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I hope the following would be clear: I'm retaking a look at the weak solution of PDEs, and now i'm at the existence and uniqueness part.So we have:

find $u : a(u,v) = f(v) ~\forall v \in V $

Now, for the existence it says that we find a solution $u$ considering the minimization problem $J (u) = \inf J(v) $, where the functional $J(v)$ is defined by:

$ J(v) = \frac{1}{2}a(v,v)-f(v) $

So my main question is, from where this functional (found in literature as energy functional, why is that name?) take the $\frac{1}{2}$ ?

Probably I'm missing something, maybe knowing the use of this energy functional more in general could be useful.

Thank you in advance

Cla
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1 Answers1

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The $\frac12$ is there because of derivatives (just like the derivative of $\tfrac12 x^2$ is $x$ in real analysis)

If we want to find the minimizer of $J$, one way is to take the derivative and set it to zero. $$ J'(u)=0 $$

This means that $J'(u)v$ is the directional derivative in direction $v$, and therefore $$ J'(u)v = 0 \;\forall v\in V $$ It can be calculated, that $$J'(u)v= \frac12 a(u,v) +\frac12 a(v,u)- f(v)$$ Because $a$ is symmetric, this yields $$ J'(u)v= a(u,v)-f(v) = 0 \forall v \in V. $$

supinf
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  • Ok thought I was missing something like deriving, thank you a lot! I don't get just one thing, why the $\frac{1}{2}$ is only ahead the bilinear form ? I mean, is that the way in which the energy operator is defined ? – Cla Nov 09 '18 at 12:29