Eigenvalue Problems

  1. Initial Value Problems.
  2. Boundary Value Problems
  3. Eigenvalue (characteristic-value) Problems Even if the boundary conditions of the differential equation are available, the solutions can only exist for some specific values of a parameter in the system. For example,

    $\displaystyle -\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}+V(x)\psi(x)=E\psi(x)
$

1 For example, standing waves are the solutions of the following differential equation for a string fixed at both ends:

$\displaystyle \frac{d^2y(x)}{dx^2}+k^2y(x)=0$    

Here $k=2\pi/\lambda$ represents the wavenumber. If the two ends of the L-length string are fixed, then corresponding boundary values are:

$\displaystyle y(0)=y(L)=0$    

Although there are sufficient boundary conditions, there are only solutions for certain $k$ values. It is impossible to satisfy these boundary conditions for other $k$ values. The standing wave solution in the string:

$\displaystyle y(x)=Asinkx+Bcoskx$    

Boundary conditions to find the coefficients A, B;

\begin{indisplay}\begin{align*}
y(0)=0 &\rightarrow& A*0+B*1&=0 &\rightarrow & B...
...&=0 &\rightarrow & kL=n\pi~ (n=1,2,\ldots) \nonumber
\end{align*}\end{indisplay}    

According to these results, there is only one set of solution as $k=\pi/L, 2\pi/L, \ldots$ values. ( $k_n=n\pi/L \rightarrow$ eigenvalue(s).)2 Another example is the Schrödinger equation in quantum mechanics. If we write it in one-dimensional space,

$\displaystyle -\frac{\hbar^2}{2m}\frac{d^2\psi(x)}{dx^2}+V(x)\psi(x)=E\psi(x)$    

Here $V(x)$ is the potential energy function of the particle, and $E$ is its total energy. Boundary conditions for the wave function are given for $x= \pm \infty$:

$\displaystyle \psi(\pm \infty) \rightarrow 0$    

Again, this differential equation has solutions satisfying the boundary conditions only for certain $E$ eigenvalues. For eigenvalue problems, the trial-and-error (shooting) method is also used.

Subsections