# Solving Heat Equation IBVP

I have a hard time understanding how to proceed with the third condition of the following IBVP: $$\begin{cases} u_t = u_{xx}, \ x\in (0, 2), \ t>0 \\[6pt] u(x, 0) = x, \ x \in [0, 2] &(i)\\[6pt] u(0, t) = 0, \ t>0 &(ii)\\[6pt] u_x(2, t) = 0, \ t>0 &(iii)\\[6pt] \end{cases}$$

I am solving this by the method of separation of variables (looking for a solution in the form $$u(x, t) = X(x)T(t)$$). After dividing it into two ODE-s, for the first one I have a solution in the form of $$X(x) = c_1\sin px + c_2 \cos px$$, where $$p$$ is the square root of the negative eigenvalue $$\lambda = -p^2$$, as it is the only one that produces a non-trivial solution.

After plugging in the second condition, I obtained $$c_1 = 0$$. From the third condition it follows that $$\cos 2p = 0$$ and $$p = \frac{\pi}{4} + \frac{\pi k}{2}$$. Hence, $$X(x) = c_2\cos((\frac{\pi}{4} + \frac{\pi k}{2})x)$$

Assuming everything above is correct, the final solution will be in the form $$u(x,t) = \sum_{mn=1}^\infty c_n\cos\left((\frac{\pi}{4} + \frac{\pi k}{2})x\right)e^{-(\frac{\pi}{4} + \frac{\pi k}{2})^2t}$$

I am looking for a hint on how to continue from here. Do I need to find $$c_m$$ as the cosine coefficients of the Fourier series of $$x$$ from 0 to 2 or do I need to perform further reductions, as the cosine part does not match the one in the series?

Thank you.

## 1 Answer

How did you get $$c_1=0$$? The boundary conditions are

$$X(0) = X'(2) = 0$$

Since $$X(0) = c_2$$ this would imply $$c_2=0$$, so you're left with

$$X(x) = \sin(px) \implies X'(2) = p\cos(2p) = 0$$

$$\cos$$ is zero at odd multiples of $$\pi/2$$, so we have

$$2p = \frac{(2n+1)\pi}{2} \implies p = \frac{(2n+1)\pi}{4}$$

and the general solution is

$$u(x,t) = \sum_{n=0}^\infty c_n\exp\left[-\left(\frac{(2n+1)\pi}{4} \right)^2t\right]\sin\left(\frac{(2n+1)\pi}{4}x\right)$$

From the initial condition

$$u(x,0) = x = \sum_{n=0}^\infty c_n\sin\left(\frac{(2n+1)\pi}{4}x\right)$$

Using orthogonality we can determine

$$c_n = \frac{\int_0^2 x \sin\left(\frac{(2n+1)\pi}{4}x\right)\ dx}{\int_0^2 \sin^2\left(\frac{(2n+1)\pi}{4}x\right)\ dx}$$

The result is similar to a Fourier series, but not a conventional one. Here's a similar answer where I went a bit further on orthogonality.

• Thank you! I did get the sine version at first, but then got mixed up with the notations. I did not know this could be solved by using orthogonality, thank you for the help. – eliott Feb 21 at 18:44