5 7.28, §5 Solve the PDE

So what is the procedure for solving the original problem for the pressure $u$ having found the eigenfunctions $X_n$? It is to write everything in terms of eigenfunctions. And if I say everything, I mean everything.

We first write our solution $u(x,t)$ in terms of the eigenfunctions:

$$u(x,t) = \sum_{n=1}^\infty u_n(t) X_n(x)$$

The coefficients $u_n(t)$ are called the “Fourier coefficients” of $u$. The sum is called the “Fourier series” for $u$.

We know our eigenfunctions $X_n(x)$, but not yet our Fourier coefficients $u_n(t)$. In fact, the $u_n(t)$ are what is still missing; if we know the $u_n(t)$, we can find the pressure $u$ we want by doing the sum above. On a computer probably, if we want to get high accuracy. Or just the first few terms by hand, if we accept some numerical error.

Second, we also write our PDE, $u_{tt} = a^2 u_{xx}$, in terms of the eigenfunctions:

$$\sum_{n=1}^\infty \ddot u_n(t) X_n(x) = a^2 \sum_{n=1}^\infty u_n(t) X_n''(x)$$

This PDE will always simplify; that is how the method of separation of variables works. Look up the Sturm-Liouville problem for $X_n$ in the previous section; it was $X_n''(x) = - \lambda_n X_n(x)$. So we can get rid of the $x$-derivatives in the PDE:

$$\sum_{n=1}^\infty \ddot u_n(t) X_n(x) = a^2 \sum_{n=1}^\infty \left(- \lambda_n u_n(t)\right) X_n(x)$$

Now if two functions are equal, all their Fourier coefficients must be equal, so we have, for any value of $n$,

$$\ddot u_n(t) = - a^2 \lambda_n u_n(t) \qquad (n=1,2,3,\ldots)$$

That no longer contains $x$ at all: the PDE has become a set of ODE in $t$ only. And we (hopefully) know how to solve those! Getting rid of $x$ is really what the method of separation variables does for us.

We can solve the ODE above easily. It is a constant coefficient one, with a characteristic equation $k^2 = -a^2 \lambda_n$, hence $k = \pm i a \sqrt\lambda_n$, giving

$$u_n(t) = C_{1n} e^{ i a \sqrt\lambda_n t} + C_{2n} e^{ -i a \sqrt\lambda_n t}$$

or after cleaning up,

$$u_n(t) = D_{1n} \cos\left(a \sqrt\lambda_n t\right) + D_{2n} \sin\left(a \sqrt\lambda_n t\right)$$

So, we have already found our pressure a bit more precisely:

$$u(x,t)= \sum_{n=1}^\infty \left[ D_{1n} \cos\left(a \sqrt... ... + D_{2n} \sin\left(a \sqrt\lambda_n t\right) \right] X_n(x)$$

but we still need to figure out what the integration constants $D_{1n}$ and $D_{2n}$ are.