Derivative of Functional

Jul 2009
1
0
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''?
 

Drexel28

MHF Hall of Honor
Nov 2009
4,563
1,566
Berkeley, California
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.
 

Jester

MHF Helper
Dec 2008
2,470
1,255
Conway AR
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)