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2024年2月29日 星期四

Solution to Partial Differential Equations: Methods and Applications (Robert McOwen) Section 4.2

Solution to Partial Differential Equations: Methods and Applications (Robert McOwen) Section 4.2

    1. If n=2 and a=1, show that (44) is equivalent to

      u(r,θ)=1r22π2π0g(ϕ)1+r22rcos(θϕ)dϕ.

    2. Use (a) and Exercise 1 in Section 4.1 to verify the formula

      rkcoskθ=1r22π2π0cos(kϕ)1+r22rcos(θϕ)dϕ,

      where k is an integer and 0r<1.
  1. Solution
    1. When n=2 and a=1, the scircumference of the unit disk is 2π, and (44) becomes

      u(ξ1,ξ2)=1(ξ21+ξ22)2πx2+y2=1g(x,y)(xξ1)2+(yξ2)2dSx,y.

      Using the polar coordinate x=cosϕ and y=sinϕ for 0ϕ<2π; ξ1=rcosθ and ξ2=rsinθ for 0r1 and 0θ<2π, we define ˜u(r,θ)=u(rcosθ,rsinθ)=u(ξ1,ξ2) and ˜g(ϕ)=g(cosϕ,sinϕ)=g(x,y). Then we have dSx,y=dϕ and

      ˜u(r,θ)=1r22π2π0˜g(ϕ)(cosϕrcosθ)2+(sinϕrsinθ)2dϕ=1r22π2π0˜g(ϕ)1+r22rcosθcosϕ2rsinθsinϕdϕ=1r22π2π0˜g(ϕ)1+r22rcos(θϕ)dϕ.

      The proof is complete.
    2. Consider the following Dirichlet problem

      {Δu=0in B1(0),u(1,θ)=cos(kθ)for 0θ<2π.

      Then using Exercise 1 in Section 4.1, we find

      u(r,θ)=12π2π0g(s)ds+1πn=1rn[cos(nθ)2π0g(s)cos(ns)ds+sin(nθ)2π0g(s)sin(ns)ds]=12π2π0cos(ks)ds+1πn=1rn[cos(nθ)2π0cos(ks)cos(ns)ds+sin(nθ)2π0cos(ks)sin(ns)ds]=rkcos(kθ).

      Here we have used the fact that 2π0cos(ks)ds=2π0cos(ks)sin(ns)ds=0 and 2π0cos(ks)cos(ns)ds=δks for kN, where δkn=1 if k=n; 0 if kn. On the other hand, by (a), we obtain the formula

      rkcoskθ=u(r,θ)=1r22π2π0cos(kϕ)1+r22rcos(θϕ)dϕ.

      The proof is complete.

  2. Let Ω be a bounded domain and fCk(ˉΩ). Show that in fact the domain potential (29a) satisfies uCk+1(Ω). Conclude that fC(ˉΩ) implies uC(Ω). [In Section 8.2 we show that fC(ˉΩ) implies uC(ˉΩ), provided Ω is C.]
  3. SolutionClearly, the statement holds true for k=1 by (iii) of Proposition at page 114 in the textbook. We firstly show that the statement holds true for k=2, i.e., fC2(ˉΩ) implies uC3(Ω). For any i{1,,n}, from the argument of Proposition at page 114, we have

    iu(x)=Ω(iK)(xy)f(y)dyfor xΩ.

    It suffices to show iuC2(Ω). To achieve this, we consider the Poisson equation Δϕ=if in Ω. Since ifC1(ˉΩ) (from fC2(ˉΩ)), then by the Proposition at page 114, we know the solution ϕC2(Ω) and

    ϕ(x)=ΩK(xy)(if)(y)dy.

    Using the integration by parts, we have

    ϕ(x)=ΩK(xy)f(y)νi(y)dSy+Ω(iK)(xy)f(y)dy.

    For any yΩ, the function ΩK(y)f(y)νi(y)dSy is smooth. Hence, we find

    ui(x)=ϕ(x)ΩK(xy)f(y)νi(y)dSyfor xΩ,

    which means iuC2(Ω) and uC3(Ω).

    Now we prove the statement by mathematical induction. Suppose the statement holds true for some positive integer m2, i.e., fCm(ˉΩ) implies uCm+1(Ω). Since fCm+1(ˉΩ), we have fiCm for any i{1,,n}. Note that iuCm(Ω) and satisfies Δ(iu)=if in Ω. Here we have used the fact that m2. Then by the inductive hypothesis, iuCm+1(Ω). Due to the arbitrariness of i, we know uCm+2(Ω). Therefore, by mathematical induction, the statement holds true for all nN.

    When fC(ˉΩ), we have fCm(ˉΩ) for all mN, which gives uCm+1(Ω) for all mN. Thus, we know uC(Ω). The proof is complete.

  4. The symmetry of the Green's function [i.e., G(x,ξ)=G(ξ,x) for all x,ξΩ] is an important fact connected with the self-adjointness of Δ.
    1. Verify the symmetry of G(x,ξ) by direct calculation when Ω=Rn+ and Ω=Ba(0).
    2. Prove the symmetry of G(x,ξ) when Ω is any smooth, bounded domain.
  5. Solution
    1. Recall that K(x)={12πlog|x|if n=2;1(2n)ωn|x|2nif n3. (cf. (23)).
      • When Ω=Rn+, the Green's function G is defined in (35), i.e.,

        G(x,ξ)=K(xξ)K(xξ)={log|xξ|log|xξ|2πif n=2;|xξ|2n|xξ|2n(2n)ωnif n3.

        Here ξ=(ξ1,,ξn1,ξn) denotes the reflection point of ξ=(ξ1,,ξn1,ξn) with respect to the plane Rn+. Clearly, |xξ|=|ξx| and

        |xξ|=n1i=1(xiξi)2+[xn(ξn)]2=n1i=1(ξixi)2+[ξn(xn)]2=|ξx|,

        where x=(x1,,xn1,xn) denotes the reflection of x with respect to the plane Rn+. Therefore, we verify that

        G(x,ξ)=K(xξ)K(xξ)=K(ξx)K(ξx)=G(ξ,x) for all x,ξRn+ and xξ.

      • When Ω=Ba(0), the Green's function G is defined in (42), i.e.,

        G(x,ξ)=K(xξ)K(|ξ|a(xξ))={12π[log|xξ|log(|ξ|a|xξ|)]if n=2;1(2n)ωn[|xξ|2n(|ξ|a|xξ|)2n]if n=3.

        Here ξ=a2ξ|ξ|2 denotes the reflection of ξ0 with respect to the sphere Ba(0). Clearly, |xξ|=|ξx| and

        ||ξ|a(xξ)|2=ni=1ξ2ia2(ni=1x2i2a2ni=1xiξinj=1ξ2j+a4ni=1ξ2i(nj=1ξ2j)2)=ni=1x2ia2(ni=1ξ2i2a2ni=1xiξinj=1x2j+a4ni=1x2i(nj=1x2i)2)=||x|a(ξx)|2,

        where x=a2x|x|2 for x0. Therefore, we verify that

        G(x,ξ)=K(xξ)K(|ξ|a(xξ))=K(ξx)K(|x|a(ξx))=G(ξ,x)for all x,ξBa(0){0} and xξ.

    2. Recall that the notion of Green's function G is defined at page 117: G(x,ξ)=K(xξ)+ω(x) for xξ. Hence G satisfies ΔxG(x,ξ)=0 for xΩ{ξ} with G(x,ξ)=0 for xΩ (cf. (32)). Fix x,ξΩ arbitrarily with xξ. We want to prove G(x,ξ)=G(ξ,x). For this purpose, we define u:Ω{x}R and v:Ω{ξ}R by

      u(z)=G(z,x)for zΩ{x},v(z)=G(z,ξ)for zΩ{ξ}.

      It suffices to show u(ξ)=v(x). Since x and ξ are fixed, there exists a positive constant ϵ>0 such that Bϵ(x)Ω, Bϵ(ξ)Ω and ¯Bϵ(x)¯Bϵ(ξ)=. Let Ωϵ=Ω(Bϵ(x)Bϵ(ξ)). By the definition of Green's function, Δu(z)=0 for zΩ{x} and Δv(z)=0 for zΩ{ξ}, which implies

      Ωϵ(vΔuuΔv)dz=0.

      Using the Green's second identity (7) at page 107, we have

      0=Ωϵ(vΔuuΔv)dz=Ωϵ(vuνuvν)dSz=(Ω+Bϵ(x)+Bϵ(ξ))(vuνuvν)dSz.

      In addition, we note the fact that u(z)=G(z,x)=0 for zΩ and v(z)=G(z,ξ)=0 for zΩ (from (32)). Hence we get

      Bϵ(x)(vuνuvν)dSz=Bϵ(ξ)(uvνvuν)dSz,

      where ν is the unit vector pointing the inward on Bϵ(x)Bϵ(ξ). Since v is smooth near x and |u(z)|=|G(z,x)||K(zx)|+|ω(z)|, we find

      |Bϵ(x)uvνdSz|nωnϵn1max.

      Here we have used the fact that

      \displaystyle\lim_{\epsilon\to0^+}\epsilon^{n-1}K(|\epsilon|)=\begin{cases}\displaystyle\lim_{\epsilon\to0^+}\frac{\epsilon\log\epsilon}{2\pi}=0&\text{if}~n=2;\\\displaystyle\lim_{\epsilon\to0^+}\frac{\epsilon^{n-1}\epsilon^{2-n}}{(2-n)\omega_n}=0&\text{if}~n\geq3.\end{cases}

      On the other hand, by the smoothness of \omega and (22)-(23), we have

      \displaystyle\lim_{\epsilon\to0^+}\int_{\partial B_\epsilon(x)}\!v\frac{\partial u}{\partial\nu}\,\mathrm dS_z=\lim_{\epsilon\to0^+}\int_{\partial B_\epsilon(x)}v(z)\frac{\partial K}{\partial\nu}(z-x)\,\mathrm dS_z=v(x).

      Hence we see \displaystyle\lim_{\epsilon\to0^+}\int_{\partial B_\epsilon(x)}\left(v\frac{\partial u}{\partial\nu}-u\frac{\partial v}{\partial\nu}\right)\,\mathrm dS_z=v(x). Similarly, one may check that \displaystyle\lim_{\epsilon\to0^+}\int_{\partial B_\epsilon(\xi)}\left(u\frac{\partial v}{\partial\nu}-v\frac{\partial u}{\partial\nu}\right)\,\mathrm dS_z=u(\xi). Therefore, we arrive at u(\xi)=v(x) and complete the proof.


    1. Use the weak maximum principle (16) to prove that G(x,\xi)\leq0 for x,\xi\in\Omega with x\neq\xi.
    2. Use the strong maximum principle (15) to prove that G(x,\xi)<0 for x,\xi\in\Omega with x\neq\xi.
  6. Solution
    1. Fix \xi\in\Omega. From the definition of Green's function G at page 117, we write G(x,\xi)=K(x-\xi)+\omega(x), where \omega satisfies \Delta\omega=0 in \Omega and \omega(x)=-K(x-\xi) for x\in\partial\Omega. Since \omega is bounded in \bar\Omega and \displaystyle\lim_{x\to\xi}K(x-\xi)=-\infty, there exists \epsilon>0 such that \overline{B_\epsilon(\xi)}\subsetneq\Omega and G(x,\xi)<0 for x\in\overline{B_\epsilon(\xi)} and x\neq\xi. Let \Omega_{\epsilon,\xi}=\Omega-\overline{B_\epsilon(\xi)}. Then we find

      \begin{aligned} &\Delta_x(x,\xi)=0\quad\text{in}~\Omega_{\epsilon,\xi},\\&G(x,\xi)\leq0\quad\text{for}~x\in\partial\Omega_{\epsilon,\xi}=\partial\Omega\cup\partial B_\epsilon(\xi).\end{aligned}

      Then by the weak maximum principle (16), we obtain

      \displaystyle\max_{x\in\bar\Omega_{\epsilon,\xi}}G(x,\xi)=\max_{x\in\partial\Omega_{\epsilon,\xi}}G(x,\xi)=\max\left\{\max_{x\in\partial\Omega}G(x,\xi),\max_{x\in\partial B_\epsilon(\xi)}G(x,\xi)\right\}=0,

      which implies G(x,\xi)\leq0 for x\in\bar\Omega_{\epsilon,\xi}. Since G(x,\xi)<0 for x\in\overline{B_\epsilon(\xi)}-\{\xi\}, we arrive at G(x,\xi)<0 for x\neq\xi.
    2. Following the process of part (a), G is not constant because G(x,\xi)=0 for x\in\partial\Omega and G(x,\xi)<0 for x\in\partial B_\epsilon(\xi). Then by Theorem 3 (Maximun Principle) at page 109, we obtain

      \displaystyle G(x,\xi)<\sup_{z\in\Omega}G(z,\xi)=\max_{z\in\bar\Omega}G(z,\xi)=0\quad\text{for}~x\in\Omega~\text{but}~x\neq\xi.

      The proof is complete.

  7. Use (33) to prove (38).
  8. SolutionLet \Omega_R=B_R(0)\cap\mathbb R_+^n=\{x=(x_1,\dots,x_n)\in\mathbb R^n\,:\,|x|<R,\,x_n>0\} for any R>0 and u\equiv1 in \bar{\mathbb R}_+^n. Using (31) and Green's function G defined by (35), we have

    \begin{aligned}1&=\int_{\Omega_R}\!G(x,\xi)\Delta u\,\mathrm dx+\int_{\partial\Omega_R}\!\left(u(x)\frac{\partial G(x,\xi)}{\partial\nu_x}-G(x,\xi)\frac{\partial u(x)}{\partial\nu}\right)\,\mathrm dS_x\\&=\int_{\partial\Omega_R}\!\frac{\partial G(x,\xi)}{\partial\nu_x}\,\mathrm dS_x=\left(\int_{\partial\Omega_R\cap(\mathbb R^{n-1}\times\{0\})}+\int_{\partial\Omega_R\cap\mathbb R_+^n}\right)\frac{\partial G(x,\xi)}{\partial\nu_x}\,\mathrm dS_x\quad\text{for}~\xi\in\Omega_R.\end{aligned}

    For x\in\partial\Omega_R\cap(\mathbb R^{n-1}\times\{0\}), we have

    \displaystyle\frac{\partial G(x,\xi)}{\partial\nu_x}=\left.-\frac{\partial G}{\partial x_n}(x,\xi)\right|_{x_n=0}=\frac{2\xi_n}{\omega_n}|x'-\xi|^{-n},

    where x'=(x_1,\dots,x_{n-1})\in\mathbb R^{n-1} (cf. (36)). For x\in\partial\Omega_R\cap\mathbb R_+^n, we have

    \begin{aligned}\frac{\partial G(x,\xi)}{\partial\nu_x}&=\nabla_xG(x,\xi)\cdot\frac{x}{|x|}=\frac1{\omega_n}\left(|x-\xi|^{-n}(x-\xi)-|x-\xi^*|^{-n}(x-\xi^*)\right)\cdot\frac{x}R\\&=\frac1{\omega_nR}\left(\frac{R^2-\xi\cdot x}{|x-\xi|^n}-\frac{R^2-\xi^*\cdot x}{|x-\xi^*|^n}\right)\\&=\left(\frac{R}{\omega_n}-\sum_{i=1}^{n-1}\frac{\xi_ix_i}{\omega_nR}\right)\left(\frac1{|x-\xi|^n}-\frac1{|x-\xi^*|^n}\right)-\frac{\xi_nx_n}{\omega_nR}\left(\frac1{|x-\xi|^n}+\frac1{|x-\xi^*|^n}\right).\end{aligned}

    A direct estimate gives

    \displaystyle\left|\int_{\partial\Omega_R\cap\mathbb R_+^n}\frac{\xi_nx_n}{\omega_nR}\left(\frac1{|x-\xi|^n}+\frac1{|x-\xi^*|^n}\right)\,\mathrm dS_x\right|\leq\frac{|\xi_n|}{\omega_n}\cdot\frac2{(R-|\xi|)^n}\cdot\frac{\omega_nR^{n-1}}2=\frac{|\xi|R^{n-1}}{R-|\xi|^n}\to0\quad\text{as}~R\to\infty.

    Here we have used the fact that |x-\xi|\geq|x|-|\xi|=R-|\xi| and |x-\xi^*|\geq|x|-|\xi^*|=R-|\xi| for x\in\partial B_R(0). Moreoever, by the trinagle inequality and x\in\partial B_R(0)\cap\mathbb R_+^n, we have R-|\xi|\leq|x-\xi|\leq|x-\xi^*|\leq R+|\xi|, which implies

    \displaystyle\frac1{|x-\xi|^n}-\frac1{|x-\xi^*|^n}\leq\frac1{(R-|\xi|)^n}-\frac1{(R+|\xi|)^n}=\frac{2nR^{n-1}|\xi|+\cdots}{(R^2-|\xi|^2)^n}.

    Hence we get

    \displaystyle\left|\int_{\partial\Omega_R\cap\mathbb R_+^n}\!\left(\frac R{\omega_n}-\sum_{i=1}^{n-1}\frac{\xi_ix_i}{\omega_nR}\right)\left(\frac1{|x-\xi|^n}-\frac1{|x-\xi^*|^n}\right)\,\mathrm dS_x\right|\leq\left(\frac R{\omega_n}+\sum_{i=1}^{n-1}\frac{|\xi_i|}{\omega_n}\right)\cdot\frac{2nR^{n-1}|\xi|+\cdots}{(R^2-|\xi|^2)^n}\cdot\frac{\omega_nR^{n-1}}2

    Combining these results, it follows that \displaystyle\lim_{R\to\infty}\int_{\partial\Omega_R\cap\mathbb R_+^n}\!\frac{\partial G(x,\xi)}{\partial\nu_x}\,\mathrm dS_x=0. Therefore, we obtain

    \displaystyle1=\lim_{R\to\infty}\int_{\partial\Omega_R}\!\frac{\partial G(x,\xi)}{\partial\nu_x}\,\mathrm dS_x=\lim_{R\to\infty}\int_{\partial\Omega_R\cap(\mathbb R^{n-1}\times\{0\})}\!\frac{\partial G(x,\xi)}{\partial\nu_x}\,\mathrm dS_x=\int_{\mathbb R^{n-1}\times\{0\}}H(x',\xi)\,\mathrm dx'\quad\text{for}~x\in\mathbb R_+^n.

    The proof is complete.

    Warning: Note that we should not use (33) to prove (38) because Green's function G defined in (35) is for half-plane \mathbb R_+^n but not B_R(0)\cap\mathbb R_+^n. Particularly, G\equiv0 on \partial\Omega_R\cap(\mathbb R^{n-1}\times\{0\}) but G may not becomes zero on \partial\Omega_R\cap\mathbb R_+^n. However, since u\equiv1, the normal derivative of u becomes zero so the boundary value of function G is not important.


  9. For n=2, use the method of reflections to find the Green's function for the first quadrant \Omega=\{(x,y)\,:\,x,y>0\}.
  10. SolutionLet \xi=(\xi_1,\xi_2)\in\Omega, i.e., \xi_1 and \xi_2 are positive. Define \xi^*=(-\xi_1,\xi_2). Then \xi^*, -\xi and -\xi^* are in the second, the third and the fourth quadrant, respectively. Now we claim

    \begin{aligned}G(x,\xi)&=G(x_1,x_2,\xi_1,\xi_2)=K(x-\xi)-K\left(\frac{|x+\xi^*||x-\xi^*|}{|x+\xi|}\right)\\&=\frac{\log|x-\xi|-\log|x-\xi^*|-\log|x+\xi^*|+\log|x+\xi|}{2\pi}\\&=\frac1{2\pi}\log\frac{|x-\xi||x+\xi|}{|x-\xi^*||x+\xi^*|}\quad\text{for}~x=(x_1,x_2)\in\Omega~\text{but}~x\neq\xi\end{aligned}

    is a Green's function for the first quadrant \Omega. Since \xi^*, -\xi and -\xi^* are not in the first quadrant, \Delta_xG(x,\xi)=0 for x=(x_1,x_2)\in\Omega except \xi. In addition, for x_1=0, we have |x-\xi|=|x-\xi^*| and |x+\xi|=|x+\xi^*|; for x_2=0, we have |x-\xi|=|x+\xi^*| and |x+\xi|=|x-\xi^*|. Thus, for x\in\partial\Omega, it is necessary that \displaystyle\frac{|x-\xi||x+\xi|}{|x-\xi^*||x+\xi^*|}=1, which implies G(x,\xi)=0. Therefore, by the definition, G is a Green's function for the first quadrant \Omega.

  11. If u\in C(\Omega) satisfies the mean value property of Section 4.1.d, then u is harmonic in \Omega.
  12. SolutionLet \xi\in\Omega. Then there exists a positive constant r such that \overline{B_r(\xi)}\subsetneq\Omega. Since u\in C(\Omega) satisfies the mean value property, we have

    \displaystyle u(\xi)=\frac1{\omega_n}\int_{|x|=1}\!u(\xi+rx)\,\mathrm dS_x.

    Consider the Laplace equation \Delta v=0 in B_r(\xi) with the Dirichlet boundary condition v=u\Big|_{\partial B_r(\xi)}. Clearly, u\Big|_{\partial B_r(\xi)} is continuous on \partial B_r(\xi). Then by Theorem 4 at page 122, v can be solved by Poisson integral formula (44) and v\in C^2(B_r(\xi))\cap C(\overline{B_r(\xi)}) is harmonic. Thus, v also satisfies the mean value property. Using the Exercise 9 in Section 4.1, we obtain v\equiv u in \overline{B_r(\xi)}. Thus, u\in C^2(B_r(\xi))\cap C(\overline{B_r(\xi)}) and \Delta u=0 in B_r(\xi). Due to arbitrariness of \xi, \Delta u=0 in \Omega. The proof is complete.

  13. Let \Omega=B_a(0), \Omega_+=\Omega\cap\mathbb R_+^n, and \Omega_0=\{x\in\Omega\,:\,x_n=0\}. If u\in C^2(\Omega_+)\cap C(\Omega_+\cup\Omega_0) is harmonic in \Omega_+, and u=0 on \Omega_0, prove that u may be extended ot a harmonic function on all of \Omega. (This is called a reflection principle.)
  14. SolutionDefine the extension \tilde u:\Omega\to\mathbb R of u by

    \tilde u(x)=\tilde u(x',x_n)=\begin{cases}u(x',x_n)&\text{if}~x_n\geq0;\\-u(x',-x_n)&\text{if}~x_n<0.\end{cases}

    Here x' denotes (x_1,\dots,x_{n-1}). Since u\equiv0 on \Omega_0, function \tilde u is continuous on \Omega. By Exercise 7, it suffices to show that \tilde u satisfies the mean value property: For any \xi\in\Omega, there exists a positive constant r (which may be dependent on \xi) such that \overline{B_r(\xi)}\subsetneq\Omega=B_a(0) and

    \displaystyle\tilde u(\xi)=\frac1{\omega_n}\int_{|x|=1}\!\tilde u(\xi+rx)\,\mathrm dS_x.

    Since \tilde u\equiv u in \Omega_+, for \xi\in\Omega_+, there exists r>0 such that \overline{B_r(\xi)}\subsetneq\Omega_+\subsetneq\Omega, then we have

    \displaystyle\tilde u(\xi)=u(\xi)=\frac1{\omega_n}\int_{|x|=1}\!u(\xi+|\xi_n|x)\,\mathrm dS_x=\frac1{\omega_n}\int_{|x|=1}\!\tilde u(\xi+|\xi_n|x)\,\mathrm dS_x.

    Similarly, for \xi\in\Omega-(\Omega_+\cup\Omega_0), there exists r>0 such that \overline{B_r(\xi)}\subsetneq\Omega-(\Omega_+\cup\Omega_0)\subsetneq\Omega, we have

    \begin{aligned}\tilde u(\xi)&=-u(\xi',-\xi_n)=-\frac1{\omega_n}\int_{|x|=1}\!u((\xi',-\xi_n)+|\xi_n|x)\,\mathrm dS_x=\frac1{\omega_n}\int_{|x|=1}\!\tilde u(\xi+|\xi_n|(x',-x_n))\,\mathrm dS_x\\&=\frac1{\omega_n}\int_{|x|=1}\!\tilde u(\xi+|\xi_n|(x',-x_n))\,\mathrm dS_x.\end{aligned}

    Here we have used the symmetry of the integral over the surface |x|=1. For \xi\in\Omega_0 and 0<r<a-|\xi|, we observe that

    \begin{aligned}\frac1{\omega_n}\int_{|x|=1}\!\tilde u((\xi',0)+rx)\,\mathrm dS_x&=\frac1{\omega_n}\int_{|x|=1, x_n>0}\!u(\xi'+rx',rx_n)\,\mathrm dS_x-\frac1{\omega_n}\int_{|x|=1,\,x_n<0}\!u(\xi'+rx',-rx_n)\,\mathrm dS_x\\&=0=\tilde u(\xi',0)\quad\text{(by substitution $y_n=-x_n$, the two integrals are same)}.\end{aligned}

    Therefore, continuous function \tilde u satisfies the mean value property, which implies \tilde u is harmonic in \Omega. The proof is complete.

    Remark. If the condition is replaced by a stronger condition: u\in C^2(\Omega_+\cup\Omega_0), then we can compute the second derivatives \displaystyle\frac{\partial^2\tilde u}{\partial x_i^2}(x) at each point x\in\Omega, and then sum up to get \Delta\tilde u(x)=0 for all x\in\Omega.

    • If x\in\Omega_+, then \displaystyle\frac{\partial^2\tilde u}{\partial x_i^2}(x)=\frac{\partial^2u}{\partial x_i^2}(x), which implies \Delta\tilde u(x)=\Delta u(x)=0.
    • If x\in\Omega-(\Omega_+\cup\Omega_0), then \displaystyle\frac{\partial\tilde u}{\partial x_i}(x)=-\frac{\partial u}{\partial x_i}(x',-x_n) for i=1,\dots,n-1 and \displaystyle\frac{\partial\tilde u}{\partial x_n}(x)=\frac{\partial u}{\partial x_n}(x',-x_n). Moreover, we get \displaystyle\frac{\partial^2\tilde u}{\partial x_i^2}(x)=-\frac{\partial^2u}{\partial x_i^2}(x',-x_n) for i=1,\dots,n-1 and \displaystyle\frac{\partial^2\tilde u}{\partial x_n^2}(x)=-\frac{\partial^2u}{\partial x_n^2}(x',-x_n). Thus, we get \Delta\tilde u(x)=-\Delta u(x',-x_n)=0.
    • If x\in\Omega_0, then by \tilde u\equiv u\equiv0 on \Omega_0, we have \displaystyle\frac{\partial^2\tilde u}{\partial x_i^2}(x',0)=0 for i=1,\dots,n-1. Moreorever, we use the definition of partial derivative to find

      \begin{aligned} \frac{\partial\tilde u}{\partial x_n}(x',0^-)&=\lim_{h\to0^-}\frac{\tilde u(x',h)-\tilde u(x',0)}h=\lim_{h\to0^-}\frac{u(x',-h)}{-h}=\frac{\partial u}{\partial x_n}(x',0)\\&=\lim_{h\to0^+}\frac{u(x',h)}h=\lim_{h\to0^+}\frac{\tilde u(x',h)-\tilde u(x',0)}h=\frac{\partial\tilde u}{\partial x_n}(x',0^+).\end{aligned}

      Thus \displaystyle\frac{\partial\tilde u}{\partial x_n}(x',0) exists. Then the second derivative gives

      \begin{aligned} &\frac{\partial^2\tilde u}{\partial x_n^2}(x',0^+)=\lim_{h\to0^+}\frac{\displaystyle\frac{\partial\tilde u}{\partial x_n}(x',h)-\frac{\partial\tilde u}{\partial x_n}(x',0)}h=\lim_{h\to0^+}\frac{\displaystyle\frac{\partial u}{\partial x_n}(x',h)-\frac{\partial u}{\partial x_n}(x',0)}h=\frac{\partial^2u}{\partial x_n^2}(x',0)=0,\\&\frac{\partial^2\tilde u}{\partial x_n^2}(x',0^-)=\lim_{h\to0^-}\frac{\displaystyle\frac{\partial\tilde u}{\partial x_n}(x',h)-\frac{\partial\tilde u}{\partial x_n}(x',0)}{h}=-\lim_{h\to0^-}\frac{\displaystyle\frac{\partial u}{\partial x_n}(x',-h)-\frac{\partial u}{\partial x_n}(x',0)}{h}=-\frac{\partial^2u}{\partial x_n^2}(x',0)=0.\end{aligned}

      Thus, \displaystyle\frac{\partial^2\tilde u}{\partial x_n^2}(x',0)=0. Therefore \Delta\tilde u(x)=0.

    Warning: The original statement is that "If u\in C^2(\Omega)\cap C(\Omega_+\cup\Omega_0) ...". It should be modified as "If u\in C^2(\Omega_+)\cap C(\Omega_+\cup\Omega_0) ...".


  15. Show that the bounded solution of the Dirichlet problem in a half-space is unique. Give unbounded counterexamples.
  16. SolutionLet \tilde u and u be the bounded solutions of the Dirichlet problem in a half-space. Then v:=\tilde u-u satisfies

    \left\{\begin{aligned} &\Delta v=0\quad\text{in}~\mathbb R_+^n,\\&v=0\quad\text{on}~\partial\mathbb R_+^n.\end{aligned}\right.

    Define \tilde v:\mathbb R^n\to\mathbb R by the reflection

    \tilde v(x)=\tilde v(x',x_n)=\begin{cases}v(x',x_n)&\text{if}~x_n\geq0;\\-v(x',-x_n)&\text{if}~x_n<0.\end{cases}

    For any x\in\mathbb R^n, there exists a sufficiently large a>0 such that x\in B_a(0). Then by Exercise 8, \tilde v is harmonic in B_a(0) and hence \Delta\tilde v(x)=0. Thus, \tilde v is harmonic in \mathbb R^n. Since \tilde u and u are bounded in \mathbb R_+^n, \tilde v is also bounded. By Theorem 7 (Liouville's Theorem) at page 123, \tilde v must be a constant, and hence \tilde v(x)=\tilde v(0)=0 for all x\in\mathbb R^n. This implies \tilde u\equiv u in \mathbb R_+^n and the uniqueness of bounded solution of the Dirichlet problem in the half-space follows.

    It is clear that the function u(x',x_n)=x_n is a unbounded harmonic solution to the zero Dirichlet problem in the half-space. (The bounded harmonic solution to the zero Dirichlet problem in the half-space is identically zero.)


    1. Suppose u\in C^2(\Omega) is harmonic, u\geq0, and \overline{B_a(0)}\subset\Omega. Use (44) to show

      \displaystyle\frac{a^{n-2}(a-|\xi|)}{(a+|\xi|)^{n-1}}u(0)\leq u(\xi)\leq\frac{a^{n-2}(a+|\xi|)}{(a-|\xi|)^{n-1}}u(0)\quad\text{for}~|\xi|<a.

    2. Prove (45).
  17. Solution
    1. In B_a(0), we use the Poisson integral formula (44):

      \displaystyle u(\xi)=\frac{a^2-|\xi|^2}{a\omega_n}\int_{|x|=a}\!\frac{u(x)}{|x-\xi|^n}\,\mathrm dS_x\quad\text{for}~|\xi|<a.

      By the triangle inequality, we have a-|\xi|=|x|-|\xi|\leq|x-\xi|\leq|x|+|\xi|=a+|\xi| for |x|=a, which gives \displaystyle\frac1{(a+|\xi|)^n}\leq\frac1{|x-\xi|^n}\leq\frac1{(a-|\xi|)^n}. Then by u(x)\geq0, we have

      \begin{aligned}u(\xi)&\geq\frac{a^2-|\xi|^2}{a\omega_n}\int_{|x|=a}\!\frac{u(x)}{(a+|\xi|)^n}\,\mathrm dS_x=\frac{a^n(a^2-|\xi|^2)}{a\omega_n(a+|\xi|)^n}\int_{|x|=a}\!\frac{u(x)}{a^n}\,\mathrm dS_x\\&=\frac{a^{n-1}(a-|\xi|)}{\omega_n(a+|\xi|)^{n-1}}\int_{|x|=a}\frac{u(x)}{|x|^n}\,\mathrm dS_x=\frac{a^{n-2}(a-|\xi|)}{(a+|\xi|)^{n-1}}\cdot\frac{a}{\omega_n}\int_{|x|=a}\!\frac{u(x)}{|x|^n}\,\mathrm dS_x=\frac{a^{n-2}(a-|\xi|)}{(a+|\xi|)^{n-1}}u(0).\end{aligned}

      Simiarly, from u(x)\geq0, we have

      \begin{aligned}u(\xi)&\leq\frac{a^2-|\xi|^2}{a\omega_n}\int_{|x|=a}\!\frac{u(x)}{(a-|\xi|)^n}\,\mathrm dS_x=\frac{a^n(a^2-|\xi|^2)}{a\omega_n(a-|\xi|)^n}\int_{|x|=a}\!\frac{u(x)}{a^n}\,\mathrm dS_x\\&=\frac{a^{n-1}(a+|\xi|)}{\omega_n(a-|\xi|)^{n-1}}\int_{|x|=a}\!\frac{u(x)}{|x|^n}\,\mathrm dS_x=\frac{a^{n-2}(a+|\xi|)}{(a-|\xi|)^{n-1}}\cdot\frac{a}{\omega_n}\int_{|x|=a}\frac{u(x)}{|x|^n}\,\mathrm dS_x=\frac{a^{n-2}(a+|\xi|)}{(a-|\xi|)^{n-1}}u(0).\end{aligned}

      Therefore, we complete the proof of Harnack's inequality on a ball.
    2. Let r=\frac13\text{dist}(\bar\Omega_1,\partial\Omega). Then for any \xi,\eta\in\bar\Omega_1, and |\xi-\eta|\leq r, we use the mean value property (12) to find

      \begin{aligned}u(\xi)&=\frac n{\omega_n}\int_{|x|\leq1}\!u(\xi+2rx)\,\mathrm dx=\frac n{\omega_n(2r)^n}\int_{B_{2r}(\xi)}\!u(y)\,\mathrm dy\\&\geq\frac n{\omega_n(2r)^n}\int_{B_r(\eta)}\!u(y)\,\mathrm dy=\frac n{2^n\omega_n}\int_{|x|\leq1}\!u(\eta+rx)\,\mathrm dx=\frac{u(\eta)}{2^n}.\end{aligned}

      Since \Omega_1 is a bounded domain, \Omega_1 is connected and \bar\Omega_1 is compact. Then there exists a finite open converging \{B_i\}_{i=1}^N such that the radius of B_i is r and B_i\cap B_{i-1}\neq\emptyset. In addition, by the extreme value theorem, there exists x\in\bar\Omega_1 and y\in\bar\Omega_1 such that

      \displaystyle u(x)=\max_{\bar\Omega_1}u=\sup_{\Omega_1}u,\quad u(y)=\min_{\bar\Omega_1}u=\inf_{\Omega_1}u.

      Therefore, we have

      \displaystyle u(y)\geq\frac1{2^{nN}}u(x),

      which implies (45) and C_1 depends on N, i.e., depends on \Omega_1.

  18. Use (46) to prove Liouville's theorem.
  19. SolutionLet u be the bounded harmonic function in \mathbb R^n and \xi\in\mathbb R^n. Define v(x)=u(x+\xi) for x\in\mathbb R^n. Clearly, v is also bounded and harmonic. By (46), we have

    \displaystyle\left|\frac{\partial u}{\partial\xi_i}(\xi)\right|=\left|\frac{\partial v}{\partial\xi_i}(0)\right|\leq\frac na\max_{|x|=a}|v(x)|=\frac na\max_{|x|=a}|u(x+\xi)|\to0\quad\text{as}~a\to\infty.

    This shows \nabla u(\xi)=0 for all \xi\in\mathbb R^n, and hence u is constant.

  20. Use Stirling's formula k!\sim(2\pi k)^{1/2}k^ke^{-k} as k\to\infty to show there is a constant C so that |\alpha|^{|\alpha|}\leq C\alpha!e^{|\alpha|}n^{|\alpha|} holds for all multi-indices \alpha.
  21. SolutionThe Stirling's formula implies

    \displaystyle\lim_{k\to\infty}\frac{k^k}{k!e^kn^k}=\lim_{k\to\infty}\frac{(2\pi k)^{1/2}k^ke^{-k}}{k!}\cdot\frac{k^k}{(2\pi k)^{1/2}k^ke^{-k}e^kn^k}=1\cdot\lim_{k\to\infty}\frac1{n^k\sqrt{2\pi k}}=0.

    There exists a positive constant K such that k^k\leq k!e^kn^k for k\geq K. Hence we take

    \displaystyle C=\max\left\{1,\max_{1\leq|\alpha|<K}\frac{|\alpha|^{|\alpha|}}{\alpha!e^{|\alpha|}n^{|\alpha|}}\right\},

    which is desired.

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