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

 

[Math100]

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

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

 

[pasmith]

Hint: $y'h' = (y'h)' - y''h$

 

[Math100]

Hint: $y'h' = (y'h)' - y''h$

I still don't understand.

 

[pasmith]

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