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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|nωnϵn1maxz¯B1(x)|v(z)|(maxzBϵ(x)|K(zx)|+maxz¯B1(x)|ω(z)|)0as ϵ0.

      Here we have used the fact that

      limϵ0+ϵn1K(|ϵ|)={limϵ0+ϵlogϵ2π=0if n=2;limϵ0+ϵn1ϵ2n(2n)ωn=0if n3.

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

      limϵ0+Bϵ(x)vuνdSz=limϵ0+Bϵ(x)v(z)Kν(zx)dSz=v(x).

      Hence we see limϵ0+Bϵ(x)(vuνuvν)dSz=v(x). Similarly, one may check that limϵ0+Bϵ(ξ)(uvνvuν)dSz=u(ξ). Therefore, we arrive at u(ξ)=v(x) and complete the proof.


    1. Use the weak maximum principle (16) to prove that G(x,ξ)0 for x,ξΩ with xξ.
    2. Use the strong maximum principle (15) to prove that G(x,ξ)<0 for x,ξΩ with xξ.
  6. Solution
    1. Fix ξΩ. From the definition of Green's function G at page 117, we write G(x,ξ)=K(xξ)+ω(x), where ω satisfies Δω=0 in Ω and ω(x)=K(xξ) for xΩ. Since ω is bounded in ˉΩ and limxξK(xξ)=, there exists ϵ>0 such that ¯Bϵ(ξ)Ω and G(x,ξ)<0 for x¯Bϵ(ξ) and xξ. Let Ωϵ,ξ=Ω¯Bϵ(ξ). Then we find

      Δx(x,ξ)=0in Ωϵ,ξ,G(x,ξ)0for xΩϵ,ξ=ΩBϵ(ξ).

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

      maxxˉΩϵ,ξG(x,ξ)=maxxΩϵ,ξG(x,ξ)=max{maxxΩG(x,ξ),maxxBϵ(ξ)G(x,ξ)}=0,

      which implies G(x,ξ)0 for xˉΩϵ,ξ. Since G(x,ξ)<0 for x¯Bϵ(ξ){ξ}, we arrive at G(x,ξ)<0 for xξ.
    2. Following the process of part (a), G is not constant because G(x,ξ)=0 for xΩ and G(x,ξ)<0 for xBϵ(ξ). Then by Theorem 3 (Maximun Principle) at page 109, we obtain

      G(x,ξ)<supzΩG(z,ξ)=maxzˉΩG(z,ξ)=0for xΩ but xξ.

      The proof is complete.

  7. Use (33) to prove (38).
  8. SolutionLet ΩR=BR(0)Rn+={x=(x1,,xn)Rn:|x|<R,xn>0} for any R>0 and u1 in ˉRn+. Using (31) and Green's function G defined by (35), we have

    1=ΩRG(x,ξ)Δudx+ΩR(u(x)G(x,ξ)νxG(x,ξ)u(x)ν)dSx=ΩRG(x,ξ)νxdSx=(ΩR(Rn1×{0})+ΩRRn+)G(x,ξ)νxdSxfor ξΩR.

    For xΩR(Rn1×{0}), we have

    G(x,ξ)νx=Gxn(x,ξ)|xn=0=2ξnωn|xξ|n,

    where x=(x1,,xn1)Rn1 (cf. (36)). For xΩRRn+, we have

    G(x,ξ)νx=xG(x,ξ)x|x|=1ωn(|xξ|n(xξ)|xξ|n(xξ))xR=1ωnR(R2ξx|xξ|nR2ξx|xξ|n)=(Rωnn1i=1ξixiωnR)(1|xξ|n1|xξ|n)ξnxnωnR(1|xξ|n+1|xξ|n).

    A direct estimate gives

    |ΩRRn+ξnxnωnR(1|xξ|n+1|xξ|n)dSx||ξn|ωn2(R|ξ|)nωnRn12=|ξ|Rn1R|ξ|n0as R.

    Here we have used the fact that |xξ||x||ξ|=R|ξ| and |xξ||x||ξ|=R|ξ| for xBR(0). Moreoever, by the trinagle inequality and xBR(0)Rn+, we have R|ξ||xξ||xξ|R+|ξ|, which implies

    1|xξ|n1|xξ|n1(R|ξ|)n1(R+|ξ|)n=2nRn1|ξ|+(R2|ξ|2)n.

    Hence we get

    |ΩRRn+(Rωnn1i=1ξixiωnR)(1|xξ|n1|xξ|n)dSx|(Rωn+n1i=1|ξi|ωn)2nRn1|ξ|+(R2|ξ|2)nωnRn12

    Combining these results, it follows that limRΩRRn+G(x,ξ)νxdSx=0. Therefore, we obtain

    1=limRΩRG(x,ξ)νxdSx=limRΩR(Rn1×{0})G(x,ξ)νxdSx=Rn1×{0}H(x,ξ)dxfor xRn+.

    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 Rn+ but not BR(0)Rn+. Particularly, G0 on ΩR(Rn1×{0}) but G may not becomes zero on ΩRRn+. However, since u1, 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 Ω={(x,y):x,y>0}.
  10. SolutionLet ξ=(ξ1,ξ2)Ω, i.e., ξ1 and ξ2 are positive. Define ξ=(ξ1,ξ2). Then ξ, ξ and ξ are in the second, the third and the fourth quadrant, respectively. Now we claim

    G(x,ξ)=G(x1,x2,ξ1,ξ2)=K(xξ)K(|x+ξ||xξ||x+ξ|)=log|xξ|log|xξ|log|x+ξ|+log|x+ξ|2π=12πlog|xξ||x+ξ||xξ||x+ξ|for x=(x1,x2)Ω but xξ

    is a Green's function for the first quadrant Ω. Since ξ, ξ and ξ are not in the first quadrant, ΔxG(x,ξ)=0 for x=(x1,x2)Ω except ξ. In addition, for x1=0, we have |xξ|=|xξ| and |x+ξ|=|x+ξ|; for x2=0, we have |xξ|=|x+ξ| and |x+ξ|=|xξ|. Thus, for xΩ, it is necessary that |xξ||x+ξ||xξ||x+ξ|=1, which implies G(x,ξ)=0. Therefore, by the definition, G is a Green's function for the first quadrant Ω.

  11. If uC(Ω) satisfies the mean value property of Section 4.1.d, then u is harmonic in Ω.
  12. SolutionLet ξΩ. Then there exists a positive constant r such that ¯Br(ξ)Ω. Since uC(Ω) satisfies the mean value property, we have

    u(ξ)=1ωn|x|=1u(ξ+rx)dSx.

    Consider the Laplace equation Δv=0 in Br(ξ) with the Dirichlet boundary condition v=u|Br(ξ). Clearly, u|Br(ξ) is continuous on Br(ξ). Then by Theorem 4 at page 122, v can be solved by Poisson integral formula (44) and vC2(Br(ξ))C(¯Br(ξ)) is harmonic. Thus, v also satisfies the mean value property. Using the Exercise 9 in Section 4.1, we obtain vu in ¯Br(ξ). Thus, uC2(Br(ξ))C(¯Br(ξ)) and Δu=0 in Br(ξ). Due to arbitrariness of ξ, Δu=0 in Ω. The proof is complete.

  13. Let Ω=Ba(0), Ω+=ΩRn+, and Ω0={xΩ:xn=0}. If uC2(Ω+)C(Ω+Ω0) is harmonic in Ω+, and u=0 on Ω0, prove that u may be extended ot a harmonic function on all of Ω. (This is called a reflection principle.)
  14. SolutionDefine the extension ˜u:ΩR of u by

    ˜u(x)=˜u(x,xn)={u(x,xn)if xn0;u(x,xn)if xn<0.

    Here x denotes (x1,,xn1). Since u0 on Ω0, function ˜u is continuous on Ω. By Exercise 7, it suffices to show that ˜u satisfies the mean value property: For any ξΩ, there exists a positive constant r (which may be dependent on ξ) such that ¯Br(ξ)Ω=Ba(0) and

    ˜u(ξ)=1ωn|x|=1˜u(ξ+rx)dSx.

    Since ˜uu in Ω+, for ξΩ+, there exists r>0 such that ¯Br(ξ)Ω+Ω, then we have

    ˜u(ξ)=u(ξ)=1ωn|x|=1u(ξ+|ξn|x)dSx=1ωn|x|=1˜u(ξ+|ξn|x)dSx.

    Similarly, for ξΩ(Ω+Ω0), there exists r>0 such that ¯Br(ξ)Ω(Ω+Ω0)Ω, we have

    ˜u(ξ)=u(ξ,ξn)=1ωn|x|=1u((ξ,ξn)+|ξn|x)dSx=1ωn|x|=1˜u(ξ+|ξn|(x,xn))dSx=1ωn|x|=1˜u(ξ+|ξn|(x,xn))dSx.

    Here we have used the symmetry of the integral over the surface |x|=1. For ξΩ0 and 0<r<a|ξ|, we observe that

    1ωn|x|=1˜u((ξ,0)+rx)dSx=1ωn|x|=1,xn>0u(ξ+rx,rxn)dSx1ωn|x|=1,xn<0u(ξ+rx,rxn)dSx=0=˜u(ξ,0)(by substitution yn=xn, the two integrals are same).

    Therefore, continuous function ˜u satisfies the mean value property, which implies ˜u is harmonic in Ω. The proof is complete.

    Remark. If the condition is replaced by a stronger condition: uC2(Ω+Ω0), then we can compute the second derivatives 2˜ux2i(x) at each point xΩ, and then sum up to get Δ˜u(x)=0 for all xΩ.

    • If xΩ+, then 2˜ux2i(x)=2ux2i(x), which implies Δ˜u(x)=Δu(x)=0.
    • If xΩ(Ω+Ω0), then ˜uxi(x)=uxi(x,xn) for i=1,,n1 and ˜uxn(x)=uxn(x,xn). Moreover, we get 2˜ux2i(x)=2ux2i(x,xn) for i=1,,n1 and 2˜ux2n(x)=2ux2n(x,xn). Thus, we get Δ˜u(x)=Δu(x,xn)=0.
    • If xΩ0, then by ˜uu0 on Ω0, we have 2˜ux2i(x,0)=0 for i=1,,n1. Moreorever, we use the definition of partial derivative to find

      ˜uxn(x,0)=limh0˜u(x,h)˜u(x,0)h=limh0u(x,h)h=uxn(x,0)=limh0+u(x,h)h=limh0+˜u(x,h)˜u(x,0)h=˜uxn(x,0+).

      Thus ˜uxn(x,0) exists. Then the second derivative gives

      2˜ux2n(x,0+)=limh0+˜uxn(x,h)˜uxn(x,0)h=limh0+uxn(x,h)uxn(x,0)h=2ux2n(x,0)=0,2˜ux2n(x,0)=limh0˜uxn(x,h)˜uxn(x,0)h=limh0uxn(x,h)uxn(x,0)h=2ux2n(x,0)=0.

      Thus, 2˜ux2n(x,0)=0. Therefore Δ˜u(x)=0.

    Warning: The original statement is that "If uC2(Ω)C(Ω+Ω0) ...". It should be modified as "If uC2(Ω+)C(Ω+Ω0) ...".


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

    {Δv=0in Rn+,v=0on Rn+.

    Define ˜v:RnR by the reflection

    ˜v(x)=˜v(x,xn)={v(x,xn)if xn0;v(x,xn)if xn<0.

    For any xRn, there exists a sufficiently large a>0 such that xBa(0). Then by Exercise 8, ˜v is harmonic in Ba(0) and hence Δ˜v(x)=0. Thus, ˜v is harmonic in Rn. Since ˜u and u are bounded in Rn+, ˜v is also bounded. By Theorem 7 (Liouville's Theorem) at page 123, ˜v must be a constant, and hence ˜v(x)=˜v(0)=0 for all xRn. This implies ˜uu in Rn+ and the uniqueness of bounded solution of the Dirichlet problem in the half-space follows.

    It is clear that the function u(x,xn)=xn 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 uC2(Ω) is harmonic, u0, and ¯Ba(0)Ω. Use (44) to show

      an2(a|ξ|)(a+|ξ|)n1u(0)u(ξ)an2(a+|ξ|)(a|ξ|)n1u(0)for |ξ|<a.

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

      u(ξ)=a2|ξ|2aωn|x|=au(x)|xξ|ndSxfor |ξ|<a.

      By the triangle inequality, we have a|ξ|=|x||ξ||xξ||x|+|ξ|=a+|ξ| for |x|=a, which gives 1(a+|ξ|)n1|xξ|n1(a|ξ|)n. Then by u(x)0, we have

      u(ξ)a2|ξ|2aωn|x|=au(x)(a+|ξ|)ndSx=an(a2|ξ|2)aωn(a+|ξ|)n|x|=au(x)andSx=an1(a|ξ|)ωn(a+|ξ|)n1|x|=au(x)|x|ndSx=an2(a|ξ|)(a+|ξ|)n1aωn|x|=au(x)|x|ndSx=an2(a|ξ|)(a+|ξ|)n1u(0).

      Simiarly, from u(x)0, we have

      u(ξ)a2|ξ|2aωn|x|=au(x)(a|ξ|)ndSx=an(a2|ξ|2)aωn(a|ξ|)n|x|=au(x)andSx=an1(a+|ξ|)ωn(a|ξ|)n1|x|=au(x)|x|ndSx=an2(a+|ξ|)(a|ξ|)n1aωn|x|=au(x)|x|ndSx=an2(a+|ξ|)(a|ξ|)n1u(0).

      Therefore, we complete the proof of Harnack's inequality on a ball.
    2. Let r=13dist(ˉΩ1,Ω). Then for any ξ,ηˉΩ1, and |ξη|r, we use the mean value property (12) to find

      u(ξ)=nωn|x|1u(ξ+2rx)dx=nωn(2r)nB2r(ξ)u(y)dynωn(2r)nBr(η)u(y)dy=n2nωn|x|1u(η+rx)dx=u(η)2n.

      Since Ω1 is a bounded domain, Ω1 is connected and ˉΩ1 is compact. Then there exists a finite open converging {Bi}Ni=1 such that the radius of Bi is r and BiBi1. In addition, by the extreme value theorem, there exists xˉΩ1 and yˉΩ1 such that

      u(x)=maxˉΩ1u=supΩ1u,u(y)=minˉΩ1u=infΩ1u.

      Therefore, we have

      u(y)12nNu(x),

      which implies (45) and C1 depends on N, i.e., depends on Ω1.

  18. Use (46) to prove Liouville's theorem.
  19. SolutionLet u be the bounded harmonic function in Rn and ξRn. Define v(x)=u(x+ξ) for xRn. Clearly, v is also bounded and harmonic. By (46), we have

    |uξi(ξ)|=|vξi(0)|namax|x|=a|v(x)|=namax|x|=a|u(x+ξ)|0as a.

    This shows u(ξ)=0 for all ξRn, and hence u is constant.

  20. Use Stirling's formula k!(2πk)1/2kkek as k to show there is a constant C so that |α||α|Cα!e|α|n|α| holds for all multi-indices α.
  21. SolutionThe Stirling's formula implies

    limkkkk!eknk=limk(2πk)1/2kkekk!kk(2πk)1/2kkekeknk=1limk1nk2πk=0.

    There exists a positive constant K such that kkk!eknk for kK. Hence we take

    C=max{1,max1|α|<K|α||α|α!e|α|n|α|},

    which is desired.

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