# Ngô Quốc Anh

## April 30, 2010

### The Pohozaev identity: Toda systems and a priori estimates

Filed under: PDEs — Tags: — Ngô Quốc Anh @ 10:15

In this topic we consider the analysis of solutions of the following system entitled Toda system $\displaystyle\left\{ \begin{gathered} - \Delta {u_1} = 2{e^{{u_1}}} - {e^{{u_2}}}, \hfill \\ - \Delta {u_2} = - {e^{{u_1}}} + 2{e^{{u_2}}}. \hfill \\ \end{gathered} \right.$

Following is our main result

Lemma 1. The following identities $\displaystyle\begin{gathered} \left( {1 - \frac{n}{2}} \right)\int_{{B_r}(0)} {|\nabla {u_1}{|^2}dx} + r\left[ {\frac{1}{2}\int_{\partial {B_r}(0)} {|\nabla {u_1}{|^2}d\sigma } - \int_{\partial {B_r}(0)} {{{\left| {\frac{{\partial {u_1}}}{{\partial \nu }}} \right|}^2}d\sigma } } \right] \hfill \\ \qquad\qquad= - 2n\int_{{B_r}(0)} {{e^{{u_1}}}dx} + 2\int_{\partial {B_r}(0)} {r{e^{{u_1}}}dx} - \int_{{B_r}(0)} {{e^{{u_2}}}(x\cdot\nabla {u_1})dx} , \hfill \\ \left( {1 - \frac{n}{2}} \right)\int_{{B_r}(0)} {|\nabla {u_2}{|^2}dx} + r\left[ {\frac{1}{2}\int_{\partial {B_r}(0)} {|\nabla {u_2}{|^2}d\sigma } - \int_{\partial {B_r}(0)} {{{\left| {\frac{{\partial {u_2}}}{{\partial \nu }}} \right|}^2}d\sigma } } \right] \hfill \\ \qquad\qquad= - 2n\int_{{B_r}(0)} {{e^{{u_2}}}dx} + 2\int_{\partial {B_r}(0)} {r{e^{{u_2}}}dx} - \int_{{B_r}(0)} {{e^{{u_1}}}(x\cdot\nabla {u_2})dx} , \hfill \\ \end{gathered}$

hold.

Proof. Multiply the first equation by $x\cdot \nabla u_1$ and integrate over the ball $B_r(0)$ $\displaystyle -\int_{{B_r}(0)} {\Delta {u_1}(x \cdot \nabla {u_1})dx} = \int_{{B_r}(0)} {\left( {2{e^{{u_1}}} - {e^{{u_2}}}} \right)(x \cdot \nabla {u_1})dx}$.

With the fact that $\displaystyle 2{e^{{u_1}}}(x\cdot\nabla {u_1}) = 2\nabla {e^{{u_1}}}\cdot x$

the RHS can be estimated as follows $\displaystyle\begin{gathered} \int_{{B_r}(0)} {\left( {2{e^{{u_1}}} - {e^{{u_2}}}} \right)(x\cdot\nabla {u_1})dx} = \int_{{B_r}(0)} {2{e^{{u_1}}}(x\cdot\nabla {u_1})dx} - \int_{{B_r}(0)} {{e^{{u_2}}}(x\cdot\nabla {u_1})dx} \hfill \\ \qquad= 2\int_{{B_r}(0)} {\nabla {e^{{u_1}}}\cdot xdx} - \int_{{B_r}(0)} {{e^{{u_2}}}(x\cdot\nabla {u_1})dx} \hfill \\ \qquad= 2\left( {\int_{\partial {B_r}(0)} {{e^{{u_1}}}x\cdot\nu d\sigma} - n\int_{{B_r}(0)} {{e^{{u_1}}}dx} } \right) - \int_{{B_r}(0)} {{e^{{u_2}}}(x\cdot\nabla {u_1})dx} \hfill \\ \qquad= 2\left( {\int_{\partial {B_r}(0)} {r{e^{{u_1}}}d\sigma} - n\int_{{B_r}(0)} {{e^{{u_1}}}dx} } \right) - \int_{{B_r}(0)} {{e^{{u_2}}}(x\cdot\nabla {u_1})dx} \hfill \\ \qquad= - 2n\int_{{B_r}(0)} {{e^{{u_1}}}dx} + 2\int_{\partial {B_r}(0)} {r{e^{{u_1}}}d\sigma} - \int_{{B_r}(0)} {{e^{{u_2}}}(x\cdot\nabla {u_1})dx.} \hfill \\ \end{gathered}$

The left hand side also can be estimated, we refer the reader to this topic for details $\displaystyle - \int_{{B_r}(0)} {\Delta {u_1}(x \cdot \nabla {u_1})dx} = \left( {1 - \frac{n}{2}} \right)\int_{{B_r}(0)} {|\nabla u{|^2}dx} + r\left[ {\frac{1}{2}\int_{\partial {B_r}(0)} {|\nabla {u_1}{|^2}d\sigma } - \int_{\partial {B_r}(0)} {{{\left| {\frac{{\partial {u_1}}}{{\partial \nu }}} \right|}^2}d\sigma } } \right]$.

Thus we firstly obtain $\displaystyle\begin{gathered} \left( {1 - \frac{n}{2}} \right)\int_{{B_r}(0)} {|\nabla u_1|^2dx} + r\left[ {\frac{1}{2}\int_{\partial {B_r}(0)} {|\nabla {u_1}{|^2}d\sigma } - \int_{\partial {B_r}(0)} {{{\left| {\frac{{\partial {u_1}}}{{\partial \nu }}} \right|}^2}d\sigma } } \right] \hfill \\ \qquad\qquad= - 2n\int_{{B_r}(0)} {{e^{{u_1}}}dx} + 2\int_{\partial {B_r}(0)} {r{e^{{u_1}}}d\sigma} - \int_{{B_r}(0)} {{e^{{u_2}}}(x\cdot\nabla {u_1})dx} \hfill \\ \end{gathered}$

Similarly, we also obtain $\displaystyle\begin{gathered} \left( {1 - \frac{n}{2}} \right)\int_{{B_r}(0)} {|\nabla u_2|^2dx} + r\left[ {\frac{1}{2}\int_{\partial {B_r}(0)} {|\nabla {u_2}|^2d\sigma } - \int_{\partial {B_r}(0)} {{{\left| {\frac{{\partial {u_2}}}{{\partial \nu }}} \right|}^2}d\sigma } } \right] \hfill \\ \qquad\qquad= - 2n\int_{{B_r}(0)} {{e^{{u_2}}}dx} + 2\int_{\partial {B_r}(0)} {r{e^{{u_2}}}d\sigma} - \int_{{B_r}(0)} {{e^{{u_1}}}(x\cdot\nabla {u_2})dx} \hfill \\ \end{gathered}$

The proof follows.

Lemma 2. The following identities $\displaystyle\begin{gathered} \int_{\partial {B_r}(0)} {r\frac{{\partial {u_1}}}{{\partial \nu }}\frac{{\partial {u_2}}}{{\partial \nu }}d\sigma } + \int_{{B_r}(0)} {\nabla {u_1}\cdot\nabla {u_2}dx} + \int_{{B_r}(0)} {x \cdot \sum\limits_{j = 1}^n {\nabla({\nabla _j}{u_2}){\nabla _j}{u_1}} dx} \hfill \\ \qquad= 2\int_{{B_r}(0)} {{e^{{u_1}}}(x\cdot\nabla {u_2})dx} + n\int_{{B_r}(0)} {{e^{{u_2}}}dx} - \int_{\partial {B_r}(0)} {r{e^{{u_2}}}d\sigma } , \hfill \\ \int_{\partial {B_r}(0)} {r\frac{{\partial {u_1}}}{{\partial \nu }}\frac{{\partial {u_2}}}{{\partial \nu }}d\sigma } + \int_{{B_r}(0)} {\nabla {u_1}\cdot\nabla {u_2}dx} + \int_{{B_r}(0)} {x \cdot \sum\limits_{j = 1}^n {\nabla({\nabla _j}{u_1}){\nabla _j}{u_2}} dx} \hfill \\ \qquad= 2\int_{{B_r}(0)} {{e^{{u_2}}}(x\cdot\nabla {u_1})dx} + n\int_{{B_r}(0)} {{e^{{u_1}}}dx} - \int_{\partial {B_r}(0)} {r{e^{{u_1}}}d\sigma } , \hfill \\ \end{gathered}$

hold.

Proof. Multiply the first equation by $x\cdot \nabla u_2$ and integrate over the ball $B_r(0)$ $\displaystyle -\int_{{B_r}(0)} {\Delta {u_1}(x \cdot \nabla {u_2})dx} = \int_{{B_r}(0)} {\left( {2{e^{{u_1}}} - {e^{{u_2}}}} \right)(x \cdot \nabla {u_2})dx}$.

Regarding to the RHS, we get $\displaystyle\begin{gathered} \int_{{B_r}(0)} {\left( {2{e^{{u_1}}} - {e^{{u_2}}}} \right)(x\cdot\nabla {u_2})dx} = \int_{{B_r}(0)} {2{e^{{u_1}}}(x\cdot\nabla {u_2})dx} - \int_{{B_r}(0)} {{e^{{u_2}}}(x\cdot\nabla {u_2})dx} \hfill \\ \qquad= 2\int_{{B_r}(0)} {{e^{{u_1}}}(x\cdot\nabla {u_2})dx} - \int_{{B_r}(0)} {{e^{{u_2}}}(x\cdot\nabla {u_2})dx} \hfill \\ \qquad= 2\int_{{B_r}(0)} {{e^{{u_1}}}(x\cdot\nabla {u_2})dx} - \left( {\int_{\partial {B_r}(0)} {{e^{{u_2}}}x\cdot\nu d\sigma } - n\int_{{B_r}(0)} {{e^{{u_2}}}dx} } \right) \hfill \\ \qquad= 2\int_{{B_r}(0)} {{e^{{u_1}}}(x\cdot\nabla {u_2})dx} + n\int_{{B_r}(0)} {{e^{{u_2}}}dx} - \int_{\partial {B_r}(0)} {r{e^{{u_2}}}d\sigma } . \hfill \\ \end{gathered}$

Involving the LHS, $\displaystyle\begin{gathered} - \int_{{B_r}(0)} {\Delta {u_1}(x\cdot\nabla {u_2})dx} = - \sum\limits_{i = 1}^n {\int_{{B_r}(0)} {{{({u_1})}_{{x_i}{x_i}}}(x\cdot\nabla {u_2})dx} } \hfill \\ \qquad= - \sum\limits_{i = 1}^n {\left[ {\int_{\partial {B_r}(0)} {{{({u_1})}_{{x_i}}}(x\cdot\nabla {u_2}){\nu ^i}d\sigma } - \int_{{B_r}(0)} {{{({u_1})}_{{x_i}}}{{(x\cdot\nabla {u_2})}_{{x_i}}}dx} } \right]} \hfill \\ \qquad= - \sum\limits_{i = 1}^n {\left[ {\int_{\partial {B_r}(0)} {r{\nu ^i}{{({u_1})}_{{x_i}}}(\nu \cdot\nabla {u_2})d\sigma } - \sum\limits_{j = 1}^n {\int_{{B_r}(0)} {{{({u_1})}_{{x_i}}}{{({x_j}{{({u_2})}_{{x_j}}})}_{{x_i}}}dx} } } \right]} \hfill \\ \qquad= - \sum\limits_{i = 1}^n {\left[ {\int_{\partial {B_r}(0)} {r(\nu \cdot\nabla {u_1})(x\cdot\nabla {u_2})d\sigma } - \sum\limits_{j = 1}^n {\int_{{B_r}(0)} {{{({u_1})}_{{x_i}}}{{({x_j}{{({u_2})}_{{x_j}}})}_{{x_i}}}dx} } } \right]} \hfill \\ \qquad= \int_{\partial {B_r}(0)} {r\frac{{\partial {u_1}}}{{\partial \nu }}\frac{{\partial {u_2}}}{{\partial \nu }}d\sigma } + \int_{{B_r}(0)} {\nabla {u_1}\cdot\nabla {u_2}dx} + \int_{{B_r}(0)} {x \cdot \sum\limits_{j = 1}^n {\nabla({\nabla _j}{u_2}){\nabla _j}{u_1}} dx} . \hfill \\ \end{gathered}$

Thus $\displaystyle\begin{gathered} \int_{\partial {B_r}(0)} {r\frac{{\partial {u_1}}}{{\partial \nu }}\frac{{\partial {u_2}}}{{\partial \nu }}d\sigma } + \int_{{B_r}(0)} {\nabla {u_1}\cdot\nabla {u_2}dx} + \int_{{B_r}(0)} {x \cdot \sum\limits_{j = 1}^n {\nabla({\nabla _j}{u_2}){\nabla _j}{u_1}} dx} \hfill \\ \qquad= 2\int_{{B_r}(0)} {{e^{{u_1}}}(x\cdot\nabla {u_2})dx} + n\int_{{B_r}(0)} {{e^{{u_2}}}dx} - \int_{\partial {B_r}(0)} {r{e^{{u_2}}}d\sigma } . \hfill \\ \end{gathered}$

Similarly, $\displaystyle\begin{gathered} \int_{\partial {B_r}(0)} {r\frac{{\partial {u_1}}}{{\partial \nu }}\frac{{\partial {u_2}}}{{\partial \nu }}d\sigma } + \int_{{B_r}(0)} {\nabla {u_1}\cdot\nabla {u_2}dx} + \int_{{B_r}(0)} {x \cdot \sum\limits_{j = 1}^n {\nabla({\nabla _j}{u_1}){\nabla _j}{u_2}} dx} \hfill \\ \qquad= 2\int_{{B_r}(0)} {{e^{{u_2}}}(x\cdot\nabla {u_1})dx} + n\int_{{B_r}(0)} {{e^{{u_1}}}dx} - \int_{\partial {B_r}(0)} {r{e^{{u_1}}}d\sigma } . \hfill \\ \end{gathered}$

The proof follows.

We are now in a position to derive the Pohozaev identity for Toda system in $\mathbb R^2$.

Theorem. The following identity $\displaystyle\begin{gathered} \int_{\partial {B_r}} {r\left( {{e^{{u_1}}} + {e^{{u_2}}}} \right)d\sigma } - 2\int_{{B_r}} {\left( {{e^{{u_1}}} + {e^{{u_2}}}} \right)dx} \hfill \\ \qquad= \frac{2}{3}\left[ {\int_{\partial {B_r}} {r\left( {{{\left| {\frac{{\partial {u_1}}}{{\partial \nu }}} \right|}^2} + {{\left| {\frac{{\partial {u_2}}}{{\partial \nu }}} \right|}^2} - \frac{1}{2}{{\left| {\nabla {u_1}} \right|}^2} - \frac{1}{2}{{\left| {\nabla {u_2}} \right|}^2}} \right)d\sigma - \int_{\partial {B_r}} {r\left( {\frac{{\partial {u_1}}}{{\partial \nu }}\frac{{\partial {u_2}}}{{\partial \nu }} - \frac{1}{2}\nabla {u_1} \cdot \nabla {u_2}} \right)d\sigma } } } \right] \hfill \\ \end{gathered}$

holds.

Proof. The proof is just an application of Lemmas 1 and 2, we omit it here.

As an application to this theorem, we are going to derive a priori estimate for solutions of Toda systems in $\mathbb R^2$. Firstly, we introduce the following notations. Denote $\displaystyle {\alpha _i} = \int_{{\mathbb{R}^2}} {{e^{{u_i}}}dx} , \quad i = 1,2$.

We then have the following result.

Corollary. We have $\displaystyle \alpha _1^2 + \alpha _2^2 - {\alpha _1}{\alpha _2} = 4\pi \left( {{\alpha _1} + {\alpha _2}} \right)$.

Proof. By the theorem above and by letting $r\to\infty$ the proof follows.

Remark. In this entry, we introduce a new techique to deal with $\displaystyle\int_{{B_r}(0)} {\Delta u(x\cdot\nabla u)dx}$

so actually lots of Pohozaev identities mentioned before can be simplified a little bit.

See also:

## 1 Comment »

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