Find the Laplace equation for $f(x)=\sin^3(x), a=b=1$.

Question

I'm working on a PDE question but before jumping to that, we know that: $$D_n =\frac{2}{a \sinh\left(\frac{n\pi b}{a}\right)} \int_0^a f(x)\sin\left(\frac{n \pi x}{a}\right),$$

$$u(x,t)=\sum_{n=1}^{\infty} D_n \sin\left(\frac{n \pi x}{a}\right) \sinh\left(\frac{n \pi b}{a}\right).$$

There is a related question which seems super easy bit actually I’m having trouble with the calculation of it. The question is :

Find the Laplace equation for $f(x)=\sin^3(x), a=b=1$.

I tried integration by parts, putting $D_n$ directly in $u(x,t)$ but I cannot simplify it.

Thanks in advance.

Answer 1

After some integration by parts, you can reduce to

$$ I_1 = \int \sin (x) \sin (n \pi x) dx\\ u = \sin (n \pi x)\\ dv = \sin x dx\\ du = n \pi \cos (n \pi x) dx\\ v = - \cos x\\ I_1 = (- \sin (n \pi x) \cos x) \mid_0^1 + n \pi \int \cos x \cos (n \pi x) dx\\ = n \pi \int \cos x \cos (n \pi x) dx\\ $$

Now do integration by parts again

$$ I_2 = \int \cos x \cos (n \pi x) dx\\ u = \cos (n \pi x)\\ dv = \cos x dx\\ du = - n \pi \sin (n \pi x) dx\\ v = \sin x\\ I_2 = (\sin x \cos (n \pi x)) \mid_0^1 + n \pi \int \sin x \sin (n \pi x) dx\\ = \sin 1 \cos n \pi + n \pi I_1\\ $$

So you get

$$ I_1 = n \pi I_2\\ = n \pi (\cos (n \pi) \sin 1 + n \pi I_1)\\ I_1 = n \pi \cos (n \pi) \sin 1 + n^2 \pi^2 I_1\\ (1 - n^2 \pi^2) I_1 = n \pi \sin 1 \cos (n \pi)\\ I_1 = \frac{n \pi \sin 1 \cos (n \pi)}{1 - n^2 \pi^2}\\ $$