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Unformatted text preview: Math 445  Solutions to selected problems from HW 4 Exercise 5, page 552. Solve u tt = c 2 u xx with boundary conditions u (0 ,t ) = 0 and u x ( L,t ) = 0, and with initial conditions u ( x, 0) = f ( x ) and u t ( x, 0) = g ( x ) = 0. Solution We follow the solution given in class/textbook for the wave equation with standard boundary conditions and proceed by separation of variables: let u ( x,t ) = f ( x ) G ( t ). Then F ( x ) G 00 ( t ) = c 2 F 00 ( x ) G ( t ) and hence F 00 ( x ) F ( x ) = G 00 ( t ) c 2 G ( t ) = k , a constant. Thus we must solve two ODE’s: F 00 ( x ) kF ( x ) = 0 and G 00 ( t ) c 2 kG ( t ) = 0 . The boundary conditions give 0 = u (0 ,t ) = F (0) G ( t ) and 0 = u x ( L,t ) = F ( L ) G ( t ) = 0. Since G ( t ) = 0 leads to u ( x,t ) = 0 which is not an acceptable solution we must have F (0) = 0 = F ( L ). Now we solve the ODE for F . Depending on the sign of k we have the three cases: If k = p 2 > : F ( x ) = Ae px + Be px If k = 0 : F ( x ) = Ax + B If k = p 2 < : F ( x ) = A cos( px ) + B sin( px ) ....
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 Fall '07
 Friedlander
 Math, Sin, Boundary value problem, Partial differential equation

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