The Variational form of a biharmonic PDE

Question

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!

Answer 1

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.

Answer 2

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 $$\Delta^2 u=f, \quad \text{ in $\Omega$}$$ $$\partial_{\nu} u = g, \quad \text{on $\partial \Omega$}$$