- #1

spaghetti3451

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- 33

## \bigg(-\frac{d^2}{dx^2}+V(x)\bigg) \phi_n = \lambda_n \phi_n,\ \ \ \ \ \ \ \ \ \ \ \ \ \phi_n(0)=0,\ \phi_n(1)=0, ##

where ##0 < \lambda_1 < \lambda_2 < ...## are discrete, non-degenerate eigenvalues that are bounded from below.

I've read somewhere that we can achieve this by constructing an initial value problem with the same operator, but with different boundary conditions:

##\bigg(-\frac{d^2}{dx^2}+V(x)\bigg) u_\lambda = \lambda u_\lambda,\ \ \ \ \ \ \ \ \ \ \ \ \ u_\lambda(0)=0,\ u^{'}_\lambda(1)=0.##

I can see that this determines the function ##u_\lambda (x)## uniquely, because the second order differential equation is supplanted with two boundary conditions. Also, I understand that if ##\lambda## is to be an eigenvalue of the original problem, then it also holds that ##u_\lambda(1)=0##. That much I understand.

However, I don't understand why we need to convert the boundary value problem into an initial value problem in order to solve it, and how the condition ##u^{'}_\lambda(1)=0## comes about anyway.

Can somebody please explain? :(