What is the 'linearization' of a PDE?

Question

Specifically I am looking at the proof of Lemma 4.1 on page 9 here, where the graphical form of curve shortening flow is given, and then its 'linearization'. I am struggling to find any online resources that explain what this means, and what the relevance of using such a linearization is. Could anyone point me in the right direction or perhaps provide an explanation here? It would be even better if someone could explain this in the context of Lemma 4.1, as I think this can be read as a standalone result that doesn't require familiarity with the rest of the paper. Any help/information on this concept would be appreciated, however!

(Sorry if this should be posted on mathoverflow - whilst it is a research based question, I am guessing the concept itself is something that should be asked about here).

Answer 1

First let's look at the linearization of the ODE $\dot{x}(t) = f(x(t))$. Suppose that $x_0$ is an equilibrium point, i.e. a point for which $f(x_0) =0$. Then $x(t) = x_0$ for all $t$ is a trivial solution to the ODE. A natural question is to examine what happens to solutions that start off near $x_0$. Assume we have a solution and write $y(t) = x(t) -x_0$. Then $$ \dot{y} = \dot{x} = f(x) = f(x_0 + y) = Df(x_0) y + [f(x_0+y) - f(x_0) - Df(x_0) y]. $$ Now, if we assume that $f$ is differentiable at $x_0$ then we should have that $$ f(x_0+y) - f(x_0) - Df(x_0) y = o(|y|) $$ and so we (formally) expect the dynamics of $y$ to be dominated by the linearized ODE $\dot{y} = Df(x_0) y$. This formal reasoning can be made precise for ODEs under some assumptions on the behavior of $f$ and the resulting matrix $Df(x_0)$.

For PDEs we can play a similar game, though it's harder to write down. Let me illustrate with two simple examples. First consider $$ \partial_t u - \Delta u = u^2-1 = f(u) $$ A trivial solution to this is $u=1$, so we can try to look for solutions near $u$ by plugging in $u = 1 + v$: $$ \partial_t v - \Delta v = \partial_t u - \Delta u = u^2 -1 = 1+2v +v^2 -1 = 2v + v^2, $$ and so the linearized PDE is $\partial_t v - \Delta v = 2v$.

Let's now look at the PDE $\partial_t u + u \partial_x u =0$. A trivial solution is any constant. We plug in $u = c + v$ to get $$ 0 = \partial_t v + (c+v) \partial_x v = \partial_t v + c \partial_x v + v \partial_x v, $$ and so the linearized PDE is $\partial_t v + c \partial_x v =0$.