Ngô Quốc Anh

April 18, 2010

The Pohozaev identity: Elliptic problem with biharmonic operator

Filed under: PDEs — Tags: — Ngô Quốc Anh @ 14:19

We now consider another kind of problem involving biharmonic operator. Let us assume u>0 a solution of the equation

\displaystyle (\Delta^2u)(x)+Q(x)f(u(x))=0

in \mathbb R^n. We shall prove the following result

Theorem. The following identity

\displaystyle\begin{gathered} \frac{3}{2}\left[ {\int_{{B_r}(0)}  {uQ(x)f(u)dx} - \int_{\partial {B_r}(0)} {u\frac{{\partial \Delta  u}}{{\partial \nu }}d\sigma } + \int_{\partial {B_r}(0)}  {\frac{{\partial u}}{{\partial \nu }}\Delta ud\sigma } } \right] +  \hfill \\ \frac{1}{2}\int_{\partial {B_r}(0)} {r{{(\Delta u)}^2}d\sigma }  + \int_{{B_r}(0)} {(x\cdot\nabla (\Delta u))\Delta udx} -  \int_{\partial {B_r}(0)} {\left[ {\Delta u\frac{{\partial (x\cdot\nabla  u)}}{{\partial \nu }} - (x\cdot\nabla u)\frac{{\partial \Delta  u}}{{\partial \nu }}} \right]d\sigma } \hfill \\ \qquad\qquad=  \int_{\partial {B_r}(0)} {ruQ(x)f(u)d\sigma } - \int_{{B_r}(0)} {u\left(  {Q(x)f(u) + (x\cdot\nabla Q)f(u) + (x\cdot\nabla u)Q(x)f'(u)}  \right)dx}\hfill \\ \end{gathered}

holds.

Proof. Multiplying the PDE by x \cdot \nabla u on the both sides

\displaystyle (x \cdot \nabla u){\Delta ^2}u + (x \cdot \nabla u)Q(x)f(u) = 0

and integrating the resulting equation over the ball B_r(0), one has

\displaystyle\int_{{B_r}(0)} {(x \cdot \nabla u){\Delta ^2}udx} + \int_{{B_r}(0)} {(x \cdot \nabla u)Q(x)f(u)dx} = 0.

Clearly, by the divergence theorem, one gets

\displaystyle\begin{gathered} \int_{{B_r}(0)} {(x \cdot \nabla u)Q(x)f(u)dx} = \sum\limits_{i = 1}^n {\int_{{B_r}(0)} {{u_{{x_i}}}{x_i}Q(x)f(u)dx} } \hfill \\ \qquad= \sum\limits_{i = 1}^n {\left[ {\int_{\partial {B_r}(0)} {{\nu ^i}u{x_i}Q(x)f(u)d\sigma } - \int_{{B_r}(0)} {u{{\left( {{x_i}Q(x)f(u)} \right)}_{{x_i}}}dx} } \right]} \hfill \\ \qquad= \sum\limits_{i = 1}^n {\left[ {\int_{\partial {B_r}(0)} {\frac{{{x_i}}}{r}u{x_i}Q(x)f(u)d\sigma } - \int_{{B_r}(0)} {u\left( {Q(x)f(u) + {x_i}{Q_{{x_i}}}f(u) + {x_i}Q(x)f'(u){u_{{x_i}}}} \right)dx} } \right]} \hfill \\ \qquad= \boxed{\int_{\partial {B_r}(0)} {ruQ(x)f(u)d\sigma } - \int_{{B_r}(0)} {u\left( {Q(x)f(u) + (x \cdot \nabla Q)f(u) + (x \cdot \nabla u)Q(x)f'(u)} \right)dx}}. \hfill \\ \end{gathered}

Regarding to the first term we still use the divergence theorem, however, we are dealing with the biharmonic operator \Delta^2 u which is defined to be \Delta(\Delta u). We have

\displaystyle\begin{gathered} \int_{{B_r}(0)} {(x \cdot \nabla u){\Delta ^2}udx} = - \int_{{B_r}(0)} {\nabla (x \cdot \nabla u) \cdot \nabla (\Delta u)dx} + \int_{\partial {B_r}(0)} {(x \cdot \nabla u)\frac{{\partial \Delta u}}{{\partial \nu }}d\sigma } \hfill \\ \qquad= \int_{{B_r}(0)} {\Delta (x \cdot \nabla u)\Delta udx} - \int_{\partial {B_r}(0)} {\Delta u\frac{{\partial (x \cdot \nabla u)}}{{\partial \nu }}d\sigma } + \int_{\partial {B_r}(0)} {(x \cdot \nabla u)\frac{{\partial \Delta u}}{{\partial \nu }}d\sigma } \hfill \\ \qquad= \int_{{B_r}(0)} {\Delta (x \cdot \nabla u)\Delta udx} - \int_{\partial {B_r}(0)} {\left[ {\Delta u\frac{{\partial (x \cdot \nabla u)}}{{\partial \nu }} - (x \cdot \nabla u)\frac{{\partial \Delta u}}{{\partial \nu }}} \right]d\sigma } . \hfill \\ \end{gathered}

Note that

\displaystyle\begin{gathered} \Delta (x \cdot \nabla u) = \sum\limits_{j = 1}^n {\sum\limits_{i = 1}^n {{{\left( {{x_i}{u_{{x_i}}}} \right)}_{{x_j}{x_j}}}} } \hfill \\ \qquad= \sum\limits_{j = 1}^n {{{\left( {{u_{{x_j}}} + \sum\limits_{i = 1}^n {{x_i}{u_{{x_i}{x_j}}}} } \right)}_{{x_j}}}} \hfill \\ \qquad= \sum\limits_{j = 1}^n {\left( {{u_{{x_j}{x_j}}} + {{\left( {\sum\limits_{i = 1}^n {{x_i}{u_{{x_i}{x_j}}}} } \right)}_{{x_j}}}} \right)} \hfill \\ \qquad= \sum\limits_{j = 1}^n {\left( {{u_{{x_j}{x_j}}} + {u_{{x_j}{x_j}}} + \left( {\sum\limits_{i = 1}^n {{x_i}{u_{{x_i}{x_j}{x_j}}}} } \right)} \right)} \hfill \\ \qquad= 2\Delta u + (x \cdot \nabla (\Delta u)). \hfill \\ \end{gathered}

Therefore

\displaystyle\int_{{B_r}(0)} {\Delta (x \cdot \nabla u)\Delta udx} = 2\int_{{B_r}(0)} {{{(\Delta u)}^2}dx} + \int_{{B_r}(0)} {(x \cdot \nabla (\Delta u))\Delta udx} .

We actually can estimate further as follows

\displaystyle\begin{gathered} \int_{{B_r}(0)} {(x \cdot \nabla (\Delta u))\Delta udx} = \int_{{B_r}(0)} {\sum\limits_{i = 1}^n {({x_i}{{(\Delta u)}_{{x_i}}}\Delta u)} dx} \hfill \\ \qquad= \sum\limits_{i = 1}^n {\int_{{B_r}(0)} {({x_i}\Delta u){{(\Delta u)}_{{x_i}}}dx} } \hfill \\ \qquad= \sum\limits_{i = 1}^n {\left[ { - \int_{{B_r}(0)} {{{({x_i}\Delta u)}_{{x_i}}}(\Delta u)dx} + \int_{\partial {B_r}(0)} {{\nu ^i}({x_i}\Delta u)(\Delta u)d\sigma } } \right]} \hfill \\ \qquad= \sum\limits_{i = 1}^n {\left[ { - \int_{{B_r}(0)} {(\Delta u + {x_i}{{(\Delta u)}_{{x_i}}})(\Delta u)dx} + \int_{\partial {B_r}(0)} {\frac{{x_i^2}}{r}{{(\Delta u)}^2}d\sigma } } \right]} \hfill \\ \qquad= - \int_{{B_r}(0)} {{{(\Delta u)}^2}dx} - \int_{{B_r}(0)} {(x \cdot \nabla (\Delta u))\Delta udx} + \int_{\partial {B_r}(0)} {r{{(\Delta u)}^2}d\sigma } \hfill \\ \end{gathered}

which follows

\displaystyle\int_{{B_r}(0)} {(x \cdot \nabla (\Delta u))\Delta udx} = \frac{1}{2}\int_{\partial {B_r}(0)} {r{{(\Delta u)}^2}d\sigma } - \frac{1}{2}\int_{{B_r}(0)} {{{(\Delta u)}^2}dx} .

Hence

\displaystyle\begin{gathered} \int_{{B_r}(0)} {(x \cdot \nabla u){\Delta ^2}udx} \hfill \\ \qquad\qquad= \boxed{\frac{3}{2}\int_{{B_r}(0)} {{{(\Delta u)}^2}dx} +\frac{1}{2}\int_{\partial {B_r}(0)} {r{{(\Delta u)}^2}d\sigma }+ \int_{{B_r}(0)} {(x \cdot \nabla (\Delta u))\Delta udx} - \int_{\partial {B_r}(0)} {\left[ {\Delta u\frac{{\partial (x \cdot \nabla u)}}{{\partial \nu }} - (x \cdot \nabla u)\frac{{\partial \Delta u}}{{\partial \nu }}} \right]d\sigma}. } \hfill \\ \end{gathered}

Multiply the PDE by u on both sides and integrating it over the ball B_r(0) together with the fact that

\displaystyle - \int_{{B_r}(0)} {f\Delta gdx} = \int_{{B_r}(0)} {\nabla f \cdot \nabla gdx} - \int_{\partial {B_r}(0)} {f\frac{{\partial g}}{{\partial \nu }}d\sigma }

we obtain

\displaystyle\begin{gathered} \int_{{B_r}(0)} {{{(\Delta u)}^2}dx} = \int_{{B_r}(0)} {\Delta u\Delta udx} \hfill \\ \qquad= - \int_{{B_r}(0)} {\nabla u \cdot \nabla \Delta udx} + \int_{\partial {B_r}(0)} {\frac{{\partial u}}{{\partial \nu }}\Delta ud\sigma } \hfill \\ \qquad= \int_{{B_r}(0)} {u{\Delta ^2}udx} - \int_{\partial {B_r}(0)} {u\frac{{\partial \Delta u}}{{\partial \nu }}d\sigma } + \int_{\partial {B_r}(0)} {\frac{{\partial u}}{{\partial \nu }}\Delta ud\sigma } \hfill \\ \qquad= \boxed{\int_{{B_r}(0)} {uQ(x)f(u)dx} - \int_{\partial {B_r}(0)} {u\frac{{\partial \Delta u}}{{\partial \nu }}d\sigma } + \int_{\partial {B_r}(0)} {\frac{{\partial u}}{{\partial \nu }}\Delta ud\sigma }} . \hfill \\ \end{gathered}

Combining all gives

\displaystyle\begin{gathered} \frac{3}{2}\left[ {\int_{{B_r}(0)} {uQ(x)f(u)dx} - \int_{\partial {B_r}(0)} {u\frac{{\partial \Delta u}}{{\partial \nu }}d\sigma } + \int_{\partial {B_r}(0)} {\frac{{\partial u}}{{\partial \nu }}\Delta ud\sigma } } \right] + \hfill \\ \frac{1}{2}\int_{\partial {B_r}(0)} {r{{(\Delta u)}^2}d\sigma } + \int_{{B_r}(0)} {(x\cdot\nabla (\Delta u))\Delta udx} - \int_{\partial {B_r}(0)} {\left[ {\Delta u\frac{{\partial (x\cdot\nabla u)}}{{\partial \nu }} - (x\cdot\nabla u)\frac{{\partial \Delta u}}{{\partial \nu }}} \right]d\sigma } \hfill \\ \qquad\qquad= \int_{\partial {B_r}(0)} {ruQ(x)f(u)d\sigma } - \int_{{B_r}(0)} {u\left( {Q(x)f(u) + (x\cdot\nabla Q)f(u) + (x\cdot\nabla u)Q(x)f'(u)} \right)dx} . \hfill \\ \end{gathered}

Cf. Exact solutions of nonlinear conformally invariant integral equations in \mathbb R^3Adv. Math. 194 (2005),  no. 2, pp. 485-503.

See also:

2 Comments »

  1. Hello!

    I am interested in the Pohozaev identity but not good at it. Please say more about its applications! It seems that it is basically used to prove non-existence?

    Thanks.

    Comment by unknown — May 3, 2011 @ 21:54

    • You are right, this is a great tool for proving non-existence result. The simplest example, originally due to Pohozaev, can be found here http://wp.me/p4uAr-Nu.

      Comment by Ngô Quốc Anh — May 3, 2011 @ 22:28


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