## ghdoru11 one year ago Find the extreme functions, y(x),z(x) for the functional F[y,z]= integrate from 0 to 1 [y^2+z^2+4(y')^2-16(z')^2] dx

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1. IrishBoy123

$\triangle$

2. ghdoru11

i don't understand

3. IrishBoy123

hi @ghdoru11 that silly sign means I am watching and interested i think you you want extreme functions y(x),z(x) for: $F[y,z]= \int_{0}^{1} y^2+z^2+4(y')^2-16(z')^2 \ dx$ i could have a go but i think that might be really counter productive as i'd be guessing the relationship between x,y,z etc ie i don't recognise some of the finer lingo so i can, in the first instance, tag some people who know way better than i and some way you will get some help if that doesn't work out.... @Phi @SithsAndGiggles

4. ghdoru11

thank you for your time,i didn't know what that sign means

5. IrishBoy123

@Michele_Laino

6. Michele_Laino

hint: here you have to write the Euler-Lagrange equation, for both functions y, and z, namely: $\Large \begin{gathered} \frac{{\partial f}}{{\partial y}} - \frac{d}{{dx}}\frac{{\partial f}}{{\partial y'}} = 0 \hfill \\ \hfill \\ \frac{{\partial f}}{{\partial z}} - \frac{d}{{dx}}\frac{{\partial f}}{{\partial z'}} = 0 \hfill \\ \end{gathered}$ where: $\Large f\left( {y,y'z,z'} \right) = {y^2} + {z^2} + 4{(y')^2} - 16{(z')^2}$ furthermore, we need to know the initial conditions for $$\large y(x), \; z(x)$$ at $$\large x=0, x=1$$

7. Michele_Laino

please solve both the Euler-Lagrange differential equations, for $$\large y(x), \; z(x)$$ respectively

8. anonymous

I'm not well-versed in functional calculus, but it sounds really interesting...

9. ghdoru11

this is wrong. right?