Sturm-Liouville Problem with sine in eigenfunction - why don't we include $n = 0$? & theory questions

Question

For the following Sturm-Liouville Problem:

$\begin{matrix} & -u^{\prime \prime}(x) = \mu\, u(x) \\ & u(0) = u^{\prime}(1) = 0 \end{matrix}$

I needed to compute all the eigenvalues and corresponding eigenfunctions.

What I found was that for $\mu < 0$, $\mu = 0$, the only solution to the differential equation was the trivial solution, and so it has no negative eigenvalues nor does it have $0$ as an eigenvalue.

It does however, have positive eigenvalues:

When $\mu > 0$, write $\mu = \beta^{2}$ for $\beta$ real and nonzero. Then, the differential equation becomes:

\begin{matrix} u^{\prime\prime}(x) + \beta^{2} u(x) = 0. \end{matrix}

Then, the characteristic equation, $r^{2} + \beta^{2} = 0$ yields a pair of complex conjugate roots, $r = \pm \beta$, so the form of the general solution to the differential equation is:

\begin{matrix} u(x) = A \cos (\beta x) + B \sin (\beta x). \end{matrix}

Applying the first boundary condition, we obtain that $u(0) = A \cos(0) + B \sin (0) = 0$, which forces $A = 0$.

Applying the second boundary condition to $u^{\prime}(x) = -\beta A \sin (\beta x) + \beta B \cos (\beta x)$, we obtain that $u^{\prime}(1) = -\beta A \sin \beta + \beta B \cos \beta = 0$, and so $\beta B \cos \beta = 0$, since $A = 0$.

Now, either $B = 0$ or $\cos \beta = 0$, since $\beta$ is real and nonzero. Now, $\cos \beta = 0 $ when $\displaystyle \beta_{n} = \frac{\pi}{2}n$ for odd $n$, or $\displaystyle \beta_{n} = \frac{\pi (2n-1)}{2}$, $n \in \mathbb{Z}$.

There are infinitely many roots of $\cos \beta = 0$, and aside from when $n=0$, they occur in positive-negative pairs; i.e., $\forall$ positive root $\beta_{n}$, $\exists$ a corresponding negative root $-\beta_{n}$.

Since we began by writing $\mu = \beta^{2}$, the eigenvalues are the squares of the roots $\beta_{n}$; i.e.,

\begin{matrix} \displaystyle \mu_{n} = \left( \frac{\pi(2n-1)}{2}\right)^{2}, & n=0,1,\cdots \end{matrix}

Since it is $A$ that vanishes here, the corresponding eigenfunction is

\begin{matrix} \displaystyle u_{n}(x) = \sin \left(\frac{\pi(2n-1)}{2} \right)x, & n=0,1,2,\cdots \end{matrix}

Now, my questions are the following:

1. In my textbook, the solution is exactly the same as what I have, except that $n$ starts at $1$. Why?
2. What is the physical significance of mixed boundary conditions like what we have in this problem? In the case of an insulated wire, say, I know that Dirichlet conditions correspond to applying a zero temperature to each end and Neumann conditions correspond to insulated ends. So, in this case, are we applying a zero temperature to the left end, and insulating the right end?
3. What possible physical phenomena can we model by "changing up" a Sturm-Liouville Problem? I.e., if the general form of a Sturm-Liouville Problem is

\begin{matrix} -[p(x)u^{\prime}(x)]^{\prime} + q(x)u(x) = \mu r(x) u(x), & a<x<b\\ A_{1}u(a)+A_{2}u^{\prime}(a) = 0\\ B_{1}u(b) + B_{2}u^{\prime}(b) = 0 \end{matrix}

then what physical phenomena can we model by, say, letting $q(x) \neq 0$, or having $p(x)$, $q(x)$, $r(x)$ be non-constant, or having all $A_{1}$, $A_{2}$, $B_{1}$, $B_{2}$ non-vanishing?

Thank you. I realize I'm asking kind of an involved question here, but I'm really very curious about how Sturm-Liouville theory works and what we can use it for.

Answer 1

1. The case $n=0$ is the same case as $n = 1$, you can see that from your eigenfunction

   $$ u_1(x) = -u_0(x) = \sin \pi x/2 $$

   So there's no difference between the solutions, only a change in the phase that can be absorbed by the time-dependent part of $u(x,t)$

2. You're right. The heat flux density relates to the temperature as $\mathbf{q} = -k \nabla T$, so $u(0) = 0$ means zero temperature at one end, and $u'(1) = 0$ means zero heat transfer at the other (insulating material)

3. This example actually gives you a good hint of a physical application of mixed boundary conditions in SL problems. Imagine now that at $x=1$ you have a material that is not perfectly insulating, but has some loses depending on the temperature $u'(1) = \alpha u(1)$

   As for your last question, I am going to switch to another equation: the acoustic wave equation (note: you can argue the same for the heat equation). The wave equation in acoustic media is the result of applying two principles:

   • Newton's law $\mathbf{F} = m \mathbf{a}$:

   $$ \rho(\mathbf{x}) \frac{\partial \mathbf{v}}{\partial t} = -\nabla P \tag{1} $$

   with $\rho$ the density of the medium and $P$ the pressure field.

   • Pressure drop is proportional to the change in velocity

   $$ \frac{\partial P}{\partial t} = -K(\mathbf{x}) \nabla \cdot \mathbf{v} \tag{2} $$

   with $K$ expressing the compressibility of the material.

   In 1D this traslates to

   $$ \frac{\partial }{\partial x} \left( \frac{1}{\rho(x)} \frac{\partial P}{\partial x}\right) = \frac{1}{K(x)} \frac{\partial^2 P}{\partial t^2} \tag{3} $$ If you assume that $P(x,t) = u(x)v(t)$, you will end up with a problem that looks like this

   $$ \frac{d}{dx}\left(\frac{1}{\rho} \frac{d u}{dx}\right) = \frac{1}{K(x)}u(x) $$

   I think this will give you an idea of a physical SL problem where the coefficients are not constant.