Separation of variables

February 26th, 2025

This post is still a work in progress.

Another technique to solve PDEs is to use the technique of Separation of Variables.

When do you know to apply this technique? As a first pass, if you see something that looks like the heat equation or the wave equation, both of which are linear PDEs, you can apply SoV. However, this does not bar off the possibility that you can use it for nonlinear PDEs as well — it’ just not guaranteed.

Heat Equation

Let’s consider the 1D Heat Equation with the following form.

\[\frac{\partial u}{\partial t} = \alpha \frac{\partial^2 u}{\partial x^2}\]

where $\alpha$ the thermal diffusivity. Let’s consider a 1D slab where the temperature is held at 0 at both $x=0$ and $x=L$. These boundary conditions are referred to as homogeneous boundary conditions. We can solve for a general solution without mentioning initial conditions, which I’ll circle back to later.

Before diving into the method, let’s get some intuition for what this equation is telling us. Given a point in space and time, the temperature $u(x,t)$ changes at the rate $\frac{\partial u}{\partial t}$ defined by the $\alpha \frac{\partial^2 u}{\partial x^2}$. But what exactly is $\frac{\partial^2 u}{\partial x^2}$? It’s the curvature (or concavity) in how temperature varies over space. Imagine a hill in the temperature landscape, where the temperature rises towards the top of the hill and falls off again as you go to the other side of the hill. Such a hill is said to have negative concavity (its curvature points downwards). The negative sign on the RHS implies that the temperature at that given spot will decrease over time, which is what we’d expect for an area that’s hotter than its surroundings.

Proposing a solution with separated variables

How do we solve $u$ using SoV? The first step in SoV is to propose a solution. Yes, propose a solution out of thin air! It’ll work when it works. If it doesn’t end up working, then maybe SoV wasn’t the right approach. If you’re so willing, take this leap of faith with me! The proposed solution will take the form:

\[u(x,t) = X(x)T(t)\]

where the solution is the product of two univariate functions. If we plug this proposed back solution into the original PDE, we get:

\[\frac{dT(t)}{dt}X(x) = \alpha \frac{d^2X(x)}{dx^2}T(t)\]

Notice how instead of partial derivatives, I can now do a normal derivative. On the LHS, since $X(x)$ is not a function of $t$, it can just be treated as a coefficient and left untouched. The same argument can be made on the RHS. Before proceeding, let’s clean this equation up such that the LHS is only a function of $t$ and the RHS is only a function of $x$. We can do this by dividing everything by $X(x)T(t)$.

\[\frac{T'(t)}{T(t)} = \alpha \frac{X''(x)}{X(x)}\]

Believe it or not, our equation in this form is telling us something very profound. Let’s just pick a random value of $t$, which sets the LHS to be a certain value (it doesn’t matter what that value is right now). What value of $x$ will make the RHS match the LHS value? Based on the equality we are claiming, any value of $x$ would work! More pertinently, varying the value of $x$ leaves the LHS unchanged. The only way this is possible is that the LHS is equal to a constant. And given that the LHS and RHS are equal, that means the RHS is also equal to a constant!

\[\frac{T'(t)}{\alpha T(t)} = \frac{X''(x)}{X(x)} = \text{Constant}\]

Let’s take a look at the $X$ term and its relationship to the constant.

\[\frac{X''(x)}{X(x)} = -\lambda\]

Why the negative sign? So that when we clean up the denominator and put everything on one side:

\[X''(x) + \lambda X(x) = 0\]

it’s just a plain ol’ ODE (second order with constant coefficients) in canonical form. More specifically, it’s an ODE that describes a simple harmonic oscillator (no damping present). We know the solution to that!

\[X(x) = C_1e^{\sqrt{-\lambda} x}+C_2e^{-\sqrt{-\lambda} x}\]

Investigating allowed eigenvalues

But what exactly are the values of $\lambda$? Let’s assume they are real, negative numbers. That means we have an exponential growth and decay term. If we take into account the homogeneous boundary conditions, [add more details here], it must be the case that the $C_1$ and $C_2$ are 0. In other words, negative values of $\lambda$ have trivial contributions to the solution.

What if $\lambda$ is 0? We get a solution of the form:

\[X''(x) = 0 \implies X(x) = Ax + B\]

Once again, if we consider the boundary conditions, it implies that $A$ and $B$ are 0.

Finally, if we let $\lambda > 0$, we get a solution of the form:

\[X(x) = C_1e^{i\sqrt{\lambda}x} + C_2e^{-i\sqrt{\lambda}x}\]

which can be written in terms of sines and cosines

\[X(x) = C_3\cos{(\sqrt{\lambda}x)}+C_4\sin{(\sqrt{\lambda}x)}\]

We apply the boundary solution to the coefficient to figure out what the coefficients are and find that the cosine coefficient must be 0. Only the sin term survives. We also learn that there can be various values of $\lambda$ so long as $X(x)=0$ at $x=0$ and $x=L$.

\[\sqrt{\lambda_n} = \frac{n\pi}{L}\] \[X_n(x) = C_{x,n}\sin{(\sqrt{\lambda_n}x)}\]

Now we can figure out what the time variable is.

\[\frac{T'(t)}{\alpha T(t)} = -\lambda\] \[T'(t) + \lambda \alpha T(t) = 0\]

This is just a first order ODE with constant coefficients. The solution has the form

\[T(t) = C_{t,n}e^{-\lambda\alpha t}\]

Given that there are multiple values of $\lambda$:

\[T_n(t) = e^{-\lambda_n \alpha t}\]

Putting it all back together

We have solved the equation for separated variables and can recombine them together to get the original solution we’re interested in.

\[u_n(x,t) = X_n(x)T_n(t) = C_{n}\sin{(\sqrt{\lambda_n}x)}e^{-\lambda_n \alpha t}\]

There are just solution components, so if we want the full solution, we sum over all possible values of $n$:

\[u(x,t) = \sum_n^\infty C_{n}\sin{(\sqrt{\lambda_n}x)}e^{-\lambda_n \alpha t}\]

Lastly, how do we even figure out the coefficients $C_n$? The value of these coefficients depend on the initial conditions of the system. I’ll go into more depth on that topic in a future update.