# using PDE existence to show a map is invertible

I have an existence/uniqueness theorem for the PDE $$u_t = a(x,t)u_{xx} + b(x,t)u_x + c(x,t)u - g(x,t).$$

Now if I have a Gateaux derivative of a map $F$ at a point $p$ satisfying $$DF(p)v = v_t - f_1v_{xx} - f_2v_x - f_3v$$

(the $f_i$ are functions of $(x,t)$) then how can I use my PDE result to say that $DF(p)$ is invertible? I thought I could rearrange to get $$v_t - f_1v_{xx} - f_2v_x - f_3v - DF(p)v = 0$$ and I want to put the $f_3$ and $DF(p)$ together so that it is in the form of the PDE and I can just quote the existence result which in this cases would tell me that there is a unique $v$ satisfying this equation and hence $DF(p)$ is invertible. But all I know is the the derivative is linear and I can't put the $f_3$ and the $DF(p)$ together.

Thank you

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@user12477 you're right, sorry. – blahb Jun 15 '12 at 22:04

I think that what you've given in your question is just about there: here's an attempt to tidy up. This also needs to be fleshed out in a way that depends on the spaces $V, W$ and on the details of your existence/uniqueness result.

We have $F:V\to W$. Then $DF(p)$ is invertible if it is a bijection from $V$ to $W$. So you need to show two things:

1. For every $w\in W$, there exists $v\in V$ such that $DF(p)(v)=w$.
2. If $DF(p)(v_1)=DF(p)(v_2)\in W$, then $v_1=v_2\in V$.

But $$DF(p)(v) = v_t-f_1v_{xx}-f_2v_x-f_3v,$$ and (I guess...) your PDE existence and uniqueness result tells you that for every $w=g\in W$ there exists a unique $v\in V$ such that $$v_t-f_1v_{xx}-f_2v_x-f_3v = w.$$

Existence gives part 1 above, and uniqueness gives part 2. (The key is that $w$ plays the role of the inhomogeneity $g(x,t)$.)

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Great, thank you!! – blahb Jun 15 '12 at 22:36