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Unformatted text preview: constant, say k . Then we get the two ODEs X 00 + kX = 0 , T 00kT = 0 The characteristic equation of the X ODE is r 2 + k = 0 ⇒ r = ± √k. Since we are looking for an initial condition of sin x , we probably want sines in the solution to this ODE. This will occur when the roots are complex, which happens when k > 0. In this case the general solution is X general ( x ) = c 1 sin( √ kx ) + c 2 cos( √ kx ) . In order to get the initial condition we want, let’s take c 2 = 0 (no cosines needed), c 1 = 1, and k = 1 (to get the right coeﬃcient inside the sine). When k = 1, the T diﬀerential equation is T 00T = 0 . This has the general solution T g ( t ) = b 1 e t + b 2 et . Let’s take the solution T ( t ) = e t , so that u ( x, t ) = X ( x ) T ( t ) = e t sin x. Note that this has the initial condition required....
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 Spring '08
 Chorin
 Math, Differential Equations, Linear Algebra, Algebra, Equations, Fourier Series, Periodic function, Mathematical analysis, Partial differential equation

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