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Unformatted text preview: Week 5 Lectures, Math 716, Tanveer 1 Separation of variable method The method of separation of variable is a suitable technique for determining solutions to linear PDEs, usually with constant coefficients, when the domain is bounded in at least one of the independent variables. We illustrate this procedure for 1D wave equation and 2D heat equation for Dirichlet, Neumann and Robin boundary conditions, though the idea is equally applicable for diffusion equation and Laplace’s equation, and other constant coefficient equations. The idea of separation of variable is first to seek simple solution to the PDE in the form of a product, each term in the product depending on only one independent variable. Solutions are then constrained by boundary conditions. This results in a countably infinite set of solutions. A linear superposition of such solution is also a solution, because of the linearity of the problem. As we shall find later, such linear superposition is capable of describing all reasonable initial conditions. 1.1 Dirichlet Problem Consider u tt c 2 u xx = 0 for < x < l (1) u (0 , t ) = 0 = u ( l, t ) (2) with initial condition u ( x, 0) = φ ( x ) ; u t ( x, 0) = ψ ( x ) (3) Recall, there is a representation of this solution by doing odd extension about x = 0 to the interval ( l, 0), and then periodically extending this problem (with period 2 l ), and then using D’Alembert representation of solution, as you will see in last week notes. Uniqueness follows from application of energy method, as we saw earlier. Here, we are seeking a different form of the same solution. We seek simple solution to (1), ignoring initial conditions (3) for now, in a product form u ( x, t ) = X ( x ) T ( t ) Plugging it into (1) and dividing it the resulting equation by c 2 X ( x ) T ( t ), we obtain T 00 ( t ) c 2 T ( t ) = X 00 ( x ) X ( x ) (4) If we set t = 1, for instance, then the left side of (4) is a number independent of x . Hence the right side of (4) cannot depend on x at all. Similarly, if we set x to some specific value, the right side of (4) is a constant; hence the left side (4) must be independent of t . Therefore, we conclude both left and right side of each side of (4) is some constant λ . Therefore, T 00 ( t ) + λc 2 λT ( t ) = 0 (5) and X 00 ( x ) + λX ( x ) = 0 (6) The boundary conditions (2) imply that X (0) = 0 and X ( l ) = 0. 1 If λ < 0, then the solution to (6) that satisfies X (0) = 0 is given by X ( x ) = C sinh √ λx (7) Since the function sinh does not vanish anywhere except the origin, any nontrivial solution in the form (7) is incapable of satisfying X ( l ) = 0. Therefore, we must discard the possibility of λ < 0. If λ = 0, then the solution to (6) that satisfies X (0) = 0 is X ( x ) = Cx, (8) But, this is not capable of satisfying X ( l ) = 0, unless C = 0, which corresponds to the trivial solution. Therefore, we conclude that λ 6 = 0....
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This note was uploaded on 07/26/2011 for the course MATH 716 taught by Professor Tanveer,s during the Spring '08 term at Ohio State.
 Spring '08
 Tanveer,S
 Math

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