# Ngô Quốc Anh

## April 13, 2010

### The Pohozaev identity: Semilinear elliptic problem with exponential nonlinearity

Filed under: PDEs — Tags: — Ngô Quốc Anh @ 2:07

We know consider another type of equation, precisely, we consider the positive solution to the following $\displaystyle -\Delta u=Ke^{2u}$

in $\mathbb R^n$. Following is what we need to prove.

Theorem. The following identity $\displaystyle\begin{gathered} \left( {1 + \frac{n}{2}} \right)\int_{{B_r}(0)} {uK{e^{2u}}dx} + \int_{{B_r}(0)} {u\left( {(x \cdot \nabla K){e^{2u}} + 2(x \cdot \nabla u)K{e^{2u}}} \right)dx} \hfill \\ \qquad= \int_{\partial {B_r}(0)} {ruK{e^{2u}}d\sigma } - \int_{\partial {B_r}(0)} {\left[ {\left( {1 - \frac{n}{2}} \right)u\frac{{\partial u}}{{\partial \nu }} + \frac{1}{2}r|\nabla u{|^2} - r{{\left| {\frac{{\partial u}}{{\partial \nu }}} \right|}^2}} \right]d\sigma } \hfill \\ \end{gathered}$

holds.

Proof. We multiply the PDE by $x \cdot \nabla u$ and integrate over $B_r(0)$ to find $\displaystyle\int_{{B_r}(0)} { - \Delta u(x \cdot \nabla u)dx} = \int_{{B_r}(0)} {(x \cdot \nabla u)K{e^{2u}}dx}$.

Followed by a calculation in this topic, the LHS is already calculated which is nothing but the following $\displaystyle LHS = \underbrace {\sum\limits_{j = 1}^n {\sum\limits_{i = 1}^n {\int_{{B_r}(0)} {{u_{{x_i}}}{{({x_j}{u_{{x_j}}})}_{{x_i}}}dx} } } }_{{A_1}} - \underbrace {\sum\limits_{j = 1}^n {\sum\limits_{i = 1}^n {\int_{\partial {B_r}(0) } {{u_{{x_i}}}{\nu ^i}{x_j}{u_{{x_j}}}d\sigma } } } }_{{A_2}}$

where $\displaystyle {A_1} = \left( {1 - \frac{n}{2}} \right)\int_{{B_r}(0)} {|\nabla u{|^2}dx} + \frac{1}{2}\int_{\partial {B_r}(0)} {|\nabla u{|^2}\left( {\nu \cdot x} \right)d\sigma }$.

In this situation, $\displaystyle\nu \cdot x = \frac{x}{r} \cdot x = r$

which helps us to write down $\displaystyle A_1 = \left( {1 - \frac{n}{2}} \right)\int_{{B_r}(0)} {|\nabla u{|^2}dx} + \frac{1}{2}\int_{\partial {B_r}(0)} {r|\nabla u{|^2}d\sigma }$.

Regarding to $A_2$, clearly $\displaystyle\begin{gathered} \int_{\partial {B_r}(0)} {\nabla u \cdot \nu x \cdot \nabla ud\sigma } = \int_{\partial {B_r}(0)} {\nabla u \cdot \frac{x}{r}x \cdot \nabla ud\sigma } \hfill \\ \qquad= r\int_{\partial {B_r}(0)} {\nabla u \cdot \nu \nu \cdot \nabla ud\sigma } \hfill \\ \qquad= r\int_{\partial {B_r}(0)} {{{\left| {\nabla u \cdot \nu } \right|}^2}d\sigma } \hfill \\ \qquad= r\int_{\partial {B_r}(0)} {{{\left| {\frac{{\partial u}}{{\partial \nu }}} \right|}^2}d\sigma } \hfill \\ \end{gathered}$

so $\displaystyle LHS = \left( {1 - \frac{n}{2}} \right)\int_{{B_r}(0)} {|\nabla u{|^2}dx} + \frac{1}{2}\int_{\partial {B_r}(0)} {r|\nabla u{|^2}d\sigma } - \int_{\partial {B_r}(0)} {{{r\left| {\frac{{\partial u}}{{\partial \nu }}} \right|}^2}d\sigma }$.

Furthermore, the Green identity tells us that $\displaystyle\int_{{B_r}(0)} {|\nabla u{|^2}dx} = \int_{{B_r}(0)} {uK{e^{2u}}dx} + \int_{\partial {B_r}(0)} {u\frac{{\partial u}}{{\partial \nu }}d\sigma }$.

Thus $\displaystyle LHS = \left( {1 - \frac{n}{2}} \right)\int_{{B_r}(0)} {uK{e^{2u}}dx} + \int_{\partial {B_r}(0)} {\left[ {\left( {1 - \frac{n}{2}} \right)u\frac{{\partial u}}{{\partial \nu }} + \frac{1}{2}r|\nabla u{|^2} - r{{\left| {\frac{{\partial u}}{{\partial \nu }}} \right|}^2}} \right]d\sigma }$.

We next evaluate the RHS. Indeed, $\displaystyle\begin{gathered} \int_{{B_r}(0)} {(x \cdot \nabla u)K{e^{2u}}dx} = \sum\limits_{i = 1}^n {\int_{{B_r}(0)} {{u_{{x_i}}}{x_i}K{e^{2u}}dx} } \hfill \\ \qquad= \sum\limits_{i = 1}^n {\left( {\int_{\partial {B_r}(0)} {{\nu ^i}u{x_i}K{e^{2u}}d\sigma } - \int_{{B_r}(0)} {u{{\left( {{x_i}K{e^{2u}}} \right)}_{{x_i}}}dx} } \right)} \hfill \\\qquad = \sum\limits_{i = 1}^n {\left( {\int_{\partial {B_r}(0)} {\frac{{{x_i}}}{r}u{x_i}K{e^{2u}}d\sigma } - \int_{{B_r}(0)} {u\left( {K{e^{2u}} + {x_i}{K_{{x_i}}}{e^{2u}} + 2{x_i}K{e^{2u}}{u_{{x_i}}}} \right)dx} } \right)} \hfill \\ \qquad= \int_{\partial {B_r}(0)} {ruK{e^{2u}}d\sigma } - \int_{{B_r}(0)} {u\left( {nK{e^{2u}} + (x \cdot \nabla K){e^{2u}} + 2(x \cdot \nabla u)K{e^{2u}}} \right)dx}. \hfill \\ \end{gathered}$

Combining all gives $\displaystyle\begin{gathered} \left( {1- \frac{n}{2}} \right)\int_{{B_r}(0)} {uK{e^{2u}}dx} + \int_{{B_r}(0)} {u\left( nKe^{2u}+ {(x \cdot \nabla K){e^{2u}} + 2(x \cdot \nabla u)K{e^{2u}}} \right)dx} \hfill \\ \qquad= \int_{\partial {B_r}(0)} {ruK{e^{2u}}d\sigma } - \int_{\partial {B_r}(0)} {\left[ {\left( {1 - \frac{n}{2}} \right)u\frac{{\partial u}}{{\partial \nu }} + \frac{1}{2}r|\nabla u{|^2} - r{{\left| {\frac{{\partial u}}{{\partial \nu }}} \right|}^2}} \right]d\sigma }. \hfill \\ \end{gathered}$

In $\mathbb R^2$, the above identity is just $\displaystyle\int_{{B_r}(0)} {u\left( 2Ke^{2u}+ {(x\cdot\nabla K){e^{2u}} + 2(x\cdot\nabla u)K{e^{2u}}} \right)dx} = \int_{\partial {B_r}(0)} {ruK{e^{2u}}d\sigma } - \int_{\partial {B_r}(0)} {\left( {\frac{1}{2}r|\nabla u{|^2} - r{{\left| {\frac{{\partial u}}{{\partial \nu }}} \right|}^2}} \right)d\sigma }$.

The left hand side of the previous identity can be simplified as follows $\displaystyle\int_{{B_r}(0)} {u\left( 2Ke^{2u}+ {(x\cdot\nabla K){e^{2u}} + 2(x\cdot\nabla u)K{e^{2u}}} \right)dx} =\int_{B_r(0)} u\left( 2Ke^{2u} +x\cdot\nabla (Ke^{2u})\right)dx.$

NB: I thank Prof. Fontana Luigi for pointing out a missing term in the above derivation.

## 12 Comments »

1. Good job! Your post is excellent and useful 🙂

Comment by Hong — April 14, 2010 @ 0:01

2. Hi Quoc Anh,

Your calculations interest me. I prefer reading the details like this. Thank you for your contribution.

However I am quite stupid about this topic. Would you teach me some?

Could you tell me what the Pohozaev identity for a PDE is? Is it an integral form of the PDE? How is it applied to study the PDE?

Looking forward to hearing from you.

Best regards,

HasPas

Ps: You are very professional.

Comment by HasPas — April 16, 2010 @ 0:44

• Thanks for the comment. Roughly speaking, the Pohozaev identities are strong tools to prove some non-existence result in PDE. As can be seen from this topic, we actually have a non-existence result provided $p>\frac{n+2}{n-2}$. I do think we have no precisely definition for the Pohozaev identities. More or less, these identities have less differentiation and get involved the boundary instead of the whole domain.

Comment by Ngô Quốc Anh — April 16, 2010 @ 0:53

3. I was looking for the definition, but I have not found it.

Comment by HasPas — April 16, 2010 @ 0:47

4. Many thanks for your answer.

At first I misunderstood it would be a weak form. Then it interests me, since it is actually a strong one.

In the next entries, if possible, I would like you to go through more details. I am now following this series. I do believe I am learning much from you.

Comment by HasPas — April 16, 2010 @ 1:02

• There is another entry involving the Pohozaev identites that is going to appear shortly. The equation now involves the biharmonic operator.

Comment by Ngô Quốc Anh — April 16, 2010 @ 23:14

5. For the equation above, the Pohozaev identity seems complicated. Could you please explain how it can be applied to conclude the non-existence of positive solutions of the equation concerned? Thank you in advance.

Comment by HasPas — April 16, 2010 @ 22:00

• You was right, originally at the time when Pohozaev created his own identity, it was helped him to prove some non-existence result as I mentioned before. Now it is not really just an application to the proof of non-existence. I strongly recommend you to read the paper mentioned above. If I have time, I will run out this stuff in details. Unfortunately, I am in the mess of things.

Honestly, I have not touched these identities before although I know them. Sometimes, we need to work on these calculations in order to fully understand their meaning. I just used my time to work on these stuff last week as I need to get some feelings. As can be seen, all are quite standard and just an easy-calculation.

Anyhow, hope it helps you a little bit.

Comment by Ngô Quốc Anh — April 16, 2010 @ 23:10

6. Thank you very much for your kindness.

Surely, your entries help me much. I love the phrase of yourself “I need to get some feelings”.

I had an overlook at “Math. Ann. 308 (1997), pp. 119-139”. But it seems there are no claims regarding the Pohozaev identity for the equation concern? Please accept my apology if I had been careless.

Comment by HasPas — April 16, 2010 @ 23:26

• Thank for encouraging me. As can be seen I am trying to create several series of topics. I do think it is not easy to see the use of the Pohozaev identity in that paper. I will post some thing related to this once I have enough connection to other topics.

Obviously, this is just a blog so I cannot post something which is too large, too difficult, too complicated. So far I just recall some beautiful results and put them here or I just cut out some basic steps in some particular papers and post them here so that people are able to fully understand.

I also want to let this blog open so that once people want some stuffs in details, I can do it without any difficulties.

Comment by Ngô Quốc Anh — April 17, 2010 @ 0:28

7. Nice job! but it seems that in the 2 dimensional case you forgot the first term in the LHS. That term is also missing in the last equality for the n dimensional case. I mention the issue to warn people that might be looking for just the
2 dimensional case and are in such a hurry not to read the statement of the Theorem!
You have a very good blog. Keep doing it!

Comment by Luigi — August 29, 2015 @ 6:33

• Dear Prof. Luigi Fontana,

Thanks for your interest in my blog and for pointing out the missing term. I have updated the post and do hope it is correct now.

Best.

Comment by Ngô Quốc Anh — August 29, 2015 @ 7:49

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