SC/MATH Partial Differential Equations Fall Assignment 3 Solutions

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1 November 16, 211 SC/MATH Partial Differential Equations Fall 211 Assignment 3 Solutions (a) on page 7 in the text To determine the equilibrium (also called steady-state) heat distribution for a thin circular ring entails solving the heat equation with =, which simplifies to 2 u t 2 condition that u( L, t) = u(l, t) and =, for L x L, with the ( L, t) = (L, t). The general solution to the steady state equation 2 u = when = 2 t is u(x, t) = a + bx. And the boundary conditions force b =. So u(x, t) = a is constant. [If the initial distribution of heat is given as f(x), then in fact the constant must be a = 1 L f(x) dx.] 2L L Separation of variables u(r, θ) = ρ(r)τ(θ) in Laplace s equation yields the eigenvalue problems r r r + 1 r 2 θ =, 2 r 2 ρ + rρ λρ = and τ = λτ with τ(θ) = τ(θ + 2π). If λ, the eigenvalue problem for τ has general solution τ(θ) = A cos( λθ) + B sin( λθ), and the periodicity condition forces λ =, 1, 2,.... If λ < then no periodic solution is possible. Thus λ = n 2 for n =, 1, 2,... are the only possible eigenvalues. The corresponding equation r 2 ρ + rρ n 2 ρ = can be solved by the substitution r = e s involving the new variable s = log r. In terms of s, the equation becomes, d 2 ρ ds 2 = n2 ρ,

2 which for n > has general solution ρ(e s ) = a n e ns +b n e ns. In terms of r, this is ρ(r) = a n r n +b n r n. If n = the solution is ρ(r) = a+b log r. Since lim r + ρ(r) should be finite, this eliminates log r and r n as possible solutions, leaving ρ n (r) = a n r n where n =, 1, 2,.... Thus for each n there is a solution u n (r, θ) = r n (A n cos(nθ)+b n sin(nθ)) that is periodic in θ and hence u(r, θ) = A + r n (A n cos(nθ) + B n sin(nθ)) n=1 is the most general such solution. To satisfy the remaining boundary condition at r = a, it is required that u(a, θ) = A + a n (A n cos(nθ) + B n sin(nθ)) = f(θ). n=1 Therefore the coefficients A n and B n are given by the formulas A = 1 2π π π f(θ) dθ and for n 1, A n = 1 a n π π π f(θ) cos(nθ) dθ, and B n = 1 π f(θ) sin(nθ) dθ. a n π π (g) on page 85 Solve Laplace s equation in the rectangle x L, y H, with the boundary conditions (, y) = (L, y) = (x, H) = and y { if x > L/2 u(x, ) = 1 if x < L/2. Applying separation of variables u(x, y) = X(x)Y (y) to Laplace s equation yields the eigenvalues problems X = λx and Y = λy, for some constant λ. The boundary conditions imply that for a non-trivial solution, X () = X (L) =, and Y (H) =. The eigenvalue problem for X has non-trivial solutions of the form X n (x) = a n cos(nπx/l), n =, 1, 2,..., where λ n = (nπ/l) 2. The corresponding general solution to Y = λ n Y is Y n (y) = A n cosh(nπy/l) + B n sinh(nπy/l).

3 The boundary condition implies that Y n(h) = (nπ/l)a n sinh(nπh/l) + (nπ/l)b n cosh(nπh/l) = = B n = A n sinh(nπh/l)/ cosh(nπh/l) = cosh(nπh/l)y n = A n (cosh(nπh/l) cosh(nπy/l) sinh(nπh/l) sinh(nπy/l)) = 2A n (cosh(nπ(y H)/L). Thus, relabelling the constants, for each n =, 1, 2,..., there is a solution to Laplace s equation of the form u n (x, y) = A n cos(nπx/l) cosh(nπ(y H)/L) that satisfies the zero boundary conditions. Hence any function of the form u(x, y) = n= u n(x, y) is also a solution satisfying the zero boundary conditions. To satisfy the last boundary condition requires Thus u(x, ) = A n cosh(nπh/l) cos(nπx/l) = n= A n cosh(nπh/l) = 2 L L/2 { if x > L/2 1 if x < L/2. cos(nπx/l) dx = 2 nπ sin(nπ/2). This is non-zero only for odd values of n = 2k + 1. Thus u(x, y) = k= 2( 1) k cos((2k+1)πx/l) cosh((2k+1)π(y H)/L). (2k + 1)π cosh((2k + 1)πH/L) (b) on page 86 Solve Laplace s equation in a quarter circle of radius 1 ( θ π/2, r 1), subject to the boundary conditions (r, ) = (r, π/2) = θ θ and u(1, θ) = f(θ). Separation of variables u(r, θ) = ρ(r)τ(θ) in Laplace s equation yields the eigenvalue problems r r r + 1 r 2 θ =, 2 r 2 ρ + rρ λρ = ; and τ = λτ with τ () = τ (π/2) =. The eigenvalue problem for τ, together with its boundary conditions, has non-trivial solutions only for λ = (2n) 2, where n =, 1, 2,..., the corresponding solution being τ n (θ) = a n cos(2nθ). The corresponding equation r 2 ρ + rρ 4n 2 ρ =

4 can be solved by the substitution r = e s involving the new variable s = log r. In terms of s, the equation becomes, d 2 ρ ds 2 = 4n2 ρ, which for n 1 has general solution ρ(e s ) = a n e 2ns + b n e 2ns. In terms of r, this is ρ(r) = a n r 2n + b n r 2n. If n = the solution is ρ(r) = a + b log r. Since lim r + ρ(r) should be finite, this eliminates log r and r 2n as possible solutions, leaving ρ n (r) = a n r 2n where n =, 1, 2,.... Thus for each n there is a solution u n (r, θ) = a n r 2n cos(2nθ) that is consistent with the zero boundary conditions, and hence u(r, θ) = a n r 2n cos(2nθ) n= is the most general such solution. To satisfy the remaining boundary condition, it is required that u(1, θ) = a n cos(2nθ) = f(θ). n= Therefore the coefficients a n are given by the formulas a = 2 π π/2 f(θ) dθ and for n 1, a n = 4 π π/2 f(θ) cos(2nθ) dθ on page 279 Irrespective of the shape of the membrane, if the displacement u(x, y, t) satisfies the wave equation t 2 = c2 ( ) + 2 u 2 y 2 then, if u(x, y, t) = ϕ(x, y)t (t), it follows that ϕt = c 2 2 ϕt. This implies that 1 T c 2 T = 2 ϕ ϕ = λ, for some constant λ. In particular 2 ϕ = λϕ. (Note that the negative sign is a matter of convention; it s not necessary.) (b) on page 286 Solve the heat equation in the rectangular region < x < L, < y < H with initial condition u(x, y, ) = f(x, y) and boundary conditions (, y, t) = (L, y, t) = y (x,, t) = (x, H, t) =. y

5 Separation of time leads to the equations T = kλt and 2 ϕ = λϕ, for u(x, y, t) = ϕ(x, y)t (t). Further separating variables for ϕ(x, y) = X(x)Y (y) results in the equation X X + Y Y = λ. Therefore each of X Y = µ and = ν must be constant, with X Y µ + ν = λ. Note that the boundary conditions are symmetric with respect to x and y. The boundary conditions (, y, t) = (L, y, t) = imply that X () = X (L) =. The only non-trivial solutions to X = µx for which X () = X (L) =, are multiples of X m (x) = cos(mπx/l), where µ = m 2 and m =, 1, 2,.... By exactly the same derivation, up to multiplication by a constant, Y n (y) = cos(nπy/h) where n =, 1, 2,... (and ν = n 2 ). For each m, n the corresponding value of λ is λ = µ + ν = m 2 + n 2. The equation T = kλt therefore has solution proportional to T m,n (t) = e k(m2 +n 2 )t. Thus the most general form for u(x, y, t) consistent with the boundary conditions is u(x, y, t) = A m,n cos(mπx/l) cos(nπy/h)e k(m2 +n 2 )t. m= n= To satisfy the initial condition u(x, y, ) = f(x, y), repeated application of Fourier series yields the formula for the coefficients A m,n = 4 LH (a) on page 287 L H cos(mπx/l) cos(nπy/h)f(x, y) dxdy. Solve the wave equation t 2 = c2 ( ) + 2 u 2 y 2 in a rectangular region < x < L, < y < H with initial conditions u(x, y, ) = and (x, y, ) = f(x, y), and boundary conditions t u(, y, t) = u(l, y, t) = (x,, t) = (x, H, t) =. y y

6 As before, separation of variables u(x, y, t) = X(x)Y (y)t (t) leads to the equations T = c 2 λt, X µx and Y = νy, where λ = µ + ν, with associated boundary conditions X() = X(L) = and Y () = Y (H) =. Up to constant multiples, the non-trivial solutions are X m (x) = sin(mπx/l) with m = 1, 2, 3,..., and Y n (y) = cos(nπy/h) with n =, 1, 2,.... And T (t) = cos(c m 2 + n 2 t) or T (t) = sin(c m 2 + n 2 t). Thus the most general solution consistent with the boundary conditions is u(x, y, t) = ( sin(mπx/l) cos(nπy/h) A m,n cos(c m 2 + n 2 t) + B m,n sin(c ) m 2 + n 2 t). m=1 n= Since u(x, y, ) =, it follows that A m,n = for each m, n. Thus u(x, y, t) = m=1 n= And the second initial condition is f(x, y) = t (x, y, ) = from which it follows that and B m,n = 4 LH B m, = L 2 LH B m,n sin(mπx/l) cos(nπy/h) sin(c m 2 + n 2 t). H m=1 n= c m 2 + n 2 B m,n sin(mπx/l) cos(nπy/h), f(x, y) sin(mπx/l) cos(nπy/h) dxdy c m 2 + n 2 for n 1, L H f(x, y) sin(mπx/l) cos(nπy/h) dxdy c m 2 + n 2.