Create the Euler–Lagrange equation for the following questions (if it's necessary change the variables).




I don't have an idea about $(1)$ and $(3)$. But here it's what I have tried for $(2)$.

2) $$\int_{x_1}^{x_2}y^{3/2}\,\mathrm{d}s = \int_{x_1}^{x_2}y^{3/2}(1+y'^2)^{1/2} = \int_{y_1}^{y_2}(1+x'^2)(y^{3/2})\,\mathrm{d}y$$

So our Euler equation is:

$$\dfrac{\mathrm{d}}{\mathrm{d}y}\left(\dfrac{\partial F}{\partial x'}\right) - \dfrac{\partial F}{\partial x} = 0$$

Then I have to find $Y'$ or $X'$. But I did not take a differential equations course yet, we use Beltrami identity to calculate the extremum points.


1 Answer 1


We know that the functional

$$I[y]=\int_{x_1}^{x_2} f(x,y(x),y'(x))\,\mathrm{d}x$$

is extremized only if $y$ satisfies the Euler–Lagrange equation


Now, if $f$ is independent of $x$, then the Euler–Lagrange equation reduces to the Beltrami identity


Note that the names of the variables are immaterial. Henceforth, we use the same notation as above.

In the first case, we have $f(x,y,y')=\dfrac{y'^2}{\sqrt{y^2+y'^2}}$, which is independent of $x$. Hence we can use the Beltrami identity. Differentiation gives


Rewriting the first term gives


which is a first-order non-linear ordinary differential equation. In total, two constants will be obtained.

Of course, we could use the Euler–Lagrange equation instead, but that would be much more complicated; just consider the derivatives:


The second and thrid cases can be treated similarly; both integrands are independent of $x$. However, in the second case, note that $\mathrm{d}s=\sqrt{1+y'^2}\,\mathrm{d}x$, so $f(x,y,y')=y^{3/2}\sqrt{1+y'^2}$; this expression is independent of $x$, as it should be.

  • $\begingroup$ you explanation is perfect but when I'm trying to derivate the functional,I can't get the same answer as you. $\endgroup$
    – BySpecops.
    Commented Jun 17, 2013 at 14:11
  • $\begingroup$ $$\dfrac {\partial F} {\partial y'}=\dfrac {2y'\sqrt {y^{2}+y'^{2}}-2y'y'^{2}} {\left( y^{2}+y'^{2}\right)^{3/2} }$$ $\endgroup$
    – BySpecops.
    Commented Jun 17, 2013 at 14:36
  • $\begingroup$ @Erbil It follows from the quotient rule: $$\frac{\partial{f}}{\partial{y'}}=\frac{2y'\sqrt{y^2+y'^2}-y'^3\Big/\sqrt{y^2+y'^2}}{y^2+y'^2}=\frac{2y^2y'+y'^3}{\left(y^2+y'^2\right)^{3/2}}.$$ $\endgroup$
    – Librecoin
    Commented Jun 17, 2013 at 18:25
  • $\begingroup$ I think I have used the same rule,and I have also take a care about partial derivative.But where did you get +,and where did the 2 coefficient go on y'3? edit : There is more than these.I can't also understand where the (y^2+y'^2)^1/2 gone :( $\endgroup$
    – BySpecops.
    Commented Jun 17, 2013 at 21:11
  • 1
    $\begingroup$ @Erbil Multiply the numerator and the denominator in the first fraction by $\sqrt{y^2+y'^2}$, and the equality follows. $\endgroup$
    – Librecoin
    Commented Jun 17, 2013 at 21:39

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