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

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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