Calculate the Euler-Lagrange equations for the functional and show that the Euler-Lagrange equations have a solution

Question

Consider the functional $J_2$ of two functions y(x) and z(x) given by

$$J_2[y,z] = \int_{x0}^{x1}\sqrt{{1+(z')^{2}+z^2(y')^2}\over{C^2+z^2}}$$

where $y' = dy/dx, z'= dz/dx$, and C is a non-zero constant.

1. Calculate the Euler-Lagrange equations for the functional $J_{2}$

2. By substitution or otherwise, show that the Euler-Lagrange equations in (1) have a solution of the form

   $$y(x) = Ax + B, z(x) = D$$

   where $A, B$ and $D$ are appropriate constants and $D$ does not equal zero.

For part (1) I have worked out that there are two Euler-Lagrange equations. The first is

$$\frac{d}{dx}\frac{\partial L}{\partial y'}-\frac{\partial L}{\partial y}=0, \quad\frac{\partial L}{\partial y}=0$$

and so

$$0=\frac{d}{dx} \left(\frac{z^2y'}{(z^2+c^2)^\frac{1}{2}(z^2(y')^2+(z')^2+1)^{\frac{1}{2}}} \right)$$

The second one is

\begin{align} \frac{d}{dx}\frac{\partial L}{\partial z'}-\frac{\partial L}{\partial z} &= \frac{d}{dx} \left(\frac{z'}{(c^2+z^2)^{\frac{1}{2}}(x^2+(y')^2z^2+1)^\frac{1}{2}} \right) - \frac{\partial L}{\partial z} \\ &= \frac{d}{dx} \left(\frac{z'}{(c^2+z^2)^\frac{1}{2}(x^2+(y')^2z^2+1)^\frac{1}{2}} \right) - \frac{z(c^2(y')^2-(z')^2-1)(z)}{(z^2+c^2)^\frac{3}{2}((y')^2z^2+(z')^2+1)^\frac{1}{2}} \\ &= 0 \end{align}

I am unsure how to do the $\frac{d}{dx}$ for part (1) if anyone can help and would appreciate any help with part (2) feel free to add adapt tags if you think any don't fit.

Answer 1

1. OP's Lagrangian $$ L ~=~ \sqrt{\frac{1+\dot{z}^2+z^2\dot{y}^2}{C^2+z^2}} \tag{1}$$ has momenta $$ p_y~:=~\frac{\partial L}{\partial \dot{y}}~=~\frac{z^2\dot{y}}{\sqrt{(C^2+z^2)(1+\dot{z}^2+z^2\dot{y}^2)}}, \tag{2}$$ $$ p_z~:=~\frac{\partial L}{\partial \dot{z}}~=~\frac{\dot{z}}{\sqrt{(C^2+z^2)(1+\dot{z}^2+z^2\dot{y}^2)}}, \tag{3}$$ and energy $$E~:=~\dot{y}p_y+\dot{z}p_z-L~=~\frac{1}{\sqrt{(C^2+z^2)(1+\dot{z}^2+z^2\dot{y}^2)}}~\neq~ 0.\tag{4}$$

2. Since $L$ does not depend explicitly on $y$ and $x$, we have 2 constants of motion $p_y$ and $E$, cf. Noether's theorem.

3. Eqs. (2) simplifies to $$\frac{p_y}{E}~=~ z^2\dot{y}.\tag{5}$$ Eqs. (4) simplifies to a first-order ODE $$\frac{1}{E^2(C^2+z^2)}~=~1+\dot{z}^2+\frac{p_y^2}{E^2z^2}\tag{6}$$ for $z\neq 0$.

4. Eqs. (5) & (6) are 2 first integrals to the 2 EL equations.

5. Clearly the Ansatz $$y(x) ~=~ Ax + B,\qquad z(x) ~=~ D \tag{7}$$ are solutions to eqs. (5) & (6) for appropriate constants $A$, $B$ and $D$.