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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
    • one year ago
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    \[\triangle\]

  2. ghdoru11
    • one year ago
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    i don't understand

  3. IrishBoy123
    • one year ago
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    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
    • one year ago
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    thank you for your time,i didn't know what that sign means

  5. IrishBoy123
    • one year ago
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    @Michele_Laino

  6. Michele_Laino
    • one year ago
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    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
    • one year ago
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    please solve both the Euler-Lagrange differential equations, for \( \large y(x), \; z(x) \) respectively

  8. anonymous
    • one year ago
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    I'm not well-versed in functional calculus, but it sounds really interesting...

  9. ghdoru11
    • one year ago
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    this is wrong. right?

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