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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.

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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.

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  • $\begingroup$ 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
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    $\begingroup$ @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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