# Change of functional derivative under rigid coordinate transformation

Suppose I have a functional $$E=\int F(y_{1,1},..y_{1,n},y_{2,1}\ldots,y_{n,n})d\boldsymbol{x}\,,$$ where $\boldsymbol{y}:\mathbb{R}^{n}\to\mathbb{R}^{n},\,\boldsymbol{y}(\boldsymbol{x})=\left(y_{1}(x_{1},...x_{n}),...,y_{n}(x_{1},...x_{n})\right)$, and $y_{1,1},...,y_{n,n}$ are partial derivatives, i.e. $y_{i,j}=\dfrac{dy_{i}}{dx_{i}}$.

How the functional derivative $\dfrac{\delta E[\boldsymbol{y}]}{\delta\boldsymbol{y}(\boldsymbol{x})}$ is changed under a rigid coordinates transformation (of both the domain and the image.) $$\left(x_{1},...,x_{n}\right)^{T}=V\left(v_{1},...,v_{n}\right)^{T};\,\left(y_{1},...,y_{n}\right)^{T}=U\left(u_{1},...,u_{n}\right)^{T},$$ where $U,V$ are orthogonal $n\times n$ matrices.

In other words, given $\dfrac{\delta E[\boldsymbol{y(x)}]}{\delta\boldsymbol{y(x)}}$ what would be the expression for $\dfrac{\delta E[\boldsymbol{u(v)}]}{\delta\boldsymbol{u(v)}}$?

P.S. By " functional derivative" I mean the one that follows from the Euler Lagrange equation for $E$ that depends on partial derivatives of $n$-functions of $n$-variables i.e., $$\dfrac{\delta E[\boldsymbol{y(x)}]}{\delta\boldsymbol{y(x)}}=-\left(\underset{i=1}{\overset{n}{\sum}}\dfrac{\partial}{\partial x_{i}}\dfrac{\partial F}{\partial y_{1,i}},\ldots,\underset{i=1}{\overset{n}{\sum}}\dfrac{\partial}{\partial x_{i}}\dfrac{\partial F}{\partial y_{n,i}}\right)^{T}.$$ Any comments would be appreciated!

The question was forwarded from https://math.stackexchange.com/questions/1906398/change-of-variation-under-rigid-coordinate-transformation

(a comment) Isn't it possible to use the same formula (from the post scriptum) but replacing $u(v)$ as it follows from the definition:$u(v) = U^T y(V v)$ and compute derivatives of $u$ with respect to $v$ as for function decomposition. Then you'll obtain a formula which may take some nice form in tensor notations. Or not?
• In my case the functional $F$ is complicated in one system of coordinates and much more simple in another. This is the main reason to look for transformation of the functional derivative without evaluating $\boldsymbol{u}(\boldsymbol{v})=U^{T}\boldsymbol{y}(V\boldsymbol{y})$. . – AlexN Sep 26 '16 at 13:51
• Using this chain rule formula en.wikipedia.org/wiki/Functional_derivative#Properties leads to $$\dfrac{\delta F[\boldsymbol{u}(\boldsymbol{y})]}{\delta\boldsymbol{y}(\boldsymbol{x})}=\dfrac{\delta F[\boldsymbol{u}(\boldsymbol{y})]}{\delta\boldsymbol{u}[\boldsymbol{y}(\boldsymbol{x})]}\dfrac{d\boldsymbol{u}(\boldsymbol{y})}{d\boldsymbol{y}(\boldsymbol{x})}=\dfrac{\delta F[\boldsymbol{u}(\boldsymbol{y})]}{\delta[\boldsymbol{y}(\boldsymbol{x})]}U$$ – AlexN Sep 26 '16 at 14:00
• And now I need to transform $\dfrac{\delta F[\boldsymbol{u}(\boldsymbol{y})]}{\delta\boldsymbol{u}[\boldsymbol{y}(\boldsymbol{x})]}$ to $\dfrac{\delta F[\boldsymbol{u}(\boldsymbol{v})]}{\delta\boldsymbol{u}(\boldsymbol{v})}$ ? The annotation of functional derivatives from Wikipedia is a little bit confusing for me. I'm not sure I understood it correctly. Any comments/corrections will be appreciated ! – AlexN Sep 26 '16 at 14:02