37.9 Hints Hint 37.1 Hint 37.2 Hint 37.3 Hint 37.4 Hint 37.5 Hint 37.6 Hint 37.7 Impose the boundary conditions u(0, t) = u(2π, t), u θ (0, t) = u θ (2π, t). Hint 37.8 Apply the separation of variables u(x, y) = X(x)Y (y). Solve an eigenvalue problem for X(x). Hint 37.9 Hint 37.10 1734 Hint 37.11 Hint 37.12 There are two ways to solve the problem. For the first method, expand the solution in a series of the form u(x, t) = ∞  n=1 a n (t) sin  nπx L  . Because of the inhomogeneous boundary conditions, the convergence of the series will not be uniform. You can differentiate the series with respect to t, but not with respect to x. Multiply the partial differential equation by the eigenfunction sin(nπx/L) and integrate from x = 0 to x = L. Use integration by parts to move derivatives in x from u to the eigenfunctions. This process will yield a first order, ordinary differential equation for each of the a n ’s. For the second method: Make the change of variables v(x, t) = u(x, t) − µ(x), where µ(x) is the equilibrium temperature distribution to obtain a problem with homogeneous boundary conditions. Hint 37.13 Hint 37.14 Hint 37.15 Hint 37.16 Hint 37.17 1735 Hint 37.18 Use separation of variable s to find eigen-solutions of the partial differential equation that satisfy the homogeneous boundary conditions. There will be two eigen-solutions for each eigenvalue. Expand u(x, t) in a series of the eigen- solutions. Use the two initial conditions to determine the constants. Hint 37.19 Expand the solution in a series of eigenfunctions in x. D etermine these eigenfunctions by using separation of variables on the homogeneous partial differential equation. You will find that the answer has the form, u(x, y) = ∞  n=1 u n (y) sin  nπx a  . Substitute this series into the partial differential equation to determine ordinary differential equations for each of the u n ’s. The boundary conditions on u(x, y) will give you boundary conditions for the u n ’s. Solve these ordinary differential equations with Green functions. Hint 37.20 Solve this problem by expanding the solution in a series of eigen-solutions that satisfy the partial differential equation and the homogeneous boundary conditions. Use the initial conditions to determine the coefficients in the expansion. Hint 37.21 Use separation of variables to find eigen-solutions that satisfy the partial differential equation and the homogeneous boundary conditions. The solution is a linear combination of the eigen-solutions. The whole solution will be exponentially decaying if each of the eigen-solutions is exponentially decaying. Hint 37.22 For parts (a), (b) and (c) use separation of variables. For part (b) the eigen-solutions will involve Bessel functions. For part (c) the eigen-solutions will involve spherical Bessel functions. Part (d) is trivial. Hint 37.23 The solution is a linear combination of eigen-solutions of the partial differential equation that satisfy the homogeneous boundary conditions. Determine the coefficients in the expansion with the initial condition. 1736 Hint 37.24 The problem is u rr + 1 r u r + 1 r 2 u θθ = 0, 0 < r < 1, 0 < θ < π u(r, 0) = u(r, π) = 0, u(0, θ) = 0, u(1, θ) = 1 The solution is a linear combination of eigen-solutions that satisfy the partial differential equation an d the three homogeneous boundary conditions. Hint 37.25 Hint 37.26 Hint 37.27 Hint 37.28 Hint 37.29 Hint 37.30 Hint 37.31 Hint 37.32 1737 Hint 37.33 Hint 37.34 Hint 37.35 Hint 37.36 Hint 37.37 1738 37.10 Solutions Solution 37.1 We expand the solution in eigenfunctions in x and y which satify the boundary conditions. u = ∞  m,n=1 u mn (t) sin  mπx a  sin  nπy b  We expand the inhomogeneities in the eigenfunctions. q(x, y, t) = ∞  m,n=1 q mn (t) sin  mπx a  sin  nπy b  q mn (t) = 4 ab  a 0  b 0 q(x, y, t) sin  mπx a  sin  nπy b  dy dx f(x, y) = ∞  m,n=1 f mn sin  mπx a  sin  nπy b  f mn = 4 ab  a 0  b 0 f(x, y) sin  mπx a  sin  nπy b  dy dx 1739 We substitute the expansion of the solution into the diffusion equation and the initial condition to determine initial value problems for the coefficients in the expansion. u t − κ(u xx + u yy ) = q(x, y, t) ∞  m,n=1  u  mn (t) + κ   mπ a  2 +  nπ b  2  u mn (t)  sin  mπx a  sin  nπy b  = ∞  m,n=1 q mn (t) sin  mπx a  sin  nπy b  u  mn (t) + κ   mπ a  2 +  nπ b  2  u mn (t) = q mn (t) u(x, y, 0) = f(x, y) ∞  m,n=1 u mn (0) sin  mπx a  sin  nπy b  = ∞  m,n=1 f mn sin  mπx a  sin  nπy b  u mn (0) = f mn We solve the ordinary differential equations for the coefficients u mn (t) subject to their initial conditions. u mn (t) =  t 0 exp  −κ   mπ a  2 +  nπ b  2  (t − τ)  q mn (τ) dτ + f mn exp  −κ   mπ a  2 +  nπ b  2  t  Solution 37.2 After looking at this problem for a minute or two, it seems like the answer would have the form u = sin(x)T (t). This form satisfies the boundary conditions. We substitute it into the heat equation and the initial condition to determine 1740 T sin(x)T  = −κ sin(x)T + A sin(x), T(0) = 0 T  + κT = A, T (0) = 0 T = A κ + c e −κt T = A κ  1 − e −κt  Now we have the solution of the heat equation. u = A κ sin(x)  1 − e −κt  Solution 37.3 First we write the Laplacian in p olar coordinates. u rr + 1 r u r + 1 r 2 u θθ = 0 1. We introduce the separation of variables u(r, θ) = R(r)Θ(θ). R  Θ + 1 r R  Θ + 1 r 2 RΘ  = 0 r 2 R  R + r R  R = − Θ  Θ = λ We have a regular Sturm-Liouville problem for Θ and a differential equation for R. Θ  + λΘ = 0, Θ  (0) = Θ(π/2) = 0 (37.8) r 2 R  + rR  − λR = 0, R is bounded 1741 First we solve the problem for Θ to determine the eigenvalues and eigenfunctions. The Rayleigh quotient is λ =  π/2 0 (Θ  ) 2 dθ  π/2 0 Θ 2 dθ Immediately we see that the eigenvalues are non-negative. I f Θ  = 0, then the right boundary condition implies that Θ = 0. Thus λ = 0 is not an eigenvalue. We find the general solution of Equation 37.8 for positive λ. Θ = c 1 cos  √ λθ  + c 2 sin  √ λθ  The solution that satisfies the left boundary condition is Θ = c cos  √ λθ  . We apply the right boundary condition to determine the eigenvalues. cos  √ λ π 2  = 0 λ n = (2n − 1) 2 , Θ n = cos ((2n − 1)θ) , n ∈ Z + Now we solve the differential equation for R. Since this is an Euler equation, we make the su bstitition R = r α . r 2 R  n + rR  n − (2n − 1) 2 R n = 0 α(α − 1) + α −(2n − 1) 2 = 0 α = ±(2n − 1) R n = c 1 r 2n−1 + c 2 r 1−2n The solution which is bounded in 0 ≤ r ≤ 1 is R n = r 2n−1 . 1742 [...]... continuous and has a continuous first derivative along θ = 0 This gives us a boundary value problem for Θ and a differential equation for R Θ + λΘ = 0, Θ(0) = Θ(2π), Θ (0) = Θ (2π) 2 r R + rR − λR = 0, R is bounded The eigensolutions for Θ form the familiar Fourier series λn = n 2 , n ∈ Z0+ 1 (1) Θ0 = , Θ(1) = cos(nθ), n ∈ Z+ n 2 (2) Θn = sin(nθ), n ∈ Z+ Now we find the bounded solutions for R The equation for. .. this form into the diffusion equation and the initial condition to determine the coefficients in the series, Tn (t) First we substitute Equation 37.10 into the partial differential equation for φ to determine ordinary differential equations for the Tn φt = a2 φxx ∞ ∞ nπx nπ Tn (t) sin = −a2 l l n=1 n=1 Tn = − anπ l 1 755 2 Tn 2 Tn (t) sin nπx l Now we substitute Equation 37.10 into the initial condition for. .. 1 750 Solution 37.9 We substitute u(r, θ) = R(r)Θ(θ) into the partial differential equation ∂ 2 u 1 ∂u 1 ∂2u + + 2 2 =0 ∂r2 r ∂r r ∂θ 1 1 R Θ + R Θ + 2 RΘ = 0 r r R Θ R r2 +r =− =λ R R Θ r2 R + rR − λR = 0, Θ + λΘ = 0 We assume that u is a strong solution of the partial differential equation and is thus twice continuously differentiable, (u ∈ C 2 ) In particular, this implies that R and Θ are bounded and. .. Sturm-Liouville problem for Θ and a differential equation for R Θ + λΘ = 0, Θ (0) = Θ (π/2) = 0 r2 R + rR − λR = 0, R is bounded (37.9) First we solve the problem for Θ to determine the eigenvalues and eigenfunctions We recognize this problem as the generator of the Fourier cosine series λn = (2n)2 , n ∈ Z0+ , 1 Θ0 = , Θn = cos (2nθ) , n ∈ Z+ 2 1743 Now we solve the differential equation for R Since this is... This is the condition for the existence of a solution of the problem If this is satisfied, we can solve for the coefficients in the expansion u0 is arbitrary un = 4 π π/2 g(θ) cos (2nθ) dθ, 0 1744 n ∈ Z+ Solution 37.4 1 ut = ν(uxx + uyy ) XY T = ν(X Y T + XY T ) T X Y = + = −λ νT X Y Y X =− − λ = −µ X Y We have boundary value problems for X(x) and Y (y) and a differential equation for T (t) X + µX = 0,... r + k2 r2 = − = λ2 R R Θ Now we have an ordinary differential equation for R(r) and an eigenvalue problem for Θ(θ) 1 λ2 R + R + k 2 − 2 R = 0, R(0) is bounded, R(1) = 0, r r 2 Θ + λ Θ = 0, Θ(−π) = Θ(π), Θ (−π) = Θ (π) We compute the eigenvalues and eigenfunctions for Θ λn = n, 1 Θ0 = , Θ(1) = cos(nθ), n 2 The differential equations for the Rn are Bessel equations 1 n2 Rn + Rn + k 2 − 2 r r n ∈ Z0+ Θ(2)... (0) = X (1) = 0 Y + (λ − µ)Y = 0, Y (0) = Y (1) = 0 T = −λνT 2 The solutions for X(x) form a cosine series µm = m 2 π 2 , 1 X0 = , 2 m ∈ Z0+ , Xm = cos(mπx) The solutions for Y (y) form a sine series λmn = (m2 + n2 )π 2 , n ∈ Z+ , Yn = sin(nπx) We solve the ordinary differential equation for T (t) 2 +n2 )π 2 t Tmn = e−ν(m We expand the solution of the heat equation in a series of the eigensolutions 1 u(x,... = sin 1 757 nπx , L n ∈ N We expand the solution of the partial differential equation in terms of these eigenfunctions ∞ u(x, t) = an (t) sin n=1 nπx L Because of the inhomogeneous boundary conditions, the convergence of the series will not be uniform We can differentiate the series with respect to t, but not with respect to x We multiply the partial differential equation by an eigenfunction and integrate... Substituting u(x, y) = X(x)Y (y) into the partial differential equation yields Y X =− = −λ X Y With the homogeneous boundary conditions, we have the two problems X + λX = 0, X(0) = X(1) = 0, Y − λY = 0, Y (1) = 0 The eigenvalues and orthonormal eigenfunctions for X(x) are λn = (nπ)2 , Xn = √ 2 sin(nπx) The general solution for Y is Yn = a cosh(nπy) + b sinh(nπy) The solution for that satisfies the right homogeneous... am (t) = κ 2 (T0 − (−1)m T1 ) L L 1 758 mπx L dx = 0 mπx L dx = 0 dx = 0 sin Now we have a first order differential equation for each of the an ’s We obtain initial conditions for each of the an ’s from the initial condition for u(x, t) u(x, 0) = f (x) ∞ an (0) sin n=1 an (0) = 2 L nπx = f (x) L L f (x) sin 0 nπx L dx ≡ fn By solving the first order differential equation for an (t), we obtain an (t) = 2(T0 . exponentially decaying. Hint 37.22 For parts (a), (b) and (c) use separation of variables. For part (b) the eigen-solutions will involve Bessel functions. For part (c) the eigen-solutions will. value problems for X(x) and Y (y) and a differential equation for T(t). X  + µX = 0, X  (0) = X  (1) = 0 Y  + (λ − µ)Y = 0, Y (0) = Y (1) = 0 T  = −λνT 2. The solutions for X(x) form a cosine. solution of the partial differential equation and is thus twice continuously differentiable, (u ∈ C 2 ). In particular, this implies that R and Θ are bounded and that Θ is continuous and has a continuous