MHB Can we get to the desired relation?

[evinda]

Hi! (Wave)

I want to show that the Euler equation for the functional $J(y)= \int_a^b f(x,y) \sqrt{1+y'^2}dx$ has the form:

$$f_y-f_xy'-\frac{fy''}{1+y'^2}=0$$$$L(x,y,y')= f(x,y) \sqrt{1+y'^2} dx$$

Substituting $L_y(x,y,y')=f_y(x,y) \sqrt{1+y'^2}, \ L_{y'}(x,y,y')= f(x,y) \frac{y'}{\sqrt{1+y'^2}}$, I got the following:$$f_y(1+y'^2)-f_x y'- f y''+ \frac{f (y')^2}{(1+y'^2)}=0$$

Can we get from this relation to the desired one, or have I done something wrong? (Thinking)

 

[I like Serena]

$$f_y(1+y'^2)-f_x y'- f y''+ \frac{f (y')^2}{(1+y'^2)}=0$$

Can we get from this relation to the desired one, or have I done something wrong? (Thinking)

Hey! (Blush)

I think there is something wrong in your result. (Sadface)

I think we should have:
\begin{aligned}
L_y - \d {} x L_{y'} &= f_y\sqrt{1+y'^2} - \d {} x\left(f \cdot \frac{y'}{\sqrt{1+y'^2}}\right) \\
&= f_y\sqrt{1+y'^2} - (f_x + f_yy')\frac{y'}{\sqrt{1+y'^2}} - f \d {} x\left(\frac{y'}{\sqrt{1+y'^2}}\right) \\
&= f_y\left(\sqrt{1+y'^2} - y'\frac{y'}{\sqrt{1+y'^2}}\right) - f_x\frac{y'}{\sqrt{1+y'^2}} - f \d {} x\left(\frac{y'}{\sqrt{1+y'^2}}\right) \\
&= 0 \end{aligned}
Multiply by $\sqrt{1+y'^2}$ to get:
$$\Rightarrow
f_y\left((1+y'^2) - y'y'\right) - f_xy' - f \sqrt{1+y'^2}\d {} x\left(\frac{y'}{\sqrt{1+y'^2}}\right)
= 0 \\
\Rightarrow
f_y - f_xy' - f \sqrt{1+y'^2}\d {} x\left(\frac{y'}{\sqrt{1+y'^2}}\right)
= 0$$

Hey! That is almost what we want! (Wait)