Disclaimer: pretty long and specific (contraction semi groups involved).
I have fourth order parabolic equation
$$
u_t + \Delta^2 u = 0
$$
on $U_T = U \times [0,T]$. $U \subset \mathbb{R}^m$ is a bounded open set with smooth boundary. Boundary conditions are:
$$
u(x,0) = g \in L^2(U)
$$
and
$$
u=\frac{\partial U}{\partial n}=0 \quad \text{on } \partial U \times [0,T]
$$
I would like to prove the existence of a weak solution. From Hille Yosida theorem i deduce that weak solution exists if the operator $-\Delta^2$ generates a contraction semigroup (is this ok?):
To prove this i define (is this ok?)
$$
D(-\Delta^2) = H^4(U) \cap H_0^2 (U)
$$
Density:
$C_0^{\infty}(U) \subset H^4(U) \cap H_0^2 (U)$. $C_0^{\infty}(U)$ is dense in $L^2(U)$ and thus $D(-\Delta^2) $ is dense in $L^2(U)$ (is this ok?)
Closedness:
$\{u_k\}_k^{\infty} \subset D(-\Delta^2)$ with
\begin{align*}
u_k & \to u \\
-\Delta^2 u_k & \to f
\end{align*}
in $L^2(U)$ when $k \to \infty$. I have
$$
||u_k – u_l||_{H^2(U)} \leq C(||-\Delta^2 u_k + \Delta^2 u_l||_{L^2(U)} + ||u_k – u_l||_{L^2(U)})
$$
and then $u \in D(-\Delta^2)$ and $-\Delta^2 u =f$
$\lambda \in \mathbb{R}$ belongs to resolvent set $\rho (-\Delta^2)$ if operator $\lambda I + \Delta^2: D(-\Delta^2) \to L^2(U)$ is one-to-one and onto.
For $\lambda \in \rho (-\Delta^2)$, the resolvent operator $R_{\lambda}: L^2(U) \to L^2(U)$ is defined by $R_{\lambda}u = (\lambda I + \Delta^2)^{-1} u $.
I have to prove also
$(0,\infty) \subset \rho (-\Delta^2)$:
I show that equation $\lambda u + \Delta^2 u = f$ has unique weak solution for $\lambda > 0$ and assume that is also regular (can i do this?)
Now $\lambda I + \Delta^2$ is one-to-one and onto for $\lambda > 0$ and thus $(0,\infty) \subset \rho (-\Delta^2)$.
Last thing to prove
$||R_{\lambda}||_{L^2(U)} \leq \frac1{\lambda}$:
Weak solution satisfies
$$ \lambda \int_U uv dx + \int_U \Delta v \Delta u dx = \int_U fv dx
$$
for all $v \in H_0^2(U)$. I set $u=v$ to get
$$
\lambda \int_U u^2 dx + \int_U (\Delta u)^2 dx = \int_U fu dx
$$
From here
$$
\lambda \int_U u^2 dx = \lambda ||u||_{L^2(U)}^2 \leq \int_U fu dx \leq ||f||_{L^2(U)} ||u||_{L^2(U)}
$$
and
$$
||u||_{L^2(U)} \leq \frac{1}{\lambda} ||f||_{L^2(U)}
$$
Acknowledging
$$R_{\lambda}f = u$$
we get
$$
||R_{\lambda}|| \leq \frac{1}{\lambda}
$$
as desired.
Is this close to ok?
Help warmly accepted.
Best Answer
If $A : \mathcal{D}(A)\subseteq X\rightarrow X$ is a closed densely-defined linear operator on a Hilbert space $X$, then $A^{\star}A$ is a closed densely-defined positive selfadjoint linear operator. So start with $\Delta$ on $H^{2}_{0}$ and define $\Delta^{\star}\Delta$, which has the same domain and action as your proposed $\Delta^{2}$.