What Determines the Sign of Lambda in Separation of Variables?

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SUMMARY

The discussion focuses on determining the sign of lambda (λ) in the context of separation of variables for functions of the form U(x,y) = X(x)Y(y). It is established that the sign of λ is crucial for simplifying the solution process, particularly in relation to eigenvalues derived from either the X or Y differential equations. When λ > 0, solutions yield real exponential functions, which cannot satisfy boundary conditions such as X(0) = X(a) = 0. Conversely, λ < 0 leads to periodic solutions that can meet these boundary conditions, highlighting the importance of considering the full range of parameters based on the specific initial or boundary conditions.

PREREQUISITES
  • Understanding of separation of variables in differential equations
  • Familiarity with eigenvalues and eigenfunctions
  • Knowledge of boundary value problems
  • Basic concepts of real and complex functions
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  • Study the implications of boundary conditions on eigenvalue problems
  • Learn about the characteristics of solutions for different signs of λ
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AStaunton
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Just quick question about sep of variables..

say have function U(x,y)=X(x)Y(y)

when do separation of variables end up with some generic case that looks like:

X''/X=Y'/Y=lamda

my question is (and I think I know now the answer but would like confirmation), is what sign should the lamda be set to to make the problem easiest to solve...

And I think that it is down to which of the two, either X or Y that are going to give us the eigenvalues...
if X gives eigen values, then set lamda to the sign that allows the X differential equation easiest to solve and vise versa if Y ODE gives eigenvalues...

I'm just looking for a rule of thumb here or any tips anyone has..Thanks
 
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The sign of [tex]\lambda[/tex] usually depends on initial or boundary conditions. For example, we can consider solutions to

[tex]\frac{X''}{X} = \lambda.[/tex]

When [tex]\lambda>0[/tex], our solutions are real exponential functions [tex]e^{\pm\sqrt{\lambda} x[/tex]. It is impossible to satisfy the boundary conditions [tex]X(0)=X(a)=0[/tex] with a linear combination of these solutions.

However, if [tex]\lambda<0[/tex], we find periodic solutions [tex]e^{\pm i\sqrt{\lambda} x[/tex] (equivalent to sin and cos) for which there is a linear combination that satisfies [tex]X(0)=X(a)=0[/tex].

Therefore, you will generally need to consider the full range of parameters and decide which classes of solutions are permissible given the boundary or initial conditions.
 

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