Problem of Partial Differential Equations

Question

For this question, I get stuck when I apply the second initial equation. My answer is $θ= Ae^-(kλ^2 t)\cos λx$, where $A$ is a constant. Would anyone mind telling me how to solve it?

Answer 1

Notice that the general solution to the heat equation with Neumann boundary conditions is given by

$$ \theta(x,t) = \sum_{n=0}^{\infty} A_n\cos\bigg(\frac{nx}{2}\bigg)\exp\bigg(-k\bigg(\frac{n}{2}\bigg)^{2}t\bigg) $$

We are given

$$ \theta(x,0) = 2\pi x - x^{2} $$

But our series solution at $ t = 0$ is

$$ \theta(x,0) = \sum_{n=0}^{\infty} A_n\cos\bigg(\frac{nx}{2}\bigg) $$

These must be equivalent, hence we have

$$ \begin{align} \theta(x,0) &= \sum_{n=0}^{\infty} A_n\cos\bigg(\frac{nx}{2}\bigg) \\ &= 2\pi x - x^{2} \\ \end{align} $$

Now, for orthogonality, we multiply both sides by

$$ \cos\bigg(\frac{mx}{2}\bigg) $$

And integrate from $0 \rightarrow 2\pi$ (you will need to integrate by parts). Hence you will need to solve

$$ \int_{0}^{2\pi} (2\pi x -x^{2})\cos\bigg(\frac{mx}{2}\bigg) dx = \sum_{n=0}^{\infty} A_n \int_{0}^{2\pi}\cos\bigg(\frac{nx}{2}\bigg)\cos\bigg(\frac{mx}{2}\bigg) dx $$

Where the RHS gives

$$ A_0 \cdot 2\pi $$ for $ n=m=0 $ $$ \implies A_0 = \frac{1}{2\pi} \int_{0}^{2\pi} (2\pi x -x^{2}) dx $$

and

$$ A_n \cdot \pi $$ for $ n=m \ne 0 $ $$ \implies A_n = \frac{1}{\pi} \int_{0}^{2\pi} (2\pi x -x^{2})\cos\bigg(\frac{mx}{2}\bigg) dx $$

Solving for $A_0, A_n$ gives

$$ \theta(x,t) = \frac{2\pi^{2}}{3} + \sum_{n=1}^{\infty} \frac{8}{n^{2}}\bigg[(-1)^{n+1}-1\bigg]\cos\bigg(\frac{nx}{2}\bigg)\exp\bigg(-k\bigg(\frac{n}{2}\bigg)^{2}t\bigg) $$