Solution to Partial Differential Equations: Methods and Applications (Robert McOwen) Section 4.2
- If n=2 and a=1, show that (44) is equivalent to
u(r,θ)=1−r22π∫2π0g(ϕ)1+r2−2rcos(θ−ϕ)dϕ.
- Use (a) and Exercise 1 in Section 4.1 to verify the formula
rkcoskθ=1−r22π∫2π0cos(kϕ)1+r2−2rcos(θ−ϕ)dϕ,
where k is an integer and 0≤r<1.
- If n=2 and a=1, show that (44) is equivalent to
- When n=2 and a=1, the scircumference of the unit disk is 2π, and (44) becomes
u(ξ1,ξ2)=1−(ξ21+ξ22)2π∫x2+y2=1g(x,y)(x−ξ1)2+(y−ξ2)2dSx,y.
Using the polar coordinate x=cosϕ and y=sinϕ for 0≤ϕ<2π; ξ1=rcosθ and ξ2=rsinθ for 0≤r≤1 and 0≤θ<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)=1−r22π∫2π0˜g(ϕ)(cosϕ−rcosθ)2+(sinϕ−rsinθ)2dϕ=1−r22π∫2π0˜g(ϕ)1+r2−2rcosθcosϕ−2rsinθsinϕdϕ=1−r22π∫2π0˜g(ϕ)1+r2−2rcos(θ−ϕ)dϕ.
The proof is complete. - Consider the following Dirichlet problem
{Δu=0in B1(0),u(1,θ)=cos(kθ)for 0≤θ<2π.
Then using Exercise 1 in Section 4.1, we findu(r,θ)=12π∫2π0g(s)ds+1π∞∑n=1rn[cos(nθ)∫2π0g(s)cos(ns)ds+sin(nθ)∫2π0g(s)sin(ns)ds]=12π∫2π0cos(ks)ds+1π∞∑n=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 k∈N, where δkn=1 if k=n; 0 if k≠n. On the other hand, by (a), we obtain the formularkcoskθ=u(r,θ)=1−r22π∫2π0cos(kϕ)1+r2−2rcos(θ−ϕ)dϕ.
The proof is complete. - Let Ω be a bounded domain and f∈Ck(ˉΩ). Show that in fact the domain potential (29a) satisfies u∈Ck+1(Ω). Conclude that f∈C∞(ˉΩ) implies u∈C∞(Ω). [In Section 8.2 we show that f∈C∞(ˉΩ) implies u∈C∞(ˉΩ), provided ∂Ω is C∞.]
- The symmetry of the Green's function [i.e., G(x,ξ)=G(ξ,x) for all x,ξ∈Ω] is an important fact connected with the self-adjointness of Δ.
- Verify the symmetry of G(x,ξ) by direct calculation when Ω=Rn+ and Ω=Ba(0).
- Prove the symmetry of G(x,ξ) when Ω is any smooth, bounded domain.
- Recall that K(x)={12πlog|x|if n=2;1(2−n)ωn|x|2−nif n≥3. (cf. (23)).
- When Ω=Rn+, the Green's function G is defined in (35), i.e.,
G(x,ξ)=K(x−ξ)−K(x−ξ∗)={log|x−ξ|−log|x−ξ∗|2πif n=2;|x−ξ|2−n−|x−ξ∗|2−n(2−n)ωnif n≥3.
Here ξ∗=(ξ1,…,ξn−1,−ξn) denotes the reflection point of ξ=(ξ1,…,ξn−1,ξn) with respect to the plane ∂Rn+. Clearly, |x−ξ|=|ξ−x| and|x−ξ∗|=√n−1∑i=1(xi−ξi)2+[xn−(−ξn)]2=√n−1∑i=1(ξi−xi)2+[ξn−(−xn)]2=|ξ−x∗|,
where x∗=(x1,…,xn−1,−xn) denotes the reflection of x with respect to the plane ∂Rn+. Therefore, we verify thatG(x,ξ)=K(x−ξ)−K(x−ξ∗)=K(ξ−x)−K(ξ−x∗)=G(ξ,x) for all x,ξ∈Rn+ and x≠ξ.
- When Ω=Ba(0), the Green's function G is defined in (42), i.e.,
G(x,ξ)=K(x−ξ)−K(|ξ|a(x−ξ∗))={12π[log|x−ξ|−log(|ξ|a|x−ξ∗|)]if n=2;1(2−n)ωn[|x−ξ|2−n−(|ξ|a|x−ξ∗|)2−n]if n=3.
Here ξ∗=a2ξ|ξ|2 denotes the reflection of ξ≠0 with respect to the sphere ∂Ba(0). Clearly, |x−ξ|=|ξ−x| and||ξ|a(x−ξ∗)|2=n∑i=1ξ2ia2(n∑i=1x2i−2a2n∑i=1xiξin∑j=1ξ2j+a4n∑i=1ξ2i(n∑j=1ξ2j)2)=n∑i=1x2ia2(n∑i=1ξ2i−2a2n∑i=1xiξin∑j=1x2j+a4n∑i=1x2i(n∑j=1x2i)2)=||x|a(ξ−x∗)|2,
where x∗=a2x|x|2 for x≠0. Therefore, we verify thatG(x,ξ)=K(x−ξ)−K(|ξ|a(x−ξ∗))=K(ξ−x)−K(|x|a(ξ−x∗))=G(ξ,x)for all x,ξ∈Ba(0)−{0} and x≠ξ.
- When Ω=Rn+, the Green's function G is defined in (35), i.e.,
- Recall that the notion of Green's function G is defined at page 117: G(x,ξ)=K(x−ξ)+ω(x) for x≠ξ. Hence G satisfies ΔxG(x,ξ)=0 for x∈Ω−{ξ} with G(x,ξ)=0 for x∈∂Ω (cf. (32)). Fix x,ξ∈Ω arbitrarily with x≠ξ. 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 z∈Ω−{x},v(z)=G(z,ξ)for z∈Ω−{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 z∈Ω−{x} and Δv(z)=0 for z∈Ω−{ξ}, which implies∫Ωϵ(vΔu−uΔv)dz=0.
Using the Green's second identity (7) at page 107, we have0=∫Ωϵ(vΔu−uΔv)dz=∫∂Ωϵ(v∂u∂ν−u∂v∂ν)dSz=(∫∂Ω+∫∂Bϵ(x)+∫∂Bϵ(ξ))(v∂u∂ν−u∂v∂ν)dSz.
In addition, we note the fact that u(z)=G(z,x)=0 for z∈∂Ω and v(z)=G(z,ξ)=0 for z∈∂Ω (from (32)). Hence we get∫∂Bϵ(x)(v∂u∂ν−u∂v∂ν)dSz=∫∂Bϵ(ξ)(u∂v∂ν−v∂u∂ν)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(z−x)|+|ω(z)|, we find|∫∂Bϵ(x)u∂v∂νdSz|≤nωnϵn−1⋅maxz∈¯B1(x)|∇v(z)|⋅(maxz∈∂Bϵ(x)|K(z−x)|+maxz∈¯B1(x)|ω(z)|)→0as ϵ→0.
Here we have used the fact thatlimϵ→0+ϵn−1K(|ϵ|)={limϵ→0+ϵlogϵ2π=0if n=2;limϵ→0+ϵn−1ϵ2−n(2−n)ωn=0if n≥3.
On the other hand, by the smoothness of ω and (22)-(23), we havelimϵ→0+∫∂Bϵ(x)v∂u∂νdSz=limϵ→0+∫∂Bϵ(x)v(z)∂K∂ν(z−x)dSz=v(x).
Hence we see limϵ→0+∫∂Bϵ(x)(v∂u∂ν−u∂v∂ν)dSz=v(x). Similarly, one may check that limϵ→0+∫∂Bϵ(ξ)(u∂v∂ν−v∂u∂ν)dSz=u(ξ). Therefore, we arrive at u(ξ)=v(x) and complete the proof. - Use the weak maximum principle (16) to prove that G(x,ξ)≤0 for x,ξ∈Ω with x≠ξ.
- Use the strong maximum principle (15) to prove that G(x,ξ)<0 for x,ξ∈Ω with x≠ξ.
- Fix ξ∈Ω. From the definition of Green's function G at page 117, we write G(x,ξ)=K(x−ξ)+ω(x), where ω satisfies Δω=0 in Ω and ω(x)=−K(x−ξ) for x∈∂Ω. Since ω is bounded in ˉΩ and limx→ξK(x−ξ)=−∞, there exists ϵ>0 such that ¯Bϵ(ξ)⊊Ω and G(x,ξ)<0 for x∈¯Bϵ(ξ) and x≠ξ. Let Ωϵ,ξ=Ω−¯Bϵ(ξ). Then we find
Δx(x,ξ)=0in Ωϵ,ξ,G(x,ξ)≤0for x∈∂Ωϵ,ξ=∂Ω∪∂Bϵ(ξ).
Then by the weak maximum principle (16), we obtainmaxx∈ˉΩϵ,ξG(x,ξ)=maxx∈∂Ωϵ,ξG(x,ξ)=max{maxx∈∂ΩG(x,ξ),maxx∈∂Bϵ(ξ)G(x,ξ)}=0,
which implies G(x,ξ)≤0 for x∈ˉΩϵ,ξ. Since G(x,ξ)<0 for x∈¯Bϵ(ξ)−{ξ}, we arrive at G(x,ξ)<0 for x≠ξ. - Following the process of part (a), G is not constant because G(x,ξ)=0 for x∈∂Ω and G(x,ξ)<0 for x∈∂Bϵ(ξ). Then by Theorem 3 (Maximun Principle) at page 109, we obtain
G(x,ξ)<supz∈ΩG(z,ξ)=maxz∈ˉΩG(z,ξ)=0for x∈Ω but x≠ξ.
The proof is complete. - Use (33) to prove (38).
- For n=2, use the method of reflections to find the Green's function for the first quadrant Ω={(x,y):x,y>0}.
- If u∈C(Ω) satisfies the mean value property of Section 4.1.d, then u is harmonic in Ω.
- Let Ω=Ba(0), Ω+=Ω∩Rn+, and Ω0={x∈Ω:xn=0}. If u∈C2(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.)
- If x∈Ω+, then ∂2˜u∂x2i(x)=∂2u∂x2i(x), which implies Δ˜u(x)=Δu(x)=0.
- If x∈Ω−(Ω+∪Ω0), then ∂˜u∂xi(x)=−∂u∂xi(x′,−xn) for i=1,…,n−1 and ∂˜u∂xn(x)=∂u∂xn(x′,−xn). Moreover, we get ∂2˜u∂x2i(x)=−∂2u∂x2i(x′,−xn) for i=1,…,n−1 and ∂2˜u∂x2n(x)=−∂2u∂x2n(x′,−xn). Thus, we get Δ˜u(x)=−Δu(x′,−xn)=0.
- If x∈Ω0, then by ˜u≡u≡0 on Ω0, we have ∂2˜u∂x2i(x′,0)=0 for i=1,…,n−1. Moreorever, we use the definition of partial derivative to find
∂˜u∂xn(x′,0−)=limh→0−˜u(x′,h)−˜u(x′,0)h=limh→0−u(x′,−h)−h=∂u∂xn(x′,0)=limh→0+u(x′,h)h=limh→0+˜u(x′,h)−˜u(x′,0)h=∂˜u∂xn(x′,0+).
Thus ∂˜u∂xn(x′,0) exists. Then the second derivative gives∂2˜u∂x2n(x′,0+)=limh→0+∂˜u∂xn(x′,h)−∂˜u∂xn(x′,0)h=limh→0+∂u∂xn(x′,h)−∂u∂xn(x′,0)h=∂2u∂x2n(x′,0)=0,∂2˜u∂x2n(x′,0−)=limh→0−∂˜u∂xn(x′,h)−∂˜u∂xn(x′,0)h=−limh→0−∂u∂xn(x′,−h)−∂u∂xn(x′,0)h=−∂2u∂x2n(x′,0)=0.
Thus, ∂2˜u∂x2n(x′,0)=0. Therefore Δ˜u(x)=0. - Show that the bounded solution of the Dirichlet problem in a half-space is unique. Give unbounded counterexamples.
- Suppose u∈C2(Ω) is harmonic, u≥0, and ¯Ba(0)⊂Ω. Use (44) to show
an−2(a−|ξ|)(a+|ξ|)n−1u(0)≤u(ξ)≤an−2(a+|ξ|)(a−|ξ|)n−1u(0)for |ξ|<a.
- Prove (45).
- Suppose u∈C2(Ω) is harmonic, u≥0, and ¯Ba(0)⊂Ω. Use (44) to show
- In Ba(0), we use the Poisson integral formula (44):
u(ξ)=a2−|ξ|2aωn∫|x|=au(x)|x−ξ|ndSxfor |ξ|<a.
By the triangle inequality, we have a−|ξ|=|x|−|ξ|≤|x−ξ|≤|x|+|ξ|=a+|ξ| for |x|=a, which gives 1(a+|ξ|)n≤1|x−ξ|n≤1(a−|ξ|)n. Then by u(x)≥0, we haveu(ξ)≥a2−|ξ|2aωn∫|x|=au(x)(a+|ξ|)ndSx=an(a2−|ξ|2)aωn(a+|ξ|)n∫|x|=au(x)andSx=an−1(a−|ξ|)ωn(a+|ξ|)n−1∫|x|=au(x)|x|ndSx=an−2(a−|ξ|)(a+|ξ|)n−1⋅aωn∫|x|=au(x)|x|ndSx=an−2(a−|ξ|)(a+|ξ|)n−1u(0).
Simiarly, from u(x)≥0, we haveu(ξ)≤a2−|ξ|2aωn∫|x|=au(x)(a−|ξ|)ndSx=an(a2−|ξ|2)aωn(a−|ξ|)n∫|x|=au(x)andSx=an−1(a+|ξ|)ωn(a−|ξ|)n−1∫|x|=au(x)|x|ndSx=an−2(a+|ξ|)(a−|ξ|)n−1⋅aωn∫|x|=au(x)|x|ndSx=an−2(a+|ξ|)(a−|ξ|)n−1u(0).
Therefore, we complete the proof of Harnack's inequality on a ball. - Let r=13dist(ˉΩ1,∂Ω). Then for any ξ,η∈ˉΩ1, and |ξ−η|≤r, we use the mean value property (12) to find
u(ξ)=nωn∫|x|≤1u(ξ+2rx)dx=nωn(2r)n∫B2r(ξ)u(y)dy≥nωn(2r)n∫Br(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 Bi∩Bi−1≠∅. In addition, by the extreme value theorem, there exists x∈ˉΩ1 and y∈ˉΩ1 such thatu(x)=maxˉΩ1u=supΩ1u,u(y)=minˉΩ1u=infΩ1u.
Therefore, we haveu(y)≥12nNu(x),
which implies (45) and C1 depends on N, i.e., depends on Ω1. - Use (46) to prove Liouville's theorem.
- Use Stirling's formula k!∼(2πk)1/2kke−k as k→∞ to show there is a constant C so that |α||α|≤Cα!e|α|n|α| holds for all multi-indices α.
Solution
Solution
Clearly, 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., f∈C2(ˉΩ) implies u∈C3(Ω). For any i∈{1,…,n}, from the argument of Proposition at page 114, we have∂iu(x)=∫Ω(∂iK)(x−y)f(y)dyfor x∈Ω.
It suffices to show ∂iu∈C2(Ω). To achieve this, we consider the Poisson equation −Δϕ=∂if in Ω. Since ∂if∈C1(ˉΩ) (from f∈C2(ˉΩ)), then by the Proposition at page 114, we know the solution ϕ∈C2(Ω) andϕ(x)=∫ΩK(x−y)(∂if)(y)dy.
Using the integration by parts, we haveϕ(x)=∫∂ΩK(x−y)f(y)νi(y)dSy+∫Ω(∂iK)(x−y)f(y)dy.
For any y∈∂Ω, the function ∫∂ΩK(⋅−y)f(y)νi(y)dSy is smooth. Hence, we find∂ui(x)=ϕ(x)−∫∂ΩK(x−y)f(y)νi(y)dSyfor x∈Ω,
which means ∂iu∈C2(Ω) and u∈C3(Ω).Now we prove the statement by mathematical induction. Suppose the statement holds true for some positive integer m≥2, i.e., f∈Cm(ˉΩ) implies u∈Cm+1(Ω). Since f∈Cm+1(ˉΩ), we have ∂fi∈Cm for any i∈{1,…,n}. Note that ∂iu∈Cm(Ω) and satisfies Δ(∂iu)=∂if in Ω. Here we have used the fact that m≥2. Then by the inductive hypothesis, ∂iu∈Cm+1(Ω). Due to the arbitrariness of i, we know u∈Cm+2(Ω). Therefore, by mathematical induction, the statement holds true for all n∈N.
When f∈C∞(ˉΩ), we have f∈Cm(ˉΩ) for all m∈N, which gives u∈Cm+1(Ω) for all m∈N. Thus, we know u∈C∞(Ω). The proof is complete.Solution
Solution
Solution
Let ΩR=BR(0)∩Rn+={x=(x1,…,xn)∈Rn:|x|<R,xn>0} for any R>0 and u≡1 in ˉRn+. Using (31) and Green's function G defined by (35), we have1=∫ΩRG(x,ξ)Δudx+∫∂ΩR(u(x)∂G(x,ξ)∂νx−G(x,ξ)∂u(x)∂ν)dSx=∫∂ΩR∂G(x,ξ)∂νxdSx=(∫∂ΩR∩(Rn−1×{0})+∫∂ΩR∩Rn+)∂G(x,ξ)∂νxdSxfor ξ∈ΩR.
For x∈∂ΩR∩(Rn−1×{0}), we have∂G(x,ξ)∂νx=−∂G∂xn(x,ξ)|xn=0=2ξnωn|x′−ξ|−n,
where x′=(x1,…,xn−1)∈Rn−1 (cf. (36)). For x∈∂ΩR∩Rn+, we have∂G(x,ξ)∂νx=∇xG(x,ξ)⋅x|x|=1ωn(|x−ξ|−n(x−ξ)−|x−ξ∗|−n(x−ξ∗))⋅xR=1ωnR(R2−ξ⋅x|x−ξ|n−R2−ξ∗⋅x|x−ξ∗|n)=(Rωn−n−1∑i=1ξixiωnR)(1|x−ξ|n−1|x−ξ∗|n)−ξnxnωnR(1|x−ξ|n+1|x−ξ∗|n).
A direct estimate gives|∫∂ΩR∩Rn+ξnxnωnR(1|x−ξ|n+1|x−ξ∗|n)dSx|≤|ξn|ωn⋅2(R−|ξ|)n⋅ωnRn−12=|ξ|Rn−1R−|ξ|n→0as R→∞.
Here we have used the fact that |x−ξ|≥|x|−|ξ|=R−|ξ| and |x−ξ∗|≥|x|−|ξ∗|=R−|ξ| for x∈∂BR(0). Moreoever, by the trinagle inequality and x∈∂BR(0)∩Rn+, we have R−|ξ|≤|x−ξ|≤|x−ξ∗|≤R+|ξ|, which implies1|x−ξ|n−1|x−ξ∗|n≤1(R−|ξ|)n−1(R+|ξ|)n=2nRn−1|ξ|+⋯(R2−|ξ|2)n.
Hence we get|∫∂ΩR∩Rn+(Rωn−n−1∑i=1ξixiωnR)(1|x−ξ|n−1|x−ξ∗|n)dSx|≤(Rωn+n−1∑i=1|ξi|ωn)⋅2nRn−1|ξ|+⋯(R2−|ξ|2)n⋅ωnRn−12
Combining these results, it follows that limR→∞∫∂ΩR∩Rn+∂G(x,ξ)∂νxdSx=0. Therefore, we obtain1=limR→∞∫∂ΩR∂G(x,ξ)∂νxdSx=limR→∞∫∂ΩR∩(Rn−1×{0})∂G(x,ξ)∂νxdSx=∫Rn−1×{0}H(x′,ξ)dx′for x∈Rn+.
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, G≡0 on ∂ΩR∩(Rn−1×{0}) but G may not becomes zero on ∂ΩR∩Rn+. However, since u≡1, the normal derivative of u becomes zero so the boundary value of function G is not important.
Solution
Let 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 claimG(x,ξ)=G(x1,x2,ξ1,ξ2)=K(x−ξ)−K(|x+ξ∗||x−ξ∗||x+ξ|)=log|x−ξ|−log|x−ξ∗|−log|x+ξ∗|+log|x+ξ|2π=12πlog|x−ξ||x+ξ||x−ξ∗||x+ξ∗|for x=(x1,x2)∈Ω but x≠ξ
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 |x−ξ|=|x−ξ∗| and |x+ξ|=|x+ξ∗|; for x2=0, we have |x−ξ|=|x+ξ∗| and |x+ξ|=|x−ξ∗|. Thus, for x∈∂Ω, it is necessary that |x−ξ||x+ξ||x−ξ∗||x+ξ∗|=1, which implies G(x,ξ)=0. Therefore, by the definition, G is a Green's function for the first quadrant Ω.Solution
Let ξ∈Ω. Then there exists a positive constant r such that ¯Br(ξ)⊊Ω. Since u∈C(Ω) satisfies the mean value property, we haveu(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 v∈C2(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 v≡u in ¯Br(ξ). Thus, u∈C2(Br(ξ))∩C(¯Br(ξ)) and Δu=0 in Br(ξ). Due to arbitrariness of ξ, Δu=0 in Ω. The proof is complete.Solution
Define the extension ˜u:Ω→R of u by˜u(x)=˜u(x′,xn)={u(x′,xn)if xn≥0;−u(x′,−xn)if xn<0.
Here x′ denotes (x1,…,xn−1). Since u≡0 on Ω0, function ˜u is continuous on Ω. By Exercise 7, it suffices to show that ˜u satisfies the mean value property: For any ξ∈Ω, 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 ˜u≡u in Ω+, for ξ∈Ω+, 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 ξ∈Ω−(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 ξ∈Ω0 and 0<r<a−|ξ|, we observe that1ωn∫|x|=1˜u((ξ′,0)+rx)dSx=1ωn∫|x|=1,xn>0u(ξ′+rx′,rxn)dSx−1ωn∫|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: u∈C2(Ω+∪Ω0), then we can compute the second derivatives ∂2˜u∂x2i(x) at each point x∈Ω, and then sum up to get Δ˜u(x)=0 for all x∈Ω.
Warning: The original statement is that "If u∈C2(Ω)∩C(Ω+∪Ω0) ...". It should be modified as "If u∈C2(Ω+)∩C(Ω+∪Ω0) ...".
Solution
Let ˜u and u be the bounded solutions of the Dirichlet problem in a half-space. Then v:=˜u−u satisfies{Δv=0in Rn+,v=0on ∂Rn+.
Define ˜v:Rn→R by the reflection˜v(x)=˜v(x′,xn)={v(x′,xn)if xn≥0;−v(x′,−xn)if xn<0.
For any x∈Rn, there exists a sufficiently large a>0 such that x∈Ba(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 x∈Rn. This implies ˜u≡u 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.)
Solution
Solution
Let u be the bounded harmonic function in Rn and ξ∈Rn. Define v(x)=u(x+ξ) for x∈Rn. Clearly, v is also bounded and harmonic. By (46), we have|∂u∂ξi(ξ)|=|∂v∂ξi(0)|≤namax|x|=a|v(x)|=namax|x|=a|u(x+ξ)|→0as a→∞.
This shows ∇u(ξ)=0 for all ξ∈Rn, and hence u is constant.Solution
The Stirling's formula implieslimk→∞kkk!eknk=limk→∞(2πk)1/2kke−kk!⋅kk(2πk)1/2kke−keknk=1⋅limk→∞1nk√2πk=0.
There exists a positive constant K such that kk≤k!eknk for k≥K. Hence we takeC=max{1,max1≤|α|<K|α||α|α!e|α|n|α|},
which is desired.
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