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Unformatted text preview: ACM 100c  Methods of Applied Mathematics Week 4 April 18, 2009 1 Fully infinite domains and the Fourier transform We next turn to the problem of solving the heat equation when the domain is semiinfinite or fully infinite in space. That is we want to solve φ t = a 2 φ xx − ∞ < x < ∞ . Such a problem requires an initial condition: φ ( x, 0) = f ( x ) − ∞ < x < ∞ , and has implicit boundary conditions of the form φ ( x,t ) finite  x  → ∞ . We have solved this type of problem using Laplace transforms, but it is of interest to try to solve it using a separation of variables type approach. If we try to find a solution of the form φ ( x,t ) = α ( t ) β ( x ) we find that we can indeed separate the variables as before: α t ( t ) α ( t ) = a 2 β xx ( x ) β ( x ) , and write separated ODE’s for α and β . Following our approach for the finite domain, we get β xx β = − λ 2 α t α = − a 2 λ 2 . 1 Our choice of the separation constant is motivated by our previous choice for the heat equation in a finite domain. The solution of the time dependent part looks sensible: α ( t ) = C exp( − λ 2 a 2 t ) . For the spatial ODE we get d 2 dx 2 β ( x ) + λ 2 β ( x ) = 0 − ∞ < x < ∞ . The domain is now fully infinite and the only boundary conditions are that β is finite as  x  → ∞ . The solutions here are β ( x ) = A cos( λx ) + B sin( λx ) but both functions are bounded as  x  → ∞ as long as λ is real. When we solve this type of problem in a finite domain (say < x < L ) we have definite boundary conditions and the values of λ are the discrete eigenvalues. Here it seems that any real value of λ is an eigenvalue. The reason for this difference is that a SturmLiouville problem defined on an infinite domain is a singular problem. For these problems there is no guarantee that the eigenvalues are discrete and in this case they are not. We can restrict λ > without loss of generality but the set of eigenvalues is dense. Even though we have a dense set of eigenvalues, it turns out we can still effect a so lution. Using the idea of superposition we write a solution using integration rather than summation. The separable functions are sin( λx ) exp( − λa 2 t ) and cos( λx ) exp( − λa 2 t ) . Instead of summing these, we write a solution as an integral: φ ( x,t ) = integraldisplay ∞ [ A ( λ ) cos( λx ) + B ( λ ) sin( λx )] exp( − λ 2 a 2 t ) dλ. In order to determine the coefficient functions A ( λ ) and B ( λ ) we try to impose the initial condition φ ( x, 0) = f ( x ) = integraldisplay ∞ [ A ( λ ) cos( λx ) + B ( λ ) sin( λx )] dλ. Another more compact way to write this is to recall that sin( x ) = exp( ix ) − exp( − ix ) 2 i cos( x ) = exp( ix ) + exp( − ix ) 2 ....
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This note was uploaded on 12/12/2009 for the course ACM 95c taught by Professor Nilesa.pierce during the Spring '09 term at Caltech.
 Spring '09
 NilesA.Pierce

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