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Consider $$ u_t=D\Delta u+f(x,u,\nabla u)~~(*). $$

In a script it is said:

We first consider the variational structure which makes $(*)$ an $L^2$-gradient flow.

I have two questions to this:

1.) What is meant with variational structure?

2.) What is an $L^2$-gradient flow and in which sense is $(*)$ an $L^2$-gradient flow?

mathfemi
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1 Answers1

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Finding a variational structure here means finding a functional such that the PDE is the gradient flow of that functional.

The gradient flow of a functional $\psi :V\to \mathbb {R} $ is the differential equation

$$ \partial_t u=-\text{grad}_u\psi. $$ That is, at each point of time the gradient flow changes $ u $ infinitesimally to achieve a minimization of $\psi $ (remember that the gradient gives the direction of steepest ascent, so the negative gradient gives the direction of steepest descent)

For example, when $\psi $ is linear, the solution of the gradient flow has constant velocity, because the gradient is constant.

A more useful example is $ V $ being the Sobolev space $ H^1(\mathbb{R}^n)$ and $\psi(u)=\int |\nabla u|^2$ (nonlinear! ). You can calculate using Greens theorem that the gradient flow here is the heat equation. However, there are technicalities involved: when you say "direction of steepest descent" you imply that you are considering directions in a certain normed space. If you considered the Sobolev space here, then you would get $ \partial_t u=-u $. You get the heat equation only if you consider the space $ L^2$ as space of allowed directions and use formal calculations, but strictly speaking the functional is not defined for $ u+hv $ when $ u $ is a Sobolev function, $ v\in L^2$ and $ h> 0$.

Bananach
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  • Is it the same to say that $u_t=-\delta \varphi(u)$, where $\delta\varphi(u)$ is the first Variation of $\varphi(u)$? In other words: Isn't the Gradient of a functional its first Variation? – John_Doe Jan 03 '17 at 14:28
  • That question is fair enough, given the word "variational structure". The answer may be yes, depending on your definition of first variation Personally, I would understand "first variation" as "total derivative" instead of "gradjent" though. If you are not sure what the difference between the two is, look it up, this is not the place to discuss this. Just remember that the definition of the gradient requires an inner product to be specified – Bananach Jan 03 '17 at 18:58
  • Cannot follow you. First variation is defined as Gateaux derivative, i.e. directional derivative of the functional, not as total derivative. – John_Doe Jan 03 '17 at 21:29
  • But of course this is connected with the total derivative/ Frechet derivative. So maybe you use another def. of first variation. – John_Doe Jan 03 '17 at 21:46
  • My point was that if you define first variation as say a linear map from the tangent space to the real numbers then the differential equation $ u'=-\delta \phi (u)$ doesn't make sense because the left hand is a vector in your space and the right hand is an element of the dual space. To identify elements of the dual space with vectors in the space itself, you need an inner product (and then the Riesz representation theorem). This gives the connection between what I call "total derivative" and what I call "gradient" – Bananach Jan 04 '17 at 12:57
  • You are right. I think the right way to say it is that the gradient is the so-called variational derivative (or: functional derivative) and so $u_t$ equals this variational derivative (and not the first variation which is the Gâteaux derivative). But the first variation comes into play when determining an explicit formula of the gradient since the scalar product of the gradient of $E(u)$ with some $h$ is the first variation of $E(u)$ in $h$ and this equals the total derivative of $E(u)$ evaluated at $h$. -- So it is missleading to say that $u_t$ directly equals the first variation. - – John_Doe Jan 04 '17 at 18:36
  • Addemdum: With E, I mean a functional. – John_Doe Jan 04 '17 at 19:10
  • I hope I got what you mean. Say, we have $u(x,t)\in R^2$ and functional $E(u)\in R$, then $u_t\in R^2$ but first variation $\delta E(u)\in R$. Hence, $u_t=\delta E(u)$ makes no sense. But when replacing $\delta E(u)$ by the gradient of $E(u)$ it does make sense. – John_Doe Jan 04 '17 at 19:29
  • Agree to all you said except: it should be $ u(t)\in R^2$ (not $ u (x, t )\in R^2$) or in PDE situations $ u (t)\in V $ for some infinite dimensional space of functions $ V $. Also, in your last comment, I am still not happy with your claim $\delta E (u)\in R $. I would say $ \delta E (u) $ is a map from your space to the reals and only once you evaluate this map at some direction you get a real. But really I think from your remaining comments that you are aware of this and this is just nitpicking – Bananach Jan 05 '17 at 06:20