# College Math Teaching

## August 19, 2011

### Partial Differential Equations, Differential Equations and the Eigenvalue/Eigenfunction problem

Suppose we are trying to solve the following partial differential equation:
$\frac{\partial \psi}{\partial t} = 3 \frac{\partial ^2 \phi}{\partial x^2}$ subject to boundary conditions:
$\psi(0) = \psi(\pi) = 0, \psi(x,0) = x(x-\pi)$

It turns out that we will be using techniques from ordinary differential equations and concepts from linear algebra; these might be confusing at first.

The first thing to note is that this differential equation (the so-called heat equation) is known to satisfy a “uniqueness property” in that if one obtains a solution that meets the boundary criteria, the solution is unique. Hence we can attempt to find a solution in any way we choose; if we find it, we don’t have to wonder if there is another one lurking out there.

So one technique that is often useful is to try: let $\psi = XT$ where $X$ is a function of $x$ alone and $T$ is a function of $t$ alone. Then when we substitute into the partial differential equation we obtain:
$XT^{\prime} = 3X^{\prime\prime}T$ which leads to $\frac{T^{\prime}}{T} = 3\frac{X^{\prime\prime}}{X}$

The next step is to note that the left hand side does NOT depend on $x$; it is a function of $t$ alone. The right hand side does not depend on $t$ as it is a function of $x$ alone. But the two sides are equal; hence neither side can depend on $x$ or $t$; they must be constant.

Hence we have $\frac{T^{\prime}}{T} = 3\frac{X^{\prime\prime}}{X} = \lambda$

So far, so good. But then you are told that $\lambda$ is an eigenvalue. What is that about?

The thing to notice is that $T^{\prime} - \lambda T = 0$ and $X^{\prime\prime} - \frac{\lambda}{3}X = 0$
First, the equation in $T$ can be written as $D(T) = \lambda T$ with the operator $D$ denoting the first derivative. Then the second can be written as $D^2(X) = 3\lambda X$ where $D^2$ denotes the second derivative operator. Recall from linear algebra that these operators meet the requirements for a linear transformation if the vector space is the set of all functions that are “differentiable enough”. So what we are doing, in effect, are trying to find eigenvectors for these operators.

So in this sense, solving a homogeneous differential equation is really solving an eigenvector problem; often this is termed the “eigenfucntion” problem.

Note that the differential equations are not difficult to solve:
$T = a exp(\lambda T)$ $X = b exp(\sqrt{\frac{\lambda}{3}} x) + cexp(-\sqrt{\frac{\lambda}{3}} x)$; the real valued form of the equation in $x$ depends on whether $\lambda$ is positive, zero or negative.

But the point is that we are merely solving a constant coefficient differential equation just as we did in our elementary differential equations course with one important difference: we don’t know what the constant (the eigenvalue) is.

Now if we turn to the boundary conditions on $x$ we see that a solution of the form $A e^{bx} + Be^{-bx}$ cannot meet the zero at the boundaries conditions; we can rule out the $\lambda = 0$ condition as well.
Hence we know that $\lambda$ is negative and we get $X = a cos(\sqrt{\frac{\lambda}{3}} x) + b sin(\sqrt{\frac{\lambda}{3}} x)$ solution and then $T = d e^{\lambda t }$ solution.

But now we notice that these solutions have a $\lambda$ in them; this is what makes these ordinary differential equations into an “eigenvalue/eigenfucntion” problem.

So what values of $\lambda$ will work? We know it is negative so we say $\lambda = -w^2$ If we look at the end conditions and note that $T$ is never zero, we see that the cosine term must vanish ($a = 0$ ) and we can ensure that $\sqrt{\frac{w}{3}}\pi = k \pi$ which implies that $w = 3k^2$ So we get a whole host of functions: $\psi_k = a_k e^{-3k^2 t}sin(kx)$.

Now we still need to meet the last condition (set at $t = 0$ ) and that is where Fourier analysis comes in. Because the equation was linear, we can add the solutions and get another solution; hence the $X$ term is just obtained by taking the Fourier expansion for the function $x(x-\pi)$ in terms of sines.

The coefficients are $b_k = \frac{1}{\pi} \int^{\pi}_{-\pi} (x)(x-\pi) sin(kx) dx$ and the solution is:
$\psi(x,t) = \sum_{k=1}^{\infty} e^{-3k^2 t} b_k sin(kx)$

## 1 Comment »

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