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2024年2月6日 星期二

Solution to Partial Differential Equations: Methods and Applications (Robert McOwen) Section 3.2

Solution to Partial Differential Equations: Methods and Applications (Robert McOwen) Section 3.2

    1. If F(s) is a C2-function of the one-variable s, find a condition on the vector α=(α1,α2,α3) so that

      u(x1,x2,x3,t)=F(α1x1+α2x2+α3x3t)

      is a solution of (26). (Such solutions are called plane waves and are constant on the planes αxt=constants.)
    2. Find the relationship which must hold between the initial data g(x) and h(x) for a plane wave solution.
    3. Find all plane wave solutions of (26) with the initial condition u(x1,x2,x3,0)=x1x2+1.
  1. Solution
    1. Plugging the form of solution into (26), we have

      F(αxt)=utt(x,t)=c2Δu(x,t)=c2(α21+α22+α23)F(αxt).

      Thus, the vector α must satisfy |α|=α21+α22+α23=1/c.
    2. The initial data g and h must satify

      g(x)=F(αx),h(x)=F(αx).

      By differentiating g with respect to x, we find that g and h satisfies the following relationship:

      g(x)=xF(αx)=αF(αx)=αh(x)for xR3,

      which also means h(x)=c2αg(x) for xR3.
    3. Let g(x)=g(x1,x2,x3)=x1x2+1 for x=(x1,x2,x3)R3. Then by part (b), we get

      h(x)=h(x1,x2,x3)=c2αg(x1,x2,x3)=c2(α1,α2,α3)(1,1,0)=c2(α1α2)for (x1,x2,x3)R3,

      which is a constant function. Now we employ the Kirchhoff's formula (37), we have

      u(x,t)=14πt(t|ξ|=1g(x+ctξ)dSξ)+t4π|ξ|=1h(x+ctξ)dSξ=14π|ξ|=1g(x+ctξ)dSξ+t4π|ξ|=1cξg(x+ctξ)dSξ+t4π|ξ|=1h(x+ctξ)dSξ=14π|ξ|=1g(x+ctξ)dSξ+t4π|ξ|=1(1cαξ)h(x+ctξ)dSξ=14π|ξ|=1[x1x2+ct(ξ1ξ2)+1]dSξc2(α1α2)t4π|ξ|=1(1cα1ξ1cα2ξ2cα3ξ3)dSξ==14π|ξ|=1(x1x2+1)dSξc2(α1α2)t4π|ξ|=11dSξ=x1x2+1c2(α1α2)tfor x=(x1,x2,x3)R3,t0.

      Here we have used the symmetry of integral, i.e., |ξ|=1ξkdSξ=0 for k=1,2,3 and the surface area of unit ball in R3 is 4π.

  2. Find the solution of the initial value problem

    {utt=uxx+uyy+uzz,u(x,y,z,0)=x2+y2,ut(x,y,z,0)=0.

    1. by using (37);
    2. by using (39).
  3. Solution
    1. Let g(x,y,z)=x2+y2 and h(x,y,z)=0 for (x,y,z)R3 and c=1. Using the Kirchhoff's formula (37), we have

      u(x,y,z,t)=14πt(t|ξ|=1g(x+ctξ1,y+ctξ2,z+ctξ3)dSξ)+t4π|ξ|=1h(x+ctξ1,y+ctξ2,z+ctξ3)dSξ=14π|ξ|=1g(x+tξ1,y+tξ2,z+tξ3)dSξ+t4π|ξ|=1ξg(x+tξ1,y+tξ2,z+tξ3)dSξ=14π|ξ|=1[(x+tξ1)2+(y+tξ2)2]dSξ+t4π|ξ|=1(ξ1,ξ2,ξ3)(2x+2tξ1,2y+2tξ2,0)dSξ=14π|ξ|=1(x2+y2)dSξ+2t4π(xξ1+yξ2)dSξ+t24π|ξ|=1(ξ21+ξ22)dSξ+2t4π|ξ|=1(xξ1+yξ2)dSξ+2t24π|ξ|=1(ξ21+ξ22)dSξ=x2+y2+3t24π|ξ|=1(ξ21+ξ22)dSξ=x2+y2+3t24π23|ξ|=1(ξ21+ξ22+ξ23)dSξ=x2+y2+2t2for (x,y,z)R3,t0.

      Here we have used the symmetry of integral, i.e., |ξ|=1ξkdSξ=0 for k=1,2,3 and

      |ξ|=1ξ21dSξ=|ξ|=1ξ22dSξ=|ξ|=1ξ31dSξ=4π3.

      In addition, the integral |ξ|=1ξg(x+tξ)dSξ also can be computed by the Gauss' divergence theorem as follows.

      |ξ|=1ξg(x+tξ1,y+tξ2,z+tξ3)dSξ=|ξ|1Δξ[g(x+tξ1,y+tξ2,z+tξ3)]dSξ=t|ξ|1(Δg)(x+tξ1,y+tξ2,z+tξ3)dSξ=t|ξ|14dSξ=t44π3=16πt3.

    2. Let g(x,y)=x2+y2 and h(x,y)=0 for (x,y)R2 and c=1. Then using (39), we have

      u(x,y,t)=14πt(2tξ21+ξ22<1g(x+ctξ1,y+ctξ2)1ξ21ξ22dξ1dξ2)+t4π(2ξ21+ξ22<1h(x+ctξ1,y+ctξ2)1ξ21ξ22dξ1dξ2)=24πξ21+ξ22<1(x+tξ1)2+(y+tξ2)21ξ21ξ22dξ1dξ2+24πξ21+ξ22<12tξ1(x+tξ1)+2tξ2(y+tξ2)1ξ21ξ22dξ1dξ2=24πξ21+ξ22(x+tξ1)2+(y+tξ2)2+2t[ξ1(x+tξ1)+ξ2(y+tξ2)]1ξ21ξ22dξ1dξ2.

      Using the polar coordinate, we set ξ1=rcosθ and ξ2=rsinθ for 0r<1 and 0θ<2π. Then the double integral can be transformed and we have

      u(x,y,t)=24π2π010x2+y2+4rt(xcosθ+ysinθ)+3t2r21r2rdrdθ=10r(x2+y2)+3t2r31r2dr=(x2+y2)10r1r2dr+3t210r31r2dr=(x2+y2)(1r2)|r=1r=0+3t2((2+r2)1r23)|r=1r=0=x2+y2+3t223=x2+y2+2t2for (x,y)R2,t0.


  4. Use Duhamel's principle to find the solution of the nonhomogeneous wave equation for three space dimensions uttc2Δu=f(x,t) with initial conditions u(x,0)=0=ut(x,0). What regularity in f(x,t) is required for the solution u to be C2?
  5. SolutionConsider the following wave equation

    {Utt(x,t,s)c2ΔU(x,t,s)=0for xR3,t>0,s>0U(x,0,s)=0for xR3,s>0Ut(x,0,s)=f(x,s)for xR3,s>0.

    By Kirchhoff's formula, we have

    U(x,t,s)=t4π|ξ|=1f(x+ctξ,s)dSξfor xR3,t0,s>0.

    Then by Duhamel's principle, the solution is

    u(x,t)=t0U(x,ts,s)ds=14πt0(ts)|ξ|=1f(x+c(ts)ξ,s)dSξds

    According to this formula for solution, when f is C2 in x and C0 in t, u will be a C2-function.

  6. Let Ω={(x,y)R2:0<x<a and 0<y<b}, and use separation of variables to solve the initial/boundary value problem

    {utt=uxx+uyyfor (x,y)Ω and t>0,u(x,y,t)=0for (x,y)Ω and t>0,u(x,y,0)=sinπxasin2πyb,and ut(x,y,0)=0for (x,y)Ω.

  7. SolutionUsing the separation of variables, we suppose that the solution u(x,t) is in the form

    u(x,y,t)=m,n=1um,n(x,y,t)=m,n=1Xm(x)Yn(y)Tm,n(t)for (x,y)Ω,t>0.

    Due to the superposition principle, we firstly consider u(x,y,t)=X(x)Y(y)T(t) to fulfill the wave equation with the boundary condition, which gives

    X(x)Y(y)T(t)=X(x)Y(y)T(t)+X(x)Y(y)T(t),X(0)Y(y)T(t)=X(a)Y(y)T(t)=X(x)Y(0)T(t)=X(x)Y(b)T(t)=0.

    Surely, we want to find a general solution so X, Y and T are not identically zero and X(0)=X(a)=Y(0)=Y(b)=0. From the differential equation, we have

    T(t)T(t)=X(x)X(x)+Y(y)Y(y)for (x,y)Ω,t>0.

    Clearly, X(x)/X(x), Y(y)/Y(y) and T(t)/T(t) are all constants. By solving the eigenvalue problems for X(x)=λX(x) in (0,a) with X(0)=X(a)=0, Y(y)=μY(y) in (0,b) with Y(0)=Y(b)=0, we obtain

    Xm(x)=sinmπxafor x(0,a),Yn(x)=sinnπybfor y(0,b),

    where λm=(mπ/a)2 and μn=(nπ/b)2 for m,nN. Then T satisfies

    T(t)=(λm+μn)T(t)=(m2π2a2+n2π2b2)T(t),

    which gives

    Tm,n(t)=Am,ncos(tωm,n)+Bm,nsin(tωm,n)andωm,n=πm2a2+n2b2.

    Hence the solution has the following form:

    u(x,t)=m,n=1[Am,ncos(tωm,n)+Bm,nsin(tωm,n)]sinmπxasinnπybfor (x,y)Ω,t>0.

    From the intial condition, it is easy to know A1,2=1, Am,n=0 for (m,n)(1,2), and Bm,n=0 for all (m,n)N2. Therefore, the solution is

    u(x,y,t)=cos(πt1a2+4b2)sinπxasin2πybfor (x,y)ˉΩ,t0.


  8. Find a formula for the solution v(x,t)=v(x1,x2,t) of the Cauchy problem for the two-dimensional Klein-Gordon equation:

    {vtt=c2Δvm2vfor xR2 and t>0,v(x,0)=g(x),vt(x,0)=h(x)for xR2.

  9. SolutionDefine u:R3×ˉR+ by

    u(x,y,z,t)=cos(mz/c)v(x,y,t)for (x,y,z,t)R3×ˉR+.

    Then it can be checked that

    utt(x,y,z,t)=cos(mz/c)vtt(x,y,t)=cos(mz/c)[c2vxx(x,y,t)+c2vyy(x,y,t)m2v(x,y,t)]=c2uxx(x,y,z,t)+c2uyy(x,y,z)+c2uzz(x,y,z)=c2Δu(x,y,z).

    This means u satisfies the three-dimensional wave equation with the initial condition

    u(x,y,z,0)=cos(mz/c)g(x,y),ut(x,y,z,0)=cos(mz/c)h(x,y)for (x,y,z)R3.

    Then by Kirchhoff's formula (37), we have

    u(x,y,z,t)=14πt(t|ξ|=1cos(m(z+ctξ3)/c)g(x+ctξ1,y+ctξ2)dSξ)+t4π|ξ|=1cos(m(z+ctξ3)/c)h(x+ctξ1,y+ctξ2)dSξ.

    The soluion v is

    v(x,y,t)=u(x,y,0,t)=14πt(t|ξ|=1cos(mtξ3)g(x+ctξ1,y+ctξ2)dSξ)+t4π|ξ|=1cos(mtξ3)h(x+ctξ1,y+ctξ2)dSξ=12πt(tξ21+ξ22<1cos(mt1ξ21ξ22)g(x+ctξ1,y+ctξ2)1ξ21ξ22dξ1dξ2)+t2πξ21+ξ22<1cos(mt1ξ21ξ22)h(x+ctξ1,y+ctξ2)1ξ21ξ22dξ1dξ2=12πt(t2π010rcos(mt1r2)g(x+ctrcosθ,y+ctrsinθ)1r2drdθ)+t2π2π010rcos(mt1r2)h(x+ctrcosθ,y+ctrsinθ)1r2drdθ.


    1. For n=3, suppose that g,hC0(R3) and consider the solution of (26), (27) given by (37). Show that there is a constant C so that |u(x,t)|C/t for all xR3 and t>0.
    2. Is a similar result true for n=2?
  10. Solution
    1. Since g and h are in C0(R3), there exist two positive constants Rg and Rh such that supp(g)ˉBRg(0) and supp(h)ˉBRh(0). By the extreme value theorem, the continuous function must attain its maximum value in any given compact set, which means there exist two positive constants Mg and Mh such that |g(x)|Mg and |h(x)|Mh for xR3. Moreover, the support of |g| is a subset of one of g, and |g| is also a continuous function, which means there exists a positive constant Mg such that |g(x)|Mg for xR3. Then we use (37) and the change of variable y=x+ctξ (with dSξ=(ct)2dSy) to get

      u(x,t)=14π|ξ|=1(g(x+ctξ)+th(x+ctξ))dSξ+t4π|ξ|=1cξg(x+ctξ)dSξ=14πc2t2Bct(x)(g(y)+th(y))dSy+t4πc2t2Bct(x)yxtg(y)dSy=14πc2t2Bct(x)(g(y)+th(y))dSy+14πc2t2Bct(x)(yx)g(y)dSy.

      Now we can estimate |u(x,t)| as follows.

      |u(x,t)|14πc2t2Bct(x)(|g(y)|+t|h(y)|)dSy+14πc2t2Bct(x)|yx||g(y)|dSy=14πc2t2Bct(x)BRg(0)|g(y)|dSy+14πc2tBct(x)BRh(0)|h(y)|dSy+14πctBct(x)BRg(0)|g(y)|dSyMg4πc2t2H2(Bct(x)BRg(0))+Mh4πc2tH2(Bct(x)BRh(0))+Mg4πctH2(Bct(x)BRg(0)),

      where H2 denotes the two-dimensional Hausdorff measure in R3. Here we have used the fact that |yx|=ct for yBct(x).

      Now we claim that H2(Bct(x)BR(0))4πR2 for R>0. When ctR, it is easy to know Bct(x)BR(0)Bct(x) and

      H2(Bct(x)BR(0))H2(Bct(x))=4π(ct)24πR2.

      When ct>0, the value of H2(Bct(x)BR(0)) can be attained as Bct(x) passes the equatorial of BR(0). As the radius of Bct(x) increases, the intersection Bct(x)BR(0) would become more flat, which means H2(Bct(x)BR(0))πR2, the area of the circle formed by the equatorial of BR(0). On the other hand, as the radius of Bct(x) decreases to R and Bct(x) passes the equatorial of BR(0), the intersection Bct(x)BR(0) may become BR(0) and H2(Bct(x)BR(0))H2(BR(0))=4πR2. This completes the proof of Claim.

      If t1, we can yse Claim to get

      |u(x,t)|MgR2gc2t2+MhR2hc2t+MgR2gct=1t(MgR2gc+MhR2hc2+MgR2gc2t)1t(MgR2gc+MhR2hc2+MgR2gc2):=C1t,

      where C1=MgR2gc+MhR2hc2+MgR2gc2. On the other hand, if 0<t<1, we have t2<t<1 and

      |u(x,t)|Mg4πc2t2H2(Bct(x))+Mh4πc2tH2(Bct(x))+Mg4πctH2(Bct(x))=Mg4πc2t2(4πc2t2)+Mh4πc2t(4πc2t2)+Mg4πct(4πc2t2)<Mg4πc2t2(4πc2t)+Mh4πc2t(4πc2)+Mg4πct(4πc2)=Mg+Mh+cMgt:=C2t,

      where C2=Mg+Mh+cMg. Hence we choose C=max, which is desired.
    2. No. To find a counterexample, we consider the initial condition u(x,y,0)\equiv0 and u_t(x,y,0)=h(x,y), where h satisfies h\equiv1 in B_1(0), 0\leq h\leq1 in B_2(0)-B_1(0) and h\equiv0 in \mathbb R^2-B_2(0). The existence of function h follows from Urysohn's lemma. Suppose by contradiction that there exists a positive constant C such that |u(x,y,t)|\leq C/t for (x,y)\in\mathbb R^2. By (39), we have

      \begin{aligned}u(x,y,t)&=\frac1{4\pi}\frac\partial{\partial t}\left(2t\int_{\xi_1^2+\xi_2^2<1}\!\frac{g(x+ct\xi_1,y+ct\xi_2)}{\sqrt{1-\xi_1^2-\xi_2^2}}\,\mathrm d\xi_1\,\mathrm d\xi_2\right)+\frac t{4\pi}\left(2\int_{\xi_1^2+\xi_2^2<1}\!\frac{h(x+ct\xi_1,y+ct\xi_2)}{\sqrt{1-\xi_1^2-\xi_2^2}}\,\mathrm d\xi_1\,\mathrm d\xi_2\right)\\&=\frac t{2\pi}\int_{\xi_1^2+\xi_2^2<1}\frac{h(x+ct\xi_1,y+ct\xi_2)}{\sqrt{1-\xi_1^2-\xi_2^2}}\,\mathrm d\xi_1\,\mathrm d\xi_2=\frac1{2\pi c^2t}\int_{B_{ct}(x,y)}\!\frac{h(\eta,\mu)}{\sqrt{1-(ct)^{-2}((\eta-x)^2+(\mu-y)^2)}}\,\mathrm d\eta\,\mathrm d\mu.\end{aligned}

      It is clear that B_1(0)\subseteq B_{ct+1}(ct,0) for t\geq0. Then we have

      \begin{aligned}u(ct,0,t+c^{-1})&=\frac1{2\pi c(ct+1)}\int_{B_{ct+1}(ct,0)}\!\frac{h(\eta,\mu)}{\sqrt{1-(ct+1)^{-2}((\eta-ct)^2+\mu^2)}}\,\mathrm d\eta\,\mathrm d\mu\\&\geq\frac1{2\pi c(ct+1)}\int_{B_1(0,0)}\!\frac{h(\eta,\mu)}{\sqrt{1-(ct+1)^{-2}((\eta-ct)^2+\mu^2)}}\,\mathrm d\eta\,\mathrm d\mu\\&=\frac1{2\pi c}\int_{B_1(0,0)}\!\frac1{\sqrt{(ct+1)^2-(\eta-ct)^2-\mu^2}}\,\mathrm d\eta\,\mathrm d\mu\\&=\frac1{2\pi c}\int_0^{2\pi}\!\int_0^1\!\frac r{\sqrt{1-r^2+2ct(1+r\cos\theta)}}\,\mathrm dr\,\mathrm d\theta\\&\geq\frac1{4\pi c}\int_0^{2\pi}\!\int_0^1\!\frac r{\sqrt{1+4ct}}\,\mathrm dr\,\mathrm d\theta=\frac1{4\sqrt{1+4ct}}.\end{aligned}

      Thus, by |u(x,y,t)|\leq C/t for all (x,y)\in\mathbb R^2 and t>0, we get

      \displaystyle\frac1{4\sqrt{1+4ct}}\leq\frac Ct\quad\text{for all}~t>0,

      which is equivalent to t^2-64c^3t-16c^2\leq0 for all t>0. This leads a contradiction. Thereofre, it is impossible to find a positive constant C such that |u(x,y,t)|\leq C/t for all (x,y)\in\mathbb R^2 and t>0.

  11. Suppose n=2k+1 for a positive integer k and u satisfies (26).
    1. Show that V^x(r,t) given by (40) satisfies (32).
    2. Derive (42) from (35) and (41); note c_n=1\cdot3\cdots(n-2).
  12. Solution
    1. Firstly, we prove the following identity by mathematical induction:

      \displaystyle\frac{\mathrm d^2}{\mathrm dr^2}\left(\frac1r\frac{\mathrm d}{\mathrm dr}\right)^{k-1}(r^{2k-1}\phi(r))=\left(\frac1r\frac{\mathrm d}{\mathrm dr}\right)^k(r^{2k}\phi'(r)),

      where k\in\mathbb N and \phi\in C^{k+1}(\mathbb R). Then we claim that
      Proof of ClaimWhen k=1, it is easy to verify that

      \begin{aligned}\frac{\mathrm d^2}{\mathrm dr^2}\left(\frac1r\frac{\mathrm d}{\mathrm dr}\right)^{k-1}(r^{2k-1}\phi(r))&=\frac{\mathrm d^2}{\mathrm dr^2}(r\phi(r))=\frac{\mathrm d}{\mathrm dr}(r\phi'(r)+\phi(r))=r\phi''(r)+2\phi'(r)\\&=\frac{r^2\phi''(r)+2r\phi'(r)}r=\frac1r\frac{\mathrm d}{\mathrm dr}(r^2\phi'(r)).\end{aligned}

      This shows the identity holds true for k=1. Now we suppose the identity holds true for k=n, i.e.,

      \displaystyle\frac{\mathrm d^2}{\mathrm dr^2}\left(\frac1r\frac{\mathrm d}{\mathrm dr}\right)^{n-1}(r^{2n-1}\phi(r))=\left(\frac1r\frac{\mathrm d}{\mathrm dr}\right)^n(r^{2n}\phi'(r)).

      Then we can observe that

      \begin{aligned}\frac{\mathrm d^2}{\mathrm dr^2}\left(\frac1r\frac{\mathrm d}{\mathrm dr}\right)^n(r^{2n+1}\phi(r))&=\frac{\mathrm d^2}{\mathrm dr^2}\left(\frac1r\frac{\mathrm d}{\mathrm dr}\right)^{n-1}\left[\frac{(2n+1)r^{2n}\phi(r)+r^{2n+1}\phi'(r)}r\right]\\&=(2n+1)\frac{\mathrm d^2}{\mathrm dr^2}\left(\frac1r\frac{\mathrm d}{\mathrm dr}\right)^{n-1}(r^{2n-1}\phi(r))+\frac{\mathrm d^2}{\mathrm dr^2}\left(\frac1r\frac{\mathrm d}{\mathrm dr}\right)^{n-1}(r^{2n-1}(r\phi'(r)))\\&=(2n+1)\left(\frac1r\frac{\mathrm d}{\mathrm dr}\right)^n(r^{2n}\phi'(r))+\left(\frac1r\frac{\mathrm d}{\mathrm dr}\right)^n(r^{2n}(r\phi'(r))')\\&=\\&=\left(\frac1r\frac{\mathrm d}{\mathrm dr}\right)^n(r^{2n+1}r^{2n}\phi'(r)+r^{2n}(r\phi''(r)+\phi'(r)))\\&=\left(\frac1r\frac{\mathrm d}{\mathrm dr}\right)^n\frac1r\frac{\mathrm d}{\mathrm dr}(r^{2n+2}\phi'(r))\\&=\left(\frac1r\frac{\mathrm d}{\mathrm dr}\right)^{n+1}(r^{2n+2}\phi'(r)).\end{aligned}

      Hence by the mathematical induction, the identity holds true for all n\in\mathbb N.

      Now we can use (28), (40) and Claim to obtain

      \begin{aligned}\frac{\partial^2V^x}{\partial t^2}(r,t)&=\frac{\partial^2}{\partial t^2}\left(\frac1r\frac{\partial}{\partial r}\right)^{k-1}(r^{2k-1}M_u(x,r,t))=\left(\frac1r\frac{\partial}{\partial r}\right)^{k-1}(r^{2k-1}\frac{\partial^2M_u}{\partial t^2}(x,r,t))\quad\text{(by (28))}\\&=\left(\frac1r\frac\partial{\partial r}\right)^{k-1}\left[r^{2k-1}\cdot c^2\left(\frac{\partial^2}{\partial r^2}+\frac{n-1}r\frac\partial{\partial r}\right)M_u(x,r,t)\right]\quad\text{(by (40))}\\&=c^2\left(\frac1r\frac\partial{\partial r}\right)^{k-1}\cdot\frac1r\left(r^{n-1}\frac{\partial^2}{\partial r^2}+(n-1)r^{n-2}\frac\partial{\partial r}\right)M_u(x,r,t)\\&=c^2\left(\frac1r\frac\partial{\partial r}\right)^{k-1}\cdot\frac1r\frac\partial{\partial r}\left(r^{n-1}\frac\partial{\partial r}\right)M_u\\&=c^2\left(\frac1r\frac\partial{\partial r}\right)^k\left(r^{2k}\frac{\partial M_u}{\partial r}(x,r,t)\right)\\&=c^2\frac{\partial^2}{\partial r^2}\left(\frac1r\frac\partial{\partial r}\right)^{k-1}(r^{2k-1}M_u(x,r,t))\quad\text{(by Claim)}\\&=c^2\frac{\partial^2V^x}{\partial r^2}(x,r,t)\quad\text{(by (28))}.\end{aligned}

      The proof is complete.
    2. For any k\in\mathbb N and \phi\in C^{k-1}(\mathbb R), we prove the following identity by the mathematical induction:

      \displaystyle\left(\frac1r\frac{\mathrm d}{\mathrm dr}\right)^{k-1}(r^{2k-1}\phi(r))=\sum_{j=0}^{k-1}\beta_j^kr^{j+1}\phi^{(j)}(r),

      where the constants \beta_j^k (j=0,\dots,k-1) are independent of \phi.
      Proof of Claim 1When k=1, it is easy that

      \displaystyle\left(\frac1r\frac{\mathrm d}{\mathrm dr}\right)^{k-1}(r^{2k-1}\phi(r))=r\phi(r)=\beta_0^1\cdot r^{0+1}\phi^{(0)}(r),

      where \beta_0^1=1=c_3. This shows the claim holds true for k=1. Now we suppose the claim holds true for k=m, i.e., there exists \beta_j^m (j=0,1,\dots,m-1) such that

      \displaystyle\left(\frac1r\frac{\mathrm d}{\mathrm dr}\right)^{m-1}(r^{2m-1}\phi(r))=\sum_{j=0}^{m-1}\beta_j^mr^{j+1}\phi^{(j)}(r),

      where \beta_0^m=c_{2m+1}. Then we can observe that

      \begin{aligned}\left(\frac1r\frac{\mathrm d}{\mathrm dr}\right)^m(r^{2m+1}\phi(r))&=\frac1r\frac{\mathrm d}{\mathrm dr}\left(\frac1r\frac{\mathrm d}{\mathrm dr}\right)^{m-1}(r^{2m-1}(r^2\phi(r)))\\&=\frac1r\frac{\mathrm d}{\mathrm dr}\sum_{j=0}^{m-1}\beta_j^mr^{j+1}\frac{\mathrm d^j}{\mathrm dr^j}(r^2\phi(r))\\&=\frac1r\sum_{j=0}^{m-1}\beta_j^m\left[(j+1)r^j\frac{\mathrm d^j}{\mathrm dr^j}(r^2\phi(r))+r^{j+1}\frac{\mathrm d^{j+1}}{\mathrm dr^{j+1}}(r^2\phi(r))\right]\\&=\sum_{j=0}^{m-1}(j+1)\beta_j^mr^{j-1}\frac{\mathrm d^j}{\mathrm dr^j}(r^2\phi(r))+\sum_{j=0}^{m-1}\beta_j^mr^j\frac{\mathrm d^{j+1}}{\mathrm dr^{j+1}}(r^2\phi(r))\\&=\sum_{j=0}^{m-1}(j+1)\beta_j^mr^{j-1}\frac{\mathrm d^j}{\mathrm dr^j}(r^2\phi(r))+\sum_{j=1}^m\beta_{j-1}^mr^{j-1}\frac{\mathrm d^j}{\mathrm dr^j}(r^2\phi(r))\\&=\beta_0^mr\phi(r)+\sum_{j=1}^{m-1}[(j+1)\beta_j^m+\beta_{j-1}^m]r^{j-1}\frac{\mathrm d^j}{\mathrm dr^j}(r^2\phi(r))+\beta_{m-1}^mr^{m-1}\frac{\mathrm d^m}{\mathrm dr^m}(r^2\phi(r)).\end{aligned}

      • For j=1, \displaystyle\frac{\mathrm d^j}{\mathrm dr^j}(r^2\phi(r))=2r\phi(r)+r^2\phi'(r);
      • for j\geq2, \displaystyle\frac{\mathrm d^j}{\mathrm dr^j}(r^2\phi(r))=r^2\phi^{(j)}(r)+2jr\phi^{(j-1)}(r)+j(j-1)\phi^{(j-2)}(r).
      Thus, we have

      \begin{aligned}\left(\frac1r\frac{\mathrm d}{\mathrm dr}\right)^m(r^{2m+1}\phi(r))&=\beta_0^mr\phi(r)+(2\beta_1^m+\beta_0^m)(2r\phi(r)+r^2\phi'(r))\\&\quad+\sum_{j=2}^{m-1}[(j+1)\beta_j^m+\beta_{j-1}^m]r^{j-1}[r^2\phi^{(j)}+2jr\phi^{(j-1)}(r)+j(j-1)\phi^{(j-2)}(r)]\\&\quad+\beta_{m-1}^mr^{m-1}(r^2\phi^{(m)}+2mr\phi^{(m-1)}(r)+m(m-1)\phi^{(m-2)}(r))\\&=(3\beta_0^m+4\beta_1^m)r\phi(r)+(2\beta_1^m+\beta_0^m)r^2\phi'(r)+\sum_{j=2}^{m-1}[(j+1)\beta_j^m+\beta_{j-1}^m]r^{j+1}\phi^{(j)}\\&\quad+\sum_{j=1}^{m-2}2[(j+2)(j+1)\beta_{j+1}^m+(j+1)\beta_j^m]r^{j+1}\phi^{(j)}(r)\\&\quad+\sum_{j=0}^{m-3}[(j+3)(j+2)(j+1)\beta_{j+2}^m+(j+2)(j+1)\beta_{j+1}^m]r^{j+1}\phi^{(j)}(r)\\&\quad+m(m-1)\beta_{m-1}^mr^{m-1}\phi^{(m-2)}(r)+2m\beta_{m-1}^mr^m\phi^{(m-1)}(r)+\beta_{m-1}^mr^{m+1}\phi^{(m)}\\&=\sum_{j=0}^m\beta_j^{m+1}r^{j+1}\phi^{(j)}(r),\end{aligned}

      where \beta_j^{m+1} are given by

      \begin{aligned} &\beta_0^{m+1}=3\beta_0^m+6\beta_1^m+6\beta_2^m,\\&\beta_1^{m+1}=\beta_0^m+4\beta_1^m+18\beta_2^m+24\beta_3^m,\\&\beta_j^{m+1}=\beta_{j-1}^m+3(j+1)\beta_j^m+3(j+2)(j+1)\beta_{j+1}^m+(j+3)(j+2)(j+1)\beta_{j+2}^m\quad\text{for}~2\leq j\leq m-3,\\&\beta_{m-2}^{m+1}=\beta_{m-3}^m+2(m-1)\beta_{m-2}^m+3m(m-1)\beta_{m-1}^m,\\&\beta_{m-1}^{m+1}=3m\beta_{m-1}^m+m\beta_{m-1}^m,\\&\beta_m^{m+1}=\beta_{m-1}^m.\end{aligned}

      Therefore, we complete the proof of Claim 1.

      Then we claim that \beta_0^k=c_{2k+1}=c_n for k\in\mathbb N.
      Proof of Claim 2Clearly, the claim holds true for k=1. From Claim 1, a direct computation gives

      \begin{aligned}\sum_{j=0}^{k-1}\beta_j^kr^{j+1}\phi^{(j)}(r)=\left(\frac1r\frac{\mathrm d}{\mathrm dr}\right)^{k-1}(r^{2k-1}\phi(r))\\&=\left(\frac1r\frac{\mathrm d}{\mathrm dr}\right)^{k-2}[(2k-1)r^{2k-3}\phi(r)+r^{2k-2}\phi'(r)]\\&=\left(\frac1r\frac{\mathrm d}{\mathrm dr}\right)^{k-3}[(2k-1)(2k-3)r^{2k-5}\phi(r)+\cdots]\\&=\cdots=(2k-1)(2k-3)(2k-5)\cdots3\cdot r^1\phi(r)+\cdots.\end{aligned}

      It is clear that \beta_0^k=(2k-1)(2k-3)(2k-5)\cdots3\cdot1=c_{2k+1}=c_n. The proof is complete.

      By (40) and Claims 1 and 2, we have

      \displaystyle V^x(r,t)=\left(\frac1r\frac\partial{\partial r}\right)^{k-1}(r^{2k-1}M_u(x,r,t))=\sum_{j=0}^{k-1}\beta_j^kr^{j+1}\frac{\partial^jM_u}{\partial r^j}(x,r,t).

      Then we use (30) and (35) to get

      \begin{aligned}u(x,t)&=\lim_{r\to0^+}M_u(x,r,t)\quad\text{(by (30))}\\&=\frac1{c_n}\lim_{r\to0^+}(c_nM_u(x,r,t))\\&=\frac1{c_n}\lim_{r\to0^+}\frac1r\sum_{j=0}^{k-1}\beta_j^kr^{j+1}\frac{\partial^jM_u}{\partial r^j}(x,r,t)\\&=\frac1{c_n}\lim_{r\to0^+}\frac{V^x(r,t)}r\\&=\frac1{c_n}\lim_{r\to0^+}\left[\frac{G^x(r+ct)+G^x(r-ct)}{2r}+\frac1{2cr}\int_{r-ct}^{r+ct}\!H^x(\rho)\,\mathrm d\rho\right]\quad\text{(by (41))}\\&=\frac1{c_n}\lim_{r\to0^+}\left[\frac{(G^x)'(r+ct)+(G^x)'(r-ct)}2+\frac{H^x(r+ct)-H^x(r-ct)}{2c}\right]\quad\text{(by L'Hospital's Rule)}\\&=\frac{(G^x)'(ct)+(G^x)'(-ct)}{2c_n}+\frac{H^x(ct)+H^x(-ct)}{2c_nc}.\end{aligned}

      From (21) and (41) with replacing the variable r by t, we note that

      \begin{aligned}G^x(t)&=\left(\frac1t\frac\partial{\partial t}\right)^{k-1}(t^{2k-1}M_g(x,t))=\left(\frac1t\frac\partial{\partial t}\right)^{k-1}\left(\frac{t^{2k-1}}{\omega_n}\int_{|\xi|=1}\!g(x+t\xi)\,\mathrm dS_\xi\right)\\&=\frac1{\omega_n}=\left(\frac1t\frac\partial{\partial t}\right)^{\frac{n-3}2}t^{n-2}\int_{|\xi|=1}\!g(x+t\xi)\,\mathrm dS_\xi,\\H^x(t)&=\left(\frac1t\frac\partial{\partial t}\right)^{k-1}(t^{2k-1}M_h(x,t))=\left(\frac1t\frac\partial{\partial t}\right)^{k-1}\left(\frac{t^{2k-1}}{\omega_n}\int_{|\xi|=1}h(x+t\xi)\,\mathrm dS_\xi\right)\\&=\frac1{\omega_n}\left(\frac1t\frac\partial{\partial t}\right)^{\frac{n-3}2}t^{n-2}\int_{|\xi|=1}\!h(x+t\xi)\,\mathrm dS_\xi.\end{aligned}

      This implies

      \begin{aligned}(G^x)'(ct)&=\frac1{\omega_n}\left(\frac\partial{\partial(ct)}\left(\frac1{ct}\frac\partial{\partial(ct)}\right)^{\frac{n-3}2}(ct)^{n-2}\int_{|\xi|=1}\!g(x+ct\xi)\,\mathrm dS_\xi\right)\\&=\frac1{\omega_n}\frac\partial{\partial t}\left(\frac1t\frac\partial{\partial t}\right)^{\frac{n-3}2}t^{n-2}\int_{|\xi|=1}\!g(x+ct\xi)\,\mathrm dS_\xi,\\H^x(ct)&=\frac1{\omega_n}\left(\frac1{ct}\frac\partial{\partial(ct)}\right)^{\frac{n-3}2}(ct)^{n-2}\int_{|\xi|=1}\!h(x+ct\xi)\,\mathrm dS_\xi\\&=\frac c{\omega_n}\left(\frac1t\frac\partial{\partial t}\right)^{\frac{n-3}2}t^{n-2}\int_{|\xi|=1}\!h(x+ct\xi)\,\mathrm dS_\xi.\end{aligned}

      Also note that (G^x)'(ct)=(G^x)'(-ct) and H^x(ct)=-H^x(-ct). Therefore, we obtain (42) by combining above equations.

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