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Suppose $\Omega \subset \mathbb{R}^d$ is a $C^{1,1}$ domain. Consider the biharmonic boundary value problem (BVP): $$ \begin{cases} \Delta^2 u = f \\ \nabla u \cdot \nu = g \\ u = u_D \end{cases} $$ wherein $\Delta^2 u = \Delta\Delta u$ is the application of the Laplace operator twice.

(a) Determine appropriate Sobolev spaces within which the functions $u, f, g,$ and $u_D$ should lie, and formulate an appropriate variational problem for the BVP. Show that the two problems are equivalent.

(b) Show that there is a unique solution to the variational problem. [Hint: use the Elliptic Regularity Theorem to prove coercivity of the bilinear form.]

(c) What would be the natural boundary conditions (BCs) for this partial differential equation?

(d) For simplicity, let $u_D$ and $g$ vanish and define the energy functional $ \int \left| \Delta v(x) \right|^2 - 2f(x)v(x) \, dx. $

It is clear for me that $f \in L^2(\Omega)$, $u \in H^2(\Omega)$, $u_D \in H^{2/3}(\Omega) $ and $g \in H^{1/2}(\partial \Omega)$,

I am having a hard time understanding how to construct the variational form for this problem. So we start by integrating against $v \in H^2_0(\Omega)$, we have

$$\int_\Omega \Delta^2 u \, v \,dx = \int_\Omega f\, v \,dx.$$

Now we consider the left hand side and try to get a term that contains $g$ which then we move to the right hand side;

\begin{align} \int_\Omega \Delta^2 u \, v \, dx &= \int_\Omega \nabla \cdot \nabla \Delta u \, v dx\\ &=\int_{\partial \Omega} v \, \nabla \Delta u \cdot \vec{n}\, dS^{d-1}(x) -\int_\Omega \nabla v \cdot \nabla \Delta u \, dx \\ &=\int_{\partial \Omega} v \, \nabla \Delta u \cdot \vec{n}\, dS^{d-1}(x) -\int_{\partial \Omega}(\nabla v) (\Delta u)\cdot \, dS^{d-1}(x)+ \int_\Omega \nabla \nabla \cdot \Delta u \, dx \end{align}

I can not see how to implement $g$ here? Could you please help!

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    $\begingroup$ Something is off with your integration-by-parts computation at the end. For example, the last term in the last line doesn't even make sense. You should be getting at the end an integral of $\Delta u \Delta v$ over $\Omega$ $\endgroup$
    – Laithy
    Commented Apr 20 at 18:50
  • $\begingroup$ @Laithy I think my main problem is the simple calculations, I have no problem in understanding the main concepts. $\endgroup$
    – Mr. Proof
    Commented Apr 23 at 16:55

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Assume that $u$ and $v$ are smooth and let's do some formal computations first (we'll worry about the correct function spaces once we have an idea of what the weak formulation looks like). Observe that

$$div( v \nabla(\Delta u)) = \color{blue}{\nabla v \cdot \nabla(\Delta u)} + v \Delta^2 u = \nabla v \cdot \nabla(\Delta u) + vf$$ and $$div(\Delta u \nabla v)) = \color{blue}{\nabla v \cdot \nabla(\Delta u)} + \Delta u \Delta v.$$ Combining these two identities (and noticing that the terms in blue cancel out) we get $$div(\Delta u \nabla v - v \nabla(\Delta u)) = \Delta u \Delta v - vf$$

By the divergence theorem we then get that $$\int_{\Omega} \Delta u \Delta v - vf = \int_{\Omega} div(\Delta u \nabla v) - v \nabla(\Delta u)) = \int_{\partial \Omega} \Delta u (\nabla v \cdot \nu) - v (\nabla(\Delta u))\cdot \nu)d\sigma.$$

Our weak formulation should then read: $$\int_{\Omega} \Delta u \Delta v - vf = 0 \qquad \text{ for all } v \in C^{\infty}_c(\Omega).$$ Since the identity is stable with respect to convergence in $H^2$, we can enlarge the class of etst functions to $H^2_0 := \overline{C^{\infty}_c}^{\| \cdot \|_{H^2}}$.

Now that we have defined weak solutions, we can give the variational formulation of the problem. Consider the energy functional $$J(w) := \int_{\Omega} (\Delta w)^2 - 2fw,$$ defined over the class $$A := \{w \in H^2(\Omega) : u = u_D \text{ on } \partial \Omega, \text{ and } \partial_{\nu}u = g \text{ on } \partial \Omega\}.$$

Showing that a minimizer $u \in A$ exists is an application of the direct method in the CoV. I'll leave this to you since this looks like a homework problem. Uniqueness of the minimizer follows from the convexity of $J$ and $A$. Moreover, notice that if $v \in H^2_0$, then $u + \epsilon v$ is also an element of $A$. Using the minimality of $u$ yields a variational inequality. Repeating the argument with $-v$ in place of $v$ yields that $u$ is a weak solution in the sense of our definition above.

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  • $\begingroup$ Could you explain where the g disappear? is is going to appear on the right hand side? $\endgroup$
    – Mr. Proof
    Commented Apr 22 at 22:57
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    $\begingroup$ @Mr.Proof $g$ does not disappear, but it is encoded in the class of admissible function $A$. The same is true for $u_D$, which also doesn't show up in the weak formulation. To be precise, while the boundary conditions do not explicitly show up in the weak PDE, your definition of weak solution should be a function $u \in H^2$ that satisfies the weak PDE \emph{and} has the correct boundary conditions (in the sense of traces). $\endgroup$
    – Giovanni
    Commented Apr 22 at 23:09
  • $\begingroup$ Can you show me the details. I just want to see the logic and how to apply it for similar situations. Otherwise, can you recommend any texts? $\endgroup$
    – Mr. Proof
    Commented Apr 23 at 0:04
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    $\begingroup$ @Mr.Proof The logic is very similar to what you would do with classical solutions. If you want to solve Laplace's equation $\Delta u = 0$ with boundary values $u = u_D$, you want $u$ to satisfy two things: the equation and the BCs. (Notice that $u_D$ does not show up in the equation itself). Similarly, if you want to reframe this in terms of weak solutions, you will require that $u$ satisfies two things: the weak formulation of the PDE (where like before $u_D$ does not show up) and require that $u \in u_D + H^1_0$ (this is the Sobolev spaces equivalent of saying that $u = u_D$ on the boundary) $\endgroup$
    – Giovanni
    Commented Apr 23 at 0:47
  • $\begingroup$ Thank you very much. $\endgroup$
    – Mr. Proof
    Commented Apr 23 at 2:04
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Assume wlog that $u_D = 0$. Then let the space that $u$ lives in be $H^2_0(\Omega)$. For $u,v \in H^2_0(\Omega)$, define

$$a(u,v) := \int_{\Omega} \Delta u \Delta v - \int_{\partial \Omega} \Delta u \partial_{\nu} v + \int_{\partial \Omega} \Delta v \partial_{\nu} u$$

Then a weak solution to your PDE will be defined as a function $u$ in $ H^2_0(\Omega)$ satisfying

$$a(u,v) = \int_{\Omega} fv + \int_{\partial \Omega} g \Delta v, \qquad \text{for every $v \in H^2_{0}(\Omega)$}$$

The reason why this is the appropriate definition is because of the following. Suppose $u$ satisfies the above definition and suppose for for now that it's smooth. Then we can integrate by parts to get:

$$\int_{\Omega} (\Delta^2 u -f )v + \int_{\partial \Omega} (\partial_{\nu} u - g)\Delta v = 0, \qquad \text{for every $v \in H^2_0(\Omega)$}$$

A standard argument then implies that $$\Delta^2 u=f, \quad \text{ in $\Omega$}$$ $$\partial_{\nu} u = g, \quad \text{on $\partial \Omega$}$$

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  • $\begingroup$ But I want the solution when $u \in H^2$, that is why $g$ should appear. The problem I am facing is that the $\vec{n}$ is doted with $v$ not with $u$! $\endgroup$
    – Mr. Proof
    Commented Apr 23 at 16:56
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    $\begingroup$ $v$ is $0$ on the boundary, so that term vanishes. I edited my answer to add more details. $\endgroup$
    – Laithy
    Commented Apr 23 at 18:36
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    $\begingroup$ I think you need to do the integration-by-parts computation more carefully. The way it's written on your post is false. $\endgroup$
    – Laithy
    Commented Apr 23 at 18:44

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