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ITET A. Cannas da Silva
Analysis III
Solutions 7
HS 2011
1.change variable and obtain1. We Z 5 2 δ(x)(x+ 3)dx, 1 then the definition can be applied and gives zero, since the pointx= 0is not in the interval]1,5[. R ∞ 2. Weuse the definition of the distributional derivative and obtain−δ(x− −∞ π) cosxdx= 1. 3. Usetwice the definition of distributional derivative: Z Z ∞ ∞ 2 2 00x2x δ(x)e dx=δ(x)(4x+ 2)e dx −∞ −∞ h i 2 2x = (4x+ 2)e x=0 = 2.
4. Followingthe hint, we have √ Z Z π3/2 δ(sin(x/3)) cos(x/3)dx= 3δ(u)du √ −π−3/2 = 3.
5. Theresult here is, immediately from the definition, equal to1. R R ∞ ∞ 0 −iwx−iwx 6.δ(x)e dx=iw δ(x)e dx=iw. −∞ −∞
2.We look for a solution of ΔG(x, y;ξ, η) =δ(x−ξ, y−η) (x, y)∈D , G(x, y;ξ, η) = 0(x, y)∈∂D . The fundamental solution ofΔis 1 2 2 Γ(x, y;ξ, ηln () =x−ξ) +(y−η). 4π
Bitte wenden!
y
1. Byreflection 1 1 2 22 2 G(x, y;ξ, ηln () =x−ξ) +(y−η)−ln (x+ξ) +(y−η) 4π4π 1 1 2 22 2 −ln (x−ξ) +(y+η) +ln (x+ξ) +(y+η) 4π4π 2. Bythe reflection (ξ, η) ˜ (ξ, η)7→(ξ, η˜) := 2 |(ξ, η)| it follows that p 2 2 1 (x−ξ() +y−η) G(x, y;ξ, η) =lnr, 2 2 2π ξ η % x−2+y−2 % % where p 2 2 %=ξ+η .
3.We look after a functionu, which satisfies −→ Δu=δ , ξ for all functionsϕ.
−→ If we assume thatu=u(r), whereris the distance toξ, then we end up solving the following equaiton foruoutside the special pointr= 0:
00 0 Δu=u(r) + 2u(r) = 0,
and it was seen in the Lectures (see lecture 12) that the solution is of the following form, witha, b∈R: 1 u(r) =a+b. r To determinea, bwe proceed as follows. Start with the wanted equation for distribu-tions: −→ ∞ −→ ∀ϕ∈C(R),hΔu, ϕi=hϕδ ,i=ϕ(ξ). c ξ
Siehe nchstes Blatt!
−→ We know that our distributionuis given by a nice function except atξ, so we can use this and the definition of derivatives for the distributions to justify the following calculations: hΔu, ϕi=hu,Δϕi Z =uΔϕ 3 R Z = limuΔϕ. −→ ε→03 R\Bε(ξ) −→ Now we use the second Green identity and the fact that away from the pointξthe functionusatisfiesΔu= 0, to continue the calculation. We obtain then: Z Z ∂ϕ ∂u hΔu, ϕi= lim(Δu)ϕ−u−ϕ −→−→ ∂r ∂r ε→03 R\Bε(ξ)∂Bε(ξ) Z ∂ϕ ∂u =−limu−ϕ . −→ ε→0 ∂r ∂r ∂Bε(ξ) We now use the above formula foruvalid away fromr= 0, and which givesu= 2 a/r+b, ∂ru=−a/r. We thus have (integrating in spherical coordinates) Z Z 2π π a a 2 hΔu, ϕi=−lim +b ∂rϕ+ϕ εsinϑdϑdψ 2 ε→0 0 0ε ε Z Z 2π π −→ =−asinϑϕ(ξ)dϑdψ 0 0 −→−→ π = 2πa[cosϑ]ψ(ξ) =−4πaϕ(ξ). 0 1 This givesa=−, andbis left free for us to choose. We may chooseb= 0for 4π simplicity.
4.The calculations are exactly of the same kind as the ones done at pages 12.15-12.22 of the lecture notes.
1. Wedenote herex= (x, y, z). We look after a solution of ΔG=δxinB1(0), G= 0on∂B1(0). We now use the reflection x ˜x=. 2 |x| Therefore the wanted Green’s function onB1(0)is 1 11 1 G(x,y) =−+. 4π|x−y|4π|x||y−˜x|
Bitte wenden!
2. Wewant to solve 3 nR, ΔG=δxi+ 3 G= 0on∂R. + The following reflection ˜x: (x, y, z)7→(x, y,−z) gives 1 11 1 G(x,y) =−+. 4π|x−y|4π|y−˜x| We can then easily compute the Poisson kernel
2z1 K(x,y) =−. 3 4π|x−y|
5.We use the Theorem at page 12.11 of the lecture notes, which gives the solution of the Dirichlet problem in terms of the Green function. The formula to be used is the one withf= 0, where the first term disappears. Z Z ∞ −→−→∂G −→ −→−→ u(ξ) =g(x)∂G(ξx ,)/∂nd x=g(x) (x,0, ξη)dx. −∞∂n ∂D In our case, we found the Green function of the upper half plane in2Din the lecture notes, see page 12.18: 2 2 1 (x−ξ() +y−η) GD(x, y, ξη) =ln, 2 2 4π(x−ξ) +(y+η) and we note that∂G/∂nin our case means just−∂G/∂y. We can compute this at y= 0and we obtain the formula for the Poisson kernel: ∂G(x,0, ξη)η −=−. 2 2 ∂y π[(x−ξ) +η] It follows that Z ∞ 1g(x)η u(ξ, η) =−dx. 2 2 π−∞(x−ξ) +η
6.We have to verify that∂N/∂nDis a constant on the boundary∂Dof our domains D(see page 13.11 of the lecture notes). We will actually find out that the partial derivative in questio is zero.
Siehe nchstes Blatt!
1. Onthe half plane{x≥1}we compute the derivative in the direction perpendi-cular to the boundary line{x= 1}(see lecture notes, chapter 12) 1 2 2 Γ =ln (x−ξ) +(y−η), 4π ∂2(Γ 1x−ξ) =, 2 2 ∂x4π(x−ξ() +y−η) ∂Γ−→∂2(1Γ 1−ξ) −→ (x;ξ1) =(1, y;ξ, η) =, 2 2 ∂x ∂y4π(1−ξ) +(y−η) x=1 ∂Γ−→∂Γ 12(ξ−1) −→∗ (x;ξ(1) =, y; 2−ξ, η) =. 2 2 1 ∂x ∂y4π(ξ−(1) +y−η) x=1 −→−→ ∗ therefore the values∂yΓatξand atξcancel out, and∂N/∂nD= 0. 2. Inthe case of the upper half space, we have 1 1 Γ =−p, 4π 2 2 2 (x−ξ() +y−η) +(z−ζ) ∂Γ−→1−3/2 −→2 22 (x;ξ() =x−ξ) +(y−η) +ζ∙(−ζ), ∂z4π z=0 −→ ∂Γ 1−3/2 −→∗22 2 (x;ξ) =(x−ξ() +y−η) +ζ ζ, ∂z4π z=0
and again the two terms cancel again, as above.
