Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Hi, let $\Omega \subset R^3$ be a bounded smooth domain, consider the elliptic equation \begin{equation} -\triangle u + u^2div u = f, \quad x \in \Omega, \quad u\big|_{\partial \Omega} = 0. \end{equation} Here the force $f \in L^2$, and $div u = \sum_j \partial u/\partial x_j $. Is there any solutions of the equation in $H^1$? And how to prove it. Thanks for any hints and references.

\textbf{Edit}: In fact, the $1-d$ equation we considered is the stationary equation of the BBM equation $$ u_t - u_{txx} - u_{xx} + u^2u_x = f. $$

share|improve this question
4  
your notation is ugly. Because $u$ is scalar, what you call $divu$ is not the divergence. Are you sure of which equation you are interested in ? –  Denis Serre Oct 10 '12 at 14:26
add comment

1 Answer 1

up vote 1 down vote accepted

Firstly, turning to the $\mathbb{R}^1$ case, we note that $$ -u_{xx}+u^2u_x=-\partial_x\left(u_x-\frac{1}{3}u^3\right)=f(x) $$ and so we get a Chini equation (see Inhomogeneous Bernoulli Equation): $$ -u_x+\frac{1}{3}u^3=\int_{x_0}^xf(x')dx'+C $$ and a general solution to this equation is not known. Turning to $\mathbb{R}^3$, you can see that your equation can be written down in the form $$ \nabla\cdot {\bf v}=-f $$ being $$ {\bf v}=\left(u_x-\frac{1}{3}u^3,u_y-\frac{1}{3}u^3,u_z-\frac{1}{3}u^3\right) $$ and you have to satisfy the Gauss identity $$ \int_{\partial\Omega}{\bf v}\cdot d{\bf S}=-\int_{\Omega}dV f. $$ This means that one has to find three functions $v_1,\ v_2,\ v_3$ that satisfy three different Chini equations. Again, a general solution misses in this case. The conclusion to draw is that you need to specialize the problem to see if this becomes manageable in some case or resort to some perturbation technique.

share|improve this answer
add comment

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.