I'm doing my own research on the Euler-Lagrange equation and came across a proof in van Brunt's textbook "The Calculus of Variations". However, there is something I don't quite understand.

Here is an excerpt from the second chapter (with a little bit of paraphrasing):

Let $J : C^2[x_0, x_1] \to \mathbb{R}$ be a functional of the form $\displaystyle J(y) = \int_{x_0}^{x_1} \! f(x, y, y') \, \mathrm{d}x$, where $f$ is assumed to have at least second order partial derivatives with respect to $x, y, y'$. Assume $y$ has fixed endpoints, i.e. $y(x_0) = y_0$ and $y(x_1) = y_1$.

Now assume that $J$ has a local maximum at $y$. Then there is an $\epsilon > 0$ such that $J(\hat{y}) - J(y) \le 0$ for all $\hat{y} \in \{y \in C^2[x_0, x_1]: y(x_0) = y_0 \text{ and } y(x_1) = y_1\}$ such that $\|\hat{y} - y\| < \epsilon$.

For any $\hat{y}$ there is an $\eta$ such that $\hat{y} = y + \epsilon \eta$, and for $\epsilon$ small Taylor's theorem implies that

$\begin{align} f(x, \hat{y}, \hat{y}') &= f(x, y + \epsilon \eta, y' + \epsilon \eta')\\ &= f(x, y, y') + \epsilon \left( \eta \frac{\partial f}{\partial y} + \eta' \frac{\partial f}{\partial y'} \right) + O(\epsilon^2) \end{align}$.

I'd like to ask: Why is the Taylor series valid? I've seen Taylor series for functions of several variables, but never for a function $f$ of functions $\hat{y}(x), \hat{y}(x)'$.

The textbook states

Here, we regard $f$ as a function of three independent variables $x, y,$ and $y'$ and the partial derivatives in the above expression are all evaluated at the point $(x, y, y')$.

But is it okay to just regard functions as independent variables like this?

  • 1
    $\begingroup$ The idea is simply to fix $x$ and differentiate w.r.t. $\epsilon$ - you will obtain the very same expression for any fixed $x$, which thus hold regardless of $x$ $\endgroup$
    – Ilya
    Apr 4 '13 at 9:29
  • $\begingroup$ Do you mean differentiating $y + \epsilon \eta$ in terms of $\epsilon$? Sorry, but I'm having trouble imagining how that would work algebraically, although the idea of fixing $x$ makes sense to me. $\endgroup$
    – Sp3000
    Apr 4 '13 at 9:56

For one variable you can write Taylor series around $x=a$ as below $$f(a+\epsilon)-f(a)=\epsilon \frac{df(x=a)}{dx}+\frac{\epsilon^2}{2}\frac{d^2f(x=a)}{dx^2}+O(\epsilon^3)$$ For functional case the same can be written as $$J(y)=\int_a^bf(x,y,y')dx$$ and for weak variations it follows that $$\hat y=y+\epsilon t\Rightarrow J(\hat y)=J(y+\epsilon t)=\int_a^bf(x,y+\epsilon t,y'+\epsilon t')dx$$ If you want to have stationary $y$ below equation must be zero, since we don't want any variations for integral value around $y$ $$J(y+\epsilon t)-J(y)=\int_a^b\bigg(f(x,y+\epsilon t,y'+\epsilon t')-f(x,y,y')\bigg)dx$$ The integral is zero independent of $dx$ if below condition is satisfied for $[a,b]$ $$f(x,y+\epsilon t,y'+\epsilon t')-f(x,y,y')=0$$ If you compare it with the Taylor expansion formula you can see that they have same formulation ifyo replace $\epsilon$ by $\epsilon t$ and it follows that $$f(x,y+\epsilon t,y'+\epsilon t')-f(x,y,y')=\epsilon t \frac{\partial f(x,y,y')}{\partial y}+\epsilon t' \frac{\partial f(x,y,y')}{\partial y'}+O(\epsilon^2)=0$$

Edit: They treat the functions as independent inside $f$. Consider the optimization case with constraints. For objective function you treat every variable as independent but with constraint equations you impose some relations between independent variables. It is the same here: the variables are treated as independent but the condition is imposed by $\hat y=y+\epsilon t$ and $\hat y'=y'+\epsilon t'$

  • $\begingroup$ Thank you for the answer, but as I've only started researching this topic I'm not sure what you mean by "stationary $y$" and "optimization case with restraints".... $\endgroup$
    – Sp3000
    Apr 4 '13 at 10:01
  • $\begingroup$ A stationary point means a point where the derivative of the function is zero. A stationary curve means the same for integral. If y is a stationary curve it means that if you apply small variations to y the value of integral remains the same. If you are not familiar with optimization I suugest you to study optimization before studying variational calculus. It would help you very much in understanding the concepts. $\endgroup$
    – AnilB
    Apr 4 '13 at 13:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.