Is this the correct way to find the Euler equation (strong form)?

[Math100]

Homework Statement

Find the Euler equation (strong form) for $\int ((\mathrm{u}')^2+e^{\mathrm{u}}) \, dx$.

Relevant Equations

Euler's equation: $J(y)=\int_{a}^{b} F(x, y, y', y") \, dx$

By the Euler's equation of the functional, we have
$J(\mathrm u)=\int ((\mathrm{u})^{2}+e^{\mathrm{u}}) \, dx$.
Then $J(\mathrm{u}+\epsilon\eta)=\int ((\mathrm{u}'+\epsilon\eta')^{2}+e^{\mathrm{u}+\epsilon\eta}) \, dx=\int ((\mathrm{u})'^{2}+2\epsilon\mathrm{u}'\eta'+\epsilon^{2}(\eta')^{2}+e^{\mathrm{u}}+\epsilon e^{\mathrm{u}}\eta) \, dx$.
Note that $\frac {J(\mathrm{u}+\epsilon\eta)-J(\mathrm{u})} {\epsilon}=\int (2\mathrm{u}'\eta'+\epsilon(\eta')^{2}+e^{\mathrm{u}}\eta) \, dx$.
Consider the following limit:
$\lim_{\epsilon \rightarrow 0} \frac {J(\mathrm{u}+\epsilon\eta)-J(\mathrm{u})} {\epsilon}=\int (2\mathrm{u}'\eta'+e^{\mathrm{u}}\eta) \, dx=0$.
Applying the method of integration by parts, we obtain
$\int (2\mathrm{u}'\eta'+e^{\mathrm{u}}\eta) \, dx=(2\mathrm{u}'\eta)-\int (2\mathrm{u}''\eta+e^{\mathrm{u}}\eta) \, dx=0$.
Thus $2\mathrm{u}'\eta-2\mathrm{u}''\eta-e^{\mathrm{u}}\eta=0\implies 2\mathrm{u}'-2\mathrm{u}''-e^{\mathrm{u}}=0$.
Therefore, the Euler equation (strong form) is $2\mathrm{u}'-2\mathrm{u}''-e^{\mathrm{u}}=0$.

 

[WWGD]

Ouch, please review your Latex. For one, don't include standard English under tags.

 

[Math100]

Ouch, please review your Latex. for one, don't include standard English under tags.

Yes, I was working on it, but still doesn't seem to work. I will see what's wrong.

 

[Mark44]

It's now fixed.
@Math100, for such complicated LaTeX, simpler is better.
For example, all the \mathrm items made it harder to find the two of them missing the leading slash.
Also, instead of \mathrm {u}, you could simplify this to \mathrm u. Braces -- { } -- can be eliminated from fractions, exponents, roots, and many other places if what's in the braces consists of a single character.
Instead of \lim_{x \rightarrow 0}, it's easier to use \lim_{x \to 0}.

 

[Math100]

It's now fixed.
@Math100, for such complicated LaTeX, simpler is better.
For example, all the \mathrm items made it harder to find the two of them missing the leading slash.
Also, instead of \mathrm {u}, you could simplify this to \mathrm u. Braces -- { } -- can be eliminated from fractions, exponents, roots, and many other places if what's in the braces consists of a single character.
Instead of \lim_{x \rightarrow 0}, it's easier to use \lim_{x \to 0}.

May you check/verify the work and solution to see if it's correct/accurate?

 

[fresh_42]

Let's see. We have $J[ u ]=\int_a^b F(x,u,u') \,dx=\int_a^b \left((u')^2+e^u\right)\,dx$ and want to calculate
\begin{align*}
\left. \dfrac{d}{d\varepsilon }\right|_{\varepsilon =0} J(u+\varepsilon \eta)&=\left. \dfrac{d}{d\varepsilon }\right|_{\varepsilon =0} \int_a^b \left(\left( \left(u+\varepsilon \eta\right)'\right)^2+e^{u+\varepsilon \eta}\right)\,dx \\
&=\left. \dfrac{d}{d\varepsilon }\right|_{\varepsilon =0} \int_a^b \left(\left(u'+\varepsilon \eta'\right)^2+e^u\cdot e^{\varepsilon \eta}\right)\,dx\\
&=\left. \dfrac{d}{d\varepsilon }\right|_{\varepsilon =0} \int_a^b\left((u')^2+2u'\varepsilon \eta'+\varepsilon ^2(\eta')^2 + e^u\cdot e^{\varepsilon \eta}\right)\,dx\\
&\stackrel{\text{compact support}}{=}\int_a^b \left. \dfrac{d}{d\varepsilon }\right|_{\varepsilon =0}\left((u')^2+2u'\varepsilon \eta'+\varepsilon ^2(\eta')^2 + e^u\cdot e^{\varepsilon \eta}\right)\,dx\\
&=\int_a^b\left( 2u' \eta' +e^u\eta \right)\,dx\\
&=2\left. u'\eta \right|_a^b -2\int_a^b \eta u'' \,dx +\int_a^b \eta e^u\,dx \\
&=\eta \left(2u'(b)-2u'(a)-2u'(b)+2u'(a)+ \int_a^b e^u\,dx \right)\\
&=\eta \int_a^b e^u\,dx
\end{align*}
That's what I get, but to be honest, variational calculus is not my strength. My solution is $E(F)=e^u.$ But I might have made a mistake.

Corrected. My mistake was that I took the factor $e^u$ for a sum.

 

[Math100]

Let's see. We have $J[ u ]=\int_a^b F(x,u,u') \,dx=\int_a^b \left((u')^2+e^u\right)\,dx$ and want to calculate
\begin{align*}
\left. \dfrac{d}{d\varepsilon }\right|_{\varepsilon =0} J(u+\varepsilon \eta)&=\left. \dfrac{d}{d\varepsilon }\right|_{\varepsilon =0} \int_a^b \left(\left( \left(u+\varepsilon \eta\right)'\right)^2+e^{u+\varepsilon \eta}\right)\,dx \\
&=\left. \dfrac{d}{d\varepsilon }\right|_{\varepsilon =0} \int_a^b \left(\left(u'+\varepsilon \eta'\right)^2+e^u\cdot e^{\varepsilon \eta}\right)\,dx\\
&=\left. \dfrac{d}{d\varepsilon }\right|_{\varepsilon =0} \int_a^b\left((u')^2+2u'\varepsilon \eta'+\varepsilon ^2(\eta')^2 + e^u\cdot e^{\varepsilon \eta}\right)\,dx\\
&\stackrel{\text{compact support}}{=}\int_a^b \left. \dfrac{d}{d\varepsilon }\right|_{\varepsilon =0}\left((u')^2+2u'\varepsilon \eta'+\varepsilon ^2(\eta')^2 + e^u\cdot e^{\varepsilon \eta}\right)\,dx\\
&=\int_a^b\left( 2u' \eta' +e^u\eta \right)\,dx\\
&=2\left. u'\eta \right|_a^b -2\int_a^b \eta u'' \,dx +\int_a^b \eta e^u\,dx \\
&=\eta \left(2u'(b)-2u'(a)-2u'(b)+2u'(a)+ \int_a^b e^u\,dx \right)\\
&=\eta \int_a^b e^u\,dx
\end{align*}
That's what I get, but to be honest, variational calculus is not my strength. My solution is $E(F)=e^u.$ But I might have made a mistake.

I will wait.

 

[Math100]

What does "compact support" mean/indicate in this problem?

 

[fresh_42]

What does "compact support" mean/indicate in this problem?

IIRC then it allowed me to switch integral and differentiation. But you have the same problem. Maybe I should look up the correct theorem.

 

[Math100]

Let's see. We have $J[ u ]=\int_a^b F(x,u,u') \,dx=\int_a^b \left((u')^2+e^u\right)\,dx$ and want to calculate
\begin{align*}
\left. \dfrac{d}{d\varepsilon }\right|_{\varepsilon =0} J(u+\varepsilon \eta)&=\left. \dfrac{d}{d\varepsilon }\right|_{\varepsilon =0} \int_a^b \left(\left( \left(u+\varepsilon \eta\right)'\right)^2+e^{u+\varepsilon \eta}\right)\,dx \\
&=\left. \dfrac{d}{d\varepsilon }\right|_{\varepsilon =0} \int_a^b \left(\left(u'+\varepsilon \eta'\right)^2+e^u\cdot e^{\varepsilon \eta}\right)\,dx\\
&=\left. \dfrac{d}{d\varepsilon }\right|_{\varepsilon =0} \int_a^b\left((u')^2+2u'\varepsilon \eta'+\varepsilon ^2(\eta')^2 + e^u\cdot e^{\varepsilon \eta}\right)\,dx\\
&\stackrel{\text{compact support}}{=}\int_a^b \left. \dfrac{d}{d\varepsilon }\right|_{\varepsilon =0}\left((u')^2+2u'\varepsilon \eta'+\varepsilon ^2(\eta')^2 + e^u\cdot e^{\varepsilon \eta}\right)\,dx\\
&=\int_a^b\left( 2u' \eta' +e^u\eta \right)\,dx\\
&=2\left. u'\eta \right|_a^b -2\int_a^b \eta u'' \,dx +\int_a^b \eta e^u\,dx \\
&=\eta \left(2u'(b)-2u'(a)-2u'(b)+2u'(a)+ \int_a^b e^u\,dx \right)\\
&=\eta \int_a^b e^u\,dx
\end{align*}
That's what I get, but to be honest, variational calculus is not my strength. My solution is $E(F)=e^u.$ But I might have made a mistake.

Corrected. My mistake was that I took the factor $e^u$ for a sum.

How to differentiate $(u')^2+2u'\varepsilon \eta'+\varepsilon ^2(\eta')^2 + e^u\cdot e^{\varepsilon \eta}$?

Never mind, I got it now. It's $2u' \eta' +2\varepsilon (\eta')^2+\eta e^{u+ \varepsilon \eta}$. And at $\varepsilon=0$, it's $2u' \eta' + \eta e^{u}$.

 

[fresh_42]

In order to exchange a limit, and differentiation is a limit, with an integral we need a majorized convergence. That means, if we replace $\varepsilon$ by $1/n$ and consider a sequence of functions $f_n$ then we have the theorem:

If there is a function $h(x)$ such that $|f_n(x)| < h(x)$ and $\int h(x) <\infty$ then
$$
\lim_{n \to \infty} \int f_n(x)\,dx = \int \lim_{n \to \infty}f_n(x) \,dx
$$

With a compact interval $[a,b]$ (in tech speech: "with a compact support" for cases where the integral isn't over an interval but over some region $\Omega$) we have such an upper limit $h(x)$ whenever the functions are integrable, e.g. continuous.

I also used $\int_a^b u'' \,dx = \left[u'(x)\right]_a^b =u'(b)-u'(a).$

 

[Math100]

But where does the solution $E(F)=e^{u}$ come from?

 

[Math100]

$\eta \int_a^b e^{u} dx= \eta (e^{u(b)}-e^{u(a)})$

 

[fresh_42]

But where does the solution $E(F)=e^{u}$ come from?

Good question.

Let me have a look at the book (Olver, GTM 107, proposition 4.2 and theorem 4.4). We have without my sloppiness with $\eta$
$$
\left. \dfrac{d}{d\varepsilon }\right|_{\varepsilon =0} J(u+\varepsilon \eta)=\int_a^b \underbrace{e^{u(x)}}_{=E(F)=\delta J[ u(x) ]} \eta(x)\, dx
$$
and the theorems say that if $f=u(x)$ is an extremal of $J[ f ]$ then $E(F)=\delta J[ u(x) ]=e^{u(x)}= 0$ and that a solution has to be of the form $E(F)=0.$

But this is never true on $[a,b]$ so there is probably no smooth extremal solution

I think my mistake is the integration step. Since $\eta$ depends on $x$ we cannot simply pretend as if it was a constant. So we actually have
\begin{align*}
\left. \dfrac{d}{d\varepsilon }\right|_{\varepsilon =0} J(u+\varepsilon \eta)&= 2u'(x)\eta(x)|_a^b+\int_a^b \left(e^u -2 u''(x)\right)\cdot \eta(x)\,dx
\end{align*}
How do we bring the constant term $2u'(x)\eta(x)|_a^b$ into the form $\int_a^b \delta(J[ v(x) ])\cdot\eta(x) \,dx$?

 

[fresh_42]

I think your solution is correct, except that it should be $+ e^u$ not minus, and could be written a bit more along my way above.

 

[renormalize]

But where does the solution $E(F)=e^{u}$ come from?

Let's attack this without getting too hung up on mathematical rigor; i.e., we do it the way physicists do!

Start with the 1D action functional:$$J\left[u,u'\right]\equiv\int_{a}^{b}dx\left(\left(u'\right)^{2}+e^{u}\right)$$We want to find the function $u(x)$ which extremizes this action for fixed values of $u(a),u(b)$. So we set the first variation of the functional to zero:$$0=\delta J\left[u,u'\right]=\int_{a}^{b}dx\left(2u'\delta\left(u'\right)+e^{u}\delta u\right)$$with the condition that $\delta u(a)=\delta u(b)=0$. Thus,$$0=\int_{a}^{b}dx\left(2u'\left(\delta u\right)'+e^{u}\delta u\right)\qquad\text{ (𝛿 commutes with ´)}$$$$=\int_{a}^{b}dx\left(\left(2u'\delta u\right)'-2u''\delta u+e^{u}\delta u\right)=\left[2u'\delta u\right]_{a}^{b}+\int_{a}^{b}dx\left(-2u''\delta u+e^{u}\delta u\right)\qquad\text{ (integration by parts)}$$Because $\delta u(x)$ is zero at the integration limits, the term in brackets vanishes, leaving us with:$$0=\int_{a}^{b}dx\left(-2u''+e^{u}\right)\delta u$$And because $\delta u(x)$ is arbitrary for $a<x<b$, this equation in turn requires that the coefficient in parentheses must itself be zero, leading to the Euler-Lagrange equation:$$0=-2u\left(x\right)''+e^{u\left(x\right)}$$

 

[fresh_42]

Because $\delta u(x)$ is zero at the integration limits, the term in brackets vanishes, leaving us with:$$0=\int_{a}^{b}dx\left(-2u''+e^{u}\right)\delta u$$

That was the part I am looking for, thanks.

 

[fresh_42]

Let me summarize my calculations after all that confusion I have caused.

\begin{align*}
\left. \dfrac{d}{d\varepsilon }\right|_{\varepsilon =0} J(u+\varepsilon \eta)&=\left. \dfrac{d}{d\varepsilon }\right|_{\varepsilon =0} \int_a^b \left(\left( \left(u+\varepsilon \eta\right)'\right)^2+e^{u+\varepsilon \eta}\right)\,dx \\
&=\left. \dfrac{d}{d\varepsilon }\right|_{\varepsilon =0} \int_a^b \left(\left(u'+\varepsilon \eta'\right)^2+e^u\cdot e^{\varepsilon \eta}\right)\,dx\\
&=\left. \dfrac{d}{d\varepsilon }\right|_{\varepsilon =0} \int_a^b\left((u')^2+2u'\varepsilon \eta'+\varepsilon ^2(\eta')^2 + e^u\cdot e^{\varepsilon \eta}\right)\,dx\\
&=\left. \dfrac{d}{d\varepsilon }\right|_{\varepsilon =0}\left(\underbrace{\int_a^b (u')^2\,dx}_{\to 0} + \int_a^b(2u'\eta')\varepsilon \,dx+\int_a^b (\eta')^2\varepsilon^2 \,dx+\int_a^be^ue^{\eta \varepsilon }\,dx
\right)\\
&=\left. \dfrac{d}{d\varepsilon }\right|_{\varepsilon =0}\left( \varepsilon \cdot \underbrace{\left[2u'\eta\right]_a^b}_{=:C} -2\varepsilon \int_a^b \eta u''\,dx +\underbrace{\varepsilon^2 \int_a^b (\eta')^2\,dx}_{\to 0} +e^u \int_a^b e^{\eta \varepsilon }\,dx
\right)\\
&=C-2 \int_a^b u''\eta\,dx +\int_a^b e^u\underbrace{\left(\left. \dfrac{d}{d\varepsilon }\right|_{\varepsilon =0}e^{\varepsilon \eta}\,dx\right)}_{=\eta}\\
&=C+\int_a^b \underbrace{\left(-2u''(x)+e^{u(x)}\right)}_{=E(F)}\cdot \eta(x)\,dx
\end{align*}