# Show that the operator $Lu$ is linear, symmetric, positive and coercive operator.

Consider the following partial differential equation in the bounded domain $\Omega\subset \mathbb{R}^2$: $$\Delta^2u - \nabla\cdot(k(x,y)\nabla u) + \lambda u = f, \,in\,\Omega$$ where $\lambda >0$ is a constant and $k(x,y) >0$ is a given function of the position. Further, $f(x,y)$ is a given function. The boundary conditions are given by $$u =0, \Delta u = 0, \,on\,\,\partial\Omega$$ I need to show that this boundary value problem can be reformulated as a minimisation problem.

I know the following theorem:

Let $L:\sum(\Omega)\to\sum'(\Omega)$, where $\sum(\Omega)$ is a linear space, and suppose that $L$ is linear, self-adjoint and positive, and let $$u_0\in\sum:Lu_0 = f$$ Then $u_0\in \sum(\Omega)$ minimises $F(u) = \displaystyle\int_\Omega \dfrac{1}{2}uLu - uf d\Omega$. If $L$ is coercive on $\sum(\Omega)$ then $u_0$ is the only minimiser of $F(u)$ in $\sum(\Omega)$.

Question: How do I show that $L$ is self-adjoint, positive and coercive? (I know how to show that $L$ is linear.)

I've tried to show that $L$ is self-adjoint (symmetric), but I already got stuck there. $L$ is symmetric if we have that $$\int vLud\Omega = \int uLvd\Omega.$$ To begin, I tried to show that $\int v\Delta^2ud\Omega = \int u\Delta^2 vd\Omega$, by integration by parts and applying Gauss theorem to show that some terms cancel out as they are zero on the boundary. However, I didn't get very far.

Edit: I did find out how to show that $\int v\Delta^2ud\Omega = \int u\Delta^2 vd\Omega$. Applying integration by parts twice on $\int v\Delta^2ud\Omega$ and using Gauss theorem to cancel out terms on the boundary leads to $\int v\Delta^2ud\Omega = \int_\Omega\Delta u\Delta vd\Omega$ which is obviously symmetric. I'll update with a more detailed solution when I'm done with the rest of the exercise.