Integration of a term that is squared, cubed, etc.

1. Jul 9, 2011

hoomanya

Hi,

I was wondering if the following statement is correct:

$\int$($\frac{du}{dx}$)$^{2}$dV=$\int$$\frac{du}{dx}$dV$\int$$\frac{du}{dx}$dV

Thanks!

2. Jul 9, 2011

BrianMath

No because the left hand side has just $dV$, whereas the right hand side has $dV^2$ because you have two of them multiplying together.

3. Jul 9, 2011

pmsrw3

Counterexample: let u(x) = x, v = x.

4. Jul 9, 2011

hoomanya

Thank you both. pmsrw3 I am not bothered with solving the integral.

Can I write it as something like this then, using the divergence theorem:

$\int$($\frac{du}{dx}$)$^{2}$dV=$\frac{1}{V}$$\int$u.n dS$\int$u.n dS

Cheers!

5. Jul 9, 2011

pmsrw3

???

But the example I gives shows that your speculated relationship doesn't hold. It doesn't matter whether you want an explicit solution or not.

Wait a minute, I think this needs some clarification. Is dV a volume element? In how many dimensions, and what is x? What's the context?

6. Jul 9, 2011

hoomanya

Sorry, I don't get the counter example. Thank you for your quick responses.

V is a volume integral, x is a dimension and u is displacement. The problem is 1D and the context is solid deformation and finite element analysis.

By the way, does it depend on the context or is it a mathematical relationship (if it's correct)?

Thanks again.

7. Jul 9, 2011

pmsrw3

If u(x) = x, du/dx = 1. If v=x, dv=dx. The left-hand-side integral is therefore x+C1. The RHS integrals comes to (x+C2)(x+C3). Obviously the LHS and RHS aren't equal.

OK, then I completely don't get the question. If you're in one dimension and it's a volume integral, doesn't that mean V = x? In one dimension, the surface of a region is just two points, the left and right ends (assuming the region is contiguous), and the normal vector n is -1 at one end, +1 at the other end. In other words, the divergence theorem in one dimension is nothing but the familiar

$$\int_a^b \frac{du}{dx}dx = u(b) - u(a)$$

I think you have the idea that the fact the the integrand is the square of the derivative of something makes it special. But it doesn't, you know: ANY (well-behaved) function can be expressed as the square of the derivative of something. $\int(\frac{du}{dx})^2 dV$ is no more special than $\int u dV$.

Both could be true. The truth or falsehood of an equation will depend on what the symbols mean. For instance you can't tell me whether the inequality $x^2 \geq 0$ is true or false unless I tell you what kind of values are allowed for x.

8. Jul 9, 2011

hoomanya

Thanks a lot.If the problem was not 1D and you had to write the left hand side of the equation in terms of a product the two terms I have on the RHS(after using the divergence theorem), would there be a way? Or is there a context or a situation this or something similar would hold?

9. Jul 9, 2011

pmsrw3

Maybe you should explain where this question comes from and what you're trying to do.

10. Jul 9, 2011

hoomanya

Thanks for trying to help. It would be too complicated to explain the entire context. I posted the question hoping this is a general mathematical relations or something close. But you helped understanding that it probably isn't. So thanks for that!

11. Jul 10, 2011

hoomanya

I think I know where the first equation I wrote is coming from.... To do with finite elements ... :D