2

I know that in $\mathbf{\mathbb{R}}^n$ the definition of the dot (or scalar) product is the following:

$x.y=x^{\mathrm{T}}y$, with ''T" denoting the transpose of the vector x.

How does this definition change when working in infinite space, e.g:

$\int_{\Omega}\nabla u \cdot \nabla v \, d\Omega = 0,\, \forall v\in H^{1}_{0}\left(\Omega\right)$

am I allowed to write:

$\int_{\Omega}\left(\nabla u\right)^{\mathrm{T}}\, \nabla v\,d\Omega = 0,\, \forall v\in H^{1}_{0}\left(\Omega\right)$ ??

Thanks in advance for your answers

Wallflower
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    What you wrote down for $\Bbb R^n$, is the standard scalar product there. Whether an inifinite-dimensional space has a standard way to define dot product, very much depends on the space - usually you want square-integrability – Hagen von Eitzen Feb 02 '21 at 14:04
  • @HagenvonEitzen For the infinite dimensional space, I chose the 'Sobolev' space, so the suqare integrability is guarranteed. Suppose I discretize the domain of definition of my first integral, what transition do I have to do or taken into account so that I can apply the 'standard' scalar product? – Wallflower Feb 02 '21 at 14:07
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    @Wallflower It sounds like you're trying to apply some type of Galerkin method. – Ben Grossmann Feb 02 '21 at 15:24
  • @BenGrossmann From what I've read in the page you shared with me, the reduction part only focuses on the change of spaces (from continuous to discrete), nothing was mentioned on the part integral one to integral 2... – Wallflower Feb 02 '21 at 15:31
  • @Wallflower There are two inner products here, it's not clear from the comments which one you're trying to replace with an expression invovling $T$. If I interpret your question literally, then the answer is yes: if $\Omega \subset \Bbb R^n$ (as is typically the case), then we can replace $\operatorname{grad}(V)\cdot \operatorname{grad}(V')$ with $\operatorname{grad}(V)^T\operatorname{grad}(V')$. However, this merely is a way of rewriting an inner product over $\Bbb R^n$, a finite dimensional space. The inner product over the infinite dimensional $H^1_0$ is still expressed as an integral – Ben Grossmann Feb 02 '21 at 15:43
  • @BenGrossmann so if I understood you correctly, the first integral is already considered as a scalar product, and the second is the standard one (the one I defined at the beginning)? I am actually trying to rewrite my first integral, which is my weak formulation, in a discrete domain (a meshed geometry) and I honestly do not know how to move correctly and meticulously from the first to the second.. I hope I was clear enough for you to maybe help me – Wallflower Feb 02 '21 at 15:47
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    @Wallflower I don't understand how writing $$ \int_{\Omega}\beta,\mathrm{\mathbf{grad}}\left(V\right)^{\mathrm{T}}, \mathrm{\mathbf{grad}}\left(V'\right) d\Omega = 0 $$ (which is certainly allowed) accomplishes your goal of rewriting your weak formulation in a discrete domain – Ben Grossmann Feb 02 '21 at 16:23
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    @Wallflower I would suggest that you look into the Finite Element Method. – Ben Grossmann Feb 02 '21 at 16:30
  • @BenGrossmann It is accomplishing rewriting my weak formulation in a discrete domain because a transpose of a gradient in infinite space makes no sense...? – Wallflower Feb 02 '21 at 19:07
  • @Wallflower My problem is that I have no idea what you are actually trying to say when you say that you're looking for a "transpose of a gradient in infinite dimensional space". It's not that these words don't make sense, it's that the thing that these words refer to seems unrelated to your goal of formulating the problem over a discrete domain. The finite element method is the standard way of discretizing the weak formulation of a linear PDE, so it should be useful for whatever you have in mind. – Ben Grossmann Feb 02 '21 at 19:18
  • @BenGrossmann I, sometimes, can be really bad at expressing myself, pardon me.. Let me put this way: The first integral is my weak formulation before discretizing my space by meshing my geometry. Knowing that the dot product in $\mathbb{R}^n$ is defined as in the first equation, does it really make sense to say that I am now meshing my geometry so I am moving from continuous to discrete space, I will use this definition of the scalar product even though that of the continuous weak formulation may be different and requires a square-integrability which is not satisfied when in discrete case. – Wallflower Feb 02 '21 at 19:23
  • @BenGrossmann I feel like I am mixing 2 dot product definitions when doing so and I do not know how to go about it... – Wallflower Feb 02 '21 at 19:25
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    @Wallflower Yes, what you're saying makes sense and is exactly what is accomplished by the finite element method. I suggest that you learn about the FEM in detail. Then, if you can't find a way to apply the method to your specific domain of interest, ask a question about how to do that. – Ben Grossmann Feb 02 '21 at 19:33
  • @BenGrossmann Understood, thank you! – Wallflower Feb 02 '21 at 19:38
  • @BenGrossmann Hey I feel like [https://math.stackexchange.com/questions/476738/difference-between-dot-product-and-inner-product]-(this question) answered my question in a way. So back to your previous answer where you confirmed that I can definitely do the transition from integral1 to integral2, I just have to mention that integral one is the definition of the inner (its general definition) product whereas when moving to integral 2, the inner product reduces to the the first definition in $\mathbb{R}^n$ – Wallflower Feb 02 '21 at 21:36

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I basically found my answer from this question, where they clarified the ambiguity I had with the definition of the inner product. In fact, when working in an infinite dimension space the inner product (a general nomenclature and definitions for different spaces) is the one used and as the dimension becomes finite its definition becomes the dot product (or you can still call it an inner product). So basically, the dot product is the definition of the inner product in a finite dimension space. Me moving from integral 1 to the second integral defined in a discretized space is completely fine as long as I remind the definition of the inner product.

I confirmed this by the comments I've gotten here.

Hope I was clear enough :)

Wallflower
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