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I want to show that $$ \partial_v\left( \frac{1}{2} | \nabla u| ^2\right)=\nabla u^T \nabla v .$$

I was using the standard formula to calculate directional derivatives, i.e. $\nabla f(x) \cdot v$, but it doesn't seem to give the desired result.

Any help in helping me understand this is appreciated!

Hakim
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Hku
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  • Your formula does not seem to make sense. If $v$ is a (fixed?) vector, what does $\nabla v$ mean? – Steven Gubkin Jun 11 '14 at 12:12
  • We want to calculate the directional derivative in the direction of $v$. – Hku Jun 11 '14 at 12:13
  • The right hand side of your formula includes the term $\nabla v$ which is meaningless in this context. – Steven Gubkin Jun 11 '14 at 12:27
  • I got this problem from someone else, it could be a typo then perhaps. What would be the correct expression then? $ \nabla u^T v$? – Hku Jun 11 '14 at 12:29
  • Do you know the full multivariable chain rule, in terms of composing the total derivatives? If not I do not think I can help you. For what its worth, I compute $\partial_v (\frac{1}{2} \left|\nabla u\right|^2) = Hu(v) \cdot \nabla u$, where $Hu$ is the hessian matrix of $u$. – Steven Gubkin Jun 11 '14 at 12:43

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Okay, let me guess: $u$ is a function (probably from the Sobolev space $H^1$) and you want to differentiate the functional $E \colon H^1 \to \mathbb{R}$ defined by $E(u)=\frac{1}{2} \int |\nabla u|^2 \, dx$. Am I (essentially) right? If so, you want to compute $$ \lim_{\varepsilon \to 0} \frac{E(u+\varepsilon v) - E(u)}{\varepsilon} = \lim_{\varepsilon \to 0} \frac{1}{2\varepsilon} \left( \int 2\varepsilon (\nabla u)^T \nabla v\, dx + \varepsilon^2 \int |\nabla v|^2 \, dx \right), $$ which is similar to your expression.

Siminore
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  • Thanks for your reply, yes you are right. I feel a bit weak in dealing with these kind of things, can you recommend me some articles/books to grasp dealing with these (i.e. differentiating norms (of gradients) and stuff)? In other words, I wonder how you can justify simplifying the expression $\int |\nabla u+ \epsilon \nabla v|^2$ Never mind, I see it now writing it out as an inproduct. – Hku Jun 11 '14 at 13:54
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    I think that the classical book by Ambrosetti and Prodi, A primer of nonlinear analysis, is still a very good reference for infinite-dimensional calculus. – Siminore Jun 11 '14 at 14:18
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You should use the following facts for $F:\mathbb R^n \to \mathbb R^m,G: \mathbb R^m \to \mathbb R$, and $H = G \circ F$.

  • If $F$ is differentiable then its partial derivative at $x$ in the direction $y$, $\delta F(x,y): \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}^m$. And if $F$ is differentiable then $(\delta F(x,y))_i = \langle \nabla F_i(x), y\rangle$ with $F(x)=(F_1(x), \ldots, F_m(x))$.

  • If the directional derivatives $\delta F(x,y)$ and $\delta G(F(x),y)$ are well defined, it follows that the directional derivative of $H$ is given by $\delta H(x,y)= \delta G(F(x),\delta G(x,y))$.

Now it should be easy to compute your directional derivative.

Surb
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  • The LHS is the derivative of a scalar function. $\frac{1}{2} \left| \nabla u \right|^2$ is a real number! – Steven Gubkin Jun 11 '14 at 13:35
  • Hello Surb, thanks for your reply. however, I am sorry for the confusion. The LHS evaluated at the point $u$ is what should equal the RHS. So then we are indeed comparing apples to apples. – Hku Jun 11 '14 at 13:50
  • ah.. sorry got it wrong... |x| is the norm of x and not the component-wise absolute value... Anyway you can use the formulas I gave you to compute your directional derivative. – Surb Jun 11 '14 at 14:59