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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...

Wallflower
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    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 03 '21 at 09:18
  • @nicomezi I've just added an edit to the post, thank you for the confirmation. – Wallflower Feb 03 '21 at 09:20
  • @nicomezi While we are still on it, if you check my question https://math.stackexchange.com/questions/4009568/difference-between-the-definition-of-the-dot-product-in-infinite-and-finite-dimensions, we can safely say that when discretizing the weak formulation the inner product becomes a dot product since my functions are no longer square integrable (all discrete), right? – Wallflower Feb 03 '21 at 09:22
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    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 03 '21 at 09:28
  • @user10354138 I fully agree with you. It's just a copy paste of the Wikipedia definition... – Wallflower Feb 03 '21 at 09:29
  • @nicomezi Not a big inner product but a discretized inner product that becomes a dot product in $\mathbb{R}^n$ is what I am trying to say – Wallflower Feb 03 '21 at 09:32
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    You can probably formulate it this way but again I do not see the point. – nicomezi Feb 03 '21 at 09:33
  • @nicomezi I am just writing a report and I am asked to be as precise and concise as possible. And since it's been a while since my last math class I'm trying my best not to mistake or confuse few defintions. – Wallflower Feb 03 '21 at 09:34
  • I would definitely not write that a discretization would lead to a discrete inner product in this case. Although it seems correct I find this confusing and make a complicated problem apparently simple. – nicomezi Feb 03 '21 at 09:36
  • @nicomezi What do you suggest I write when moving from the continuous form of the weak formulation to a discrete one, when it comes to the definition of the inner product ? – Wallflower Feb 03 '21 at 09:38
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    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 03 '21 at 09:41
  • @nicomezi Understood, thank you so much :) – Wallflower Feb 03 '21 at 09:42
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    $\equiv$ 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 '21 at 01:01
  • @K.defaoite They have not defined it so it really is confusing. Thank you for your remark btw :) – Wallflower Feb 25 '21 at 10:18

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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 $u=0$.

Ellya
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