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

• Math100
In summary, the conversation discusses the Euler's equation of the functional and its application in calculating the limit of a specific equation. The result is the Euler equation (strong form) of 2u' - 2u'' - e^u = 0. The conversation also includes a solution to this equation using the method of integration by parts.
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 ##.

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

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WWGD said:
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.

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
Mark44 said:
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?

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.

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Math100
fresh_42 said:
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.

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

Math100 said:
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
fresh_42 said:
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} ##.

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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).##

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Math100
But where does the solution ## E(F)=e^{u} ## come from?

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

Math100 said:
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##?

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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.

Math100
Math100 said:
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:$$0=-2u\left(x\right)''+e^{u\left(x\right)}$$

Math100 and fresh_42
renormalize said:
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.

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*}

Math100 and renormalize

## What is the Euler equation in the context of calculus of variations?

The Euler equation, in the context of calculus of variations, is a differential equation that provides a necessary condition for a function to be an extremum of a functional. It is derived by considering the first variation of the functional and setting it to zero.

## How do you derive the Euler equation from a given functional?

To derive the Euler equation, start with a functional of the form $$J[y] = \int_{a}^{b} F(x, y, y') \, dx$$. Introduce a perturbation $$y(x) + \epsilon \eta(x)$$ and expand the functional to first order in $$\epsilon$$. Set the first variation $$\delta J = 0$$ and apply integration by parts to isolate $$\eta(x)$$. This leads to the Euler-Lagrange equation: $$\frac{\partial F}{\partial y} - \frac{d}{dx}\left(\frac{\partial F}{\partial y'}\right) = 0$$.

## Can the Euler equation be applied to functionals involving higher-order derivatives?

Yes, the Euler equation can be extended to functionals involving higher-order derivatives. For a functional $$J[y] = \int_{a}^{b} F(x, y, y', y'', \ldots, y^{(n)}) \, dx$$, the corresponding Euler-Lagrange equation is derived similarly, resulting in a higher-order differential equation: $$\frac{\partial F}{\partial y} - \frac{d}{dx}\left(\frac{\partial F}{\partial y'}\right) + \frac{d^2}{dx^2}\left(\frac{\partial F}{\partial y''}\right) - \cdots + (-1)^n \frac{d^n}{dx^n}\left(\frac{\partial F}{\partial y^{(n)}}\right) = 0$$.

## What are common mistakes to avoid when deriving the Euler equation?

Common mistakes include neglecting boundary terms when integrating by parts, incorrectly applying the chain rule, and not properly handling higher-order derivatives if present. It is crucial to carefully manage the variations and ensure all terms are accounted for in the derivation process.

## How do boundary conditions affect the Euler equation?

Boundary conditions play a crucial role in solving the Euler equation. Natural boundary conditions arise from the requirement that the first variation vanishes, leading to additional conditions on the boundary values of $$y$$ and its derivatives. These conditions must be satisfied along with the Euler equation to find the extremal function.

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