{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

10 - Separation of Variables

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

This preview shows pages 1–3. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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
This is the end of the preview. Sign up to access the rest of the document.
• Fall '10
• Vasy
• Boundary conditions, Dirichlet boundary condition, Neumann boundary condition, bn sin, general separated solution

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern