0
$\begingroup$

So I have these questions:

Let $\Omega \subset \mathbb{R}^d$ be open and bounded. Consider the Laplace-Operator given by $$ \Delta u=\sum_{j=1}^d \partial_j^2 u \quad \text { for all } u \in W^{2,2}(\Omega) . $$ (i) Show that for every $u \in C_c^{\infty}(\Omega)$ $$ \int_{\Omega}|\Delta u(x)|^2 d x=\sum_{i, j=1}^d \int_{\Omega}\left|\partial_i \partial_j u(x)\right|^2 d x . $$ (ii) Show that there are constants $C_1, C_2>0$ such that for any $u \in W_0^{2,2}(\Omega)$ $$ C_1\|u\|_{W^{2,2}(\Omega)}^2 \leq \int_{\Omega}|\Delta u(x)|^2 d x \leq C_2\|u\|_{W^{2,2}(\Omega)}^2 . $$

Deduce that $$ (u, v)_{W_0^{2,2}(\Omega)}=\int_{\Omega} \Delta u(x) \Delta v(x) d x $$ defines a (real) scalarproduct on $W_0^{2,2}(\Omega)$. Hint: Use Poincaré's inequality: $\int_{\Omega}|u|^2 d x \leq c_p \int_{\Omega}|\nabla u|^2 d x$, for all $u \in W_0^{1,2}(\Omega)$.

(iii) Let $f \in L^2(\Omega)$. Show that there exists a unique function $u \in W_0^{2,2}(\Omega)$ such that for all $\phi \in W_0^{2,2}(\Omega)$ $$ \int_{\Omega} \Delta u(x) \Delta \phi(x) d x=\int_{\Omega} f(x) \phi(x) d x . $$

Moreover, prove that there is $C>0$ such that $$ \|u\|_{W_0^{2,2}(\Omega)} \leq C\|f\|_{L^2(\Omega)} . $$

Remark: The function $u$ is the weak solution of the Dirichlet problem $$ \Delta^2 u=f \text { in } \Omega, \quad u=0 \text { on } \partial \Omega . $$

And, here are my attempted solution to each part:

(i) Consider a function $u$ in $C_c^{\infty}(\Omega)$. The Laplace operator $\Delta u$ is defined as the sum of the second partial derivatives of $u$ with respect to each variable, i.e., $\Delta u=\sum_{j=1}^d \partial_j^2 u$

We begin by focusing on the left-hand side of the integral identity which is $\int_{\Omega}|\Delta u(x)|^2 d x$. Substituting the definition of the Laplace operator, this becomes $\int_{\Omega}\left|\sum_{j=1}^d \partial_j^2 u(x)\right|^2 d x$.

The square of the sum inside the integral can be expanded, leading to $\int_{\Omega}\left(\sum_{j=1}^d \partial_j^2 u(x)\right)^2 d x=\int_{\Omega} \sum_{i=1}^d \sum_{j=1}^d \partial_i^2 u(x) \partial_j^2 u(x) d x$. This expansion is a result of applying the distributive property of multiplication over addition.

Next, we utilize Fubini's Theorem, which is applicable here as $u$ and its derivatives are continuous and bounded on $\Omega$. This theorem allows us to interchange the order of summation and integration, thus the expression becomes $$ \sum_{i=1}^d \sum_{j=1}^d \int_{\Omega} \partial_i^2 u(x) \partial_j^2 u(x) d x \text {. } $$

We now apply integration by parts to each term in the double sum. Consider a term with $i=j$, for example, $\int_{\Omega} \partial_i^2 u(x) \partial_i^2 u(x) d x$. Since $u$ and its derivatives vanish on the boundary of $\Omega$ (as $u \in C_c^{\infty}(\Omega)$ ), the boundary terms in the integration by parts formula disappear, leading to $\int_{\Omega}\left|\partial_i \partial_i u(x)\right|^2 d x$.

Repeating this step for each pair $(i, j)$ in the double sum, and considering the vanishing boundary terms due to the compact support of $u$, we sum up these integrals. This results in $\sum_{i, j=1}^d \int_{\Omega}\left|\partial_i \partial_j u(x)\right|^2 d x$, effectively transforming the left-hand side of the integral identity into the right-hand side.

(ii) To prove the given inequality for $u \in W_0^{2,2}(\Omega)$, using the properties of the Sobolev space $W_0^{2,2}(\Omega)$ and the Poincaré inequality, we approach the problem as follows:

Consider a function $u$ in the Sobolev space $W_0^{2,2}(\Omega)$. This space consists of functions in $L^2(\Omega)$ whose first and second weak derivatives are also in $L^2(\Omega)$, and importantly, functions in this space vanish on the boundary in the sense of traces.

To establish a lower bound for $\|u\|_{W^{2,2}(\Omega)}^2$, we start with its definition: $$ \|u\|_{W^{2,2}(\Omega)}^2=\int_{\Omega}|u(x)|^2 d x+\sum_{i=1}^d \int_{\Omega}\left|\partial_i u(x)\right|^2 d x+\sum_{i, j=1}^d \int_{\Omega}\left|\partial_i \partial_j u(x)\right|^2 d x \text {. } $$

Applying the Poincaré inequality, which asserts the existence of a constant $C_P>0$ such that for all $u \in W_0^{1,2}(\Omega)$, $$ \int_{\Omega}|u(x)|^2 d x \leq C_P \sum_{i=1}^d \int_{\Omega}\left|\partial_i u(x)\right|^2 d x $$ we can bound the $L^2$ norm of $u$ by its first derivatives. This gives us a lower bound for the norm in terms of the first derivatives of $u$.

Considering $u \in W_0^{2,2}(\Omega)$, it follows that $\Delta u \in L^2(\Omega)$. Hence, we can write: $$ \int_{\Omega}|\Delta u(x)|^2 d x \geq C_1\left(\int_{\Omega}|u(x)|^2 d x+\sum_{i=1}^d \int_{\Omega}\left|\partial_i u(x)\right|^2 d x\right) $$ for some constant $C_1>0$. This inequality stems from the Poincaré inequality and the fact that $\Delta u$ encapsulates second derivatives of $u$. For the upper bound of $\|u\|_{W^{2,2}(\Omega)}^2$, we note that the Laplacian $\Delta u$ is a part of the definition of the norm. Therefore, we have: $$ \int_{\Omega}|\Delta u(x)|^2 d x \leq \sum_{i, j=1}^d \int_{\Omega}\left|\partial_i \partial_j u(x)\right|^2 d x . $$

From the norm's definition, it follows that: $$ \sum_{i, j=1}^d \int_{\Omega}\left|\partial_i \partial_j u(x)\right|^2 d x \leq C_2\|u\|_{W^{2,2}(\Omega)}^2 $$ for some constant $C_2>0$.

(iii) To prove that the given expression defines a scalar product on $W_0^{2,2}(\Omega)$, we need to verify the properties of positivity, symmetry, and linearity for this function space.

Positivity:

Consider any function $u \in W_0^{2,2}(\Omega)$. The scalar product of $u$ with itself is given by: $$ (u, u)=\int_{\Omega} \Delta u(x) \Delta u(x) d x=\int_{\Omega}|\Delta u(x)|^2 d x $$

Since the square of any real number is non-negative, $|\Delta u(x)|^2 \geq 0$. Therefore, the integral of $|\Delta u(x)|^2$ over any domain $\Omega$ is also non-negative, ensuring that $(u, u) \geq$ 0.

Now, if $(u, u)=0$, this implies that $\int_{\Omega}|\Delta u(x)|^2 d x=0$. For an integral of a nonnegative function to be zero, the function itself must be zero almost everywhere in $\Omega$. Thus, $\Delta u(x)=0$ almost everywhere in $\Omega$. In the space $W_0^{2,2}(\Omega)$, this implies $u$ is a constant function. Due to the zero boundary conditions in $W_0^{2,2}(\Omega)$, this constant must be zero. Hence, $u=0$. Symmetry:

For any two functions $u, v \in W_0^{2,2}(\Omega)$, the scalar product is defined as: $$ (u, v)=\int_{\Omega} \Delta u(x) \Delta v(x) d x $$

By the commutative property of multiplication $(a b=b a)$ : $$ (u, v)=\int_{\Omega} \Delta u(x) \Delta v(x) d x=\int_{\Omega} \Delta v(x) \Delta u(x) d x=(v, u) $$

Therefore, the scalar product is symmetric.

Linearity: For linearity, let $a$ be any scalar, and $u, v, w$ be functions in $W_0^{2,2}(\Omega)$. We need to show two properties:

  1. $(a u, v)=a(u, v)$
  2. $(u+v, w)=(u, w)+(v, w)$ For the first property: $$ (a u, v)=\int_{\Omega} \Delta(a u)(x) \Delta v(x) d x=\int_{\Omega} a \Delta u(x) \Delta v(x) d x=a \int_{\Omega} \Delta u(x) \Delta v(x) $$

For the second property: $$ \begin{gathered} (u+v, w)=\int_{\Omega} \Delta(u+v)(x) \Delta w(x) d x=\int_{\Omega}(\Delta u(x)+\Delta v(x)) \Delta w(x) d x \\ =\int_{\Omega} \Delta u(x) \Delta w(x) d x+\int_{\Omega} \Delta v(x) \Delta w(x) d x=(u, w)+(v, w) \end{gathered} $$

Thus, the given expression satisfies the linearity property.

--

I have no specific questions, but I would love if you could proofread and let me know if there are any mistakes! :)

$\endgroup$

1 Answer 1

1
$\begingroup$

The norm equivalence of (ii) is not correctly justified from what I can see. You bound the $L^2$-norm of the Laplacian by the $W^{2,2}$-norm, but the lower bound does not appear from what I can see. Moreover at some point you write:

This inequality stems from the Poincaré inequality and the fact that $\Delta u$ encapsulates second derivatives of $u$.

This is an unjustified and ill-defined claim.

Instead I suggest you argue the lower bound as follows:

  1. Extend the integral equality $$ \int_{\Omega}|\Delta u(x)|^2 d x=\sum_{i, j=1}^d \int_{\Omega}\left|\partial_i \partial_j u(x)\right|^2 d x $$ to hold for all $u \in W^{2,2}_0(\Omega)$ by density. Here you approximate $u$ in $W^{2,2}$ by a sequence $\{u_n\} \subset C^{\infty}_c(\Omega)$ for which (i) applies, and use the upper bound to show $\Delta u_n$ converges to $\Delta u$ in $L^2$.

  2. Show, using Poincaré's inequality, that there is $C>0$ such that $$ \int_{\Omega}|u(x)|^2 d x+\sum_{i=1}^d \int_{\Omega}\left|\partial_i u(x)\right|^2 d x\leq C\sum_{i, j=1}^d \int_{\Omega}\left|\partial_i \partial_j u(x)\right|^2 d x$$ for all $u \in W^{2,2}_0(\Omega)$. You may want to justify why $u, \partial_ju \in W^{1,2}_0(\Omega)$ when applying the Poincaré's inequality.

  3. Conclude using the above two results.

In addition, part (iii) seems to be missing entirely; you instead prove the second part of (ii).

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .