- #1
zetafunction
- 391
- 0
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' ?
[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' ?