Finding the particular solution of a pde.

Question

I have solved a PDE up to the point of finding the particular solution. I am trying to find the constant $$C_n$$

I have the expression $$3x-x^2=\sum_{n=1}^{\infty} C_{n} \, \sin\left(\frac{\pi n x}{3}\right)$$ Now I now I need to use fourier expansion because we can not find the constant by inspection.

However I am stuck on what calculation I need to do and why?

Answer 1

You have

$$3x - x^{2} = \sum_{n = 1}^{\infty} c_{n} \sin \bigg( \frac{n \pi x}{3} \bigg)$$

Multiplying both sides by

$$\sin \bigg( \frac{m \pi x}{3} \bigg)$$

for orthogonality, we get

$$(3x - x^{2})\sin \bigg( \frac{m \pi x}{3} \bigg) = \sum_{n = 1}^{\infty} c_{n} \sin \bigg( \frac{n \pi x}{3} \bigg)\sin \bigg( \frac{m \pi x}{3} \bigg)$$

Integrating over the domain $[0, 3]$, we get

\begin{align} \int_{0}^{3} (3x - x^{2})\sin \bigg( \frac{m \pi x}{3} \bigg) dx &= \int_{0}^{3} \sum_{n = 1}^{\infty} c_{n} \sin \bigg( \frac{n \pi x}{3} \bigg)\sin \bigg( \frac{m \pi x}{3} \bigg) dx \\ &= \sum_{n = 1}^{\infty} c_{n} \int_{0}^{3} \sin \bigg( \frac{n \pi x}{3} \bigg)\sin \bigg( \frac{m \pi x}{3} \bigg) dx \ \ (1) \\ \end{align}

Now, I suggest you try and do this last integral yourself (this is the 'orthogonality condition'); Hint 1: set $m = n$, hence

\begin{align} \sin \bigg( \frac{n \pi x}{3} \bigg)\sin \bigg( \frac{m \pi x}{3} \bigg) &= \sin^{2} \bigg( \frac{m \pi x}{3} \bigg) \\ &= \frac{1}{2} \bigg[ 1 - \cos \bigg( \frac{2 m \pi x}{3} \bigg) \bigg] \ \ (2) \end{align}

and integrate $(2)$ from $[0, 3]$. If you do this integral, you will find that it equals $\frac{3}{2}$, so $(1)$ becomes

$$\frac{3 c_{m}}{2}$$

where we don't have a sum anymore because the case above was only for when $n = m$ for some $m$.

When $m \ne n$, the integral equals $0$ (why?), hence we have

$$\sum_{n = 1}^{\infty} c_{n} \int_{0}^{3} \sin \bigg( \frac{n \pi x}{3} \bigg)\sin \bigg( \frac{m \pi x}{3} \bigg) dx = \begin{cases} 0 & n \ne m \\ \frac{3 c_{m}}{2} & n = m \\ \end{cases}$$

But from $(1)$ we also know

$$\sum_{n = 1}^{\infty} c_{n} \int_{0}^{3} \sin \bigg( \frac{n \pi x}{3} \bigg)\sin \bigg( \frac{m \pi x}{3} \bigg) dx = \int_{0}^{3} (3x - x^{2})\sin \bigg( \frac{m \pi x}{3} \bigg) dx$$

Hence

\begin{align} \frac{3 c_{m}}{2} &= \int_{0}^{3} (3x - x^{2})\sin \bigg( \frac{m \pi x}{3} \bigg) dx \\ \implies c_{m} &= \frac{2}{3} \int_{0}^{3} (3x - x^{2})\sin \bigg( \frac{m \pi x}{3} \bigg) dx \ \ (3) \end{align}

So all we need to do is solve $(3)$ and we will get the coefficients of our Fourier Series (Hint 2: Use integration by parts).

Answer 2

Using the integral \begin{align} \int_{0}^{L} \sin\left( \frac{m \pi x}{L} \right) \, \sin\left(\frac{n \pi x}{L} \right) \, dx = \frac{L}{2} \, \delta_{n,m} \end{align} then \begin{align} \int_{0}^{L} (3x - x^{2}) \, \sin\left( \frac{m \pi x}{L} \right) \, dx &= \sum_{n=1}^{\infty} B_{n} \, \int_{0}^{L} \sin\left(\frac{m \pi x}{L}\right) \, \sin\left(\frac{n \pi x}{L}\right) \, dx = \sum_{n=1}^{\infty} \frac{L}{2} B_{n} \, \delta_{n,m} = \frac{L \, B_{m}}{2}. \end{align} Now, \begin{align} \int_{0}^{L} x \, \sin\left(\frac{m \pi x}{L}\right) \, dx &= \frac{L^{2} (-1)^{m+1}}{\pi} \\ \int_{0}^{L} x^{2} \, \sin\left(\frac{m \pi x}{L}\right) \, dx &= \frac{L^{3}}{m^{3} \, \pi^{3}} \left( (2-\pi^{2} m^{2}) (-1)^{m} - 2 \right) \end{align} for which the series becomes \begin{align} B_{m} = \frac{2}{L} \left[ 3 \, \frac{L^{2} (-1)^{m+1}}{\pi} - \frac{L^{3}}{m^{3} \, \pi^{3}} \left( (2-\pi^{2} m^{2}) (-1)^{m} - 2 \right) \right] \end{align}