# Analyzing the Laplace Operator in $W_0^{2,2}(\Omega)$: Integral Identities, Inequalities, and Weak Solution

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.

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

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).