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}<br /> S[y] &amp;= y(0) + \frac12 \int_0^1 2y&#039;^2 + x^2 y^2 \,dx \\<br /> S[y + \epsilon h] &amp;= y(0) + \epsilon h(0) + \frac12 \int_0^1 2(y&#039; + \epsilon h&#039;)^2 + x^2(y + \epsilon h)^2\,dx \\<br /> \frac{d}{d\epsilon}S[y + \epsilon h] &amp;= h(0) + \frac12 \int_0^1 4(y&#039; + \epsilon h&#039;)h&#039; + 2x^2(y + \epsilon h)h\,dx<br /> \end{split} so \begin{split}\Delta S[y,h] &amp;= h(0) + \frac12 \int_0^1 4y&#039;h&#039; + 2x^2yh\,dx\\<br /> &amp;= h(0) + \int_0^1 2y&#039;h&#039; + 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}<br /> S[y] &amp;= y(0) + \frac12 \int_0^1 2y&#039;^2 + x^2 y^2 \,dx \\<br /> S[y + \epsilon h] &amp;= y(0) + \epsilon h(0) + \frac12 \int_0^1 2(y&#039; + \epsilon h&#039;)^2 + x^2(y + \epsilon h)^2\,dx \\<br /> \frac{d}{d\epsilon}S[y + \epsilon h] &amp;= h(0) + \frac12 \int_0^1 4(y&#039; + \epsilon h&#039;)h&#039; + 2x^2(y + \epsilon h)h\,dx<br /> \end{split} so \begin{split}\Delta S[y,h] &amp;= h(0) + \frac12 \int_0^1 4y&#039;h&#039; + 2x^2yh\,dx\\<br /> &amp;= h(0) + \int_0^1 2y&#039;h&#039; + 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&#039;h&#039; = (y&#039;h)&#039; - y&#039;&#039;h

 

[Math100]

Hint: y&#039;h&#039; = (y&#039;h)&#039; - y&#039;&#039;h

I still don't understand.

 

[pasmith]

For S[y] = \int_0^1 F(y,y&#039;,x)\,dx the Gateux derivative is \begin{split}<br /> \Delta S[y,h] &amp;= \lim_{\epsilon \to 0} \frac{d}{d\epsilon}S[y + \epsilon h] \\<br /> &amp;= \int_0^1 \frac{\partial F}{\partial y}h + \frac{\partial F}{\partial y&#039;}h&#039;\,dx \\<br /> &amp;= \int_0^1 \frac{\partial F}{\partial y}h + \frac{d}{dx}\left( \frac{\partial F}{\partial y&#039;} h \right)- \frac{d}{dx}\left(\frac{\partial F}{\partial y&#039;}\right)h\,dx \\<br /> &amp;= \left[ \frac{\partial F}{\partial y&#039;} h\right]_0^1 <br /> + \int_0^1 \left( \frac{\partial F}{\partial y} - \frac{d}{dx}\left( \frac{\partial F}{\partial y&#039;} \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 <br /> \begin{split}<br /> \frac{\partial F}{\partial y} - \frac{d}{dx}\left( \frac{\partial F}{\partial y&#039;} \right) &amp;= 0 \\<br /> \frac{\partial F}{\partial y&#039;}(y(0),y&#039;(0),0)h(0) &amp;= 0 \\<br /> \frac{\partial F}{\partial y&#039;}(y(1),y&#039;(1),1)h(1) &amp;= 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 <br /> \frac{\partial F}{\partial y&#039;}(y(c),y&#039;(c),c) = 0.<br /> 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&#039;}h&#039; 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 <br /> S[y] = y(0) + \int_0^1 F(y,y&#039;,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.