Consider the heat equation:

\begin{equation} \frac{\partial u}{\partial t} = K \frac{\partial^{2} u}{\partial x^2} \end{equation}

with initial condition:

$ u(x,0) = sin(\pi ^2 - x^2) $

and Neumann boundary conditions:

$ \frac{\partial u}{\partial x}(\pm \pi, t) = 0 $.

I have attempted using separation of variables, where:

\begin{equation} u(x,t) = X(x)T(t)\\ \Rightarrow \frac{\partial T}{\partial t} = -K\lambda T, \\ \Rightarrow \frac{\partial^2 X}{\partial x^2} = -\lambda X \end{equation} and $X'(-\pi) = X'(\pi) = 0$.

This leads to the solution of $T$:

\begin{equation} T = A exp(-K \lambda t) \end{equation}

However, I am stuck from this point onward in obtaining the solution for $X(x)$.

  • 1
    $\begingroup$ I would look for the $X$ solutions first; they should involve sines and cosines. The boundary conditions then imply restrictions on $\lambda$. Then, express your $T$ solution in terms of the allowable $\lambda$s. $\endgroup$ Dec 31, 2020 at 13:05

1 Answer 1


As per Mark's hint, you should start with separating and solving for $X(x)$

  1. Suppose $u(t,x) = T(t)X(x)$ and apply separation of variables:

\begin{align*} \frac{1}{K} \frac{T'(t)}{T(t)} = \frac{X''(x)}{X(x)} = -\lambda \leq 0. \end{align*}

  1. Solve for $X(x) = A\cos(\sqrt{\lambda}(x+\pi)) + B\sin(\sqrt{\lambda}(x+\pi))$ with boundary conditions $X'(\pm \pi)=0$ (and $X(x) \equiv C$ for $\lambda =0$). Argue that a non-trivial eigenfunction in terms of $\cosh$ and $\sinh$ cannot satisfy the boundary condition.

  2. We must have $B=0$ as $X'(-\pi)=0$. Deduce $X'(\pi) = -2A\sqrt{\lambda}\sin(\sqrt{\lambda}2\pi) = 0 \implies \lambda = \left (\frac{n}{2} \right )^{2}$. So you will get eigenfunctions of form $X_{n}(x) = \cos \left (\frac{n}{2} (x+\pi) \right )$.

  3. Solve for $T_{n}(t) = e^{-\left (\frac{n}{2}\right)^{2}Kt}$ and express as a sum

\begin{align*} u(t,x) &= \sum_{n=0}^{\infty} C_{n}T_{n}(t)X_{n}(x) \\ &= \sum_{n=0}^{\infty} C_{n} e^{-n^{2}Kt/4} \cos \left (\frac{n}{2} (x+\pi) \right ). \end{align*}

  1. Solve for $C_{n} = \frac{1}{\pi} \int_{-\pi}^{\pi} u(0,x)\cos \left (\frac{n}{2} (x+\pi) \right )\, dx$ for $n \geq 1$ and $C_{0} = \frac{1}{2\pi} \int_{-\pi}^{\pi} u(0,x)\, dx$. These values may not be integrable in terms of elementary functions.

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .