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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 semi-infinite 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 Sturm-Liouville 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