6 7.28, §6 Satisfy the IC

In the previous section, we found the general solution of the PDE with the given boundary conditions to be

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

Now we want to find the coefficients $D_{1n}$ and $D_{2n}$ for all values of $n$ from the initial conditions. That will fully determine the solution.

To do so, we must first write our initial condition $u(x,0)=f(x)$ and $u_t(x,0)=g(x)$ in terms of the eigenfunctions. Writing the Fourier series for the two functions as

$$f(x) = \sum_{n=1}^\infty f_n X_n(x) \qquad g(x) = \sum_{n=1}^\infty g_n X_n(x).$$

and using the Fourier series for $u$ above, the two initial conditions become

$$\sum_{n=1}^\infty D_{1n} X_n(x) = \sum_{n=1}^\infty f_n X_n(x)$$

$$\sum_{n=1}^\infty a \sqrt\lambda_n D_{2n} X_n(x) = \sum_{n=1}^\infty g_n X_n(x).$$

The Fourier coefficients must again be equal, so we conclude that the coefficients we are looking for are

$$D_{1n} = f_n \qquad D_{2n} = \frac{g_n}{a \sqrt\lambda_n}$$

The Fourier series for $u$ becomes now

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

where

$$\lambda_n = \frac{(2n-1)^2 \pi^2}{4\ell^2} \quad \qquad X_n = \cos\left(\frac{(2n-1) \pi x}{2\ell}\right)$$

So, if we can find the Fourier coefficients $f_n$ and $g_n$ of functions $f(x)$ and $g(x)$, we are done.

Now $f(x)$ and $g(x)$ are, supposedly, given functions, but how do we find their Fourier coefficients? The answer is the following important formula:

$$\fbox{$ \displaystyle f_n = \frac{\int_0^l f(x) X_n(x)\/ {\rm d}x}{\int_0^l X_n(x)^2\/ {\rm d}x} $}$$

which is called the “orthogonality relation”. Even if $f(x) = 1$, say, we still need to do those integrals. The same for $g$ of course:

$$g_n = \frac{\int_0^l g(x) X_n(x)\/ {\rm d}x}{\int_0^l X_n(x)^2\/ {\rm d}x}$$

We are done! Or at least, we have done as much as we can do until someone tells us the actual functions $f(x)$ and $g(x)$. If they do, we just do the integrals above to find all the $f_n$ and $g_n$, (maybe analytically or on a computer), and then we can sum the expression for $u(x,t)$ for any $x$ and $t$ that strikes our fancy.

Note that we did not have to do anything with the boundary conditions $u_x(0,t)=0$ and $u(\ell,t)=0$; since every eigenfunction $X_n$ satisfies them, the expression for $u$ above automatically also satisfies these homogeneous boundary conditions.