# Confusing dot product and inner product in a weak formulation

I have been struggling with this for a while. Here, as you can see, they define the weak formulation of the Poisson equation as:

$$-\int_{\Omega}\nabla u\cdot\nabla v\,ds = \int_{\Omega}fv\,ds \equiv -\phi(u,v)$$ (not sure what the $$\equiv$$ symbol means here(?)), where they specify that the ''." denotes the dot product, $$v$$ is a test function, $$u$$ the unknown and $$\phi$$ an operator (or map). Few lines later; they explain: "Such functions are (weakly) once differentiable and it turns out that the symmetric bilinear map $$\phi$$ then defines an inner product which turns $$H_0^1(0,1)$$ into a Hilbert space"

Does this mean that the weak formulation in this case is by default an inner product that uses the dot product?

EDIT: I checked this pdf file that studies the weak formulation of Poisson's equation, apparently the weak formulation defines an inner product.

I just need a confirmation if what I am doing is correct...

• Not sure I fully understand your last sentence but I think you got it. The weak formulation of the Poisson's equation defines a bilinear map $\phi$ which happens to be an inner product over $H_0^1$. Also the use of the sign $\equiv$ is not clear to me as well. – nicomezi Feb 3 at 9:18
• The $\equiv$ sign is a (terrible) way to indicate a definition (terrible because $X\equiv Y$ doesn't tell you whether $X$ is defined by $Y$ or $Y$ is defined as $X$ in the abstract). I'd much prefer $=:$ here. – user10354138 Feb 3 at 9:28
• You can probably formulate it this way but again I do not see the point. – nicomezi Feb 3 at 9:33
• It depends on what is intended with the discretization of the weak formulation. I guess it is for numerical computations ? In this case I would rather speak about a weighted/rescaled sum (depending on your quadrature rule). – nicomezi Feb 3 at 9:41
• $\equiv$ is sometimes used to state not just an equality, but a definition, e.g $$e\equiv \lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n$$ – K.defaoite Feb 25 at 1:01

The reason it's an inner product is that the inner product axioms are satisfied for $$\phi(u,v)=\int_\Omega\nabla u\cdot\nabla v.$$
The easy ones to show are symmetry and linearity, i.e., $$\phi(u,v)=\phi(v,u)$$ and $$\phi(au,v)=a\phi(u,v)$$ ($$a$$ constant). One has to also show that $$\phi(u,u)\ge0$$ for all $$u$$, and $$\phi(u,u)=0$$ if and only if $$u=0$$.
One has that $$\phi(u,u)=\int_\Omega|\nabla u|^2\ge0$$. Then, if $$u=0$$, then $$\phi(u,u)=\int_\Omega0=0$$, and if $$\phi(u,u)=0$$, then $$\int_\Omega|\nabla u|^2=0$$, and so $$u$$ must be constant, but also $$u\in H^1_0(\Omega)$$, so the constant must in fact be zero, so $$u=0$$.