Euler-Lagrange equation in Optimal Control book by Kirk

[mad mathematician]

Homework Statement

Question 4.14 in the book asks the following:
Find the extremal curves for the functional: $J(x)=\int_0^{t_f}[\frac{\sqrt{1+\dot{x}(t)^2}}{x(t)}dt$ $x(0)=0$, and $x(t_f)$ must lie on the line $\theta(t)=t-5$.

Relevant Equations

Euler Lagrange equation.
Transervality condition which in the book of Kirk on page 140 equation (4.2-72).

According to the SM which can be found in google or any other search engine the EL can be simplified to:
$$x\ddot{x}+\dot{x}^2=-1$$.

But I don't see how can I arrive at this ode.
I get the following:
$$-1=\dot{x}^2+\ddot{x}x+\ddot{x}x\dot{x}^2-\dot{x}^2x$$

What do you get here?

Thanks!

 

[renormalize]

$$x\ddot{x}+\dot{x}^2=-1$$But I don't see how can I arrive at this ode.

That Euler-Lagrange equation is indeed correct. For the Lagrangian: $$L\left(x,\dot{x}\right)=\frac{\sqrt{1+\dot{x}^{2}}}{x}$$a straightforward calculation gives:$$\frac{\partial L}{\partial x}-\frac{d}{dt}\left(\frac{\partial L}{\partial\dot{x}}\right)=-\left(\frac{x\ddot{x}+\dot{x}^{2}+1}{\text{stuff}}\right)$$

 

[mad mathematician]

@renormalize I get the following:
$$-d/dt(\frac{\dot{x}}{\sqrt{1+\dot{x}^2}x})-\frac{\sqrt{1+\dot{x}^2}}{x^2}=0$$

How to get at what you wrote?
I don't see it.

 

[renormalize]

@renormalize I get the following:
$$-d/dt(\frac{\dot{x}}{\sqrt{1+\dot{x}^2}x})-\frac{\sqrt{1+\dot{x}^2}}{x^2}=0$$

How to get at what you wrote?
I don't see it.

What expression do you get when you perform the $d/dt$ differentiation?

 

[mad mathematician]

I might be wrong, but I get:
$$\frac{\ddot{x}}{x\sqrt{1+\dot{x}^2}}-\dot{x}^2[1/(\sqrt{1+\dot{x}^2}x^2)+1/(x(\sqrt{1+\dot{x}^2})^3)$$

 

[Orodruin]

Just at one glance that does not look correct to me. The last term is presumably from the derivative of $1/\sqrt{1+\dot x^2}$ yet contains no $\ddot x$.

Note that $d(\dot x^2)/{dt} = 2 \dot x \ddot x$.

 

[Orodruin]

On top of the above, I’d like to add that the integrand does not depend explicitly on $t$. Because of this, the Beltrami identity -which is a first integral of the EL in that case- applies:
$$\frac{\partial L}{\partial \dot x}\dot x - L = E$$ where $E$ is the integration constant. This usually saves a lot of differentiating, particularly when square roots are involved.

 

[mad mathematician]

Just at one glance that does not look correct to me. The last term is presumably from the derivative of $1/\sqrt{1+\dot x^2}$ yet contains no $\ddot x$.

Note that $d(\dot x^2)/{dt} = 2 \dot x \ddot x$.

Yes, spot on.
Thanks!
I guess no one is perfect...