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

  1. Jul 9, 2011 #1
    Hi,

    I was wondering if the following statement is correct:

    [itex]\int[/itex]([itex]\frac{du}{dx}[/itex])[itex]^{2}[/itex]dV=[itex]\int[/itex][itex]\frac{du}{dx}[/itex]dV[itex]\int[/itex][itex]\frac{du}{dx}[/itex]dV

    Please help!

    Thanks!
     
  2. jcsd
  3. Jul 9, 2011 #2
    No because the left hand side has just [itex]dV[/itex], whereas the right hand side has [itex]dV^2[/itex] because you have two of them multiplying together.
     
  4. Jul 9, 2011 #3
    Counterexample: let u(x) = x, v = x.
     
  5. Jul 9, 2011 #4
    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:

    [itex]\int[/itex]([itex]\frac{du}{dx}[/itex])[itex]^{2}[/itex]dV=[itex]\frac{1}{V}[/itex][itex]\int[/itex]u.n dS[itex][/itex][itex]\int[/itex]u.n dS

    Cheers!
     
  6. Jul 9, 2011 #5
    ???

    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?
     
  7. Jul 9, 2011 #6
    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.
     
  8. Jul 9, 2011 #7
    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

    [tex]\int_a^b \frac{du}{dx}dx = u(b) - u(a)[/tex]

    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. [itex]\int(\frac{du}{dx})^2 dV[/itex] is no more special than [itex]\int u dV[/itex].

    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 [itex]x^2 \geq 0[/itex] is true or false unless I tell you what kind of values are allowed for x.
     
  9. Jul 9, 2011 #8
    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?
     
  10. Jul 9, 2011 #9
    Maybe you should explain where this question comes from and what you're trying to do.
     
  11. Jul 9, 2011 #10
    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!
     
  12. Jul 10, 2011 #11
    I think I know where the first equation I wrote is coming from.... To do with finite elements ... :D
     
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