How to derive the associated Euler-Lagrange equation?

[Math100]

Homework Statement

For each of the following functionals, find the Gateaux differential and use it to derive the associated Euler-Lagrange equation. In each case, be sure to specify all boundary conditions that the stationary path must satisfy.
a) ## S[y]=\int_{1}^{2}(3y'^2-2y^2)dx, y(1)=1, y(2)=3 ##.
b) ## S[y]=y(0)+\frac{1}{2}\int_{0}^{1}(2y'^2+x^2y^2)dx, y(1)=2 ##.

Relevant Equations

Gateaux differential: ## \bigtriangleup S[y, h]=\displaystyle \lim_{\epsilon \to 0}\frac{d}{d\epsilon}S[y+\epsilon h] ## is defined on admissible paths ## y+\epsilon h ## for all ## \epsilon ## in the neighborhood of zero.

A path ## y(x) ## is a stationary path of a functional if the Gateaux differential ## \bigtriangleup S[y, h] ## is zero for all admissible paths ## y+\epsilon h ##.

For the functional ## S[y]=\int_{a}^{b}F(x, y, y')dx, y(a)=A, y(b)=B ##, the Euler-Lagrange equation is ## \frac{d}{dx}(\frac{\partial F}{\partial y'})-\frac{\partial F}{\partial y}=0, y(a)=A, y(b)=B ##.

a) We have ## S[y+\epsilon h]=\int_{1}^{2}[3(y'+\epsilon h')^2-2(y+\epsilon h)^2]dx ##.
Note that ## \frac{d}{d\epsilon}S[y+\epsilon h]=\int_{1}^{2}[6(y'+\epsilon h')h'-4(y+\epsilon h)h]dx=2\int_{1}^{2}[3(y'+\epsilon h')h'-2(y+\epsilon h)]dx ##.
Then ## \bigtriangleup S[y, h]=\frac{d}{d\epsilon}S[y+\epsilon h]\rvert_{\epsilon=0}=2\int_{1}^{2}(3y'h'-2yh)dx ##.
Thus, the Gateaux differential is ## \bigtriangleup S[y, h]=2\int_{1}^{2}(3y'h'-2yh)dx ##.
Consider the functional ## S[y]=\int_{1}^{2}(3y'^2-2y^2)dx, y(1)=1, y(2)=3 ##.
Let ## F(x, y, y')=3y'^2-2y^2 ##.
Then ## \frac{\partial F}{\partial y'}=6y' ## and ## \frac{\partial F}{\partial y}=-4y ##.
Observe that ## \frac{d}{dx}(\frac{\partial F}{\partial y'})-\frac{\partial F}{\partial y}=0\implies 6\frac{d^2y}{dx^2}+4y=0 ##.
Therefore, the Euler-Lagrange equation is ## 6\frac{d^2y}{dx^2}+4y=0, y(1)=1, y(2)=3 ##.

b) We have ## S[y+\epsilon h]=y(0)+\frac{1}{2}\int_{0}^{1}(2(y'+\epsilon h')^2+x^2(y+\epsilon h)^2)dx ##.
Note that ## \frac{d}{d\epsilon}S[y+\epsilon h]=y(0)+\frac{1}{2}\int_{0}^{1}[4(y'+\epsilon h')h'+2x^2(y+\epsilon h)h]dx=y(0)+\int_{0}^{1}[2(y'+\epsilon h')h'+x^2(y+\epsilon h)h]dx ##.
Thus, the Gateaux differential is ## \bigtriangleup S[y, h]=\int_{0}^{1}(2y'h'+x^2yh)dx ##.
Consider the functional ## S[y]=y(0)+\frac{1}{2}\int_{0}^{1}(2y'^2+x^2y^2)dx, y(1)=2 ##.
Let ## F(x, y, y')=2y'^2+x^2y^2 ##.
Then ## \frac{\partial F}{\partial y'}=4y' ## and ## \frac{\partial F}{\partial y}=2x^2y ##.
Observe that ## \frac{d}{dx}(\frac{\partial F}{\partial y'})-\frac{\partial F}{\partial y}=0\implies 4\frac{d^2y}{dx^2}-2x^2y=0 ##.
Therefore, the Euler-Lagrange equation is ## 4\frac{d^2y}{dx^2}-2x^2y=0, y(1)=2 ##.

 

[fresh_42]

a) We have ## S[y+\epsilon h]=\int_{1}^{2}[3(y'+\epsilon h')^2-2(y+\epsilon h)^2]dx ##.
Note that ## \frac{d}{d\epsilon}S[y+\epsilon h]=\int_{1}^{2}[6(y'+\epsilon h')h'-4(y+\epsilon h)h]dx=2\int_{1}^{2}[3(y'+\epsilon h')h'-2(y+\epsilon h)]dx ##.

There is a factor ##h## missing (only here at the end).

Then ## \bigtriangleup S[y, h]=\frac{d}{d\epsilon}S[y+\epsilon h]\rvert_{\epsilon=0}=2\int_{1}^{2}(3y'h'-2yh)dx ##.
Thus, the Gateaux differential is ## \bigtriangleup S[y, h]=2\int_{1}^{2}(3y'h'-2yh)dx ##.
Consider the functional ## S[y]=\int_{1}^{2}(3y'^2-2y^2)dx, y(1)=1, y(2)=3 ##.
Let ## F(x, y, y')=3y'^2-2y^2 ##.
Then ## \frac{\partial F}{\partial y'}=6y' ## and ## \frac{\partial F}{\partial y}=-4y ##.
Observe that ## \frac{d}{dx}(\frac{\partial F}{\partial y'})-\frac{\partial F}{\partial y}=0\implies 6\frac{d^2y}{dx^2}+4y=0 ##.
Therefore, the Euler-Lagrange equation is ## 6\frac{d^2y}{dx^2}+4y=0, y(1)=1, y(2)=3 ##.

Are you sure you shouldn't solve this equation? What else should be the initial values for? At least you should divide by ##6.##

b) We have ## S[y+\epsilon h]=y(0)+\frac{1}{2}\int_{0}^{1}(2(y'+\epsilon h')^2+x^2(y+\epsilon h)^2)dx ##.
Note that ## \frac{d}{d\epsilon}S[y+\epsilon h]=y(0)+\frac{1}{2}\int_{0}^{1}[4(y'+\epsilon h')h'+2x^2(y+\epsilon h)h]dx=y(0)+\int_{0}^{1}[2(y'+\epsilon h')h'+x^2(y+\epsilon h)h]dx ##.

Note that ##\frac{d}{d\epsilon} y(0)=0.##

Thus, the Gateaux differential is ## \bigtriangleup S[y, h]=\int_{0}^{1}(2y'h'+x^2yh)dx ##.
Consider the functional ## S[y]=y(0)+\frac{1}{2}\int_{0}^{1}(2y'^2+x^2y^2)dx, y(1)=2 ##.
Let ## F(x, y, y')=2y'^2+x^2y^2 ##.
Then ## \frac{\partial F}{\partial y'}=4y' ## and ## \frac{\partial F}{\partial y}=2x^2y ##.
Observe that ## \frac{d}{dx}(\frac{\partial F}{\partial y'})-\frac{\partial F}{\partial y}=0\implies 4\frac{d^2y}{dx^2}-2x^2y=0 ##.
Therefore, the Euler-Lagrange equation is ## 4\frac{d^2y}{dx^2}-2x^2y=0, y(1)=2 ##.

This is definitely harder to solve, so maybe all you actually need to do is note the Lagrangian. Divide it by ##4##.

It looks okay apart from the typo (missing ##h##), ##y(0)## (vanishes under the differentiation along ##\epsilon##) and the missing divisions (to make ##y''## with a coefficient ##1## in the Lagrangian) and the possible missing solution of the differential equations. At least from what I can tell from re-reading
https://www.physicsforums.com/insights/pantheon-derivatives-part-ii/
which I wrote more than six years ago.

 

[pasmith]

For (b), [tex]\begin{split}
S[y] &= y(0) + \frac12 \int_0^1 2y'^2 + x^2 y^2 \,dx \\
S[y + \epsilon h] &= y(0) + \epsilon h(0) + \frac12 \int_0^1 2(y' + \epsilon h')^2 + x^2(y + \epsilon h)^2\,dx \\
\frac{d}{d\epsilon}S[y + \epsilon h] &= h(0) + \frac12 \int_0^1 4(y' + \epsilon h')h' + 2x^2(y + \epsilon h)h\,dx
\end{split}[/tex] so [tex]\begin{split}\Delta S[y,h] &= h(0) + \frac12 \int_0^1 4y'h' + 2x^2yh\,dx\\
&= h(0) + \int_0^1 2y'h' + x^2 yh\,dx.\end{split}[/tex] Note the extra [itex]h(0)[/itex] term, which is due to the fact that the variation of [itex]y(0)[/itex] is [itex]y(0) + \epsilon h(0)[/itex]. You are given that [itex]y(1) = 2[/itex], so [itex]h(1) = 0[/itex] is required; you are not given a condition on [itex]y[/itex] at [itex]x = 0[/itex] so you can't assume [itex]h(0) = 0[/itex]. What condition will you impose on [itex]y[/itex] at [itex]x = 0[/itex] to guarantee a unique solution?

 

[Math100]

For (b), [tex]\begin{split}
S[y] &= y(0) + \frac12 \int_0^1 2y'^2 + x^2 y^2 \,dx \\
S[y + \epsilon h] &= y(0) + \epsilon h(0) + \frac12 \int_0^1 2(y' + \epsilon h')^2 + x^2(y + \epsilon h)^2\,dx \\
\frac{d}{d\epsilon}S[y + \epsilon h] &= h(0) + \frac12 \int_0^1 4(y' + \epsilon h')h' + 2x^2(y + \epsilon h)h\,dx
\end{split}[/tex] so [tex]\begin{split}\Delta S[y,h] &= h(0) + \frac12 \int_0^1 4y'h' + 2x^2yh\,dx\\
&= h(0) + \int_0^1 2y'h' + x^2 yh\,dx.\end{split}[/tex] Note the extra [itex]h(0)[/itex] term, which is due to the fact that the variation of [itex]y(0)[/itex] is [itex]y(0) + \epsilon h(0)[/itex]. You are given that [itex]y(1) = 2[/itex], so [itex]h(1) = 0[/itex] is required; you are not given a condition on [itex]y[/itex] at [itex]x = 0[/itex] so you can't assume [itex]h(0) = 0[/itex]. What condition will you impose on [itex]y[/itex] at [itex]x = 0[/itex] to guarantee a unique solution?

I don't know. What will the condition be?

 

[pasmith]

Hint: [itex]y'h' = (y'h)' - y''h[/itex]

 

[Math100]

Hint: [itex]y'h' = (y'h)' - y''h[/itex]

I still don't understand.

 

[pasmith]

For [tex]S[y] = \int_0^1 F(y,y',x)\,dx[/tex] the Gateux derivative is [tex]\begin{split}
\Delta S[y,h] &= \lim_{\epsilon \to 0} \frac{d}{d\epsilon}S[y + \epsilon h] \\
&= \int_0^1 \frac{\partial F}{\partial y}h + \frac{\partial F}{\partial y'}h'\,dx \\
&= \int_0^1 \frac{\partial F}{\partial y}h + \frac{d}{dx}\left( \frac{\partial F}{\partial y'} h \right)- \frac{d}{dx}\left(\frac{\partial F}{\partial y'}\right)h\,dx \\
&= \left[ \frac{\partial F}{\partial y'} h\right]_0^1
+ \int_0^1 \left( \frac{\partial F}{\partial y} - \frac{d}{dx}\left( \frac{\partial F}{\partial y'} \right)\right)h\,dx. \end{split}[/tex] To determine the optimum solution [tex]y : \forall \mbox{ admissible $h$} : \Delta S[y,h] = 0[/tex] we impose the conditions [tex]
\begin{split}
\frac{\partial F}{\partial y} - \frac{d}{dx}\left( \frac{\partial F}{\partial y'} \right) &= 0 \\
\frac{\partial F}{\partial y'}(y(0),y'(0),0)h(0) &= 0 \\
\frac{\partial F}{\partial y'}(y(1),y'(1),1)h(1) &= 0. \end{split}[/tex] This has the form of a second order ODE (the Euler-Lagrange equation) for [itex]y[/itex] subject to two boundary conditions: at [itex]x = c \in \{0,1\}[/itex] either [itex]y(c)[/itex] is specified in which case [itex]h(c) = 0[/itex] so that the corresponding condition is satisfied, or else [itex]y(c)[/itex] is not specified so that [itex]h(c)[/itex] is arbitrary and we must impose [tex]
\frac{\partial F}{\partial y'}(y(c),y'(c),c) = 0.
[/tex] I'm sure your text or your lecturer will have explained this derivation, and in particular the step of integrating [itex]\dfrac{\partial F}{\partial y'}h'[/itex] by parts. In your working, however, instead of following it through completely to obtain not just the Euler-Lagrange equation but also the boundary terms, you reach the second line of my derivation of [itex]\Delta S[y,h][/itex] and then just substitute [itex]F[/itex] into the Euler-Lagrange equation, which you quote in your Relevant Equations for the case where [itex]y[/itex] is specified on both boundaries. That is not the point of this exercise. In part (a) [itex]y[/itex] is specified at both boundaries so that approach works, but part (b) is an extension to the case [tex]
S[y] = y(0) + \int_0^1 F(y,y',x)\,dx[/tex] where the Euler-Lagrange equation does not change but the boundary terms do, and you need to obtain those terms because [itex]y(0)[/itex] is not specified.