Let $\Omega = B(0,1) \subset \mathbb{R}^n$, $n \geqslant 3$ and $f \in
L^n(\Omega)$. Show the existence and uniqueness of a weak solution of\begin{align} \Delta u+x\cdot \nabla u – \frac{u}{3} &= \frac{f}{|x|}
\\ u &= 0 \quad \text{on } \partial \Omega \end{align}
I think I managed to show the boundedness of $B[u,v] = \int( \Delta u v + x \nabla u v – \frac{1}{3}uv)$. We have
\begin{align}
\lvert B(u,v) \rvert &= \bigg \lvert -\int \bigg( \nabla u \nabla v + x \nabla u v – \frac{u}{3}v \bigg) \bigg\rvert \\
&\leqslant \int \lvert \nabla u \nabla v \rvert + \int \lvert x \nabla u v \rvert + \int \lvert \frac{u}{3}v \rvert \\
&\leqslant \| \nabla u \|_{L_2} \cdot \| \nabla v \|_{L_2} + \sup \lvert x \rvert \cdot \| \nabla u \|_{L_2} \cdot \| v \|_{L_2} + \frac{1}{3} \cdot \| u \|_{L_2} \cdot \| v \|_{L_2} \\
&\leqslant \| \nabla u \|_{L_2} \cdot \| \nabla v \|_{L_2} + \| \nabla u \|_{L_2} \cdot \| v \|_{L_2} + \frac{C}{3} \cdot \| \nabla u \|_{L_2} \cdot \| v \|_{L_2} \\
&= C \| \nabla u \|_{L_2} \| v \|_{L_2}
\end{align}
I was using here respectively triangle inequality, Cauchy-Schwarz inequality and Poincare inequality.
But then I tried to show coercivity and I got
\begin{align}
B[u,u] &= – \int \nabla u \nabla u+ \int x \nabla u u – \int \frac{u^2}{3} \\
&= -\| \nabla u \|^2_{L_2} + \int( x \cdot \nabla u u) – \frac{1}{3} \| u \|^2_{L_2}
\end{align}
Now I'm a bit confused, because I get two negative components, and I want to have $B[u,u] \geqslant \theta ||u||^2_{L_2}$. So I'm asking for a little hint.
Best Answer
The bilinear form which is coercive is $$ B[u,v]=\int\nabla u\cdot\nabla v-vx\cdot\nabla u+\frac 1{3}uv $$
The first and the second terms are easy to handle. Regarding the middle term, observe that $$ -u\nabla u\cdot x=\frac 1{2}\left(nu^2-\nabla\cdot (u^2x)\right) $$ and then $$ -\int\limits_B ux\cdot\nabla u\,dV=\frac 1{2}\int\limits_B\left(nu^2-\nabla\cdot (u^2x)\right)\,dV=\frac n{2}\|u\|_{L^2}-\frac 1{2}\int\limits_B\nabla\cdot (u^2x)\,dV $$ By Stokes Thm, the second term is the flux of $u^2 x$ through the boundary of $B$, but $u=0$ there so that term disappears.You end up with a combination of the squares of the norms of $\nabla u$ and $u$ in $L^2$, bounded below by the square of the $H^1$-norm.