0
$\begingroup$

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

$$\tag{1}\int_{y_1}^{y_2}\dfrac{x'^2}{\sqrt{x'^2+x^2}}\,\mathrm{d}y$$

$$\tag{2}\int_{x_1}^{x_2}y^{3/2}\,\mathrm{d}s$$

$$\tag{3}\int\dfrac{y\cdot{y'}}{1+yy'}\,\mathrm{d}x$$

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.

$\endgroup$
2
$\begingroup$

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

$$\frac{\partial{f}}{\partial{y}}-\frac{\mathrm{d}}{\mathrm{d}x}\frac{\partial{f}}{\partial{y'}}=0.$$

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

$$f-y'\frac{\partial{f}}{\partial{y'}}=C=\text{constant}.$$

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

$$\frac{\partial{f}}{\partial{y'}}=\frac{2y^2y'+y'^3}{\left(y^2+y'^2\right)^{3/2}}\implies\frac{y'^2}{\sqrt{y^2+y'^2}}-\frac{2y^2y'^2+y'^4}{\left(y^2+y'^2\right)^{3/2}}=C.$$

Rewriting the first term gives

$$\frac{y^2y'^2+y'^4}{\left(y^2+y'^2\right)^{3/2}}-\frac{2y^2y'^2+y'^4}{\left(y^2+y'^2\right)^{3/2}}=C\implies-\frac{y^2y'^2}{\left(y^2+y'^2\right)^{3/2}}=C,$$

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:

$$\frac{\partial{f}}{\partial{y}}=-\frac{yy'^2}{\left(y^2+y'^2\right)^{3/2}},\,\frac{\partial{f}}{\partial{y'}}=\frac{2y^2y'+y'^3}{\left(y^2+y'^2\right)^{3/2}},\,\frac{\mathrm{d}}{\mathrm{d}x}\frac{\partial{f}}{\partial{y'}}=\frac{y(2y^2-y'^2)(yy''-y'^2)}{\left(y^2+y'^2\right)^{5/2}}.$$

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.

$\endgroup$
  • $\begingroup$ you explanation is perfect but when I'm trying to derivate the functional,I can't get the same answer as you. $\endgroup$ – Erbil Jun 17 '13 at 14:11
  • $\begingroup$ $$\dfrac {\partial F} {\partial y'}=\dfrac {2y'\sqrt {y^{2}+y'^{2}}-2y'y'^{2}} {\left( y^{2}+y'^{2}\right)^{3/2} }$$ $\endgroup$ – Erbil Jun 17 '13 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 Jun 17 '13 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 :( $\endgroup$ – Erbil Jun 17 '13 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 Jun 17 '13 at 21:39

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.