Processing math: 100%

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{C1,C2}, which is desired.
    2. No. To find a counterexample, we consider the initial condition u(x,y,0)0 and ut(x,y,0)=h(x,y), where h satisfies h1 in B1(0), 0h1 in B2(0)B1(0) and h0 in R2B2(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)|C/t for (x,y)R2. By (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)=t2πξ21+ξ22<1h(x+ctξ1,y+ctξ2)1ξ21ξ22dξ1dξ2=12πc2tBct(x,y)h(η,μ)1(ct)2((ηx)2+(μy)2)dηdμ.

      It is clear that B1(0)Bct+1(ct,0) for t0. Then we have

      u(ct,0,t+c1)=12πc(ct+1)Bct+1(ct,0)h(η,μ)1(ct+1)2((ηct)2+μ2)dηdμ12πc(ct+1)B1(0,0)h(η,μ)1(ct+1)2((ηct)2+μ2)dηdμ=12πcB1(0,0)1(ct+1)2(ηct)2μ2dηdμ=12πc2π010r1r2+2ct(1+rcosθ)drdθ14πc2π010r1+4ctdrdθ=141+4ct.

      Thus, by |u(x,y,t)|C/t for all (x,y)R2 and t>0, we get

      141+4ctCtfor all t>0,

      which is equivalent to t264c3t16c20 for all t>0. This leads a contradiction. Thereofre, it is impossible to find a positive constant C such that |u(x,y,t)|C/t for all (x,y)R2 and t>0.

  11. Suppose n=2k+1 for a positive integer k and u satisfies (26).
    1. Show that Vx(r,t) given by (40) satisfies (32).
    2. Derive (42) from (35) and (41); note cn=13(n2).
  12. Solution
    1. Firstly, we prove the following identity by mathematical induction:

      d2dr2(1rddr)k1(r2k1ϕ(r))=(1rddr)k(r2kϕ(r)),

      where kN and ϕCk+1(R). Then we claim that
      Proof of ClaimWhen k=1, it is easy to verify that

      d2dr2(1rddr)k1(r2k1ϕ(r))=d2dr2(rϕ(r))=ddr(rϕ(r)+ϕ(r))=rϕ(r)+2ϕ(r)=r2ϕ(r)+2rϕ(r)r=1rddr(r2ϕ(r)).

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

      d2dr2(1rddr)n1(r2n1ϕ(r))=(1rddr)n(r2nϕ(r)).

      Then we can observe that

      d2dr2(1rddr)n(r2n+1ϕ(r))=d2dr2(1rddr)n1[(2n+1)r2nϕ(r)+r2n+1ϕ(r)r]=(2n+1)d2dr2(1rddr)n1(r2n1ϕ(r))+d2dr2(1rddr)n1(r2n1(rϕ(r)))=(2n+1)(1rddr)n(r2nϕ(r))+(1rddr)n(r2n(rϕ(r)))==(1rddr)n(r2n+1r2nϕ(r)+r2n(rϕ(r)+ϕ(r)))=(1rddr)n1rddr(r2n+2ϕ(r))=(1rddr)n+1(r2n+2ϕ(r)).

      Hence by the mathematical induction, the identity holds true for all nN.

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

      2Vxt2(r,t)=2t2(1rr)k1(r2k1Mu(x,r,t))=(1rr)k1(r2k12Mut2(x,r,t))(by (28))=(1rr)k1[r2k1c2(2r2+n1rr)Mu(x,r,t)](by (40))=c2(1rr)k11r(rn12r2+(n1)rn2r)Mu(x,r,t)=c2(1rr)k11rr(rn1r)Mu=c2(1rr)k(r2kMur(x,r,t))=c22r2(1rr)k1(r2k1Mu(x,r,t))(by Claim)=c22Vxr2(x,r,t)(by (28)).

      The proof is complete.
    2. For any kN and ϕCk1(R), we prove the following identity by the mathematical induction:

      (1rddr)k1(r2k1ϕ(r))=k1j=0βkjrj+1ϕ(j)(r),

      where the constants βkj (j=0,,k1) are independent of ϕ.
      Proof of Claim 1When k=1, it is easy that

      (1rddr)k1(r2k1ϕ(r))=rϕ(r)=β10r0+1ϕ(0)(r),

      where β10=1=c3. This shows the claim holds true for k=1. Now we suppose the claim holds true for k=m, i.e., there exists βmj (j=0,1,,m1) such that

      (1rddr)m1(r2m1ϕ(r))=m1j=0βmjrj+1ϕ(j)(r),

      where βm0=c2m+1. Then we can observe that

      (1rddr)m(r2m+1ϕ(r))=1rddr(1rddr)m1(r2m1(r2ϕ(r)))=1rddrm1j=0βmjrj+1djdrj(r2ϕ(r))=1rm1j=0βmj[(j+1)rjdjdrj(r2ϕ(r))+rj+1dj+1drj+1(r2ϕ(r))]=m1j=0(j+1)βmjrj1djdrj(r2ϕ(r))+m1j=0βmjrjdj+1drj+1(r2ϕ(r))=m1j=0(j+1)βmjrj1djdrj(r2ϕ(r))+mj=1βmj1rj1djdrj(r2ϕ(r))=βm0rϕ(r)+m1j=1[(j+1)βmj+βmj1]rj1djdrj(r2ϕ(r))+βmm1rm1dmdrm(r2ϕ(r)).

      • For j=1, djdrj(r2ϕ(r))=2rϕ(r)+r2ϕ(r);
      • for j2, djdrj(r2ϕ(r))=r2ϕ(j)(r)+2jrϕ(j1)(r)+j(j1)ϕ(j2)(r).
      Thus, we have

      (1rddr)m(r2m+1ϕ(r))=βm0rϕ(r)+(2βm1+βm0)(2rϕ(r)+r2ϕ(r))+m1j=2[(j+1)βmj+βmj1]rj1[r2ϕ(j)+2jrϕ(j1)(r)+j(j1)ϕ(j2)(r)]+βmm1rm1(r2ϕ(m)+2mrϕ(m1)(r)+m(m1)ϕ(m2)(r))=(3βm0+4βm1)rϕ(r)+(2βm1+βm0)r2ϕ(r)+m1j=2[(j+1)βmj+βmj1]rj+1ϕ(j)+m2j=12[(j+2)(j+1)βmj+1+(j+1)βmj]rj+1ϕ(j)(r)+m3j=0[(j+3)(j+2)(j+1)βmj+2+(j+2)(j+1)βmj+1]rj+1ϕ(j)(r)+m(m1)βmm1rm1ϕ(m2)(r)+2mβmm1rmϕ(m1)(r)+βmm1rm+1ϕ(m)=mj=0βm+1jrj+1ϕ(j)(r),

      where βm+1j are given by

      βm+10=3βm0+6βm1+6βm2,βm+11=βm0+4βm1+18βm2+24βm3,βm+1j=βmj1+3(j+1)βmj+3(j+2)(j+1)βmj+1+(j+3)(j+2)(j+1)βmj+2for 2jm3,βm+1m2=βmm3+2(m1)βmm2+3m(m1)βmm1,βm+1m1=3mβmm1+mβmm1,βm+1m=βmm1.

      Therefore, we complete the proof of Claim 1.

      Then we claim that βk0=c2k+1=cn for kN.
      Proof of Claim 2Clearly, the claim holds true for k=1. From Claim 1, a direct computation gives

      k1j=0βkjrj+1ϕ(j)(r)=(1rddr)k1(r2k1ϕ(r))=(1rddr)k2[(2k1)r2k3ϕ(r)+r2k2ϕ(r)]=(1rddr)k3[(2k1)(2k3)r2k5ϕ(r)+]==(2k1)(2k3)(2k5)3r1ϕ(r)+.

      It is clear that βk0=(2k1)(2k3)(2k5)31=c2k+1=cn. The proof is complete.

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

      Vx(r,t)=(1rr)k1(r2k1Mu(x,r,t))=k1j=0βkjrj+1jMurj(x,r,t).

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

      u(x,t)=limr0+Mu(x,r,t)(by (30))=1cnlimr0+(cnMu(x,r,t))=1cnlimr0+1rk1j=0βkjrj+1jMurj(x,r,t)=1cnlimr0+Vx(r,t)r=1cnlimr0+[Gx(r+ct)+Gx(rct)2r+12crr+ctrctHx(ρ)dρ](by (41))=1cnlimr0+[(Gx)(r+ct)+(Gx)(rct)2+Hx(r+ct)Hx(rct)2c](by L'Hospital's Rule)=(Gx)(ct)+(Gx)(ct)2cn+Hx(ct)+Hx(ct)2cnc.

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

      Gx(t)=(1tt)k1(t2k1Mg(x,t))=(1tt)k1(t2k1ωn|ξ|=1g(x+tξ)dSξ)=1ωn=(1tt)n32tn2|ξ|=1g(x+tξ)dSξ,Hx(t)=(1tt)k1(t2k1Mh(x,t))=(1tt)k1(t2k1ωn|ξ|=1h(x+tξ)dSξ)=1ωn(1tt)n32tn2|ξ|=1h(x+tξ)dSξ.

      This implies

      (Gx)(ct)=1ωn((ct)(1ct(ct))n32(ct)n2|ξ|=1g(x+ctξ)dSξ)=1ωnt(1tt)n32tn2|ξ|=1g(x+ctξ)dSξ,Hx(ct)=1ωn(1ct(ct))n32(ct)n2|ξ|=1h(x+ctξ)dSξ=cωn(1tt)n32tn2|ξ|=1h(x+ctξ)dSξ.

      Also note that (Gx)(ct)=(Gx)(ct) and Hx(ct)=Hx(ct). Therefore, we obtain (42) by combining above equations.

沒有留言:

張貼留言