Math 412-501Theory of Partial Differential EquationsLecture 2-4:Laplace’s equation in polar coordinates.Laplace’s equation in a rectangleChunk of an annulusIn polar coordinates: r1< r < r2, 0 < θ <π2Laplace’s equation in polar coordinatesIn Cartesian coordinates (x, y),∇2u =∂2u∂x2+∂2u∂y2= 0.Polar coordinates (r , θ). Transition formulas:x = r cos θ, y = r sin θ.Jacobian matrix: ∂x∂r∂x∂θ∂y∂r∂y∂θ!=cos θ −r sin θsin θ r cos θJacobian determinant:∂x∂r∂x∂θ∂y∂r∂y∂θ= rInverse matrix:a bc d−1=1ad − bcd −b−c aInverse Jacobian matrix: ∂r∂x∂r∂y∂θ∂x∂θ∂y!= ∂x∂r∂x∂θ∂y∂r∂y∂θ!−1=1rr cos θ r sin θ−sin θ cos θLet u : R2→ R be an arbitrary smooth function.∂u∂x=∂u∂r∂r∂x+∂u∂θ∂θ∂x= cos θ ·∂u∂r−1rsin θ ·∂u∂θ∂u∂y=∂u∂r∂r∂y+∂u∂θ∂θ∂y= sin θ ·∂u∂r+1rcos θ ·∂u∂θ∂2u∂x2=cos θ ·∂∂r−1rsin θ ·∂∂θcos θ ·∂u∂r−1rsin θ ·∂u∂θ= cos2θ ·∂2u∂r2+1r2cos θ sin θ ·∂u∂θ−1rcos θ sin θ ·∂2u∂r ∂θ+1rsin2θ ·∂u∂r−1rsin θ cos θ ·∂2u∂θ ∂r+1r2sin θ cos θ ·∂u∂θ+1r2sin2θ ·∂2u∂θ2.∂2u∂y2=sin θ ·∂∂r+1rcos θ ·∂∂θsin θ ·∂u∂r+1rcos θ ·∂u∂θ= sin2θ ·∂2u∂r2−1r2sin θ cos θ ·∂u∂θ+1rsin θ cos θ ·∂2u∂r ∂θ+1rcos2θ ·∂u∂r+1rcos θ sin θ ·∂2u∂θ ∂r−1r2cos θ sin θ ·∂u∂θ+1r2cos2θ ·∂2u∂θ2.∇2u =∂2u∂r2+1r·∂u∂r+1r2·∂2u∂θ2=1r·∂∂rr∂u∂r+1r2·∂2u∂θ2Laplace’s equation in polar coordinates:1r∂∂rr∂u∂r+1r2∂2u∂θ2= 0or∂2u∂r2+1r·∂u∂r+1r2·∂2u∂θ2= 0Separation of variables: u(r, θ) = h(r)φ(θ).Substitute this into Laplace’s equation:d2hdr2φ(θ) +1rdhdrφ(θ) +1r2h(r)d2φdθ2= 0.Divide both sides by r−2h(r)φ(θ ) = r−2u(r , θ):1h·r2d2hdr2+ rdhdr= −1φ·d2φdθ2.It follows that1h·r2d2hdr2+ rdhdr= −1φ·d2φdθ2= λ = const.The variables have been separated:r2d2hdr2+ rdhdr= λh,d2φdθ2= −λφ.Proposition Suppose h and φ are solutions of theabove ODEs for the same value of λ. Thenu(r , θ) = h(r)φ(θ) is a solution of Laplace’sequation.Euler’s (or equidimensional) equationr2d2hdr2+ rdhdr− λh = 0 (r > 0)Suppose h(r) = rp, p ∈ R. Thenh′(r) = prp−1, h′′(r) = p(p − 1)rp−2.Hencer2d2hdr2+ rdhdr=p(p − 1) + prp= p2rp.λ > 0 =⇒ h(r) = C1rp+ C2r−p(λ = p2, p > 0)λ = 0 =⇒ r2h′′(r) + rh′(r) = 0 =⇒ r(rh′)′= 0=⇒ rh′(r) = C2=⇒ h(r) = C1+ C2log r .Chunk of an annulusBoundary value problem1r∂∂rr∂u∂r+1r2∂2u∂θ2= 0 (r1< r < r2, 0 < θ < L),u(r , 0) = u(r, L) = 0 (r1< r < r2),u(r1, θ) = 0, u(r2, θ) = f (θ) (0 < θ < L).It is assumed that r1> 0, L < 2π.If r1= 0 then the chunk (annular sector) becomes awedge (circular sector).We are looking for a solution u(r, θ) = h(r)φ(θ)to Laplace’s equation that satisfies the threehomogeneous boundary conditions.PDE holds ifr2d2hdr2+ rdhdr= λh,d2φdθ2= −λφ.for the same constant λ.Boundary conditions u(r , 0) = u(r, L) = 0 hold ifφ(0) = φ(L) = 0.Boundary condition u(r1, θ) = 0 holds ifh(r1) = 0.Eigenvalue problem: φ′′= −λφ, φ(0) = φ(L) = 0.Eigenvalues: λn= (nπL)2, n = 1, 2, . . .Eigenfunctions: φn(θ) = sinnπθL.Dependence on r :r2h′′+ rh′= λh, h(r1) = 0.=⇒ h(r) = C0(rr1)p− (r1r)p(p =√λ)Solution of Laplace’s equation:u(r , θ) =(rr1)nπ/L− (r1r)nπ/LsinnπθL, n = 1, 2, . . .We are looking for the solution of the reducedboundary value problem as a superposition ofsolutions with separated variables.u(r , θ) =X∞n=1Cnrr1nπ/L−r1rnπ/LsinnπθLBoundary condition u(r2, θ) = f (θ) is satisfied iff (θ) =X∞n=1Cnr2r1nπ/L−r1r2nπ/LsinnπθLHow do we solve the boundary value problem?1r∂∂rr∂u∂r+1r2∂2u∂θ2= 0 (r1< r < r2, 0 < θ < L),u(r2, θ) = f (θ), u(r, 0) = u(r , L) = u(r1, θ) = 0.• Expand f into the Fourier sine series:f (θ) =X∞n=1ansinnπθL.• Write the solution:u(r , θ) =X∞n=1Cn(rr1)nπ/L− (r1r)nπ/LsinnπθL,where
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