Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am attempting to solve some problems from Evans , I need some help with the following question. If $u\in H^2_0\Omega$, where $\Omega$ is open, bounded subset of $\mathbb{R}^n$. How can I solve the biharmonic equation

$\Delta^2u=f$ in $\Omega$, $u =\frac {\partial u } {\partial n }=0$ on $\partial \Omega$, $n$ is the normal vector .

provided $\int _\Omega \Delta u \Delta v \, \,dx =\int _\Omega fv $ for all $v\in H^2_0\Omega$. Given $f \in L^2(\Omega)$ , and prove that the weak solution is unique. Any kind of help would be great .

share|cite|improve this question
$u=\frac{\partial u}{\partial n} = 0$ on $\partial \Omega$? Instead of $u \frac{\partial u}{\partial n}$? – user20266 Jun 10 '12 at 7:44
@Thomas : Thank you for pointing out the mistake . – Theorem Jun 10 '12 at 7:58
Do you still need an answer for this question? – Shuhao Cao Jun 18 '13 at 21:35

5 Answers 5

Suppose that $u \in C_0^\infty(\Omega)$.

Then $$\int_\Omega |D^2 u|^2 \, dx = \int_\Omega \sum_{j,k=1}^n (u_{x_jx_k})^2 \, dx = \sum_{j,k=1}^n \int_\Omega u_{x_jx_k} u_{x_j x_k} \, dx.$$ You can integrate by parts twice to get $$\int_\Omega u_{x_jx_k} u_{x_jx_k} \, dx = - \int_\Omega u_{x_jx_kx_j}u_{x_k} \, dx = \int_\Omega u_{x_j x_j}u_{x_kx_k}\, dx$$ taking into account that $u$ is smooth and vanishes near the boundary of $\Omega$. Thus $$\int_\Omega |D^2 u|^2 \, dx = \sum_{j,k=1}^n \int_\Omega u_{x_j x_j}u_{x_kx_k}\, dx = \int_\Omega \left( \sum_{j=1}^n u_{x_jx_j} \right) \left( \sum_{k=1}^n u_{x_k x_k} \right) \, dx = \int_\Omega |\Delta u|^2 \, dx.$$ Thus $\|D^2 u\|_2^2 = \|\Delta u\|_2^2$. You can use the Poincare inequality to find a constant $C = C(n,\Omega)$ with the property that $\|u\|_2^2 \le C \|Du\|_2^2.$ On the other hand, for any $\epsilon > 0$ you have $$\|Du\|_2^2 = \int_\Omega |Du|^2 \, dx = \int_\Omega Du \cdot Du \, dx = - \int_\Omega u (\Delta u) \, dx \le \frac \epsilon 2 \|u\|_2^2 + \frac 1{2\epsilon} \|\Delta u\|_2^2$$ by Young's inequality. Thus $$\|Du\|_2^2 \le \frac{\epsilon C}{2} \|D u\|_2^2 + \frac{1}{2\epsilon} \|\Delta u\|_2^2.$$ With e.g. $\epsilon = \dfrac 1 C$ it follows that $\|Du\|_2^2 \le \dfrac{C}{2} \|\Delta u\|_2^2$ and consequently $\|u\|_2^2 \le \dfrac{C^2}{2} \|\Delta u\|_2^2$.

Finally we obtain $$ \|u\|_2^2 + \|Du\|_2^2 + \|D^2u\|_2^2 \le \left(1 + \frac C2 + \frac{C^2}{2} \right) \|\Delta u\|_2^2.$$ This can be extended to $u \in H_0^2(\Omega)$ using the density of $C_0^\infty(\Omega)$ in that space.

share|cite|improve this answer
It seems this answer is trying to prove that $u \in H_0^2(\Omega)$, rather than asserting the existence of a weak solution, which was what the question asked. – Cookie Feb 1 at 20:20
IIRC this answer was in response to a bounty asking why the operator $$B[u,v] = \int_\Omega \Delta u \Delta v \, dx$$ is coercive. – Umberto P. Feb 1 at 20:27
@UmbertoP. Could you help me with the density argument? Let $u\in H^2_0$. Then there is $(u_n)$ in $C_c^\infty$ such that $u_n\to u$ in $H^2$. By the proved part, $\|u_n\|_{H^2}\leq \|\Delta u_n\|_{L^2}$ for all $n\in\mathbb{N}$. So, we have to pass to the limit in the last inequality. I see that $\lim\|u_n\|_{H^2}=\|u\|_{H^2}$. But, since $\|\Delta u_n\|$ isn't exactly a term of the $H^2$-norm, how can we conclude that $\lim\|\Delta u_n\|=\|\Delta u\|$? – Pedro Oct 26 at 23:20
You have $$|\|\Delta u_n\|_2 - \|\Delta u\|_2| \le \|\Delta u_n - \Delta u\|_2 \le \|u - u_n\|_{H^2}.$$ – Umberto P. Oct 27 at 10:36

We will use the Lax-Milgram Theorem. A weak solution of your problem is a $u\in H^2(\Omega)$ such that

\begin{eqnarray} \Delta^2u=f & \Rightarrow & [\Delta^2u]\varphi=f\varphi\\ & \Rightarrow & \int_{\Omega}\Delta(\Delta u)\varphi=\int_{\Omega} f\varphi\\ & \Rightarrow & -\int_{\Omega}\nabla(\Delta u)\nabla\varphi=\int_{\Omega} f\varphi\\ & \Rightarrow & \int_{\Omega}\nabla u\nabla\varphi=\int_{\Omega} f\varphi, \end{eqnarray} for all $\varphi\in H^2_0(\Omega)$. Define the bilinear operator $B:H^2_0(\Omega)\times H^2_0(\Omega)\rightarrow\mathbb{R}$, $$B(u,\varphi)=\int_{\Omega}\Delta u\Delta\varphi.$$ Statement 1 This bilinear operator is continuos.

In fact,

\begin{eqnarray} |B(u,\varphi)| & \leq & \int_{\Omega}|\Delta u||\Delta\varphi|\\ & \leq & \|\Delta u\|^2_{L^2(\Omega)}\|\Delta \varphi\|^2_{L^2(\Omega)}\\ & \leq & C\|u\|^2_{H^2_0(\Omega)}\|\varphi\|^2_{H^2_0(\Omega)} \end{eqnarray} You can prove easily this last inequality.

Statemant 2 The bilinear operator is coercive.

In fact, $$B(u,u)=\int|\Delta u|^2=\|\Delta u\|^2_{L^2(\Omega)}\geq C\|u\|^2_{H^2_0(\Omega)}.$$ We used that $\|\Delta u\|_{L^2(\Omega)}$ defines a norm on $H^2_0(\Omega)$ equivalent to the usual norm.

Then, by the Lax-Milgram Theorem, for each $f\in H^2_0(\Omega)$, exists an unique function $u\in H^2_0(\Omega)$ such that $$B(u,\varphi)=\int_{\Omega}\Delta u\Delta\varphi=\int_{\Omega}f\varphi,$$ for all $\varphi\in H^2_0(\Omega)$.

I hope I help you.

share|cite|improve this answer
Proof of $\|\Delta u\|_{L^2} \geq \|u\|_{H^2}$ is totally non-trivial! You might wanna address that problem. For you used $\|\Delta u\|_{L^2} \geq \|D^2 u\|_{L^2}$, $\|\Delta u\|_{L^2} \geq \|\nabla u\|_{L^2} \leq \|u\|_{L^2}$. And a suggestion on notation is: there is no such term as "$H^2_0(\Omega)$-norm" in your subscript for norms, only $H^2$-norm. – Shuhao Cao Jun 16 '13 at 7:40
Thank you for this observations. You are right. I knew that the proof of this inequality is non-trivial, but as I dont know to prove this, I wrote just what I did in my exercise. – José Carlos Jun 17 '13 at 4:12

Four hints:

i) what kind of functional is $v\mapsto \int fv$? It's obviously linear, but is it continuous?

ii) Assume $u,\bar{u}$ solve the problem. Then $\int\Delta(u-\bar{u})\Delta v dx =0$ for all admissible $v$. What is the image of $\Delta$ when applied to admissible $v$? I.e. for which $\phi$ can you solve $\Delta v = \phi$? All these $\phi$ are admissible test functions. What does that tell you about $u-\bar{u}$?

iii) What does ii) tell you about $(u,v)\mapsto \langle u, v\rangle := \int\Delta u \Delta v dx$ ? Could this possibly be a scalar product? If yes, on which space?

iv) Now try to combine i) and iii). Does any representation theorem for linear functionals apply?

This comes without any kind of warranty, I do explicitly state that I did not check the details, it's just the roadmap I'd try first (with quite some confidence, I'd like to add, though). But you asked for assistance not for a solution :-)

share|cite|improve this answer
Thank you very much :) . First I will try with the hints. – Theorem Jun 10 '12 at 8:53
I have been trying to get some idea but i am not able at all . If you could help me with detailed solution it would be great . – Theorem Jun 10 '12 at 10:20

The Lax-Milgram theorem

Given a Hilbert space $V$ with scalar product $(.,.)_V$ and corresponding norm $\|\cdot\|_V$, a continuous and coercive bilinear form $a(.,.)$ on $V \times V$ and a continuous linear functional $L$ on $V$, there exists unique $u \in V$ s.t.

$a(u,v) = L(v)\ \ \ \ \forall v \in V$.

Some of the proofs required if you want to use this:

  • $a(.,.)$ is symmetric, ie, it holds that $a(u,v) = a(v,u) \ \ \ \forall u,v \in V$
  • $a(.,.)$ is continuous, ie, there exists a constant $C>0$ s.t. $|a(u,v)|\leq C\|u\|_V\|v\|_V \ \ \forall u,v \in V $
  • $a(.,.)$ is coercive or V-elliptic, ie, there exists a constant $\alpha>0$ s.t. $a(u,u) \geq \alpha \|u\|^2_V \ \ \ \forall u \in V$
  • $L$ is continuous, ie, there exists a constant $\Lambda>0$ s.t. $|L(v)| \leq \lambda \|v\|_V \ \ \ \forall v \in V$

I hope this gets you started.

share|cite|improve this answer

I have some ideas about the first question.You can use the method that we find the solution of the Possion's Equation.

First, you can find a spherical symmetry solution of the biharmonic equation. Thus, you will get a fundamental solution. Then you can get solution of $\Delta^2u=f$ by Green's Identity.

share|cite|improve this answer
For the second question,you can use the representation theorem in Hilbert Space.For more details,you can read Page 118 in the PDE by Fritz John. – Juntao Huang Jun 10 '12 at 8:34

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.