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Let F[y] be a functional defined like so: F[y] = $\int y(x)^2 + (y'(x))^2 dx$.

I'm trying to find the function y which maximizes the value of F, and because the Euler Lagrange equation specifies a necessary condition for an extremal y, this looks like the place to start: $$\frac{\partial{[y(x)^2 + (y'(x))^2]}}{\partial{y(x)}} - \frac{d}{dx} [ \frac{\partial{[y(x)^2 + (y'(x))^2]}}{\partial{y'(x)}}] = 0$$

In the process of producing a differential equation where I can solve for y(x), I apply this transformation:

$$ 2 y(x) + \frac{\partial{[y'(x)^2]}}{\partial{y(x)}} - 2 y''(x) - \frac{d}{dx} [\frac{\partial{[y(x)^2]}}{\partial{y'(x)}}] = 0$$

Now I'm stuck. What does $ \frac{\partial{[y'(x)^2]}}{\partial{y(x)}} $ mean? What does $ \frac{d}{dx} [\frac{\partial{[y(x)^2]}}{\partial{y'(x)}}] $ mean? What theorems/rules allow me to simplify those expressions? Or am I going down the wrong path?

seewalker
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  • maybe you mean $$\frac { \partial \left[ y^{ \prime }\left( x \right) \right] }{ \partial x } $$ ? – haqnatural Jul 23 '15 at 22:14
  • hmm, i think i mean what i typed. It's my understanding that a partial derivative with respect to x does not appear anywhere in the Euler Lagrange equations. – seewalker Jul 23 '15 at 23:41

1 Answers1

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When applying the Euler-Lagrange equations, you treat $y$ and $y'$ as separate entities, and therefore, both of the derivatives about which you inquire are zero. However, there is a mistake in your evaluation. It should read

$$2y(x)+\frac{\partial [(y'(x))^2]}{\partial y}-2y''(x)-\frac{d}{dx}\frac{\partial [y(x)^2]}{dy'}=0.$$

You forgot to apply the $x$ derivative to $y'$.

Plutoro
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  • Treating y and y' as separate entities so that dy/dy' and dy'/dy are zero-valued the same way dn/dm = 0 = dm/dn seems like it would not work in the case where y(x) = 0.5 x^2 and y'(x) = x, because dy/dy' would have value 'x'. – seewalker Jul 23 '15 at 23:22
  • thanks for noticing my arithmetic error, the original question is edited with the correction. – seewalker Jul 23 '15 at 23:42
  • The reason these derivatives are 0 is that these are partial derivatives. Though they are related the partial derivative of y with respect to y' is zero. And vice-versa. – Plutoro Jul 23 '15 at 23:49