1. ## Derivative of Functional

hi there, i am actually studying about functional equation.
I got stucked with some derivatives problem,
and where i could find nowhere to refer or study from,
because it seems it is out of university book level.

my question is this :

what does it means by taking derivative with respect to partial derivative?
can anyone visualize this idea to me?
because i couldn't figure out the term with respect to partial derivative,
when it comes to a functional equation,
of which is a differential equations.

For e.g : F(x,y(x),y'(x),y''(x)) , find $\displaystyle \frac{d}{dx}$ of $\displaystyle \partial$ F(x,y(x),y'(x),y''(x)) / $\displaystyle \partial$ y' .

How can we write the full solution with partial derivative respect to y' ? and how bout y''?

2. Originally Posted by danong
hi there, i am actually studying about functional equation.
I got stucked with some derivatives problem,
and where i could find nowhere to refer or study from,
because it seems it is out of university book level.

my question is this :

what does it means by taking derivative with respect to partial derivative?
can anyone visualize this idea to me?
because i couldn't figure out the term with respect to partial derivative,
when it comes to a functional equation,
of which is a differential equations.

For e.g : F(x,y(x),y'(x),y''(x)) , find $\displaystyle \frac{d}{dx}$ of $\displaystyle \partial$ F(x,y(x),y'(x),y''(x)) / $\displaystyle \partial$ y' .

How can we write the full solution with partial derivative respect to y' ? and how bout y''?
I assume you're talking about a standard Calculus of Variations problem with a functional of the form $\displaystyle J=\int_a^b F(x,y,y',y'')dx$. Now, for this specific case the variational derivative is of the form $\displaystyle F_y-\frac{\partial }{\partial x}F_{y'}=0$. Now, I agree this can be confusing at first but what that really means is call $\displaystyle F(x,y,y',y'')=F(t,u,v,w)$. So, for example if $\displaystyle F(x,y,y',y'')=x^2y+y'+\frac{y'}{y''}$ you would get $\displaystyle F(t,u,v,w)=t^2u+v+\frac{v}{w}$. Now, the Euler-Lagrange equation then says take $\displaystyle F_u$ so in this case $\displaystyle F_u=t^2$ and $\displaystyle F_v=1+\frac{1}{v}$. But! You have to sub back in what that really means, i.e. $\displaystyle F_y=x^2,F_{y'}=1+\frac{1}{y''}$. Now, treating these again as actual functions of $\displaystyle x$ you compute $\displaystyle F_y-\frac{d}{dx}F_{y'}$. The point is that $\displaystyle F_y,F_{y'}$ intend you to treat $\displaystyle F$ as a function of the independent variables $\displaystyle x,y,y',y''$ and differentiate with respect to the specified one. But! Once done with that the $\displaystyle \frac{d}{dx}$ intends you to treat $\displaystyle F_y,F_{y'}$ again as a function of functions of $\displaystyle x$.

Write back if you need more clatification.

3. Originally Posted by danong
hi there, i am actually studying about functional equation.
I got stucked with some derivatives problem,
and where i could find nowhere to refer or study from,
because it seems it is out of university book level.

my question is this :

what does it means by taking derivative with respect to partial derivative?
can anyone visualize this idea to me?
because i couldn't figure out the term with respect to partial derivative,
when it comes to a functional equation,
of which is a differential equations.

For e.g : F(x,y(x),y'(x),y''(x)) , find $\displaystyle \frac{d}{dx}$ of $\displaystyle \partial$ F(x,y(x),y'(x),y''(x)) / $\displaystyle \partial$ y' .

How can we write the full solution with partial derivative respect to y' ? and how bout y''?
It might be best to look at an example. Consider

$\displaystyle L(x,y,y') = \frac{\sqrt{1 + y'^2}}{\sqrt{y}}$.

Then

$\displaystyle \frac{\partial L}{\partial y'} = \frac{1}{\sqrt{y}} \frac{y'}{\sqrt{1+y'^2}}$ (treat $\displaystyle x,y$ and $\displaystyle y'$ as all independent variables). Then

$\displaystyle \frac{d}{dx} \left( \frac{\partial L}{\partial y'} \right) = \frac{d}{dx} \left(\frac{1}{\sqrt{y}} \frac{y'}{\sqrt{1+y'^2} } \right)$

$\displaystyle = \frac{\partial}{\partial x} \underbrace{ \left(\frac{1}{\sqrt{y}} \frac{y'}{\sqrt{1+y'^2 }}\right)}_{\text{this will} = 0\, \text{- no}\,x }$ $\displaystyle + \frac{\partial}{\partial y} \underbrace{ \left(\frac{1}{\sqrt{y}} \frac{y'}{\sqrt{1+y'^2 }}\right)}_{\text{hold}\, y'\, \text{fixed} } y'$ $\displaystyle + \frac{\partial}{\partial y'} \underbrace{ \left( \frac{1}{\sqrt{y}} \frac{y'}{\sqrt{1+y'^2}} \right)}_{\text{hold}\, y\, \text{fixed} } y''$.

For the last two derivatives, treat $\displaystyle y$ and $\displaystyle y'$ and independent.

(A little slow)