Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $E:M\to\mathbb{R}\cup\{+\infty\}$ be an energy functional of the form \begin{equation} E[u]=\int_\Omega L(x,u,\nabla u)dx, \end{equation} where $M$ is a subset of $W^{1,p}(\Omega)$ $(1< p<\infty)$, and $L=L(x,z,v)$ is a nice function.

Using variational method, the minimizer of this functional is associated with the Euler-Lagrange equation \begin{equation} \sum \partial_jL_{v_j}(x,u,\nabla u)=L_z(x,u,\nabla u). \end{equation} All this makes sense, but I got confused by the definition of a weak solution, which is defined to be $u\in M$ such that \begin{equation} \int_\Omega \sum L_{v_j}(x,u,\nabla u)\phi_{x_j}+L_z(x,u,\nabla u)\phi=0 \end{equation} for all $\phi\in W^{1,p}(\Omega)$.

I am in particular confused by the choice of $W^{1,p}$. Usually, when we talk about a weak solution in a general Banach space $X$, we use the dual space $X'$. That is $x\in X$ is a solution if and only if $(x,\ell)=0$ for all $\ell\in X'$. But here, we choose the 'test' functions from the same space $W^{1,p}$, not the dual space $W^{1,p'}$.

Is there a specific reason for this choice?

Thanks!

share|improve this question

3 Answers 3

The choice of $p$ depends on $L$, and its nonlinear (in general) nature. If $L=|\nabla u|^2$, then the most convenient choice is $p=2$. If $L=|\nabla u|^s$, then $p=s/(s-1)$.

share|improve this answer

The E-L equations which arrived at by using the variational method is a second order PDE. By a weak solution we generally mean that the solution should have at least one order lesser regularity than the order of the PDE. Hence we seek a solution to this PDE in a space of one order lesser than the order of the PDE which is 2 in the current case.

share|improve this answer
    
That is why we look the space W^1,p instead of W^2,p. –  Alexander Jan 27 at 18:06
up vote 0 down vote accepted

As mentioned by Yiorgos S. Smyrlis, the choice of $p$ does depend on the Lagrangian $L$. But I guess a more detailed explanation might be helpful here.

First of all, there is a very natural growth rate control on $L:\Omega\times\mathbb{R}\times\mathbb{R}^d$. Since we want the energy to be at least finite for all $u\in W^{1,p}(\Omega)$, we want to impose on $L$ \begin{equation} |L(x,z,v)|\le C(1+|z|^p+|v|^p). \end{equation} In this way, $|L(x,u,\nabla u)|\le C(1+|u|^p+|\nabla u|^p)$ will be in $\mathcal{L}^1(\Omega)$.

However, once we impose this control on $L$, the followings become natural \begin{equation} |L_{z}(x,z,v)|\le C(1+|z|^{p-1}+|v|^{p-1}), \end{equation} and \begin{equation} |L_{v}(x,z,v)|\le C(1+|z|^{p-1}+|v|^{p-1}) \end{equation} To see this just note that the growth rate of a function is one order higher than its derivative.

But once we have this bound on the derivatives of $L$ and note that our 'test function' is to be paired up with the derivatives of $L$, then we actually need our test function to be in $W^{1,q}$, where $q$ is the conjugate of $p/(p-1)$ (because we will end up with terms like $|u|^{p-1}\phi$, then we need to use Holder to bound this term and raise $|u|$ to the $p$th power). Then a direct computation shows \begin{equation} q=p. \end{equation}

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.