Solving the Heat Equation (Sect. 10.5).

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1 Solving the Heat Equation Sect Review: The Stationary Heat Equation. The Heat Equation. The Initial-Boundary Value Problem. The separation of variables method. An example of separation of variables. Review: The Stationary Heat Equation. Review: The Stationary Heat Equation describes the temperature distribution in a solid material in thermal equilibrium. The temperature is time-independent. Problem: The time-independent temperature, T, of a bar of length with insulated horizontal sides and vertical extremes kept at fixed temperatures T, T, is the solution of the BVP: T x =, x,, T = T, T = T, y insulation T T x x z insulation Remark: The heat transfer occurs only along the x-axis.

2 Solving the Heat Equation Sect Review: The Stationary Heat Equation. The Heat Equation. The Initial-Boundary Value Problem. The separation of variables method. An example of separation of variables. The Heat Equation. Remarks: The unknown of the problem is ut, x, the temperature of the bar at the time t and position x. The temperature does not depend on y or z. The one-dimensional Heat Equation is: t ut, x = k x ut, x, where k > is the heat conductivity, units: [k] = distance. time The Heat Equation is a Partial Differential Equation, PDE. u u = u < u > t t t ut,x t is held constant. x

3 Solving the Heat Equation Sect Review: The Stationary Heat Equation. The Heat Equation. The Initial-Boundary Value Problem. The separation of variables method. An example of separation of variables. The Initial-Boundary Value Problem. Definition The IBVP for the one-dimensional Heat Equation is the following: Given a constant k > and a function f : [, ] R with f = f =, find u : [, [, ] R solution of t ut, x = k x ut, x, I.C.: u, x = f x, B.C.: ut, =, ut, =. t u t, = u t, = d u = k d u t x u, x = f x x

4 Solving the Heat Equation Sect Review: The Stationary Heat Equation. The Heat Equation. The Initial-Boundary Value Problem. The separation of variables method. An example of separation of variables. The separation of variables method. Summary: IBVP for the Heat Equation. Propose: where ut, x = c n v n t w n x. v n : Solution of an IVP. w n : Solution of a BVP, an eigenvalue-eigenfunction problem. c n : Fourier Series coefficients. Remark: The separation of variables method does not work for every PDE.

5 The separation of variables method. Summary: The idea is to transform the PDE into infinitely many ODEs. We describe this method in 6 steps. Step 1: One looks for solutions u given by an infinite series of simpler functions, u n, that is, ut, x = c n u n t, x, where u n is simpler than u is the sense, u n t, x = v n t w n x. Here c n are constants, n = 1,,. The separation of variables method. Step : Introduce the series expansion for u into the Heat Equation, t u k x [ u = c n t u n k x ] u n =. A sufficient condition for the equation above is: To find u n, for n = 1,,, solutions of t u n k x u n =. Step 3: Find u n t, x = v n t w n x solution of the IBVP t u n k x u n =. I.C.: u n, x = w n x, B.C.: u n t, =, u n t, =.

6 The separation of variables method. Step 4: Key step. Transform the IBVP for u n into: a IVP for v n ; b BVP for w n. Notice: t u n t, x = t [ vn t w n x ] = w n x dv n dt t. x u n t, x = x [ vn t w n x ] = v n t d w n dx x. Therefore, the equation t u n = k x u n is given by w n x dv n dt t = k v nt d w n dx x 1 k v n t dv n dt t = 1 w n x d w n dx x. Depends only on t = Depends only on x. The separation of variables method. Recall: 1 k v n t dv n dt t = 1 w n x d w n dx x. Depends only on t = Depends only on x. The Heat Equation has the following property: The left-hand side depends only on t, while the right-hand side depends only on x. When this happens in a PDE, one can use the separation of variables method on that PDE. We conclude that for appropriate constants λ m holds 1 k v n t dv n dt t = λ n, 1 w n x d w n dx x = λ n. We have transformed the original PDE into infinitely many ODEs parametrized by n, positive integer.

7 The separation of variables method. Summary Step 4: The original IBVP for the Heat Equation, PDE, can transformed into: a We choose to solve the following IVP for v n, 1 k v n t dv n dt t = λ n, I.C.: v n = 1. Remark: This choice of I.C. simplifies the problem. b The BVP for w n, Step 5: 1 w n x a Solve the IVP for v n. b Solve the BVP for w n. d w n dx x = λ n, B.C.: w n =, w n =. The separation of variables method. Step 5a: Solving the IVP for v n. v nt + kλ n v n t =, I.C.: v n = 1. The integrating factor method implies that µt = e kλ nt. e kλ nt v nt + kλ n e kλ nt v n t = [ e kλ nt v n t] =. e kλ nt v n t = c n v n t = c n e kλ nt. 1 = v n = c v n t = e kλ nt.

8 The separation of variables method. Step 5a: Recall: v n t = e kλ nt. Step 5b: Eigenvalue-eigenvector problem for w n : Find the eigenvalues λ n and the non-zero eigenfunctions w n solutions of the BVP w n x + λ n w n x = B.C.: w n =, w n =. We know that this problem has solution only for λ n >. Denote: λ n = µ n. Proposing w n x = e r nx, we get that pr n = r n + µ n = r n± = ±µ n i The real-valued general solution is w n x = c 1 cosµ n x + c sinµ n x. The separation of variables method. Recall: v n t = e kλ nt, w n x = c 1 cosµ n x + c sinµ n x. The boundary conditions imply, = w n = c 1 w n x = c sinµ n x. = w n = c sinµ n, c, sinµ n =. µ n = nπ µ n = nπ nπ. λ n = nπx Choosing c = 1, we get w n x = sin. We conclude that: u n t, x = e k nπ nπx t sin, n = 1,,.

9 The separation of variables method. Step 6: Recall: u n t, x = e k nπ nπx t sin. Compute the solution to the IBVP for the Heat Equation, ut, x = c n u n t, x. ut, x = c n e k nπ nπx t sin By construction, this solution satisfies the boundary conditions, ut, =, ut, =. Given a function f with f = f =, the solution u above satisfies the initial condition f x = u, x iff holds nπx f x = c n sin.. The separation of variables method. Recall: ut, x = c n e k nπ nπx t sin, f x = nπx c n sin. This is a Sine Series for f. The coefficients c n are computed in the usual way. Recall the orthogonality relation nπx mπx, m n, sin sin dx =, m = n. Multiply the equation for u by sin mπx nd integrate, nπx mπx mπx c n sin sin dx = f x sin dx. c n = nπx f x sin dx, ut, x = c n e k nπ nπx t sin.

10 The separation of variables method. Summary: IBVP for the Heat Equation. Propose: where ut, x = c n v n t w n x. v n : Solution of an IVP. w n : Solution of a BVP, an eigenvalue-eigenfunction problem. c n : Fourier Series coefficients. Remark: The separation of variables method does not work for every PDE. Solving the Heat Equation Sect Review: The Stationary Heat Equation. The Heat Equation. The Initial-Boundary Value Problem. The separation of variables method. An example of separation of variables.

11 An example of separation of variables. Example Find the solution to the IBVP 4 t u = x u, t >, x [, ], u, x = 3 sinπx/, ut, =, ut, =. Solution: et u n t, x = v n t w n x. Then 4w n x dv dt t = v nt d w dx x 4v nt v n t = w n x w n x = λ n. The equations for v n and w n are v nt + λ n 4 v nt =, w n x + λ n w n x =. We solve for v n with the initial condition v n = 1. e λ n 4 t v nt + λ n 4 e λn 4 t v n t = [ e λ n 4 t v n t ] =. An example of separation of variables. Example Find the solution to the IBVP 4 t u = x u, t >, x [, ], u, x = 3 sinπx/, ut, =, ut, =. Solution: Recall: [ e λn 4 t v n t ] =. Therefore, v n t = c e λ n 4 t, 1 = v n = c v n t = e λ n 4 t. Next the BVP: w n x + λ n w n x =, with w n = w n =. Since λ n >, introduce λ n = µ n. The characteristic polynomial is pr = r + µ n = r n± = ±µ n i. The general solution, w n x = c 1 cosµ n x + c sinµ n x. The boundary conditions imply = w n = c 1, w n x = c sinµ n x.

12 An example of separation of variables. Example Find the solution to the IBVP 4 t u = x u, t >, x [, ], u, x = 3 sinπx/, ut, =, ut, =. Solution: Recall: v n t = e λ n 4 t, and w n x = c sinµ n x. = w n = c sinµ n, c, sinµ n =. Then, µ n = nπ, that is, µ n = nπ. Choosing c = 1, we conclude, nπ λ m = ut, x =, nπx wn x = sin c n e nπ nπx 4 t sin.. An example of separation of variables. Example Find the solution to the IBVP 4 t u = x u, t >, x [, ], u, x = 3 sinπx/, ut, =, ut, =. Solution: Recall: ut, x = c n e nπ nπx 4 t sin. πx The initial condition is 3 sin = nπx c n sin. The orthogonality of the sine functions implies πx mπx nπx 3 sin sin dx = sin mπx sin dx. If m 1, then = c m, that is, c m = for m 1. Therefore, πx πx 3 sin = c 1 sin c 1 = 3.

13 An example of separation of variables. Example Find the solution to the IBVP 4 t u = x u, t >, x [, ], u, x = 3 sinπx/, ut, =, ut, =. Solution: We conclude that ut, x = 3 e π πx 4 t sin.

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