Solutions to Math 4233 Second Exam: Heat Conduction and Laplace's Equation - Prof. Birne B, Exams of Differential Equations

Download Solutions to Math 4233 Second Exam: Heat Conduction and Laplace's Equation - Prof. Birne B and more Exams Differential Equations in PDF only on Docsity! Math 4233 SOLUTIONS TO SECOND EXAM 2:30 – 3:20 pm, October 31, 2008 1. Find the solution of the following heat conduction problem. Explain the steps you take in solving this problem in as much detail as possible. 4ut − uxx = 0 , 0 < x < 2 , t > 0(1a) u (0, t) = 0(1b) u (2, t) = 0(1c) u (x, 0) = 4 sin (2πx)(1d) • We first use separation of variables to find a suitable family of solutions of the heat equation satisfying the first two boundary conditions. Thus, we look for functions of the form (1e) u (x, t) = X (x)Y (t) that will satisfy (1a), (1b) and (1c). Substituting (1e) into (1a) and dividing the result by X (x)Y (t) we obtain 4 Ẏ Y = X ′′ X Since the left hand side does not depend on x and the right hand side does not depend on t, we conclude that both sides must be equal to a constant, which we shall denote by −λ2. We are thus led to X ′′ = −λ2X =⇒ X (x) = A sin (λx+ δ) Ẏ = −λ 2 4 Y =⇒ Y (t) = Ce−λ 2 4 t Imposing the boundary conditions at x = 0 on the expression (1e) we find 0 = u (0, t) = ACe− λ2 4 t sin (δ) =⇒ δ = 0 (for non-trivial solutions) Taking then δ = 0 and imposing the boundary condition at x = 2 we find 0 = u (2, t) = ACe− λ2 4 t sin (2λ) =⇒ λ = nπ 2 , n = 1, 2, . . . (for non-trivial solutions) Thus, any function of the form φn (x, t) = e− 1 4 (nπ2 ) 2 t sin (nπ 2 x ) = e−( nπ 4 ) 2 t sin (nπ 2 x ) , n = 1, 2, . . . will satisfy equations (1a) - (1c). Moreover, any linear combination of the functions φn will continue to satisfy equations (1a) - (1c). We thus set (1f) u (x, t) = ∞∑ n=1 ane −(nπ4 ) 2 t sin (nπ 2 x ) = and try to choose the coefficients an so that the final boundary condition (1d) is satisfied. Plugging (1f) into (1d) we obtain 4 sin (2πx) = ∞∑ n=1 ane −(nπ4 ) 2 0 sin (nπ 2 x ) = ∞∑ n=1 an sin (nπ 2 x ) From this we conclude that the coefficients an should coincide with coefficients of sin ( nπ 2 x ) in the Fourier-sine expansion of the function on the right hand side on the interval [0, 2]. Thus, we set an = 2 L ∫ L 0 4 sin (2πx) sin (nπ L x ) dx = 4 ∫ 2 0 sin (2πx) sin (nπ 2 x ) dx 1 2 By the orthogonality properties of the Fourier-sine functions 2 L ∫ L 0 sin (mπ L x ) sin (nπ L x ) dx = { 1 if n = m 0 if n 6= m we find an = { 4 if n = 4 0 otherwise Thus, u (x, t) = 4e−( 4π 4 ) 2 t sin (2πx) = 4e−π 2t sin (2πx) 5 3. Find the solution of Laplace’s equation (3a) uxx + uyy = 0 satisfying the boundary conditions u (x, 0) = 0 , u (x, b) = g (x)(3b) u (0, y) = 0 , u (a, y) = 0(3c) • We set (3d) u (x, y) = X (x)Y (y) and substitute (3d) into (3a) and then divide the resulting equation by X (x)Y (y). This yields 1 X (x) d2X dx2 (x) + 1 Y (y) d2Y dy2 = 0 The usual Separation of Variables argument now tells us that 1 X (x) d2X dx2 (x) = C = − 1 Y (y) d2Y dy2 where C is a constant. This leads us to the following pair of ODEs X ′′ = CX(3e) Y ′′ = −CY(3f) Next we plug (3d) into the boundary conditions (3c). This lead us to 0 = X (0)Y (y) , ∀ y ∈ [0, b] =⇒ X (0) = 0 0 = X (a)Y (y) , ∀ y ∈ [0, b] =⇒ X (a) = 0 otherwise, we’d be forced to set Y (y) = 0 for all y ∈ [0, b] and the solution (3d) would be identically zero for all x and y. Now X ′′ = CX has two different kinds of solutions, depending on whether C is postive or negative. If C is positive, say C = λ2, then the general soltuion of (3e) will be of the form X (x) = c1 cosh (λx) + c2 sinh (λx) To satisfy X (0) = 0 we’d have to take c1 = 0, but then we could not also satisfy 0 = (a) = c2 sinh (λa) unless c2 were also zero. We conclude that C can not be positive. So we take C = −λ2 < 0. Now X ′′ = −λ2C =⇒ X = A sin (λx+ δ) for some A ∈ R, some δ ∈ [0, 2π) Imposing the first boundary condition 0 = X (0) = A sin (δ) =⇒ δ = 0 because setting A = 0 would otherwise trivialize the solution. Setting δ = 0 and imposing the second boundary conditions leads to 0 = X (a) = A sin (λa) =⇒ λa = nπ =⇒ λ = nπ a , n = 1, 2, 3, . . . What we have so far is that C = −λ2 = −n 2π2 a2 , X (x) = A sin (nπ a x ) We now impose the boundary condition u (x, 0) = 0 on our ansatz (3d): 0 = u (x, 0) = X (x)Y (0) =⇒ Y (0) = 0 But Y (y) is also to satisfy (3f) Y ′′ = −CY = n 2π2 a2 Y =⇒ Y = c1 cosh (nπ a y ) + c2 sinh (nπ a y ) 6 The boundary condition 0 = Y (0) forces us to take c1 = 0. The preceding arguments have furnished us the following family of solutions of (3a), (3c) and the first condition of (3b): φn (x, y) = sin (nπ a x ) sinh (nπ a y ) , n = 1, 2, 3, . . . Any linear combination of these functions will continuation to satisfy the PDE and the first three boundary conditions, and so we set u (x, t) = ∞∑ n=1 cn sin (nπ a x ) sinh (nπ a y ) and try to impose the last boundary condition: (3g) g (x) = u (x, b) = ∞∑ n=1 cn sin (nπ a x ) sinh (nπ a b ) Multiplying both sides of (3g) by a2 sin ( mπ a x ) and then integrating over [0, a] yields 2 a ∫ a 0 sin (mπ a x ) g (x) dx = ∞∑ n=1 2 a ∫ a 0 cn sin (mπ a x ) sin (nπ a x ) sinh (nπ a b ) dx = ∞∑ n=1 cn sinh (nπ a b ) δm,n = cn sinh (mπ b b ) We conclude if the constants cn are chosen so that cn = 2 a sinh ( nπ a b ) ∫ a 0 sin (mπ a x ) g (x) dx then u (x, t) = ∞∑ n=1 cn sin (nπ a x ) sinh (nπ a y ) will satisfy Laplace’s equation and all four boundary conditions. 7 4. Apply Separation of Variables to reduce the problem of finding a solution the following PDE to that of solving a pair of ODEs. (4a) ∂2φ ∂r2 + 1 r ∂φ ∂r + 1 r2 ∂2φ ∂θ2 = 0 • Suppose (4b) φ (r, θ) = R (r)T (θ) Then substituting (4b) into (4a) yields (4c) T (θ) d2R dr2 (r) + 1 r T (θ) dR dr (r) + 1 r2 R (r) d2T dθ2 (θ) = 0 Multiplying both sides of (4c) by r2/ (T (θ)R (r)) yields r2 R (r) d2R dr2 (r) + r R (r) dR dr (r) + 1 T (θ) d2T dθ2 (θ) = 0 or (4d) r2 R (r) d2R dr2 (r) + r R (r) dR dr (r) = − 1 T (θ) d2T dθ2 (θ) Since the right hand side of (4d) is independent of r so must be the left hand side; and since the left hand side of is independent of θ so must be the right hand side. So both sides are independent of r and θ; hence both sides equal a constant. Call this constant C. We then have r2 R (r) d2R dr2 (r) + r R (r) dR dr (r) = C = − 1 T (θ) d2T dθ2 (θ) or r2 R (r) d2R dr2 (r) + r R (r) dR dr (r) = C(4e) − 1 T (θ) d2T dθ2 (θ) = C(4f) Multiplying (4e) by R (r) and (4f) by T (θ) we obtain r2 d2R dr2 + r dR dr − CR = 0 ,(4g) d2T dθ2 + CT = 0 ;(4h) a pair of ordinary differential equations for R (r) and T (θ).