Example 1.8.G(x,y)=Γ(x−y)−Γ(R∣x∣x−R∣x∣y) for x,y∈BR(a),
G(x,y)=Γ(x−y)−Γ(x~−y) for x,y∈R+n={xn>0} where x~=(x1,⋯,xn−1,−xn).
Proposition 1.9. (1) u is harmonic in BR(0), then u(x)=∫∂BR(0)ωnR∣x−y∣nR2−∣x∣2u(y)dy.
(2) u is harmonic in R+n, then u(x)=∫∂R+nnωn∣x−y∣n2xnu(y)dy.
Lemma 1.10. (Hopf) B is an open ball and x0∈∂B, Lu≥0 with c≤0 in B. If u(x0)>u(x),∀x∈B and u(x0)≥0, then for any outward direction v at x0 (v⋅n(x0)>0) we have t→0+liminftu(x0)−u(x0−tv)>0.
2Heat equation
{∂tu−Δu=0,u(0,x)=φ(x),(t,x)∈(0,+∞)×Rnx∈Rn
Proposition 2.1. A solution is u(t,x)=(4πt)2n1∫Rnφ(y)e−4t∣x−y∣2dy,and is unique provided that ∣u(t,x)∣≤CeA∣x∣2. Plus, all solutions are smooth.
Proposition 2.2. Write ΩT=(0,T]×Ω and ΓT={0}×U∪[0,T]×∂U. u attains its maximum and minimum on ΩT only at ΓT unless u is constant.
Proposition 2.3. Write E(x,t;r)={(y,s)∈Rn+1∣s≤t,(4π(t−s))2n1e−4(t−s)∣x−y∣2≥rn1}, then u(t,x)=4rn1∫E(x,t;r)u(y,s)(t−s)2∣x−y∣2dyds.
Consider the boundary-value problem{∂tu−Δu=f(t,x),u(0,x)=φ(x),∂tu(0,x)=ψ(x),(t,x)∈(0,T)×Ω,x∈Ω.LetE(t)=21∫Ωu2(t,x).We haveE′(t)=∫Ωu∂tu=∫ΩuΔu=−∫Ω∣∇u∣2≤0.
Proposition 3.1. The solution is u(t,x)=21(φ(x+t)+φ(x−t))+21∫x−tx+tψ(y)dywhen n=1.
Proposition 3.2. The solution is u(t,x)=∂t(t−∫∂Bt(x)φ)+t−∫∂B(t,x)ψ=−∫∂Bt(x)φ(y)+Dφ(y)⋅(y−x)+tψ(y)dy.when n=3.
Proposition 3.3. The solution is u(t,x)=∂t(t−∫∂Bt(x)21−∣ty−x∣2φdy)+t−∫∂B(t,x)21−∣ty−x∣2ψdy=−∫∂Bt(x)21−∣ty−x∣2φ(y)+Dφ(y)⋅(y−x)+tψ(y)dy.when n=2.
Consider the initial&boundary-value problem⎩⎨⎧∂t2u−Δu=f(t,x),u(0,x)=φ(x),∂tu(0,x)=ψ(x),u(t,x)=0,(t,x)∈(0,T)×Ω,x∈Ω,(t,x)∈(0,T)×∂Ω.LetE(t)=21∫Ω(∂tu)2+∣∇u∣2.We haveE′(t)=∫Ω∂tu(Δu+f)+∇u⋅∂t∇u=∫Ωdiv(∂tu∇u)+∂tuf=∫Ω∂tufby divergence theorem and ∂tu∣(0,T)×∂Ω=0≤E(t)+21∫Ωf2,so by GronwallE(t)≤C(∥ψ∥L22+∥∇φ∥L22+∥f∥L2((0,T)×Ω)2).
Consider the Cauchy problem when n=3.
For fixed (t0,x0), letE(t)F(t)=21∫Bt0−t(x0)(∂tu)2+∣∇u∣2,=21∫0t∫∂Bt0−s∣n∂tu−∇u∣2.We haveE′(t)=21(−∫∂Bt0−t(x0)(∂tu)2+∣∇u∣2+∫Bt0−t(x0)∂t((∂tu)2+∣∇u∣2))=21(−∫∂Bt0−t(x0)(∂tu)2+∣∇u∣2+2∫Bt0−t(x0)div(∂tu∇u))as ∂t2u=Δu=21(−∫∂Bt0−t(x0)(∂tu)2+∣∇u∣2−2∂tu∂n∂u)=−F′(t),soE(t)≤E(t)+F(t)=E(0).