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If $u(x)$ and $v(x)$ satisfy $u(0)=1$, $v(0)=-1$, $u(\pi/2) =0$, $v(\pi/2) =0$ on extremals of the functional

$$ \int_0^{\pi/2}\left[\big({\frac{du}{dx}\big)^2 +\big(\frac{dv}{dx}\big)^2 + 2 \,u v }\,\right] dx $$

then which of the following is correct?

1) $u({\pi/4}) + v(\pi/4) = 0$

2) $u({\pi/3}) - v(\pi/3) = 0$

3) $u({\pi/4}) - v(\pi/4) = 1$

4) $u({\pi/3}) + v(\pi/3) = 0$

Rodrigo de Azevedo
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zafran
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1 Answers1

1

HINT: You have functional of the form

$$ I[u,v] = \int_0^{\frac{\pi}{2}} \mathcal{L}\left(u,v,u',v'\right)dx, \ \ \text{ where } \ \mathcal{L}\left(u,v,u',v'\right) = \left(u'\right)^2+\left(v'\right)^2 + 2uv. $$

Using the Euler-Lagrange equation for several functions depending on one variable, you can write the system of differential equations

$$ \left\{ \begin{aligned} \frac{\partial \mathcal{L}}{\partial u} - \frac{\mathrm{d} }{\mathrm{d} x} \bigg( \frac{\partial \mathcal{L}}{\partial u'} \bigg) = 0 \\ \frac{\partial \mathcal{L}}{\partial v} - \frac{\mathrm{d} }{\mathrm{d} x} \bigg( \frac{\partial \mathcal{L}}{\partial v'} \bigg) = 0 \end{aligned} \right. \iff \left\{ \begin{aligned} 2v - \frac{\mathrm{d} }{\mathrm{d} x} \left( 2u' \right) = 0 \\ 2u - \frac{\mathrm{d} }{\mathrm{d} x} \left( 2v' \right) = 0 \end{aligned} \right. \iff \left\{ \begin{aligned}v-u'' = 0\\ u - v'' = 0 \end{aligned} \right. $$ Solving these equations for the boundary conditions for $u$ and $v$ at $0$ and $\frac{\pi}{2}$, you will find critical points $(u_0,v_0)$ of the original functional $I[u,v]$ by solving these equations.

As soon as you get these solutions, you can instantly plug in values and determine which answer is correct.

Vlad
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    Thank you very much.I am learning calculus of variation on my own and only help is internet. – zafran May 13 '15 at 17:50