# linearized operator for ODE system

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?

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.
• By $(\tilde{U})'$, do you mean $F(U_0+\tilde{U})-F(U_0)$? Oct 2 at 14:33
• 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 $U' = (\widetilde{U})'$ Oct 2 at 15:52
• 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 $U$ Oct 2 at 16:00
• $U' = (U_0 + \widetilde{U})' = (\widetilde{U})'$, and also $U' = F(U) = F(U_0+\widetilde{U}) \approx F'(U_0)\widetilde{U}$. Equating these two yields the result. Oct 2 at 17:18