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OK, using the calculus of variations, I want to find a function $f$ that maximizes:

$$J = \int_0^n L(x,f(x)) \text{d}x$$.

But $L$ has multiple integrals in it (for example, $\displaystyle \int_0^n y f(y) \text{d}y$). Can I still use the Euler-Lagrange equation to find f?

Here's the full equation: I want to maximize

$$J = \frac{\int_0^n x f(x) \text{d}x}{\sqrt{\int_0^n \left( y - \int_0^n z f(z) \text{dz} \right)^2 f(y) \text{d}y}} = \int_0^n \frac{x f(x)}{\sqrt{\int_0^n \left( y - \int_0^n z f(z) \text{d}z \right)^2 f(y) dy}} \text{dx}$$

Thanks in advance!

Edit: Actually, I should say, I don't care whether I use the calculus of variations or not, I just want to find $f$. I assumed calculus of variations is the way to do that.

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What are the constraints on $f$? If $f$ is unsigned, the denominator could possibly be 0... – Willie Wong Jun 24 '11 at 23:00
f is a probability distribution, so it is between zero and one. (I'm not sure how to enforce that, however....) – usul Jun 24 '11 at 23:40
chapter 7 of… might help – yoyo Jun 25 '11 at 0:57
Thanks. Actually I'm thinking now that the denominator can be zero. Does that mean I'm screwed? (I'd still like to be able to find function that produces the "max" of infinity if possible....) – usul Jun 25 '11 at 4:33
If I understand correctly, you're just dividing the mean by the standard deviation. You can make the standard deviation arbitrarily small by making $f$ arbitrarily peaked; as $f$ tends towards a delta distribution around any value $x_0\neq0$, $J$ grows without limits. – joriki Jun 25 '11 at 5:07

1 Answer 1

Let us assume that the answer is indeed bounded and there exists an optimal value.

That is given

$$ \int_{0}^{n} \frac{x f(x) }{\sqrt{\int _0^n \left[y - \int_{0}^{n} \left[ z f(z)\right] dz\right] dy}} dx $$

Now since you particularly wanted a Maximizer it suffices that we must make:

$$\sqrt{\int _0^n \left[y - \int_{0}^{n} \left[ z f(z)\right] dz\right] dy} $$

As small as possible Thus we need to make:

$$\int_{0}^{n} \left[ z f(z)\right] dz$$

As big as possible

Not however this is trivially unbounded as we can make $f(z)$ an arbitrarily large constant function. We can declare that this is equal to the value $E_1$ whereas $E_1$ can get as big as you want it to.

If however we want the square root:

$$\sqrt{\int _0^n \left[y - \int_{0}^{n} \left[ z f(z)\right] dz\right] dy} $$

To make any sense in terms of real values (I assume that's your perogative given the word "maximize" then it must be that

$$ \int _0^n \left[y - \int_{0}^{n} \left[ z f(z)\right] dz\right] dy \ge 0$$


$$ \int _0^n \left[y -E_1\right] dy \ge 0$$

Clearly if we take $E_1 = \frac{n}{2}$ that'll do. Returning us the value $0$ for this integral.

Now the thing to be maximized is

$$ \int_{0}^{n} \left[ \frac{x f(x) }{E_2 } \right] dx $$

Where we have shown, $E_2$ can be made arbitrarily small and therefore this is arbitrarily large! Also even if we ignore $E_2$ the analysis on the Z integral (which was identical to this) shows we can STILL make this arbitrarily large, so that's no good either. In short, this problem without ANY additional COnstraints is trivial to optimize to infinity.

So for the sake of completeness lets say you added in constraints that ensure this can't be made arbitrarily large/Small, it's still worthwhile to hit this with Euler Lagrange and see if that does any better

We note that our Z integral has euler Langrage equations

$$ \frac{\partial E}{\partial f(z)} - \frac{d}{dx}\frac{\partial E}{\partial f'(z)} = 0 $$


$$ E = zf(z) $$

Which yields us

$$ z = 0 $$

The function here is basically only defined at $z=0$ elsewhere it is undefined, or what this is really saying: unbounded.

And if you work your way up for the Y integral and then X integral, it'll be a similar story.

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