extremals no. 1

**wik_chick88** · September 6th 2009, 05:09 PM

Find the extremals of $\int^{2}_{0}[\dot{x}^2 + 2\dot{x}] dt$ with $x(0) = 0$ and $x(2) = 1$ subject to the constraint $\int^{2}_{0}x dt = 2$

**luobo** · September 6th 2009, 07:26 PM

Originally Posted by wik_chick88

Find the extremals of $\int^{2}_{0}[\dot{x}^2 + 2\dot{x}] dt$ with $x(0) = 0$ and $x(2) = 1$ subject to the constraint $\int^{2}_{0}x dt = 2$

$\text{A problem of Hamilton's Variational Principle.}$

$L(\dot{x},x,t) = \dot{x}^2 + 2\dot{x} +\lambda (x-1)$

- $\frac{d}{dt} \frac{\partial L}{\partial \dot{x}} - \frac{\partial L}{\partial x} =0 \Rightarrow 2\ddot{x} - \lambda = 0 \Rightarrow x(t)=\frac{1}{4} \lambda \, t^2+\mu t + \gamma$

- $x(0)=0 \Rightarrow \gamma=0$
$x(2)=1 \Rightarrow \lambda+2\mu=1 \Rightarrow \mu=\frac{1-\lambda}{2}$
$\int^{2}_{0}x \, dt = 2 \Rightarrow \lambda=-3$
$x(t)=-\frac{3}{4} t^2+2 t \Rightarrow \dot{x}(t)=-\frac{3}{2} t+2$
$\int^{2}_{0}(\dot{x}^2 + 2\dot{x}) \, dt=\int^{2}_{0}(\tfrac{9}{4}t^2 -9t + 8) \, dt = 4$

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