Prove that this $ \langle u,v\rangle = \int_\Omega [ \bigtriangledown u \cdot\bigtriangledown v +uv ] $ is an inner product in $H^1_0 $

Question

Information needed

$$I = (a,b)$$

$$ H_0^1 (I) = \{ v \in H^1(I) : v(a)= v(b) = 0 \} $$

$$ H^1 (I) = \{ v: v , v' \in \mathbb{L}_2(I) \} $$

$$ \mathbb{L}_2(I) = \{ v:v \text{ it's defined at } I \text{ and } \int v^2\,dx < \infty \} $$

Prove that this is a scalar product in $H^1_0 $

Using the notation $$ \langle u,v\rangle = \int_\Omega [ \bigtriangledown u \cdot\bigtriangledown v +uv ] $$

where $\Omega$ is a bounded domain.

In the book numerical solutions of PDE by the finite element method by Claes there is this passage and I could not see clearly this statement.

Thanks to any help !

Answer 1

Once you establish that the expression is well defined, you just need to verify the conditions for an inner product.

1. $\langle u,v \rangle = \langle v,u \rangle$.
2. $\langle a u, v\rangle = a \langle u,b\rangle$ and $\langle u+v,w\rangle = \langle u,w\rangle+\langle v,w\rangle$.
3. $\langle u,u\rangle\ge 0$ and $\langle u,u\rangle=0 \Leftrightarrow u = 0$

The third property is the trickiest. You need to show that if $u \in H^1_0$ and $$ \int_I (|\nabla u|^2 + u^2) = 0 $$ then $u=0$ almost everywhere in $I$.