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>



Change of functional derivative 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?


