# PDE weak solution problem

My professor grades really strictly (details). I would be very happy if you could help me with this problem:

Let $U \subset R^n$ be a bounded set. Consider $\Delta^2 u = f$ on $U$ and $u=\frac{\partial u}{\partial n}=0$ on $\partial U$.

Part (a): Prove Poincare inequality for $H_0^2(U)$, i.e. $\int_U u^2 dx \leq C\int_U |\Delta u|^2 dx$ for all $u \in H_0^2(U)$.

Part (b): Define the notion of weak solution.

Part (c): Let $f \in L^2(U)$. Show that the problem has a unique weak solution in an appropriately chosen Sobolev space.

Part (d): Consider the same problem with boundary conditions $u=\Delta u = 0$ on $\partial U$ and prove the existence of a weak solution in the appropriate Sobolev space.

My work:

(a): $u \in H_0^2 (U) \subset H_0^1(U)$ and therefore we can use the Poincare inequality for $H_0^1$ $(\int_U u^2 dx \leq \widetilde{C} \int_U |Du|^2 dx)$ twice: $$\int_U u^2 dx \leq \widetilde{C} \int_U |Du|^2 dx \leq \widetilde{C} \widetilde{\widetilde{C}} \int_U |D^2u|^2 dx$$ since $Du \in H_0^1 (U)$.

(b): Multiply $\Delta^2 u = f$ on $U$ by a test function $v \in H_0^2(U)$ and integrate over $U$. Green's identity for operator $L= \Delta^2$: $$\int_U v \Delta^2 u dx= \int_{\partial U} (v \frac{\partial \Delta u}{\partial n} - \frac{\partial v}{\partial n}\Delta u)dS + \int_U \Delta v \Delta u dx = \int_U \Delta v \Delta u dx.$$ since $v$ is a test function and therefore the integral on the boundary vanishes. (i assumed $\frac{\partial v}{\partial n}= 0$. is this ok?) We can now define weak solution in the following way. Function $u \in H_0^2(U)$ is a weak solution if $$B[u,v]:= \int_U \Delta v \Delta u dx = \int_U fv dx \textrm{ for any } v \in H_0^2(U).$$ The weak solution of our fourth order problem will lie in the Hilbert space $H_0^2(U)$. Do you agree?

(c): Define an inner product on the Hilbert space $H_0^2(U)$: $$<u,v>_2 = \int_U \Delta u \Delta v dx$$ which implies the norm $$||u||_2 = (\int_U |\Delta u|^2 dx)^{\frac{1}{2}} = (\int_U |D^2 u|^2 dx)^{\frac{1}{2}}.$$ Norm $||\cdot||_2$ is equivalent to the standard norm on $H_0^2$: $||u||_1 := \big( \int_U (u^2 + |Du|^2 + |D^2u|^2)dx \big)^{\frac{1}{2}}$. It is obvious that $||u||_2 \leq ||u||_1$. On the other hand, \begin{align*} ||u||_1^2 &= \int_U (u^2 + |Du|^2 + |D^2u|^2 )dx\\ & = \int_U u^2dx + \int_U |Du|^2dx + \int_U |D^2u|^2 dx\\ & \leq C_1 \int_U |D^2u|^2dx + C_2 \int_U |D^2u|^2dx + \int_U |D^2u|^2 dx\\ &= (1+C_1+C_2)||u||_2^2 \end{align*} using Poincare inequalities as in (a).

Define a linear functional $F_f \in H_0^{2*}(U)$: $$F_f(v) = \int_U fv dx \quad (=<f,v>_{L^2}).$$ $F_f$ is bounded which follows from Cauchy-Schwarz inequality and Poincare inequality (is this ok?): $$|\int_U fv dx | \leq ||f||_{L^2}||v||_{L^2} \leq ||f||_{L^2}||v||_{H_0^2}.$$ Linearity is easy.

Equip $H_0^2$ with the inner product $<,>_2$ .Apply Riesz representation theorem and you have weak solution. But is not that already a strong solution? Confused here

(d): Denote $-\Delta u = t$. Since $\Delta u = 0$ on $\partial U$, $t=0$ on $\partial U$. We know existence of weak solution $t \in H_0^1(U)$ for $-\Delta t = f$, $f \in L^2(U)$ from previous homework. Remains $-\Delta u = t$ on $U$ and $u=0$ on $\partial U$. We can find unique $u \in H_0^1(U)$ that is a weak solution (if $t \in L^2(U)$ but here only $t \in H_0^1(U) ?)$

Would be grateful if someone could shed some light on me. Is at least the general idea of the solution ok?

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Didn't read the whole thing but the first proof has a problem, using Poincare inequality twice leads to the Hessian matrix, not the Laplacian. – Shuhao Cao Apr 2 '13 at 17:37
Hmm. Do you see another way of proving it? The hint for part (a) was: $\int_U |\Delta u|^2 dx = \int |D^2 u|^2 dx$ – UrošSlovenija Apr 2 '13 at 18:22
Yes, but you need to prove that. It's not difficult - it's basically integration by parts, but it does require proof. Basically $\int u_{ij} u_{ij} = \int u_{ii} u_{jj}$ after two integrations by parts, for functions which do not leave boundary terms. – Ray Yang Apr 2 '13 at 18:26
@RayYang in the answer below you used Fourier transformation instead. But wouldn't summing over $i,j$ would be enough. – simon Apr 14 '14 at 9:48

• (a) I think we first have to justify the usage of Poincare inequality for $Du$, especially the boundary condition: since $u=0$ on the boundary, $Du\cdot t = 0$ where $t$ is an $(n-1)$-dimensional tangent vector. Together with the normal derivative equaling 0: $\frac{\partial u}{\partial n} = Du\cdot n=0$, we have $Du = 0$ on the boundary. Then the exploiting of Poincare inequality leads to $$\int_U u^2 \leq \int_U (D^2u)^2 .$$ Yet $\Delta u = \mathrm{trace}(D^2 u)$, which says $\|\Delta u \|_{L^2(U)}\leq \|D^2 u \|_{L^2(U)}$. To prove the reverse, we could use Fourier transform to prove (left for you as exercise): $$\|\widehat{D^{\alpha} u} \|_{L^2(\mathbb{R}^n)}\leq \|\widehat{\Delta u}\|_{L^2(\mathbb{R}^n)},$$ for any multi-index $|\alpha|=2$. This basically takes advantage of the "diagonalization" property of the Fourier transform. Then Fourier transform being an isometry implies: $$\|D^{\alpha} u\|_{L^2(U)}\leq \|\Delta u\|_{L^2(U)}.$$ Therefore: $$\|D^2 u\|_{L^2(U)}^2 = \sum_{|\alpha|=2}\|D^{\alpha} u\|_{L^2(U)}^2\leq C\|\Delta u\|_{L^2(U)}^2.$$ And $$\int_U u^2 \leq \int_U (D^2u)^2 \leq C \int_U (\Delta u)^2$$ follows.

• (b) Yes, assuming $v=0$ and $\frac{\partial v}{\partial n}=0$ is appropriate.

• (c) Again I think we have to use the result in (a) $\|D^2 u\|_{L^2(U)}^2 \leq C\|\Delta u\|_{L^2(U)}^2$ again. Normally we denote the Laplacian norm as a semi-norm: $$|u|_2 := \|\Delta u\|_{L^2(U)}.$$ Using the Poincare inequality we could get the semi-norm is equivalent to the full $H^2$-norm you gave there. Right hand side is a bounded linear functional as you reasoned. Then using Riesz gives the answer like you did there. Strong solution requires the existence of derivative up to 4th order, and this not strong solution. However, if your domain is smooth or convex, elliptic regularity kicks in and your weak solution is not only strong, but also smooth.

• (d) The Sobolev space for the original problem is: $$H^2_0(U) = \{v\in H^2: v=0 , \frac{\partial v}{\partial n}=0 \text{ on }\partial \Omega\}.$$ Now let's look at the integration by parts result: $$\int_U v\, \Delta^2 u = \int_{\partial U} (v \frac{\partial \Delta u}{\partial n} - \frac{\partial v}{\partial n}\Delta u)dS + \int_U \Delta v \Delta u .$$ For the new problem in this question, if $\Delta u$ vanishes on boundary, then only assuming $v=0$ is enough for test function, i.e., choosing test function space as $H^1_0(U)\cap H^2(U)$ looks appropriate. For your last question, notice $t\in H^1_0$ implies $t\in L^2$, hence the weak solution $u\in H^1_0$. Then proving Poincare inequality for $H^1_0(U)\cap H^2(U)$, and repeat what you have done for (b) and (c).

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Thanks, it is quite helpful. Some additional questions: can we assume $\frac{\partial v}{\partial n} = 0$ if we have $v=0$ (test function) on boundary?, part (d) is my reasoning wrong?, is $L^2(U) \subset H_0^1(U)?$ – UrošSlovenija Apr 2 '13 at 18:50
@UrošSlovenija $H^1_0\subset L^2$, therefore if $t \in H^1_0$, then $t$ has to be in $L^2$. – Shuhao Cao Apr 2 '13 at 18:55
Thanks again. The only thing troubling me is if i really can assume $\frac{\partial v}{\partial n}=0$ and why. – UrošSlovenija Apr 2 '13 at 19:07
@UrošSlovenija if we assume the test function satisfying Homogeneous Neumann $\frac{\partial v}{\partial n} = 0$, then the weak solution also satisfies $\frac{\partial u}{\partial n} = 0$. Now for a particular $H^2$-function, say the true solution $u$, when $\Delta u = 0$ on the boundary, $\frac{\partial u}{\partial n}$ is not necessarily 0. Just think $-\Delta u = f$ Dirichlet boundary problem, if we have the boundary condition $u= g$ on boundary, then $u$ is uniquely determined, imposing an extra $\frac{\partial u}{\partial n}= g_N$ will make the equation overdetermined. – Shuhao Cao Apr 2 '13 at 19:20
@UrošSlovenija $\langle\cdot,\cdot\rangle_2$ you defined is indeed an inner product in $H^2_0$: the $u|_{\partial \Omega} = 0$ rules out the possibility that $u$ may be a non-zero number. $\partial u/\partial n=0$ rules out the possibility that $\nabla u$ may be a non-zero constant vector. – Shuhao Cao Apr 4 '13 at 17:19