Solving 2D heat equation with separation of variables

Question

Solve:

$$ \begin{cases} u_t = Ku_{xx}\\ u(0,y,t) = u(\pi,y,t) = 0\\ u_y(x,0,t) = u_y(x,\pi,t) = 0\\ u(x,y,0) = 1 \end{cases}$$

for $0 < x < \pi$, $0 < y < \pi$, $t>0$:

This was the problem given to me, but I don't believe it has a nontrivial solution (correct me if I'm wrong). Instead, I think the problem was meant to say: $$u_t = K(u_{xx} + u_{yy})$$ In this case,

Let $u(x,y,t) = f(x)g(y)h(t)$: $$\frac{h'(t)}{h(t)} = \frac{f"(x)}{f(x)} + \frac{g"(y)}{g(y)}$$ Let: $$\frac{f"(x)}{f(x)} = -n^2$$ $$\frac{g"(y)}{g(y)} = -m^2$$ $$\frac{h'(t)}{h(t)} = -(m^2 + n^2)$$ This leads to the seaparted solutions: $$f(x) = A\cos(nx) + B\sin(mx)$$ $$g(y) = C\cos(my) + D\sin(my)$$ $$h(t) = Ee^{-(n^2 + m^2)t}$$

$$u(o,y,t) = u(\pi,y,t) = 0 \space\text{ implies } \space A = 0$$ $$u_y(x,0,t) = u_y(x,\pi,t) = 0 \space\text{ implies } \space D = 0$$ Thus, $$u(x,y,t) = B\sin(nx)\cos(my)e^{-(n^2 + m^2)t}$$ To satisfy the initial value, we can exploit the superposition principle:

$$u(x,y,t) = \sum_{m=0}^\infty \sum_{n=0}^\infty B_{nm}\sin(nx)\cos(my)e^{-(n^2 + m^2)t}$$ $$u(x,y,0) = \sum_{m=0}^\infty \sum_{n=0}^\infty B_{nm}\sin(nx)\cos(my) = 1$$

This was as far as I was able to get. I'm unsure how to satisfy the initial condition given this double sum. Some help would be appreciated!

Answer 1

You can find m and n using boundary conditions. Instead of calling your constant n or m, call them k or $\lambda$. m and n are used frequently for natural numbers.

$$u(x=\pi) = 0 \Rightarrow Bsin(\lambda\pi) = 0 $$

And we want a non trivial solution, so $B\ne 0$. Therefore $sin(\lambda \pi)=0$

$\lambda \pi = \pi n \Rightarrow \lambda = n $

I didn't see you use the BVs so I'm not sure if you did. that step.

Now all that's left is to find the coefficient $B_{nm}$ using the orthogonality properties of your eigenfunctions.

$$B_{nm} = \frac{4}{\pi^2}\int_0^\pi\int_0^\pi f(x,y)*sin(n'x)*sin(m'y)dxdy$$

where f(x) is 1 in this case.

You don't even have to memorize the integral above to find the coefficient in the future.

You can simply multiply both sides by sin(n'x)sin(m'y) and integrate on the domain. Using properties of Kroneckers delta, only when m' = m and n'=n will get something that isn't zero. The left hand side will simply always be:

$$ \\B_{nm}\int_0^\pi sin^2(nx)dx\int_0^\pi sin^2(my)dy =B_{nm}\frac{\pi}{2}*\frac{\pi}{2} = B_{nm}\frac{\pi^2}{4}$$

Now we have:

$$B_{nm}\frac{\pi^2}{4} = \int_0^\pi\int_0^\pi 1*sin(nx)*sin(my)dxdy$$