Deriving solution to calculus of variation problem

Question

Please help me out. I don't know how the solution $y(x)$ (on the last line) was derived from solving the previous line before the last. Thanks in advance.

Answer 1

Spent a little bit of time trying to simplify the last bit, but it is pretty messy and not the expression I come up with when I approach the same functional. What I did is as follows: $$ S[y(x),x] =\int_a^b y(x) \sqrt{1+y'(x)} \: \textrm{d}x $$ Setting up the Euler-Lagrange equations as follows (where $\mathcal{L}$ is the integrand): $$ \frac{\partial \mathcal{L}}{\partial y(x) } =\sqrt{1+y'(x)} $$ $$\frac{\partial \mathcal{L}}{\partial y'(x) } = \frac{y(x) y'(x)}{\sqrt{y'(x)^2+1}}$$ $$\frac{d}{dx}\left( \frac{\partial \mathcal{L}}{\partial y'(x) } \right) = \frac{y'(x)^2}{\sqrt{y'(x)^2+1}}-\frac{y(x) y'(x)^2 y''(x)}{\left(y'(x)^2+1\right)^{3/2}}+\frac{y(x) y''(x)}{\sqrt{y'(x)^2+1}}$$ The equation we need to solve is: $$\frac{\partial \mathcal{L}}{\partial y(x) } - \frac{d}{dx}\left( \frac{\partial \mathcal{L}}{\partial y'(x) } \right) =0 $$ That is: $$ -\frac{y'(x)^2}{\sqrt{y'(x)^2+1}}+\sqrt{y'(x)^2+1}+\frac{y(x) y'(x)^2 y''(x)}{\left(y'(x)^2+1\right)^{3/2}}-\frac{y(x) y''(x)}{\sqrt{y'(x)^2+1}} =0 $$ Simplifying this we have: $$ - y(x) y''(x)+y'(x)^2+1=0 $$ (Clarification to the simplifications):

(1) Multiply the above expression by $\left(y'(x)^2+1\right)^{3/2}$: $$\left(y'(x)^2+1\right)^{3/2} \left(-\frac{y'(x)^2}{\sqrt{y'(x)^2+1}}+\sqrt{y'(x)^2+1}+\frac{y(x) y'(x)^2 y''(x)}{\left(y'(x)^2+1\right)^{3/2}}-\frac{y(x) y''(x)}{\sqrt{y'(x)^2+1}}\right) =0 $$ Which will give you - by cancelling terms: $$ - y(x) y''(x)+y'(x)^2+1=0 $$ Let $z(y) = y'(x)$, then the equation becomes: $$ - y(x) \frac{dv(y)}{dx} + z(y)^2 +1 = - y(x) \left( \frac{dz(y)}{dy} \frac{dy}{dx}\right) + z(y)^2 +1 = - y(x) z'(y) z(y) + z(y)^2 + 1 = 0 $$ Which gives: $$z(y)\frac{dz}{dy} = \frac{z(y)^2 + 1 }{y(x)}$$ So we perform the usual integration procedure: $$ \int \frac{\frac{dz}{dy}z(y)}{z(y)^2 + 1 } \textrm{ d}y = \int \frac{1}{y} \textrm{ d}y$$ $$ \frac{1}{2}\ln(z(y)^2 +1) = \ln(y(x)) + c_1 $$ Solving for $z(y)$ we get two solutions: $$z(y) = \pm \sqrt{e^{2c_1} y^2(x) -1 } $$ Where we recall that $z(y) = y'(x)$ and redefine $c_1 = e^{2c_1}$. Thus: $$y'(x) =\pm \sqrt{c_1 y^2(x) -1 } $$ Doing the same integration procedure after dividing: $$\int \frac{y'(x)}{\sqrt{c_1 y^2(x) -1 }} \textrm{ d}y = \int \pm 1 \textrm{ d}x $$ $$ \frac{\ln(\sqrt{c_1} \sqrt{c_1 y^2(x)-1} + c_1 y(x))}{\sqrt{c_1}} = \pm x + c_2 $$ The isolate $y(x)$ to get the solutions: $$y(x) = \frac{e^{-\sqrt{c_1}(x+c_2)}}{2c_1 }\left(c_1 + e^{2(x+c_2)} \right)$$ $$y(x) =\frac{e^{-\sqrt{c_1}(x+c_2)}}{2c_1 }\left(c_1e^{2\sqrt{c_1}x} +e^{2\sqrt{c_2}c_2} \right) $$