Maths Biology - linear instability

[Auron87]

1. Homework Statement

Given the linear diffusion equation with a linear source term,

$$\frac{\partial u}{\partial t} =\frac{\partial^2u}{\partial x^2} + au$$

where a is a positive constant and inital data u(x,0) = u0(x) use a linear instability analysis to show u identically equal to 0 is always unstable no matter how small a > 0.

2. Homework Equations

3. The Attempt at a Solution

To be honest I really don't know where to start. We've seen a similar example without the 'au' term and I'm struggling to follow through that. We've been told that the differential equation should come to

{sigma}*u = -k^2*u + au

but I'm not sure how to get to there. From this I can tell that sigma could be positive or negative whereas in his example, sigma was always negative which meant that u identically equal to 0 was always stable. I'm genuinely unsure of where to go with this problem.
Thanks for any help.

 

[mighty2000]

Thank you for bringing up this interesting problem. I would like to provide some guidance on how to approach this question.

Firstly, let's review the linear instability analysis method. This method is used to determine the stability of a solution to a differential equation by considering small perturbations around the solution. If the perturbations grow with time, the solution is considered unstable.

In this case, we have a linear diffusion equation with a linear source term. Let's start by considering a small perturbation to the solution u(x,t):

u(x,t) = u0(x) + \epsilon(x,t)

where \epsilon is a small perturbation. Substituting this into the diffusion equation, we get:

\frac{\partial}{\partial t}(u0 + \epsilon) = \frac{\partial^2}{\partial x^2}(u0 + \epsilon) + a(u0 + \epsilon)

Expanding and keeping only the first-order terms in \epsilon, we get:

\frac{\partial \epsilon}{\partial t} = \frac{\partial^2 \epsilon}{\partial x^2} + au0

Note that we have used the fact that the initial data is u0(x,0) = u0(x). Now, let's assume that the solution u0(x) is identically equal to 0 (meaning u0(x) = 0 for all x). In this case, the above equation reduces to:

\frac{\partial \epsilon}{\partial t} = \frac{\partial^2 \epsilon}{\partial x^2}

This is the same equation as the one we have seen in the previous example without the 'au' term. We also know that this equation has a negative eigenvalue \sigma = -k^2, which leads to a stable solution.

However, in this case, we have the additional term 'au0' in the equation. This term introduces a source of energy into the system, which can cause the perturbations to grow with time. To see this, let's consider the Fourier transform of the above equation:

\frac{\partial \hat{\epsilon}}{\partial t} = -k^2 \hat{\epsilon} + a\hat{u0}

where \hat{\epsilon} and \hat{u0} are the Fourier transforms of \epsilon and u0, respectively. Note that the Fourier