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**[Lucas' Notes for Lecture 3](/SummerCourseHeatWave.pdf)**
Below are answers to some questions that came up during the lecture. If anyone remembers other unanswered questions please post them here as well, and hopefully they'll get answered.
# Heat Equation With Nonperiodic Boundary Conditions
Suppose instead of having a metal ring we had a metal rod of length $L$, and we kept the ends of the rod at constant temperatures $u_0$ and $u_L$. How might we solve the heat equation in this context?
There is one obvious solution that satisfies these boundary conditions, namely the time-independent or steady state solution
$$ g(x,t) = u_0 + \frac{u_L- u_0}{L}x $$
This satisfies the heat equation for trivial reasons since it is time-independent and its second spatial derivative is zero, hence both sides of the heat equation are zero independently of one another.
Now suppose that $h(x,t)$ is another solution satisfying the same boundary conditions. Then the function
$$ u(x,t) = g(x,t) - h(x,t)$$
also satisfies the heat equation, but it is zero at both endpoints.
$$ u(x,0) = u(x,L) = 0 $$
To solve for $h$, it is clear that we only need to solve for $u$. First we'll use what's called the ``separation of variables'' trick to generate a lot of nice solutions, then hope and pray that any other solution can be expressed as a linear combination of these.
Here's how the separation of variables trick works. We seek a solution of the form
$$u(x,t) = a(t)b(x)$$
for some functions $a$ and $b$. Then the heat equation tells us:
$$ \frac{da}{dt}b = a \frac{d^2b}{dx^2} $$
Rearranging, we get:
$$ \frac{da}{dt}/a = \frac{d^2b}{dx^2}/b $$
The left hand side only depends on $t$ and the right hand side depends only on $x$. Therefore, neither side depends on either $x$ or $t$. We conclude that both sides are equal to a constant, which for convenience we'll write as $-\lambda^2$.
First we solve for $a$. It satisfies the equation
$$ \frac{da}{dt} = - \lambda^2 a$$
and therefore (up to a constant multiple) it is given by:
$$ a(t) = e^{-\lambda^2 t} $$
Next we solve for $b$. It satisfies the equation
$$ \frac{d^2b}{dx^2} = -\lambda b $$
However we have to be a bit more careful in picking our solutions because $b$ is supposed to satisfy the boundary conditions
$$ b(0) = b(L) = 0$$
To satisfy $b(0) = 0$, we must take $b$ to be (a constant multiple of) a sine function:
$$ b(x) = \sin(\lambda x) $$
and to satisfy $b(L) = 0$, we must impose a constraint on $\lambda$:
$$ \lambda = \frac{\pi n}{L} $$
So, the most general solution we can generate in this manner is:
$$ u(x,t) = \sum_{n = 1}^{\infty} c_n e^{-(\frac{\pi n}{L})^2t} \sin(\frac{\pi n x}{L}) $$
We would like to assert that any solution takes this form. One way to prove this assertion would be to show that any function $f:[0,L] \to \R$ satisfying $f(0) = f(L) = 0$ has a unique ``[Fourier sine expansion](http://mathworld.wolfram.com/FourierSineSeries.html)'':
$$ f(x) = \sum_{n = 1}^{\infty} c_n \sin(\frac{\pi n x}{L}) $$
One could then allow the coefficients $c_n$ to vary with $t$ and apply the same method of solution that we used in the case of periodic boundary conditions.
In fact, every function of the kind described above does have a Fourier sine expansion. The link above contains a hint of how to do it. First you extend the function $f(x)$ to a certain odd, periodic function $\tilde{f}(x)$ defined on the interval $[-L,L]$. Then you can use convergence of the usual Fourier series for $\tilde{f}(x)$.
## Convergence for not-so-nice Fourier series.
How do we know that the Fourier series of a square wave or sawtooth function converge?
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