left\{ \begin{aligned} & \Delta^2 U = f_{\gamma}(U)\quad\text{in}\; {\bf B} \\[6pt] & U = \frac{\partial U}{\partial r} = 0\quad\text{on} \; \partial {\bf B}, \end{aligned} \right. \end{equation} (4.2)

    and prove the following result:

    Proposition 4.2. For every \gamma > 0 , the unique strong solution U_{\gamma} \in H^{2}_{0}({\bf B}) \cap L^\infty({\bf B}) of (4.2) (which is radially symmetric) is analytic and can be written as

    \begin{equation} U_{\gamma}(r) = \sum\limits_{k = 0}^{\infty} A_{k}(\gamma) r^{2k} = A_{0}(\gamma) + A_{1}(\gamma) r^{2} + A_{2}(\gamma) r^{4} + A_{3}(\gamma) r^{6} + ... \qquad \forall r \in [0,1], \end{equation} (4.3)

    for some coefficients \{ A_{k}(\gamma) \}_{k \in \mathbb{N}} \subset \mathbb{R} that satisfy the following properties:

    \left\{ \begin{aligned} & A_{0}(\gamma) = \sum\limits_{k = 2}^{\infty} (k-1) A_{k}(\gamma) \, , \qquad \quad A_{1}(\gamma) = - \sum\limits_{k = 2}^{\infty} k A_{k}(\gamma) \, , \\[6pt] & A_{2}(\gamma) = \dfrac{r_{0}^{4}}{192}(|v(r_{0})| - \gamma A_{0}(\gamma)) \, , \qquad A_{k}(\gamma) = - \dfrac{\gamma r_{0}^{4}}{16 k^{2}(k^{2}-1)} A_{k-2}(\gamma) \qquad \forall k \geq 3 \, . \end{aligned} \right.

    Moreover, there exists \gamma_{0} > 0 such that, for every \gamma \in (0, \gamma_{0}) , the function U_{\gamma} is positive in [0, 1) and strictly decreasing in the radial variable.

    Proof. For every \gamma > 0 we notice that the condition (2.2) is satisfied. Then, Theorem 2.1 guarantees the existence of at most one strong solution U_{\gamma} \in H^{2}_{0}({\bf B}) \cap L^\infty({\bf B}) to problem (4.2) which is, moreover, radially symmetric and analytic in \overline{ {\bf B}} . Problem (4.2) in radial coordinates reads:

    \begin{equation} \left\{ \begin{aligned} & U_{\gamma}^{(4)}(r) + \dfrac{6}{r} U_{\gamma}'''(r) + \dfrac{3}{r^{2}} U_{\gamma}''(r) - \dfrac{3}{r^{3}} U_{\gamma}'(r) = r_{0}^{4} \left(|v(r_{0})|-\gamma U_{\gamma}(r) \right) \qquad \forall r\in(0,1) \\[6pt] & U_{\gamma}(1) = U_{\gamma}'(1) = 0\, . \end{aligned} \right. \end{equation} (4.4)

    Upon substitution of (4.3) into (4.4) _1 we obtain

    192A_{2}(\gamma)+16\sum\limits_{k = 3}^{\infty}A_{k}(\gamma) k^{2}(k^{2}-1) r^{2k - 4} \\ = r_{0}^{4} (|v(r_{0})| - \gamma A_{0}(\gamma)) - \gamma r_{0}^{4} \sum\limits_{k = 3}^{\infty} A_{k-2}(\gamma) r^{2k - 4} \qquad \forall r \in (0,1),

    which yields the identities

    \begin{equation} A_{2}(\gamma) = \dfrac{r_{0}^{4}}{192}(|v(r_{0})| - \gamma A_{0}(\gamma)) \, , \qquad A_{k}(\gamma) = - \dfrac{\gamma r_{0}^{4}}{16 k^{2}(k^{2}-1)} A_{k-2}(\gamma) \qquad \forall k \geq 3 \, . \end{equation} (4.5)

    Moreover, the boundary conditions in (4.4) _2 imply that

    \sum\limits_{k = 0}^{\infty} A_{k}(\gamma) = 0 \, , \ \sum\limits_{k = 1}^{\infty} k A_{k}(\gamma) = 0\ \Longrightarrow\ A_1(\gamma) = -\sum\limits_{k = 2}^{\infty}kA_k(\gamma)\, , \ A_0(\gamma) = \sum\limits_{k = 2}^{\infty}(k-1) A_k(\gamma)\, .

    In connection with (3.23) we define

    U_{*}(r) \doteq \sigma_{0} (1 - r^{2})^{2} \doteq \dfrac{r_{0}^{4} |v(r_{0})|}{64} (1 - r^{2})^{2} \qquad \forall r \in [0,1].

    By Corollary (3.4) and by continuous dependence, we have that

    U_{\gamma} \to U_{*} \quad \text{in} \ \ \mathcal{C}^{3}(\overline{ {\bf B}}) \ \ \text{as} \ \ \gamma \to 0^{+} \, ;

    in fact, the convergence holds in a stronger norm. Notice that

    U''_{*}(0) = -4 \sigma_{0} < 0 \, , \qquad U''_{*}(1) = 8 \sigma_{0} > 0 \, .

    Thus, by uniform convergence of U_\gamma'' and U_\gamma''' , there exist \delta_{0} \in (0, 1) and \gamma_{0} > 0 such that:

    U''_{\gamma}(r) \leq - 2 \sigma_{0} \quad \text{if} \quad r \in [0, \delta_{0}], \qquad U''_{\gamma}(r) \geq 4 \sigma_{0} \quad \text{if} \quad r \in [1-\delta_{0},1], \qquad \forall \gamma \in (0, \gamma_{0}).

    Since U'_{\gamma}(0) = U'_{\gamma}(1) = 0 for every \gamma > 0 , this implies that

    U'_{\gamma}(r) \leq - 2 \sigma_{0}r < 0 \quad \text{if} \quad r \in (0, \delta_{0}], \\ U'_{\gamma}(r) \leq 4 \sigma_{0}(r-1) < 0 \quad \text{if} \quad r \in [1-\delta_{0},1), \qquad \forall \gamma \in (0, \gamma_{0}).

    Moreover, by uniform convergence of U'_{\gamma} , there exists \gamma_{1} > 0 such that U'_{\gamma}(r) < 0 if r \in [\delta_{0}, 1-\delta_{0}] , for every \gamma \in (0, \gamma_{1}) . Therefore U'_{\gamma} < 0 in (0, 1) whenever \gamma < \min \{ \gamma_{0} \, , \gamma_{1} \} , which also implies that U_{\gamma} is strictly positive in [0, 1) , since U_{\gamma}(1) = 0 .

    In fact, the function U_{\gamma} is also unique within the class of weak solutions. Proposition 4.2 does not come unexpected, since the solution of (4.4) is closely related to a Bessel function, see [16]. Therefore, the number of critical points of U_{\gamma} is increasing with respect to \gamma and becomes arbitrarily large as \gamma \to \infty , whereas there are no critical points for any sufficiently small \gamma . By taking \gamma = 1 , we numerically obtain the solution displayed in the right picture of Figure 7, which represents the function u in (4.1). Several numerical simulations were performed in order to yield a radially decreasing solution of problem (4.2). The obtained results may be summarized as follows:

    ● if \gamma \leq 0.317 , then the associated solution U_{\gamma} of (4.2) is radially symmetric and strictly decreasing in the radial variable in the interval [0, 1] (and therefore, positive), see Figure 8;

    Figure 8.  Plots of U_{\gamma} in the intervals [0, 0.01] and [0, 1] , for \gamma = 0.317 .

    ● if \gamma \geq 0.318 , then the associated solution U_{\gamma} of (4.2) is radially symmetric but not strictly decreasing in the interval [0, 1] , see Figure 9.

    Figure 9.  Plots of U_{\gamma} in the intervals [0, 0.05] and [0, 1] , for \gamma = 0.318 .

    We numerically determined the position of the maximum point r_{\gamma} \in [0, 1] of U_{\gamma} as a function of \gamma \in [0, 1] , see the Table and Figure below. From these data we deduce that the map \gamma \in [0, 1] \mapsto r_{\gamma} is increasing (but not continuous!), giving thus consistency to the discussion following Proposition 4.1. Finally, in connection with the condition (3.25) given in Propositions 3.7 and 4.1, we observe that, for the unique radial solution U_{\gamma} \in H^{2}_{0}({\bf B}) of problem (4.2), the following inequality holds:

    f_{\gamma}(U_{\gamma}(r)) \geq r_{0}^{4} \left[|v(r_{0})|-\gamma \max\limits_{t \in [0,1]} |U_{\gamma}(t)| \right] \qquad \forall r \in [0,1].

    From the Table and Figure displayed below we further deduce that

    \begin{equation} f_{\gamma}(U_{\gamma}(r)) \geq r_{0}^{4} \left( |v(r_{0})|-\gamma U_{\gamma}(0) \right) \qquad \forall r \in [0,1], \quad \forall \gamma \in [0,0.317]; \end{equation} (4.6)

    recall that U_{\gamma} is nonnegative in [0, 1] for every \gamma \in [0, 0.317] . Numerical experiments yield the value of \gamma = 0.057 as a threshold for ensuring the positivity of the right-hand side of (4.6), in the sense that

    f_{0.057}(U_{0.057}(r)) \geq 0.014 \qquad \text{and} \qquad f_{0.058}(U_{0.058}(r)) \geq -0.092 \qquad \forall r \in [0,1].

    The first Author is supported by the PRIN project Direct and inverse problems for partial differential equations: theoretical aspects and applications and by INdAM. The second Author is supported by the Research Programme PRIMUS/19/SCI/01, by the program GJ19-11707Y of the Czech National Grant Agency GA\v{C}R, and by the University Centre UNCE/SCI/023 of the Charles University in Prague.

    The authors declare no conflict of interest.



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