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?)


$\{u_k\}_k^{\infty} \subset D(-\Delta^2)$ 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.

  • 1
    $\begingroup$ For closedness of $-\Delta^2$, you first need to check that $u \in \mathcal{D}(-\Delta^2)$. To see that $[0,\infty[ \subset \rho(-\Delta^2)$, note that this reduces to solving, for arbitrary $f \in L^2(U)$, the elliptic problem $\lambda u + \Delta^2 u = f$ with Dirichlet boundary conditions on $U$. $\endgroup$ Commented May 17, 2013 at 10:33
  • $\begingroup$ @ZulfiqarIII Is it enough to say that $-\Delta^2 u = 0$? And from here that $u \in D(-\Delta^2)$? Thanks for helping $\endgroup$ Commented May 17, 2013 at 16:09
  • $\begingroup$ No, you cannot talk about $\Delta^2 u$ unless you already know that $u \in \mathcal{D}(\Delta^2)$. $\endgroup$ Commented May 17, 2013 at 18:03
  • 1
    $\begingroup$ I would use that, by elliptic regularity (here we need $\partial U$ to be sufficiently smooth), the graph norm is equivalent to the $H^4$ norm on $\mathcal{D}(\Delta^2)$. $\endgroup$ Commented May 17, 2013 at 20:32
  • 1
    $\begingroup$ An alternative is to prove that $\Delta^2$ with the boundary conditions given above has a complete system of orthonormal (with respect to $L^2$) eigenfunctions $\varphi_i$ with eigenvalues $\lambda_i > 0$. Then $u(t,x) = \sum_i c_i e^{-\lambda_i t} \varphi(x)$ with $c_i = \langle g, \varphi_i\rangle$. The convergence is in $L^\infty(0,\infty;L^2)$ and locally uniformly in all $C^k(U \times (0,\infty))$. That approach is a bit more concrete than the general Hille-Yosida approach, and it uses more of the given structure. $\endgroup$ Commented May 25, 2013 at 13:39

1 Answer 1


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 $\Delta^{2}$.


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