Solving Boundary Value Problems: Are Eigenvalues Equal?

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SUMMARY

The discussion centers on the comparison of eigenvalues from two boundary value problems defined by the equations -D²y(x) + f(x)y(x) = λₙy(x) with boundary conditions y(0) = 0 and y(∞) = 0, and -D²y(x) + f(x)y(x) = βₙy(x) with y(-∞) = 0 and y(∞) = 0. It is established that while both problems are solvable, the eigenvalues λₙ and βₙ are generally not equal due to differences in their characteristic equations. The necessity of applying boundary conditions after determining eigenvalues is emphasized, indicating that the eigenvalues depend on the specific formulation of the boundary value problem.

PREREQUISITES
  • Understanding of boundary value problems in differential equations
  • Familiarity with eigenvalues and eigenfunctions
  • Knowledge of characteristic equations in the context of ODEs
  • Proficiency in applying boundary conditions to differential equations
NEXT STEPS
  • Study the derivation of eigenvalues from boundary value problems
  • Learn about the impact of boundary conditions on eigenvalue solutions
  • Explore the differences between homogeneous and non-homogeneous ODEs
  • Investigate the role of the characteristic equation in determining eigenvalues
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Mathematicians, physicists, and engineers dealing with differential equations, particularly those focused on boundary value problems and eigenvalue analysis.

zetafunction
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let be the two boundary value problem

[tex]-D^{2}y(x)+f(x)y(x)= \lambda _{n} y(x)[/tex]

with [tex]y(0)=0=y(\infty)[/tex]

and the same problem [tex]-D^{2}y(x)+f(x)y(x)= \beta _{n} y(x)[/tex]

with [tex]y(-\infty)=0=y(\infty)[/tex]

i assume that in both cases the problem is SOLVABLE , so my question is , are the eigenvalues in both cases equal ? , i mean [tex]\lambda _{n} = \beta _{n}[/tex] , or have the same dependence on parameter 'n' ?
 
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I don't understand your notation. You first homogenize your ODE, then you find the eigenvalues of your characteristic equation. Then you apply BCs to extract constants of integration. So you find eigenvalues before applying BCs. So, in general, yes.

However, the way you have described your equations, your characteristic equation will be different, so different eigenvalues. So if I don't understand your query, then my answer is probably wrong.
 

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