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10 - Separation of Variables

10 - Separation of Variables - MATH 220 SEPARATION OF...

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MATH 220: SEPARATION OF VARIABLES Separation of variables is a method to solve certain PDEs which have a ‘warped product’ structure. The general idea is the following: suppose we have a linear PDE Lu = 0 on a space M x × N y . We look for solutions u ( x,y ) = X ( x ) Y ( y ). In general, there are no non-trivial solutions (the identically 0 function being trivial), but in special cases we might be able to find some. We cannot expect even then that all solutions of the PDE are of this form. However, if we have a family u n ( x,y ) = X n ( x ) Y n ( y ) , n ∈ I , of separated solutions, where I is some index set (e.g. the positive integers), then, this being a linear PDE, u ( x,y ) = summationdisplay n ∈I c n u n ( x,y ) = summationdisplay n ∈I c n X n ( x ) Y n ( y ) solves the PDE as well in a distributional sense for any constants c n C , n ∈ I , provided the sum converges in distributions, and we may be able to choose the constants so that this in fact gives an arbitrary solution of the PDE. We emphasize that our endeavor, in general, is very unreasonable. Thus, we may make assumptions as we find it fit – we need to justify our results after we derive them. As an example, consider the wave equation u tt c 2 Δ x u = 0 on M x × R t , where M is the space – for instance, M is R n , or a cube [ a,b ] n or a ball B n = { x R n : | x | ≤ 1 } . A separated solution is one of the form u ( x,t ) = X ( x ) T ( t ). Substituting into the PDE yields X ( x ) T ′′ ( t ) c 2 T ( t )(Δ x X )( x ) = 0 . Rearranging, and assuming T and X do not vanish, T ′′ ( t ) c 2 T ( t ) = Δ x X ( x ) X ( x ) . Now, the left hand side is a function independent of x , the right hand side is a function independent of t , so they are both equal to a constant, λ , namely pick your favorite value of x 0 and t 0 , and then for any x and t , RHS( x ) = LHS( t 0 ) = RHS( x 0 ) = LHS( t ) , so the constant in question is LHS( t 0 ). Thus, we get two ODEs: T ′′ ( t ) = λc 2 T ( t ) , x X )( x ) = λX ( x ) . Now typically one has additional conditions. For instance, one has boundary conditions at ∂M : u | ∂M × R = 0 (DBC) or ∂u ∂n | ∂M × R = 0 (NBC) . 1
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2 Then X ( x ) T ( t ) has to satisfy these conditions for all x ∂M and all t R . Taking some t for which T ( t ) does not vanish, we deduce that the analogous boundary condition is satisfied, namely X | ∂M = 0 (DBC) or ∂X ∂n | ∂M = 0 (NBC) . We also have initial conditions, such as u ( x, 0) = φ ( x ) , u t ( x, 0) = ψ ( x ) , but as these are not homogeneous, we do not impose these at this point, and hope that we will have sufficient flexibility from the c n to match these. We start by solving the ODE for T , which is easy: T ( t ) = A cos( λct ) + B sin( λct ) , λ negationslash = 0 , and T ( t ) = A + Bt if λ = 0. (We could have used complex exponentials instead. If λ is not positive, the trigonometric functions should be thought of as given by the corresponding complex exponentials.) Now, in general the spatial equation, Δ X = λX, X | ∂M = 0 (DBC) or ∂X ∂n | ∂M = 0 (NBC) (1) is impossible to solve explicitly. However, we point out that it is an eigenvalue equation for Δ: the statement is that X is an eigenfunction of Δ with eigenvalue λ , in the strong sense that it also satisfies the boundary condition. If we let
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  • Fall '10
  • Vasy
  • Boundary conditions, Dirichlet boundary condition, Neumann boundary condition, bn sin, general separated solution

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