Math 512, Winter 2012, Midterm 2, Solutions

# Math 512, Winter 2012, Midterm 2, Solutions - M ATH 512-M...

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M ATH 512-M IDTERM 2 February 24, 2012, 9:30 AM - 10:18 AM No books, notes or calculators. Two pages of formulas are provided. Answer all questions. Show work for all problems. Simplify your answers. NAME: P ROBLEM Y OUR S CORE O UT OF 1 20 2 20 3 20 T OTAL 60

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1. (20 points) Consider the following initial-boundary-value problem: u tt = u xx for 0 < x < π, t > 0 u x (0 , t )=0 , u x ( π, t )=0 , for t > 0 u ( x, 0)= x, u t ( x, 0)=0 , for 0 < x < π. By separating variables, derive a two-point boundary-value problem for an ordinary differential equation, and determine the unique solution. Solution (similar to example 1, Section 3.6): We look for solutions of the form u ( x, t )= X ( x ) T ( t ) . We substitute into the equation to get T ′′ ( t ) X ( x )= X ′′ ( x ) T ( t ) , and separate the variables T ′′ ( t ) T ( t ) = X ′′ ( x ) X ( x ) = k. We therefore have the fol- lowing ODEs: X ′′ ( x ) kX ( x )=0 , T ′′ ( t ) kT ( t )=0 . Before we solve them, we need boundary conditions. We use u ( x, t )= X ( x ) T ( t ) and substitute it into the boundary conditions of the PDE. We get X (0) T ( t )=0= X ( π ) T ( t ) , if T ( t )=0 we have the trivial solution u ( x, t )=0 which is not acceptable, so X (0)= X ( π )=0 . We first solve the ODE for X ( · ): if k =0: the solution of the ODE is X ( x )= Ax + B. We use the boundary conditions X ( x ) = A, X (0) = 0 = X ( π ) therefore A = 0 . Notice that we do not have information for B, so let B = 1 and the solution for k = 0
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