I would like to solve the following coupled system of linear PDEs by separation of variables, where a and b are constants:

${\partial{u}\over\partial{t}} = {b-a \over a+b}u + (b+a)^2v + {\partial{u}^2\over\partial{x^2}} $

${\partial{v}\over\partial{t}} = {-2b \over a+b}u + -(b+a)^2v + d{\partial{v}^2\over\partial{x^2}} $

Or, in matrix form:

$\vec w_t = A \vec w + D \vec w_{xx}$

where $ \vec w = \begin{pmatrix}u\\v\\\end{pmatrix}, A = \begin{pmatrix} {b-a \over a+b} & (b+a)^2\\ {-2b \over a+b} & -(b+a)^2 \\\end{pmatrix}, D = \begin{pmatrix} 1 & 0\\ 0 & d \\\end{pmatrix} $

Since this is a linear equation, one should be able to apply the method of Separation of Variables. I've gone through the exercise of separating variables for single variable models in one or more spatial dimensions many times. However, I've never seen a completely worked example where there are two coupled equations. My intuition tells me to start with:

$\vec w = \begin{pmatrix}u(x,t)\\v(x,t)\\\end{pmatrix} = \begin{pmatrix}\phi_1(x)h_1(t)\\\phi_2(x)h_2(t)\\\end{pmatrix} $ Differentiating with respect to x and t, I obtain:

$\begin{pmatrix}\phi_1(x)h_1'(t)\\\phi_2(x)h_2'(t)\\\end{pmatrix} = A\begin{pmatrix}\phi_1(x)h_1(t)\\\phi_2(x)h_2(t)\\\end{pmatrix} + D\begin{pmatrix}\phi_1''(x)h_1(t)\\\phi_2''(x)h_2(t)\\\end{pmatrix} $

But at this point I get stuck, because expanding this seems to make the problem more difficult.

Textbooks/papers in mathematical biology solve this problem, which is the linearized form of a reaction-diffusion model, by immediately assuming particular solutions of the form:

$\vec w = \begin{pmatrix}u(x,t)\\v(x,t)\\\end{pmatrix} = \begin{pmatrix}\alpha_1\\\alpha_2\\\end{pmatrix}cos(kx)e^{\lambda t} $, where k is the wavenumber and $\lambda $ is the eigenvalue.

I find this really unsatisfying, because it seems to fall out of nowhere.

My questions are:

  1. Is there a general method for separating variables in coupled linear PDEs or PDEs written in vector form?
  2. Is there a book/paper/tutorial that I can use to help me work this out in all of its details?
  3. Is there some deeper theory that I need to first understand?

1 Answer 1


I do not think that a separation of variables is convenient in this case. There are two problems that have to be considered. First there is the time evolution$$ \partial _{t}\mathbf{w}=(\mathbf{A}+\mathbf{D}\partial _{x}^{2})\mathbf{w}, $$ which leads to$$ \mathbf{w}(\mathbf{x},t)=\exp [(\mathbf{A}+\mathbf{D}\partial _{x}^{2})t]% \mathbf{w}(0). $$ Now it remains to find $\exp [(\mathbf{A}+\mathbf{D}\partial _{x}^{2})t]$. This is unpleasant in view of the differential operator. But switching to Fourier transforms this becomes$$ \mathbf{\tilde{w}}(\mathbf{k},t)=\exp [(\mathbf{A}-\mathbf{D}k^{2})t]\mathbf{% \tilde{w}}(0). $$ Now $\mathbf{A}-\mathbf{D}k^{2}$ is a matrix of which one can determine the eigenvalues and eigenfunctions$$ \mathbf{A}-\mathbf{D}k^{2}=\sum_{n}\lambda _{n}|\mathbf{p}_{n}><\mathbf{q} _{n}| $$ so$$ \exp [(\mathbf{A}-\mathbf{D}k^{2})t]=\sum_{n}\exp [\lambda _{n}t]|\mathbf{p}% _{n}><\mathbf{q}_{n}|. $$ and$$ |\mathbf{\tilde{w}}(\mathbf{k},t)>=\sum_{n}\exp [\lambda _{n}t]|\mathbf{p}% _{n}><\mathbf{q}_{n}|\mathbf{\tilde{w}}(0)>. $$ The Fourier inverse then gives the solution in coordinate space.

  • $\begingroup$ Thank you for answering my question. I don't understand all of the details, though this is along the lines of what I was looking for. I'm guessing you can consider the time evolution part first/separately because the eq. is linear? Do you know of any books that will take me step-by-step through this process and help me gain some experience dealing with these types of problems. These linearized systems of PDEs come up very often in mathematical biology, and I'd like to know a rigorous way that justifies my saying that solution is a sum of exponentials/cosines. Thank you again. $\endgroup$
    – razalfuhl
    Commented Nov 14, 2014 at 18:05
  • $\begingroup$ I can provide you with more information. This could be conveniently done by a personal message or by e-mail but I do not know if Stack Exchange has the option of sending personal messages. $\endgroup$
    – Urgje
    Commented Nov 16, 2014 at 12:44
  • $\begingroup$ Is this technique common for solving these sorts of problems? $\endgroup$
    – Cuhrazatee
    Commented Jun 26, 2018 at 19:31

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