Math 412-501Theory of Partial Differential EquationsLecture 2-3: Separation of variablesfor the one-dimensional wave equation.Laplace’s equation in a rectangle.Separation of variables: wave equation∂2u∂t2= c2∂2u∂x2Suppose u(x, t) = φ(x)G (t). Then∂2u∂t2= φ(x)d2Gdt2,∂2u∂x2=d2φdx2G (t).Henceφ(x)d2Gdt2= c2d2φdx2G (t).Divide both sides by c2· φ(x) · G (t) = c2· u(x, t):1c2G·d2Gdt2=1φ·d2φdx2.It follows that1c2G·d2Gdt2=1φ·d2φdx2= −λ = const.The variables have been separated:d2φdx2= −λφ,d2Gdt2= −λc2G .Proposition Suppose φ and G are solutions of theabove ODEs for the same value of λ. Thenu(x, t) = φ(x)G (t) is a solution of the waveequation.Example. u(x, t) = cos ct · sin x. (standing wave)Finite string with fixed ends∂2u∂t2= c2∂2u∂x2, 0 ≤ x ≤ L,u(0, t) = u(L, t) = 0.We are looking for solutions u(x, t) = φ(x)G (t).PDE holds ifd2φdx2= −λφ,d2Gdt2= −λc2Gfor the same constant λ.Boundary conditions hold ifφ(0) = φ(L) = 0.Eigenvalue problem: φ′′= −λφ, φ(0) = φ(L) = 0.Eigenvalues: λn= (nπL)2, n = 1, 2, . . .Eigenfunctions: φn(x) = sinnπxL.Dependence on time:G′′= −λc2G=⇒ G (t) = C1cos(c√λt) + C2sin(c√λt)Solution of the heat equation: u(x, t) = φ(x)G (t).Theorem For n = 1, 2, . . . and arbitrary constantsC1, C2, the functionu(x, t) = φn(x) ·C1cos(c√λnt) + C2sin(c√λnt)= sinnπxL·C1cosnπctL+ C2sinnπctLis a solution of the following boundary valueproblem for the wave equation:∂2u∂t2= c2∂2u∂x2, u(0, t) = u(L, t) = 0.Normal modes (a.k.a. harmonics)Natural frequencies: nc/(2L), n = 1, 2, . . .Initial-boundary value problem∂2u∂t2= c2∂2u∂x2, 0 ≤ x ≤ L,u(x, 0) = f (x),∂u∂t(x, 0) = g (x), u(0, t) = u(L, t) = 0.Principle of superposition: the solution is asuperposition of normal modes.u(x, t) =X∞n=1sinnπxLCncosnπctL+ DnsinnπctLInitial conditions are satisfied iff (x) =X∞n=1CnsinnπxLg(x) =X∞n=1DnnπcLsinnπxLHow do we solve the initial-boundary value problem?∂2u∂t2= c2∂2u∂x2, 0 ≤ x ≤ L,u(x, 0) = f (x),∂u∂t(x, 0) = g (x), u(0, t) = u(L, t) = 0.• Expand f and g into Fourier sine series:f (x) =X∞n=1ansinnπxL,g(x) =X∞n=1bnsinnπxL.• Write the solution:u(x, t) =X∞n=1sinnπxLCncosnπctL+ DnsinnπctL,where Cn= an, Dn=Lnπcbn.The solutionu(x, t) =X∞n=1sinnπxLCncosnπctL+ DnsinnπctLis defined in the whole plane.It satisfies initial conditionsu(x, 0) = F (x),∂u∂t(x, 0) = G (x), −∞ < x < ∞,where F and G are the sums of Fourier sine seriesof f and g, respectively.F and G are odd 2L-periodic extensions of f and g .F and G are odd with respect to 0 and L.Separation of variables: Laplace’s equation∂2u∂x2+∂2u∂y2= 0Suppose u(x, y) = φ(x)h(y). Then∂2u∂x2=d2φdx2h(y),∂2u∂y2= φ(x)d2hdy2.Henced2φdx2h(y) + φ(x)d2hdy2= 0.Divide both sides by φ(x)h(y) = u(x, y):1φ·d2φdx2= −1h·d2hdy2.It follows that1φ·d2φdx2= −1h·d2hdy2= −λ = const.The variables have been separated:d2φdx2= −λφ,d2hdy2= λh.Proposition Suppose φ and h are solutions of theabove ODEs for the same value of λ. Thenu(x, t) = φ(x)h(y) is a solution of Laplace’sequation.Example. u(x, y) = eysin x.Laplace’s equation inside a rectangle∂2u∂x2+∂2u∂y2= 0 (0 < x < L, 0 < y < H)Boundary conditions:u(0, y) = g1(y)u(L, y) = g2(y)u(x, 0) = f1(x)u(x, H) = f2(x)Principle of superposition:u = u1+ u2+ u3+ u4,where∇2u1= ∇2u2= ∇2u3= ∇2u4= 0,u1(x, 0) = f1(x), u1(0, y) = u1(L, y) = u1(x, H) = 0;u2(L, y) = g2(y), u2(0, y) = u2(x, 0) = u2(x, H) = 0;u3(x, H) = f2(x), u3(0, y) = u3(L, y) = u3(x, 0) = 0;u4(0, y) = g1(y), u4(L, y) = u4(x, 0) = u4(x, H) = 0.Reduced boundary value problem∂2u∂x2+∂2u∂y2= 0 (0 < x < L, 0 < y < H)Boundary conditions:u(0, y) = 0u(L, y) = 0u(x, 0) = f1(x)u(x, H) = 0Separation of variablesWe are looking for a solution u(x, y) = φ(x)h(y).PDE holds ifd2φdx2= −λφ,d2hdy2= λhfor the same constant λ.Boundary conditions u(0, y ) = u(L, y) = 0 hold ifφ(0) = φ(L) = 0.Boundary condition u(x, H) = 0 holds ifh(H) = 0.Eigenvalue problem: φ′′= −λφ, φ(0) = φ(L) = 0.Eigenvalues: λn= (nπL)2, n = 1, 2, . . .Eigenfunctions: φn(x) = sinnπxL.Dependence on y:h′′= λh, h(H) = 0.=⇒ h(y) = C0sinh√λ(y − H)Solution of Laplace’s equation:u(x, y) = sinnπxLsinhnπ(y −H)L, n = 1, 2, . . .We are looking for the solution of the reducedboundary value problem as a superposition ofsolutions with separated variables.u(x, y) =X∞n=1CnsinnπxLsinhnπ(y −H)LBoundary condition u(x, 0) = f1(x) is satisfied iff (x) = −X∞n=1CnsinhnπHLsinnπxLHow do we solve the reduced boundary value problem?∂2u∂x2+∂2u∂y2= 0 (0 < x < L, 0 < y < H),u(x, 0) = f1(x), u(x, H) = u(0, y) = u(L, y) = 0.• Expand f1into the Fourier sine series:f1(x) =X∞n=1ansinnπxL.• Write the solution:u(x, y) =X∞n=1CnsinnπxLsinhnπ(y −H)L,where Cn=
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