3 7.28, §3 Eigenfunctions

To find the solution in the form,

$$u(x,t) = \sum_n u_n(t) X_n(x)$$

we try substituting an individual term of this general form into the PDE. In particular, we substitute a trial solution $u=T(t) X(x)$ into the homogeneous P.D.E. $u_{tt} = a^2 u_{xx}$. This gives:

$$T''(t)X(x) = a^2 T(t) X''(x)$$

Now we can take the terms depending on $t$ only to one side of the equation, and the ones depending on $x$ only to the other side:

$$\frac{T''(t)}{a^2 T(t)} = \frac{X''(x)}{X(x)}$$

This trick is why this solution procedure is called the “method of separation of variables.”

While the right hand side, $X''(x)/X(x)$, does not depend on $t$, you would think that it would depend on the position $x$; both $X$ and $X''$ change when $x$ changes. But actually, $X''/X$ does not change with $x$; after all, if we change $x$, it does nothing to $t$, so the left hand side does not change. And since the right hand side is the same, it too does not change. So the right hand side does not depend on either $x$ or $t$; it must be a constant. By convention, we call the constant $-\lambda$:

$$\frac{T''}{a^2 T} = \frac{X''}{X} = \hbox{ constant } = -\lambda$$

If we also require $X$ to satisfy the same homogeneous boundary conditions as $u$, i.e., that at $x=0$, its $x$-derivative is zero, and that at $x=\ell$, $X$ itself is zero, we get the following problem for $X$:

$$X'' + \lambda X = 0 \qquad X'(0) = 0 \qquad X(\ell)=0$$

This is an ordinary differential equation boundary value problem.

Note that this problem is completely homogeneous: $X(x)=0$ satisfies both the PDE and the boundary conditions. This is similar to the eigenvalue problem for vectors $A\vec v = \lambda \vec v$, which is certainly always true when $\vec v=0$. But for the eigenvalue problem, we are interested in nonzero vectors $\vec v$ so that $A\vec v=\lambda v$, which only occurs for special values $\lambda_1, \lambda_2, \ldots$ of $\lambda$.

Similarly, we are interested only in nonzero solutions X(x) of the above ODE and boundary conditions. Eigenvalue problems for functions such as the one above are called “Sturm-Liouville problems.” The biggest differences from matrix eigenvalue problems are:

• There are infinitely many eigenvalues $\lambda_1, \lambda_2, \ldots$ and corresponding eigenfunctions $X_1(x)$, $X_2(x)$, $\ldots$ rather than just $n$ eigenvalues and eigenvectors.
• We cannot write a determinant to find the eigenvalues. Instead we must solve the problem using our methods for solving ODE.

Fortunately, the above ODE is simple: it is a constant coefficient one, so we write its characteristic polynomial:

$$k^2 + \lambda = 0 \quad\Longrightarrow\quad k = \pm \sqrt{-\lambda} = \pm {\rm i} \sqrt{\lambda}$$

We must now find all possible eigenvalues $\lambda$ and corresponding eigenfunctions that satisfy the required boundary conditions. We must look at all possibilities, one at a time.

Case $\lambda < 0$:

Since $k = \pm \sqrt{-\lambda}$

$$X = A e^{\sqrt{-\lambda} x} + B e^{-\sqrt{-\lambda} x}$$

We try to satisfy the boundary conditions:

$$X'(0) = 0 = A \sqrt{-\lambda} - B \sqrt{-\lambda} \quad\Longrightarrow\quad B = A$$

$$X(\ell) = 0 = A \left( e^{\sqrt{-\lambda} \ell} + e^{-\sqrt{-\lambda} \ell} \right) \quad\Longrightarrow\quad A = 0$$

So $A=B=0$; there are no nontrivial solutions for $\lambda < 0$.

Case $\lambda = 0$:

Since $k_1 = k_2 = 0$ we have a multiple root of the characteristic equation, and the solution is

$$X = A e^{0x} + B x e^{0x} = A + B x$$

We try to satisfy the boundary conditions again:

$$X'(0) = 0 = B \qquad X(\ell) = 0 = A$$

So $A=B=0$; there are again no nontrivial solutions.

Case $\lambda > 0$:

Since $k = \pm \sqrt{-\lambda} = \pm {\rm i} \sqrt{\lambda}$, the solution of the ODE is after cleanup:

$$X = A \sin\left(\sqrt{\lambda} x\right) + B \cos\left(\sqrt{\lambda} x\right)$$

We try to satisfy the first boundary condition:

$$X'(0) = 0 = A \sqrt{\lambda}$$

Since we are looking at the case $\lambda > 0$, this can only be true if $A=0$. So, we need

$$X = B \cos\left(\sqrt{\lambda} x\right)$$

We now try to also satisfy the second boundary condition:

$$X(\ell) = 0 = B \cos\left(\sqrt{\lambda} \ell\right) = 0$$

For a nonzero solution, $B$ may not be zero, so the cosine must be zero. For positive argument, a cosine is zero at $\frac12\pi, \frac32\pi,\ldots$, so that our eigenvalues are

$$\sqrt{\lambda_1} = \frac{\pi}{2\ell}, \sqrt{\lambda_2} = \... ...{3\pi}{2\ell}, \sqrt{\lambda_3} = \frac{5\pi}{2\ell}, \ldots$$

The same as for eigenvectors, for our eigenfunctions we must choose the one undetermined parameter $B$. Choosing each $B=1$, we get the eigenfunctions:

$$X_1 = \cos\left(\frac{\pi x}{2\ell}\right), X_2 = \cos\lef... ...\right), X_3 = \cos\left(\frac{5\pi x}{2\ell}\right), \ldots$$

Total:

The only eigenvalues for this problem are the positive ones above, with the corresponding eigenfunctions. If we want to evaluate them on a computer, we need a general formula for them. You can check that it is:

$$\lambda_n = \frac{(2n-1)^2 \pi^2}{4\ell^2} \qquad X_n = \c... ...frac{(2n-1) \pi x}{2\ell}\right) \qquad (n = 1, 2, 3, \ldots)$$

Just try a few values for $n$. We have finished finding the eigenfunctions.