How can I solve for coefficients Bn in the sol. of heat equation / Fourier sine series?

I've discovered the solution:

$$\sum_{n=1}^{\infty} u_n(x,t)=\sum_{n=1}^{\infty} B_n \sin \bigg(\frac{n \pi x}{L} \bigg)e^{-k \bigg( \frac{n \pi}{L} \bigg)^2t}$$

and with the IC (I'm given $u(x,0)=x$):

$$u(x,0)=\sum_{n=1}^{\infty} B_n \sin \bigg(\frac{n \pi x}{L} \bigg)=x$$

How can I solve for $B_n$ in an easy to explain way?

I've done some googling (obviously), but noticed that a lot of solutions seem to rely on "exchange order of integral with infinite summation", which I find non-trivial (or at least I cannot understand how can I prove that it's possible).

Another way suggested "assume the sum is twice-differentiable", again something that I find hard to chew.

Or is it truly that solving for $B_n$s is non-trivial?

But surely there must be some explanation as to why the "exchange order of integral with infinite sum" is legit?


1 Answer 1


if your original pde was the heat equation with zero temperature at finite ends then you should have.

$$ \begin{align}\begin{cases} \frac{\partial u}{\partial t} = k\frac{\partial^{2} u}{\partial x^{2}} & 0 < x < L, t > 0 \\ u(0,t) =0 \\ u(L,t) = 0 \\ u(x,0) = f(x) \end{cases} \end{align} \tag{1}$$

which would give you

$$ u(x,t) = B \sin(\sqrt{\lambda}x) e^{-k \lambda t} \tag{2} $$

this becomes

$$ u(x,t) = \sum_{n=1}^{\infty}B_{n} \sin(\sqrt{\lambda}x) e^{-k \lambda t} \tag{3} $$

to solve for the coefficients, we note that

$$\int_{0}^{L} \sin(\frac{n \pi x}{L})\sin(\frac{m \pi x}{L}) dx = \begin{align}\begin{cases} 0 & m \neq n \\ \frac{L}{2} & m =n \end{cases} \end{align} \tag{4} $$

then we do the following. This is called Fouriers Trick. We multiply $f(x)$ by $\sin(\frac{m \pi x}{L})$ and

$$ f(x) \sin(\frac{m \pi x}{L}) = \sum_{n=1}^{\infty} B_{n} \sin(\frac{n \pi x}{L}) \sin(\frac{m \pi x}{L}) \tag{5} $$

The question is about here

When you go from step $5$ on. You take the integral from $\int_{0}^{L}$ of both sides.

$$ \int_{0}^{L} f(x) \sin(\frac{m \pi x}{L}) dx = \int_{0}^{L} \sum_{n=1}^{\infty} B_{n} \sin(\frac{n \pi x}{L})\sin(\frac{m \pi x}{L}) dx \tag{6}$$

there is a theorem on this page

Theorem: if $ \{ f_{n} \}_{n} $ is a positive sequence of integrable functions and $f = \sum_{n} f_{n} $ then

$$\int f = \sum_{n} \int f_{n} \tag{7} $$

this comes from the monotone convergence theorem.

$$ \int_{0}^{L} f(x) \sin(\frac{m \pi x}{L}) dx = \sum_{n=1}^{\infty} B_{n} \int_{0}^{L} \sin(\frac{n \pi x}{L})\sin(\frac{m \pi x}{L}) dx \tag{8}$$

now when $ m \neq n$ that is $0$ so we have

$$ \int_{0}^{L} f(x) \sin(\frac{m \pi x}{L}) dx = B_{m} \sum_{n=1}^{\infty} \int_{0}^{L} \sin^{2}(\frac{m\pi x}{L}) dx \tag{9}$$

which then yields

$$ B_{m} = \frac{\int_{0}^{L} f(x) \sin(\frac{m \pi x}{L}) dx}{ \int_{0}^{L} \sin^{2}(\frac{m\pi x}{L}) dx} = \frac{2}{L} \int_{0}^{L} f(x) \sin(\frac{m \pi x}{L}) dx \tag{10}$$

  • $\begingroup$ But are you not exchanging the integral with the summation at $(6)$. Why can you do that? $\endgroup$
    – mavavilj
    Commented Oct 31, 2018 at 23:17
  • $\begingroup$ Again that is something that's done in plentiful of other "solutions", like math.stackexchange.com/questions/705096/…. But no explanation given as to why it's legit. $\endgroup$
    – mavavilj
    Commented Oct 31, 2018 at 23:19
  • $\begingroup$ you can exchange the integral and series because the inner integral is a finite sum I believe. Is that what you're concerned about. $\endgroup$
    – user3417
    Commented Oct 31, 2018 at 23:29
  • $\begingroup$ That's arguing backwards. So you claim that $(5)->integral->(6)$ because in $(6)$ you have definite integral inside. But why did you get it inside the sum? The question asks for "why is the fourier series integrable term-by-term". $\endgroup$
    – mavavilj
    Commented Oct 31, 2018 at 23:30
  • $\begingroup$ I read the title of your question and the beginning of it and started typing, which was how to solve for $B_{n}$ and you said it was non-trivial. I imagine it has to do with this. math.stackexchange.com/questions/83721/… $\endgroup$
    – user3417
    Commented Oct 31, 2018 at 23:47

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