Integration by parts with a dxdy

[bugatti79]

Homework Statement

The functional $I(u,v)=\int_\Omega F(x,y,u,v,u_x,u_y, v_x,v_y)dxdy$

The partial variation of a functional is given as

$\displaystyle \delta I =\int_\Omega (\frac{\partial F}{\partial u} \delta u+\frac{\partial F}{\partial u_x} \delta u_x+\frac{\partial F}{\partial u_y} \delta u_y+\frac{\partial F}{\partial v} \delta v+\frac{\partial F}{\partial v_x} \delta v_x+\frac{\partial F}{\partial v_y} \delta v_y)dxdy$

Homework Equations

The Attempt at a Solution

The next step in developing this equation is to integrate by parts the 2nd, 3rd, 5th and 6th terms.

1) Why do we not integrate the 1st term since F and $\delta u$ are both functions of x and y, ie a product..? Similarly for the 4th term?

2) For the second term I let $U= \frac{\partial F}{\partial u_x}$ and $\displaystyle dV=\frac{\partial \delta u}{\partial x}$ Therefore

$\displaystyle \int_\Omega (\frac{\partial F}{\partial u_x} \frac{\partial \delta u_x}{\partial x} )dxdy=\frac{ \partial F}{\partial u_x} \delta u - \int_\Omega [\frac{\partial }{\partial x}(\frac{\partial F}{\partial u_x}) \delta u]dxdy=\int_\Omega [\frac{\partial}{\partial x}(\frac{ \partial F}{\partial u_x} \delta u) - \frac{\partial }{\partial x}(\frac{\partial F}{\partial u_x}) \delta u]dxdy$

We seem to be integrating wrt x, how do we justify not doing anything with dy...? Becasue F and u are also functions of y..? Thanks

 

[klondike]

You might not have *fully* understood the derivation of Euler-Lagrange equation for the classic example of
S(t)=\int_{a}^{b}L(t,x,\dot{x})dt. I'm using the classical mechanics convention here where x is function of t, and \dot{x}=dx/dt, L is Lagrangian.

The integration by part of the selected terms does the following two good things:
1) Make use of the nice boundary conditions of the perturbation function (vanished at both ends)
2) Leave the rest parts to be handled by the Fundamental Lemma of Calculus Of Variations -- Which gives you the Oiler-Lagrange Equation.

Homework Statement

The functional $I(u,v)=\int_\Omega F(x,y,u,v,u_x,u_y, v_x,v_y)dxdy$

The partial variation of a functional is given as

$\displaystyle \delta I =\int_\Omega (\frac{\partial F}{\partial u} \delta u+\frac{\partial F}{\partial u_x} \delta u_x+\frac{\partial F}{\partial u_y} \delta u_y+\frac{\partial F}{\partial v} \delta v+\frac{\partial F}{\partial v_x} \delta v_x+\frac{\partial F}{\partial v_y} \delta v_y)dxdy$

Homework Equations

The Attempt at a Solution

The next step in developing this equation is to integrate by parts the 2nd, 3rd, 5th and 6th terms.

1) Why do we not integrate the 1st term since F and $\delta u$ are both functions of x and y, ie a product..? Similarly for the 4th term?

2) For the second term I let $U= \frac{\partial F}{\partial u_x}$ and $\displaystyle dV=\frac{\partial \delta u}{\partial x}$ Therefore

$\displaystyle \int_\Omega (\frac{\partial F}{\partial u_x} \frac{\partial \delta u_x}{\partial x} )dxdy=\frac{ \partial F}{\partial u_x} \delta u - \int_\Omega [\frac{\partial }{\partial x}(\frac{\partial F}{\partial u_x}) \delta u]dxdy=\int_\Omega [\frac{\partial}{\partial x}(\frac{ \partial F}{\partial u_x} \delta u) - \frac{\partial }{\partial x}(\frac{\partial F}{\partial u_x}) \delta u]dxdy$

We seem to be integrating wrt x, how do we justify not doing anything with dy...? Becasue F and u are also functions of y..? Thanks

 

[bugatti79]

You might not have *fully* understood the derivation of Euler-Lagrange equation for the classic example of
S(t)=\int_{a}^{b}L(t,x,\dot{x})dt. I'm using the classical mechanics convention here where x is function of t, and \dot{x}=dx/dt, L is Lagrangian.

The integration by part of the selected terms does the following two good things:
1) Make use of the nice boundary conditions of the perturbation function (vanished at both ends)

2) Leave the rest parts to be handled by the Fundamental Lemma of Calculus Of Variations -- Which gives you the Oiler-Lagrange Equation.

Homework Statement

The functional $I(u,v)=\int_\Omega F(x,y,u,v,u_x,u_y, v_x,v_y)dxdy$

The partial variation of a functional is given as

$\displaystyle \delta I =\int_\Omega (\frac{\partial F}{\partial u} \delta u+\frac{\partial F}{\partial u_x} \delta u_x+\frac{\partial F}{\partial u_y} \delta u_y+\frac{\partial F}{\partial v} \delta v+\frac{\partial F}{\partial v_x} \delta v_x+\frac{\partial F}{\partial v_y} \delta v_y)dxdy$

Homework Equations

The Attempt at a Solution

The next step in developing this equation is to integrate by parts the 2nd, 3rd, 5th and 6th terms.

1) Why do we not integrate the 1st term since F and $\delta u$ are both functions of x and y, ie a product..? Similarly for the 4th term?

I see, if I had of read on further I would have spotted this thanks.

2) For the second term I let $U= \frac{\partial F}{\partial u_x}$ and $\displaystyle dV=\frac{\partial \delta u}{\partial x}$ Therefore

$\displaystyle \int_\Omega (\frac{\partial F}{\partial u_x} \frac{\partial \delta u_x}{\partial x} )dxdy=\frac{ \partial F}{\partial u_x} \delta u - \int_\Omega [\frac{\partial }{\partial x}(\frac{\partial F}{\partial u_x}) \delta u]dxdy=\int_\Omega [\frac{\partial}{\partial x}(\frac{ \partial F}{\partial u_x} \delta u) - \frac{\partial }{\partial x}(\frac{\partial F}{\partial u_x}) \delta u]dxdy$

We seem to be integrating wrt x, how do we justify not doing anything with dy...? Becasue F and u are also functions of y..? Thanks

Just on a technical note regarding integrating parts. How do we treat this dxdy arrangement since we do not have a double integral such that we can integrate the inner and then the outer...? We know u is a function of both x and y according to post 1 but we seem to have done nothing with dy...? thanks

 

[klondike]

I see, if I had of read on further I would have spotted this thanks.

You are welcome! This happens a lot in self-studying.

Just on a technical note regarding integrating parts. How do we treat this dxdy arrangement since we do not have a double integral such that we can integrate the inner and then the outer...?

We don't actually integrate the functional because u,v, and their partial derivatives are unknown. Or did I miss your question?

We know u is a function of both x and y according to post 1 but we seem to have done nothing with dy...? thanks

We are dealing with the terms related to first variation of \frac{\partial u}{\partial x}. y is treated as constant.