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Determine the Gâteaux and the Fréchet derivate of $$ f\colon\mathbb{R}^2\to\mathbb{R}^2, z=\begin{pmatrix}x\\y\end{pmatrix}\mapsto\begin{pmatrix}\sin x\cosh y\\\cos x\sinh y\end{pmatrix}. $$

I started with the Gâteaux derivate with any direction $h=(h_1,h_2)$:

$$ \lim\limits_{t\to 0}\frac{f(z+th)-f(z)}{t}=\lim\limits_{t\to 0}\frac{1}{t}\begin{pmatrix}\sin(x+th_1)\cosh(y+th_2)-\sin(x)\cosh(y)\\\cos(x+th_1)\sinh(y+th_2)-\cos(x)\sinh(y)\end{pmatrix} $$ This is the differential quotient, isn't it? Is the Gâteaux derivate therefore given by $$ Df(x)[h]=\begin{pmatrix}(\sin(x)\cosh(y))'\\(\cos(x)\sinh(y))'\end{pmatrix}? $$

Let me determine the Fréchet derivate later.

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No. You didn't differentiate correctly. For example, you need to differentiate $t \mapsto \sin(x+th_1)\cosh(y+th_2)$ with respect to $t$, not $x$, so that terms with $h_1, h_2$ appear.

If a function has a Fréchet derivative $Df(x)$, then the Gâteaux derivative can be easily computed using the chain rule. Let $\phi_{x,h}(t) = x+t h$, and consider the function $f \circ \phi_{x,h}$. The Gâteaux derivative is then $df(x,h) = (f \circ \phi_{x,h})'(0)$, and it follows from the chain rule that $df(x,h) = Df(x)h$.

In this example, $Df(x) = \begin{bmatrix} \cos x \cosh y & \sin x \sinh y \\ -\sin x \sinh y & \cos x \cosh y \end{bmatrix}$, hence $df(x,h) = \begin{bmatrix} h_1 \cos x \cosh y + h_2 \sin x \sinh y \\ -h_1 \sin x \sinh y + h_2 \cos x \cosh y \end{bmatrix}$, where $h = \binom{h_1 } {h_2}$.

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That is how the Gâteaux derivative is defined. See your first limit in the question, you are taking the limit with respect to $t$. Basically you are turning a higher dimensional function into a one dimensional one. –  copper.hat Feb 12 '13 at 17:19
    
Is the Fréchet-derivate always identical with the Jacobian matrix in finite dimensional case?! And does the Fréchet-derivate always exists then when all partial derivates exists? –  math12 Feb 12 '13 at 17:55
    
Essentially yes. The Jacobian is the matrix of partials, hence is the matrix representation of the Fréchet derivative in the standard basis. However, just having the partials does not ensure the existence of the Fréchet derivative. One sufficient condition is if the partials exist and are continuous, then the Fréchet derivative exists. –  copper.hat Feb 12 '13 at 18:00
    
I ask because: Here you suggested to calculate the Fréchet derivate first (what here is: the Jacobian matrix). But to do so, one has to know that the Fréchet derivate exists. And this is the case here, because all partials exist and they are continious? –  math12 Feb 12 '13 at 18:02
    
Yes, the functions here are smooth, hence Fréchet differentiable. As you are probably aware, the Gâteaux derivative can exist when the Fréchet derivatve does not. –  copper.hat Feb 12 '13 at 18:07
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