Web Bonus Problem to Week 7 (#2)

[Victor Ivrii]

Problem 4.2.7 http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter4/S4.2.P.html#problem-4.2.P.7

[Chi Ma]

Rewrite the problem as follows:
\begin{align} &\frac{u_{xx}}{u} = - (H + \lambda) \qquad & |x|\le L \\
&\frac{u_{xx}}{u} = - \lambda \qquad & |x| > L
\end{align}

For $\lambda \in (−H,0)$, $−\lambda$ is positive and $-(H + \lambda)$ is negative. The solution is in exponential form outside the rectangular well and sinusoidal within the well. All eigenvalues therefore belong to the interval $−H < \lambda < 0$.

Consider even and odd eigenfunctions separately. Even eigenfunctions take the following form:
\begin{align} &u = A \cos\omega x \qquad & |x|\le L \\
&u = Be^{-\gamma |x|} \qquad & |x|> L
\end{align}
where $\omega^2 = H + \lambda$ and $\gamma^2 = -\lambda$.

Continuity and continuous first derivative at $L$ imply
\begin{align} A \cos\omega L &= Be^{-\gamma L} \\
A \omega \sin\omega L &= \gamma Be^{-\gamma L}
\end{align}

The corresponding eigenvalues are solutions to the following:
\begin{equation}\sqrt{H + \lambda} \tan(\sqrt{H + \lambda} L) = \sqrt{-\lambda}  \qquad \qquad −H < \lambda < 0 \end{equation}


Similarly, odd eigenfunctions take the following form:
\begin{align} &u = A \sin\omega x \qquad &|x|\le L \\
&u = Be^{-\gamma x} \qquad &x > L \\
&u = -Be^{\gamma x} \qquad &x < -L
\end{align}

Continuity and continuous first derivative at $L$ imply
\begin{equation} \omega \cot\omega L = -\gamma \end{equation}

The corresponding eigenvalues are solutions to the following:
\begin{equation}\sqrt{H + \lambda} \cot(\sqrt{H + \lambda} L )= -\sqrt{-\lambda}  \qquad \qquad −H < \lambda < 0 \end{equation}

[Victor Ivrii]

Correct. Why $\lambda >0$ is impossible? Because then $u$ is "sinusoidal" outside the well and thus is not square integrable.

Why $\lambda<-H$ is impossible? The simplest way to prove:
\begin{equation}
\int_{-\infty}^\infty  (-u''-+V) u\,dx =\int_{-\infty}^\infty  (u^{\prime\,2}+Vu^2)\,dx\ge -H \int_{-\infty}^\infty u^2\,dx
\end{equation}
(since $V$ is real one can consider real and imaginary parts of $u$ separately)