I would like to solve the following coupled system of linear PDEs by separation of variables, where

Ryan Reynolds

Ryan Reynolds

Answered question

2022-05-27

I would like to solve the following coupled system of linear PDEs by separation of variables, where a and b are constants:
u t = b a a + b u + ( b + a ) 2 v + u 2 x 2
v t = 2 b a + b u + ( b + a ) 2 v + d v 2 x 2
Or, in matrix form:
w t = A w + D w x x
where w = ( u v ) , A = ( b a a + b ( b + a ) 2 2 b a + b ( b + a ) 2 ) , D = ( 1 0 0 d )
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:
w = ( u ( x , t ) v ( x , t ) ) = ( ϕ 1 ( x ) h 1 ( t ) ϕ 2 ( x ) h 2 ( t ) ) Differentiating with respect to x and t, I obtain:
( ϕ 1 ( x ) h 1 ( t ) ϕ 2 ( x ) h 2 ( t ) ) = A ( ϕ 1 ( x ) h 1 ( t ) ϕ 2 ( x ) h 2 ( t ) ) + D ( ϕ 1 ( x ) h 1 ( t ) ϕ 2 ( x ) h 2 ( t ) )
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:
w = ( u ( x , t ) v ( x , t ) ) = ( α 1 α 2 ) c o s ( k x ) e λ t , where k is the wavenumber and λ 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?

Answer & Explanation

Jarrett Reyes

Jarrett Reyes

Beginner2022-05-28Added 8 answers

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
t w = ( A + D x 2 ) w ,
which leads to
w ( x , t ) = exp [ ( A + D x 2 ) t ] w ( 0 ) .
Now it remains to find exp [ ( A + D x 2 ) t ]. This is unpleasant in view of the differential operator. But switching to Fourier transforms this becomes
w ~ ( k , t ) = exp [ ( A D k 2 ) t ] w ~ ( 0 ) .
Now A D k 2 is a matrix of which one can determine the eigenvalues and eigenfunctions
A D k 2 = n λ n | p n >< q n |
so
exp [ ( A D k 2 ) t ] = n exp [ λ n t ] | p n >< q n | .
and
| w ~ ( k , t ) >= n exp [ λ n t ] | p n >< q n | w ~ ( 0 ) > .
The Fourier inverse then gives the solution in coordinate space.

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