# Separation of Variables in PDEs

1. Mar 28, 2015

### Husaaved

I am using the text by Farlow to study elementary methods of solving PDEs, and there is a point in his illustration of separation of variables where I am not seeing something. I am clear on everything that comes after and before this point, but after having returned to a certain step a few times I still don't see its justification.

PDE: $$u_t = \alpha^2u_{xx}$$ $$0 < x < 1$$ $$0 < t < \infty$$

BCs: $$u(0, t) = 0, u(1, t) = 0$$
$$0 < t < \infty$$

IC: $$u(x, 0) = \phi(x)$$
$$0 \leq x \leq 1$$

We assume a solution to the PDE of the form $$u(x, t) = X(x)T(t)$$ and ultimately reduce our original second order partial differential equation into two ordinary differential equations, one first order and one second order:

$$T' - k\alpha^2T = 0$$
$$X'' - kX = 0$$

Here, Farlow says "[w]e now make an important observation, namely, that we want the separation constant k to be negative (or else the T(t) factor doesn't go to zero as t → ∞). With this in mind, it is general practice to rename $$k = -\lambda^2$$ where lambda is nonzero." (pp35-36)

I understand that since k is an arbitrary constant we can manipulate it as we see fit depending on what would be most convenient in the physical system under study. In this case we are studying a heat diffusion problem through a rod of fixed length with heat sinks at either end and some initial temperature φ(x).

What I'm confused about is why we should want the T(t) factor to go to zero as t → ∞ -- is this somehow required by the boundary or initial conditions?

I'm clear on how to proceed from this point and how to arrive at the solution which is a Fourier sine expansion, but the reasoning/justification behind that particular step of changing the constant wasn't clear to me.

Last edited: Mar 28, 2015
2. Mar 28, 2015

### Orodruin

Staff Emeritus
No, it is not really required from the T solution itself, although the boundary conditions will make it so. Instead, solve the X part first. Given the BCs, you will find that k has to be negative in order to allow a nontrivial solution.

3. Mar 30, 2015

### HallsofIvy

Specifically, with equation $u_t= \alpha u_{xx}$, and boundary conditions, $u(0, t)= u(1, t)= 0$ "separating the variables" as you do, we get the equation for X, $X''- kX= 0$ with boundary conditions X(0)= X(1)= 0.

Now look at three cases:

k= 0. Then the differential equation is $X''= 0$ which, by integrating twice, has general solution X= Cx+ D. X(0)= D= 0 and then X(1)= C= 0. The only solution is the "trivial" solution which will not allow us to satisfy the initial condition.

k> 0. Write $k= \alpha^2$. The differential equation is $X''- \alpha^2X= 0$ which has characteristic equation $r^2- \alpha^2= 0$ , with roots $\pm\alpha$, so general solution $X= Ce^{\alpha x}+ De^{-\alpha x}$. $X(0)= C= 0$ so that $X(1)= De^{-\alpha}= 0$. But $e^{-\alpha}$ is never 0 so we must have D= 0. Again, we have only the "trivial" solution.

k< 0. Write $k= -\lambda^2$. The differential equation is $X''+ \lambda^2 X= 0$ which has characteristic equation $r^2+ \lambda^2= 0$ with roots $\pm\lambda i$ so general solution $X= C cos(\lambda x)+ D sin(\lambda x)$. $X(0)= C= 0$ so that $X(1)= D sin(\lambda)= 0$ which requires that either D= 0, which gives the "trivial" solution or $\lambda= n\pi$ for some integer n.

We would say that $X''- kX= 0$ is an eigenvalue problem which has all numbers of the form $n\pi$ as eigenvalues with corresponding eigevectors multiples of $sin(n\pi x)$.

That is how I would look at it. In terms of T, the solution to $T'= k\alpha^2 T$ is of the form $T= Ce^{k\alpha^2 t}$ which, if k> 0 will go to infinity as t goes to infinity- in other words, is unbounded, which this author is rejecting as "unphysical".