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

calculus-of-variationseuler-lagrange-equationordinary differential equationspartial differential equations

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.

Best Answer

  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 $D$.

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