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:

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