# Parabolic PDEs: Boundary conditions

I'm working through Pinchover and Rubinstein's "Introduction to Partial Differential Equations" and am trying to understand the motivation for studying Sturm Liouville problems. To this end, I am following the process of taking a general parabolic PDE:

$$u_{t}-\frac{1}{r\left(x\right)m\left(t\right)}\left(\left(p\left(x\right)u_{x}\right)_{x}+q\left(x\right)u\right) = 0, \ x\in\left(a,b\right),t\in\left(0,\infty\right)$$ with boundary conditions $$B_{a}\left[u\right]:=\alpha u\left(a,t,\right)+\beta u_{x}\left(a,t\right) = 0, \ t\in\left[0,\infty\right)$$ $$B_{b}\left[u\right]:=\gamma u\left(b,t\right)+\delta u_{x}\left(b,t\right) = 0, \ t\in\left[0,\infty\right)$$ and initial condition $$u\left(x,0\right) = f\left(x\right), \ x\in\left[a,b\right].$$

Let us consider the parabolic problem in more detail. Just as in the case for the heat equation, we seek non-trivial solutions of the form $$u\left(x,t\right)=X\left(x\right)T\left(t\right).$$

Differentiating and substituting, noting that $\left(pX_{x}T\right)_{x}=T\left(pX_{x}\right)$, \begin{eqnarray} XT_{t}-\frac{1}{rm}\left(\left(pX_{x}T\right)_{x}+qXT\right) & = & 0\\ \frac{T_{t}m}{T} & = & \frac{\left(pX_{x}\right)_{x}+qX}{rX} \end{eqnarray} where the LHS is a function only of $t$ and the RHS a function only of $x$, so that $$\frac{T_{t}m}{T}=\frac{\left(pX_{x}\right)_{x}+qX}{rX}=-\lambda.$$ Hence, we have the system of ordinary differential equations

\begin{array}{cccc} \frac{d}{dx}\left(pX\right)+qX+\lambda rX & = & 0, & x\in\left(a,b\right),\\ m\frac{dT}{dt}+\lambda T & = & 0, & t\in\left(0,\infty\right). \end{array}

All fine so far by me. However, the authors then go on to say that since $u$ satisfies the boundary conditions $B_a [u]$ and $B_b [u]$, it follows that

$$B_a [X] = 0, \ B_b [X] =0.$$

I'm afraid I don't see why this is. Can anybody explain?

• Well, $B_a[u(x,t)] = T(t) B_a[X]$ right? Now $T(t)B_a[X] = 0$ for all $t$, hence $B_a[X] = 0$. May 6, 2013 at 21:54

## 1 Answer

The boundary conditions are formulated in a way that is only dependent on $t$ - recall that $a$ and $b$ in e.g. $B_a[u] = \alpha u(a,t) + \beta u_t(a,t) = 0$ are constants (namely the start and end values of the spatial domain).

However, $B_a[u] = 0$ should hold for all times. If we insert the ansatz, we find

$$B_a[u] = \alpha X(a)T(t) + \beta X'(a)T(t) = \left[\alpha X(a) + \beta X'(a)\right]T(t) = B_a[X]\cdot T(t) = 0$$

Since $T(t)\equiv 0 \Longleftrightarrow u \equiv 0$ is a very uninteresting solution, we cannot require this in general. Therefore, we must instead have $B_a[X] = 0$. The exact same argument can be applied to the boundary at $x=b$.