hw11_solution - Homework#11 Solutions 10.7#1 a Given a2 uxx...

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Homework #11 Solutions 10 . 7 #1 a) Given: a 2 u xx = u tt , u (0 , t ) = u ( L, t ) = 0, u t ( x, 0) = 0, u ( x, 0) = f ( x ), where f ( x ) = ( 2 x/L, 0 x L/ 2 2( L - x ) /L, L/ 2 x L We start by assuming a solution of the form u ( x, t ) = X ( x ) T ( t ). Plugging this into the PDE yields a 2 X 00 T = XT 00 Where the primes denote differentiation with respect to the argument of that function. Separating variables, we get X 00 X = T 00 a 2 T = - λ Since each side depends on a different variable, the only way for this equation to hold is if both sides are equal to a constant. We will assume that the constant is negative here ( - λ < 0), as it is the only case to have a nontrivial solution with the boundary conditions. This gives us the following two ODEs: X 00 + λX = 0 T 00 + a 2 λT = 0 For λ > 0, the general solution to these equations are: X ( x ) = c 1 sin( λx ) + c 2 cos( λx ) T ( t ) = c 3 sin( a λt ) + c 4 cos( a λt ) Applying the boundary condition u (0 , t ) = X (0) T ( t ) = 0 gives us X (0) = c 1 (0) + c 2 (1) = c 2 = 0 Applying the boundary condition u ( L, t ) = X ( L ) T ( t ) = 0 gives us X ( L ) = c 1 sin( λL ) = 0 To obtain a nontrivial solution, we set sin( λL ) = 0 and get λ n = n 2 π 2 L 2 . Applying the boundary condition u t ( x, 0) = X ( x ) T 0 (0) = 0 gives us T 0 (0) = a λc 3 (1) + a λc 4 (0) = a λc 3 = 0 Thus, ignoring constants of proportionality, we obtain the fundamental solutions: u n ( x, t ) = X ( x ) T ( t ) = sin( nπx L ) cos( anπt L ) 1
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The general solution is therefore u ( x, t ) = X n =1 c n u n = X n =1 c n sin( nπx L ) cos( anπt L ) We use the remaining initial condition to solve for the constants c n : u ( x, 0) = X n =1 c n sin( nπx L ) = f ( x ) The c n are the coefficients of the sine Fourier series of f ( x ), and can be found using the appropriate Euler- Fourier formula: c n = 2 L Z L 0 f ( x ) sin( nπx L ) dx = 2 L Z L/ 2 0 2 x L sin( nπx L ) dx + 2 L Z L L/ 2 2( L - x ) L sin( nπx L ) dx (Note that 2 L R L L/ 2 2( L - x ) L sin( nπx L ) dx = 2 L R L L/ 2 2 sin( nπx L ) dx + 2 L R L L/ 2 - 2 x L sin( nπx L ) dx )
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