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Remarks on radial symmetry and monotonicity for solutions of semilinear higher order elliptic equations

  • Received: 15 July 2021 Accepted: 30 September 2021 Published: 12 October 2021
  • Half a century after the appearance of the celebrated paper by Serrin about overdetermined boundary value problems in potential theory and related symmetry properties, we reconsider semilinear polyharmonic equations under Dirichlet boundary conditions in the unit ball of Rn. We discuss radial properties (symmetry and monotonicity) of positive solutions of such equations and we show that, in conformal dimensions, the associated Green function satisfies elegant reflection and symmetry properties related to a suitable Kelvin transform (inversion about a sphere). This yields an alternative formula for computing the partial derivatives of solutions of polyharmonic problems. Moreover, it gives some hints on how to modify a counterexample by Sweers where radial monotonicity fails: we numerically recover strict radial monotonicity for the biharmonic equation in the unit ball of R4.

    Citation: Filippo Gazzola, Gianmarco Sperone. Remarks on radial symmetry and monotonicity for solutions of semilinear higher order elliptic equations[J]. Mathematics in Engineering, 2022, 4(5): 1-24. doi: 10.3934/mine.2022040

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  • Half a century after the appearance of the celebrated paper by Serrin about overdetermined boundary value problems in potential theory and related symmetry properties, we reconsider semilinear polyharmonic equations under Dirichlet boundary conditions in the unit ball of Rn. We discuss radial properties (symmetry and monotonicity) of positive solutions of such equations and we show that, in conformal dimensions, the associated Green function satisfies elegant reflection and symmetry properties related to a suitable Kelvin transform (inversion about a sphere). This yields an alternative formula for computing the partial derivatives of solutions of polyharmonic problems. Moreover, it gives some hints on how to modify a counterexample by Sweers where radial monotonicity fails: we numerically recover strict radial monotonicity for the biharmonic equation in the unit ball of R4.



    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

    H(s,t)=ts0zm1(z+1)mdz=ts0(11z+1)m11z+1dz=1ss+t1ξ(1ξ)m1dξ=1ss+t[1ξ+m1k=1(1)k(m1k)ξk1]dξ=log(1+ts)+m1k=1(1)kk(m1k)(1(ss+t)k)s>0, t0.

    Therefore, by setting κmkm2m, for every x,yB we have

    G(x,y)=κm2[log(1+θ(x,y)|xy|2)+m1k=1(1)kk(m1k)(1(|xy|2|xy|2+θ(x,y))k)]. (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=(x1,x)B with x1>0, we define

    Rx1|x|22x1,yx(x1+Rx,x),Cx{yB;|yyx|>Rx}, (3.7)

    and the set

    Sx{yRn;(1|x|2)(x1y1)+x1|xy|2=0}, (3.8)

    which is a sphere with center at yx and radius Rx (see the proof of Theorem 3.2 below and also Figure 4). We then denote by Px:Rn{yx}Rn the inversion in Sx, given by the expression

    Px(y)=yx+R2xyyx|yyx|2yRn{yx}.

    Finally, the Kelvin transform of a function h:CxR with respect to Sx is defined as

    Kx(h)(y)=h(Px(y))yCx.

    Then we prove

    Theorem 3.2. For any x=(x1,x)BR2m with x1>0 and any uH2m(B)Hm0(B) we have

    1|x|2κmx1ux1(x)=Cx|yyx|2R2x|xy|2(θ(x,y)|xy|2+θ(x,y))m(Δy)m[Kx(u)(y)u(y)]dy. (3.9)

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

    Ht(s,t)=tm1(t+s)m,Hs(s,t)=tms(t+s)m,2Hst(s,t)=mtm1(t+s)m+1, (3.10)

    see also [3]. Moreover, for every x,yB we have

    Gx1(x,y)=κm[(x1y1)Hs(|xy|2,θ(x,y))x1(1|y|2)Ht(|xy|2,θ(x,y))]. (3.11)

    We then prove the announced symmetry property, namely that the zero level sets of x1G(x,) are spherical caps. Let us denote the zero level set by

    Lx{yB;Gx1(x,y)=0}.

    By (3.10) and by (3.11) we have

    1κmGx1(x,y)=x1(1|y|2)Ht(|xy|2,θ(x,y))(x1y1)Hs(|xy|2,θ(x,y))=x1(1|y|2)θ(x,y)m1(|xy|2+θ(x,y))m+(x1y1)θ(x,y)m|xy|2(|xy|2+θ(x,y))m=θ(x,y)m1|xy|2(|xy|2+θ(x,y))m(1|y|2)[x1|xy|2+(1|x|2)(x1y1)]. (3.12)

    Therefore, x1G(x,y) vanishes if and only if yB belongs to the set Sx defined in (3.8). Then we notice that yBSx if and only if

    0=1x1[(1|x|2)(x1y1)+x1|xy|2]=1|x|2x1(x1y1)+(x1y1)2+|xy|2=2Rx(x1y1)+(x1y1)2+|xy|2=(Rx+x1y1)2+|xy|2R2x=|yxy|2R2x. (3.13)

    Hence, Sx is a sphere with center at yx and radius Rx and the set of negativity of x1G(x,y) is the cap Cx defined in (3.7).

    Next, we notice that, by definition of yx and Rx, we have

    |yx|2=(x1+1|x|22x1)2+|x|2=x21+(1|x|2)+R2x+|x|2=1+R2x.

    Then we compute

    |Px(y)|2=|Px(y)yx+yx|2=|Px(y)yx|2+2(Px(y)yx)yx+|yx|2=R4x|yyx|2+2R2xyyx|yyx|2yx+|yx|2=R2x|yyx|2(R2x+2(yyx)yx+|yx|2|yyx|2R2x)=R2x|yyx|2((R2x|yyx|2)+(|yyx|2+2(yyx)yx+|yx|2)|yx|2+|yx|2|yyx|2R2x)=R2x|yyx|2((R2x|yyx|2)+|y|2|yx|2+|yx|2|yyx|2R2x)=R2x|yyx|2((R2x+|y|2|yx|2)+(|yx|2R2x)|yyx|2R2x)=R2x|yyx|2((|y|21)+|yyx|2R2x)=1R2x|yyx|2(1|y|2).

    This computation has two main consequences. First, that the unit sphere B is invariant under the inversion in Sx, that is,

    yB  Px(y)B.

    Second, that

    θ(x,Px(y))=R2x|yxy|2θ(x,y)yB. (3.14)

    We next claim that if Cx is the cap defined in (3.7), then B=CxPx(Cx)Lx. To this end, it suffices to show that

    Px(Cx)={yB;|yyx|<Rx}.

    Indeed, let yCx, so that Px(y)Px(Cx) and |yyx|>Rx. By definition of Px we have

    |Px(y)yx|=R2x|yyx|<Rx,

    thus proving that Px(Cx){yB;|yyx|<Rx}. On the other hand, let yB be such that |yyx|<Rx. Since the function Px is bijective, there exists a point y0Rn such that Px(y0)=y. By definition, we then have

    |y0yx|=R2x|yyx|>Rx,

    implying y0Cx, and subsequently, that yPx(Cx) and {yB;|yyx|<Rx}Px(Cx).

    Since B=CxPx(Cx)Lx, by Proposition 3.1, we have

    ux1(x)=CxGx1(x,y)(Δy)mu(y)dy+Px(Cx)Gx1(x,y)(Δy)mu(y)dy=CxGx1(x,y)(Δy)mu(y)dy+Cx(Rx|yyx|)2nGx1(x,Px(y))((Δy)mu)(Px(y))dy=CxGx1(x,y)(Δy)mu(y)dy+Cx(Rx|yyx|)4mGx1(x,Px(y))((Δy)mu)(Px(y))dy,

    where we used that the Jacobian matrix Px satisfies

    |det(Px(y))|=(Rx|yyx|)2nyB.

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

    1κm1|x|2x1Gx1(x,y)=|yxy|2R2x|xy|2(θ(x,y)θ(x,y)+|xy|2)myB, (3.15)

    so that

    1κm1|x|2x1ux1(x)=I1(x)+I2(x), (3.16)

    where

    To compute I2(x), we first notice that

    |yxPx(y)|2R2x=R4x|yxy|2R2x=R2x|yxy|2(|yxy|2R2x)<0yCx, (3.17)

    whereas, for every yB we have

    |xPx(y)|2=|xyx|2+2(xyx)(yxPx(y))+|yxPx(y)|2=|xyx|2+2R2x|yxy|2(xyx)(yxy)+R4x|yyx|2=R2x|yxy|2(|yxy|2R2x|xyx|2+2(xyx)(yxy)+R2x)=R2x|yxy|2(|yxy|2+2(xyx)(yxy)+|xyx|2)=R2x|yxy|2|xy|2. (3.18)

    After plugging (3.14), (3.17) and (3.18) into I2(x) we get

    I2(x)=Cx|yyx|2R2x|xy|2(θ(x,y)|xy|2+θ(x,y))m(Rx|yyx|)4m((Δy)mu)(Px(y))dy

    and, after recalling (3.16), we deduce

    1κm1|x|2x1ux1(x)=Cx|yyx|2R2x|xy|2(θ(x,y)|xy|2+θ(x,y))m[(Δy)mu(y)(Rx|yyx|)4m((Δy)mu)(Px(y))]dy.

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

    (Δy)mKx(u)(y)=(Rx|yyx|)4m((Δy)mu)(Px(y))yCx, (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 n2, n2m. This can be seen after computing

    Hs(s,t)=(nm2)smn21ts0zm1(z+1)n/2dztms(t+s)n/2s>0, t0,

    so that the determination of the level sets of x1G(x,) as in (3.12) becomes a delicate task. Is it possible to obtain a representation formula similar to (3.9) in any dimension n2?

    Clearly, a representation formula similar to (3.9) holds for any other directional derivative of u. Note that (3.7) may be rewritten as yx=x+(Rx,0), which implies that yx is an "horizontal translation" of xB. Moreover, Rx0 if xB while Rx+ if x10: the limit of the map xRx does not exist when x approaches the equator B{x1=0}. The level surfaces of xRx are spheres:

    Rx=k(0,)  (x1+k)2+|x|2=1+k2.

    The center of the level-surface sphere is then (x1,x)=(k,0) and its radius is 1+k2. For a fixed k>0, when x runs over the level surface Rx=k, the corresponding yx, defined by (3.7), runs over a portion of the sphere centered in the origin with the same radius 1+k2, see Figure 3.

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

    Note also that Cx, the domain of negativity of x1G(x,y), is the intersection between two balls. In Figure 4 we sketch the mutual position of the sphere Sx and the ball B. In particular, we emphasize that

    |yx|2=(x1+1|x|22x1)2+|x|2=1+(1|x|2)24x21>1xB(x1>0)

    which, again, tells us that yx is exterior to B and it approaches B as x approaches B, while it goes to infinity if x approaches the plane x1=0.

    Figure 4.  Representation of the sphere Sx for different positions of xB.

    Theorem 3.2 has several relevant consequences. Firstly, we deduce

    Corollary 3.4. Let m1. For every xBR2m such that x1>0, the following identities hold

    (1|x|2)m=1πm(2m1)!(m1)!Cx|yyx|2R2x|xy|2(θ(x,y)|xy|2+θ(x,y))m[1(Rx|yyx|)4m]dy, (3.20)
    Kmπm(m1)!(2m1)!=Cx|yyx|2R2x|xy|2(1|y|2|xy|2+θ(x,y))m[1(Rx|yyx|)4m]dy. (3.21)

    Proof. We start by noticing that

    Δr2m=2m(2m+n2)r2m2,Δ2r2m=2m(2m2)(2m+n2)(2m+n4)r2m4,Δ3r2m=2m(2m2)(2m4)(2m+n2)(2m+n4)(2m+n6)r2m6,

    where n=2m. By induction we then obtain

    Δmr2m=(2m)!!(2m+n2)!!(n2)!!=(2m)!!(4m2)!!(2m2)!!=22m1(2m)!.

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

    {(Δ)mu=1in Bu=ur==m1urm1=0on B. (3.22)

    A direct application of the binomial expansion yields

    (Δ)m(1r2)m=(Δ)m[mk=0(mk)(1)kr2k]=Δmr2m=22m1(2m)!,

    that is

    (Δ)mUm(r)=1withUm(r)(1r2)m22m1(2m)!r[0,1]. (3.23)

    Since Um satisfies the Dirichlet boundary conditions in (3.22)2, it is a strong C(¯B)-solution of (3.22). Applying identity (3.9) to the function Um 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 θ(x,y).

    It is remarkable that the right hand side of (3.21) does not depend on xB. 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 BR2m be the unit ball. Suppose that fC(R;R) and that uHm0(B)L(B) is a strong solution of (1.2). Then, for every xB such that x1>0, the following formula holds:

    |x|21κmx1ux1(x)=Cx|yyx|2R2x|xy|2[θ(x,y)|xy|2+θ(x,y)]m[f(u(y))(Rx|yyx|)4mf(u(Px(y)))]dy. (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

    ux1(x)<0for all xB  with  x1>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 Cx is determined by the behavior of the quantity

    Φx(y)f(u(y))(Rx|yyx|)4mf(u(Px(y)))yCx.

    Fixing xB such that x1>0 and uHm0(B), notice that Φx(y)=0 for all yCxB, while for yCxB (so that Px(y)B) we have that Φx(y)=0 if f(0)=0, and Φx(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 x1u(x)<0 for all xB located far away from B, with x1ε for some ε>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 B in which the strict negativity of x1u(x) holds.

    To this end, we define

    1ΓinfvC2m(¯B)Hm0(B){0}ΔmvL(B)vL(B),

    and we state:

    Proposition 3.7. Assume that fW1,(R+;R+) satisfies

    f(s)Ms0,  for some  M>0, (3.25)

    and let uHm0(B)L(B) be a strong solution of (1.2). For any point x=(x1,x)B with x1>0 we have

    |x|21κmx1ux1(x)(M2ΓfL(R+)fL(R+))Km(1|x|2)m2ΓfL(R+)fL(R+)1|x|2κmx1Cx(Rx|yyx|)4mGx1(x,Px(y))dy,

    where Km>0 is defined in (3.21).

    Proof. In view of Corollary 3.4, by the Maximum Principle and assumption (3.25), any strong solution uHm0(B)L(B) of (1.2) satisfies

    u(x)M22m1(2m)!(1|x|2)mxB.

    By embedding theorems and elliptic regularity we also have that

    uL(B)ΓfL(R+).

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

    |u(Px(y))u(y)|ΓfL(R+)|Px(y)y|yCx.

    From this we deduce that

    |f(u(Px(y)))f(u(y))|ΓfL(R+)fL(R+)|Px(y)y|yCx,

    thus yielding

    f(u(Px(y)))f(u(y))+ΓfL(R+)fL(R+)|Px(y)y|yCx

    and, by using (3.25), we finally obtain

    f(u(y))(Rx|yyx|)4mf(u(Px(y)))[1(Rx|yyx|)4m]M(Rx|yyx|)4mΓfL(R+)fL(R+)|Px(y)y|.

    By inserting this bound within (3.24), we get

    |x|21κmx1ux1(x)I1(x)I2(x), (3.26)

    where

    From Corollary 3.4 we infer that

    I1(x)=MKm(1|x|2)m. (3.27)

    In order to bound I2(x), we notice that

    |Px(y)y|<2andRx|yyx|<1yCx.

    Hence,

    I2(x)2ΓfL(R+)fL(R+)Cx|yyx|2R2x|xy|2[θ(x,y)|xy|2+θ(x,y)]mdy. (3.28)

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

    Cx|yyx|2R2x|xy|2[θ(x,y)|xy|2+θ(x,y)]mdy=1κm1|x|2x1CxGx1(x,y)dy=1κm1|x|2x1(BGx1(x,y)dyPx(Cx)Gx1(x,y)dy)=1κm1|x|2x1[122m1(2m)!x1((1|x|2)m)Cx(Rx|yyx|)4mGx1(x,Px(y))dy]=Km(1|x|2)m+1|x|2κmx1Cx(Rx|yyx|)4mGx1(x,Px(y))dy.

    After plugging this into (3.28) we deduce that

    I2(x)2ΓfL(R+)fL(R+)[Km(1|x|2)m+1|x|2κmx1Cx(Rx|yyx|)4mGx1(x,Px(y))dy]. (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 fL(R+) is small. This assumption is not needed if m=1 (see Theorem 1.1), while for m2, 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(R+) small. This issue is tackled numerically in the next section, for the biharmonic equation in BR4.

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

    Proposition 4.1. Assume that fW1,(R+;R+) satisfies (3.25), and let uHm0(B)L(B) be a strong solution of (1.2). Then, there exists γf>0 such that

    xu(x)<0for all xB such that γf<|x|<1.

    Guido Sweers [16] provided an explicit example of a non-decreasing radial solution of a linear biharmonic problem in BR2 (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 γf can be made arbitrarily small (possibly zero) provided that (f)L(R+) is sufficiently small.

    Consider the following radial function defined in R4 (the conformal dimension for the biharmonic operator):

    v(x)=k=0akr4k+2k=0(1)k24k+2r4k+2(2k+2)!(2k+1)!xR4,

    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

    Δ2ψ(r)=ψ(4)(r)+6rψ(r)+3r2ψ(r)3r3ψ(r)r(0,1).

    After noticing that

    ak+1=164(k+2)(k+1)(2k+3)2akkN,

    one readily sees that

    Δ2v=vinR4.

    Let r09.2218 be the first nonzero local minimum of v; numerically, we find that v(r0)14.8388. Then define the function

    u(r)v(r0r)v(r0)r[0,1], (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 γ=1.

    By defining

    f(w)=r40(|v(r0)|w)w0,

    we observe that f(w)r40<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:

    {Δ2u=f(u)inBu=ur=0onB.

    The solution is radially symmetric and a simple computation shows that it coincides with u defined in (4.1). Note that u(0)=|v(r0)| and u(0.5)>16.5>|v(r0)|, so that f(u) is sign-changing in [0,1].

    We now modify f in order to reduce (f)L(R+). We take

    fγ(w)r40(|v(r0)|γw)w0,

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

    {Δ2U=fγ(U)inBU=Ur=0onB, (4.2)

    and prove the following result:

    Proposition 4.2. For every γ>0, the unique strong solution UγH20(B)L(B) of (4.2) (which is radially symmetric) is analytic and can be written as

    Uγ(r)=k=0Ak(γ)r2k=A0(γ)+A1(γ)r2+A2(γ)r4+A3(γ)r6+...r[0,1], (4.3)

    for some coefficients {Ak(γ)}kNR that satisfy the following properties:

    {A0(γ)=k=2(k1)Ak(γ),A1(γ)=k=2kAk(γ),A2(γ)=r40192(|v(r0)|γA0(γ)),Ak(γ)=γr4016k2(k21)Ak2(γ)k3.

    Moreover, there exists γ0>0 such that, for every γ(0,γ0), the function Uγ is positive in [0,1) and strictly decreasing in the radial variable.

    Proof. For every γ>0 we notice that the condition (2.2) is satisfied. Then, Theorem 2.1 guarantees the existence of at most one strong solution UγH20(B)L(B) to problem (4.2) which is, moreover, radially symmetric and analytic in ¯B. Problem (4.2) in radial coordinates reads:

    {U(4)γ(r)+6rUγ(r)+3r2Uγ(r)3r3Uγ(r)=r40(|v(r0)|γUγ(r))r(0,1)Uγ(1)=Uγ(1)=0. (4.4)

    Upon substitution of (4.3) into (4.4)1 we obtain

    192A2(γ)+16k=3Ak(γ)k2(k21)r2k4=r40(|v(r0)|γA0(γ))γr40k=3Ak2(γ)r2k4r(0,1),

    which yields the identities

    A2(γ)=r40192(|v(r0)|γA0(γ)),Ak(γ)=γr4016k2(k21)Ak2(γ)k3. (4.5)

    Moreover, the boundary conditions in (4.4)2 imply that

    k=0Ak(γ)=0, k=1kAk(γ)=0  A1(γ)=k=2kAk(γ), A0(γ)=k=2(k1)Ak(γ).

    In connection with (3.23) we define

    U(r)σ0(1r2)2r40|v(r0)|64(1r2)2r[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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