Consider a system of reaction-diffusion equations where we write as $$ \begin{cases} u_t=\Delta u+f(u,v),\\ v_t=\Delta v+g(u,v) \end{cases} $$ In vector form, we also have $U_t=F(U)$ where $U=\begin{bmatrix}u \\ v\end{bmatrix}$ and $F$ is the abstract operator that gives the right-hand side. The stationary solution is just $F(U_0)=0$. But why the linearized operator is claimed to be $F'(U_0)$?

I am thinking of doing the Taylor expansion where

$$ F(U_0)=F(U)+F'(U)(U_0-U)=U_t+F'(U)(U_0-U)=0 $$ which gives $$ U_t=F'(U)(U-U_0) $$ But again I still don't see a direct implication of linearity here.

A follow-up question:

If we let $\mathbf{u}=\mathbf{u_0}+\epsilon \mathbf{U}$, then the linearized operator is $$ \frac{d\mathbf{U}}{dt}=\Delta\mathbf{U}+\mathbf{JU} $$ where $\mathbf{JU}$ is the Jacobian evaluated at the stationary solution. I trie to derive this expression by writing $\mathbf{U}=\frac{1}{\epsilon}(\mathbf{u}-\mathbf{u_0})$ and differentiate. But how did we get the Jacobian for stationary solution?


1 Answer 1


When you linearize about a point $U_0$, you typically also make a change of variable to something like $\widetilde{U} = U - U_0$ so that the new variables $\widetilde{u}$ and $\widetilde{v}$ have an equilibrium value at the origin. Indeed, $$F(U_0 + \widetilde{U}) = F(U_0) + F'(U_0)\widetilde{U} + \mathcal{O}(\|\widetilde{U}\|^2).$$ This combined with the fact that $U' = (U_0 + \widetilde{U})' = (\widetilde{U})'$ yields $$(\widetilde{U})' = F(U_0 + \widetilde{U}) \approx F'(U_0)(U-U_0)= F'(U_0)\widetilde{U},$$ which is a linear system in $\widetilde{U}$, as desired. The approximation is made assuming $\widetilde{U}$ is small enough that a linear Taylor series approximation is good enough. This can be done because linear systems are scale-invariant so we can take $\widetilde{U}$ as small as we want and still get the same linearized system.

  • $\begingroup$ By $(\tilde{U})'$, do you mean $F(U_0+\tilde{U})-F(U_0)$? $\endgroup$
    – 79999
    Commented Oct 2, 2023 at 14:33
  • $\begingroup$ I have added some more to the answer to make the expression for $(\widetilde{U})'$ more clear. Recall that $U_0$ is constant in time, so its derivative is zero and therefore $U' = (\widetilde{U})'$ $\endgroup$
    – whpowell96
    Commented Oct 2, 2023 at 15:52
  • $\begingroup$ I still don't see how the first equality comes in in your second equation. From the Taylor expansion, you are basically saying $F(U_0-\tilde{U})=(\tilde{U})'$. I gues I am confused about how the derivation on $U$ relates to the operator applies to $U$ $\endgroup$
    – 79999
    Commented Oct 2, 2023 at 16:00
  • $\begingroup$ $U' = (U_0 + \widetilde{U})' = (\widetilde{U})'$, and also $U' = F(U) = F(U_0+\widetilde{U}) \approx F'(U_0)\widetilde{U}$. Equating these two yields the result. $\endgroup$
    – whpowell96
    Commented Oct 2, 2023 at 17:18
  • 1
    $\begingroup$ Oh, I see! So in the follow-up question, the Jacobian also comes from this linearization. Thanks! $\endgroup$
    – 79999
    Commented Oct 2, 2023 at 18:54

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .