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Is This Line Integral Differentiable?

  1. Jan 17, 2013 #1
    1. The problem statement, all variables and given/known data
    [itex]\vec { F } \left( x,y \right) =u\left( x,y \right) \hat { i } +v\left( x,y \right) \hat { j }[/itex]
    [itex]u\left( x,y \right) , v\left( x,y \right)[/itex] are continuous on ℝ²
    [itex]\Gamma[/itex] is piecewise smooth.
    Is [itex]\psi (x,y){ =\int { \vec { F } \left( x,y \right) \cdot \vec { dr } } }[/itex] differentiable? How can I show this if it is?
    ψ is meant to be the line integral over Γ.

    2. Relevant equations
    Definition of differentiability; definition of partial derivatives; formula for calculating line integrals under a parametrization.


    3. The attempt at a solution
    I wanted to show that the partial derivatives of ψ are continuous, which would imply differentiability. I can't deal with ψ as it's defined, only with the formula, and I think differentiability must be independent of some arbitrary parametrization..
     
    Last edited: Jan 17, 2013
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  3. Jan 17, 2013 #2

    Zondrina

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    So what happens if you parametrize your curve? Remember that it's piecewise smooth, so you might have to break it up a bit. Plug in what you know and remember to choose the interval.

    Think about the continuity of u and v.
     
  4. Jan 17, 2013 #3
    Could you be a bit more specific?
    Parametrizing my curve allows me to use the Line Integral formula, which is an integral of 't' alone (t being a parameter). I still can't see anything as far as proving continuity of partial derivatives goes..

    In the original problem, I was to find a formula for ∂ψ/∂x for a given Γ. When I tried to solve it, I did exactly what you said - plugged in all I know etc - but from what I noticed, I only used continuity to guarantee integrability.. The formula I arrived at required ψ be differentiable (used the chain rule).
     
  5. Jan 17, 2013 #4

    Zondrina

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    Could you show me your work?
     
  6. Jan 17, 2013 #5

    Dick

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    If you think about the integral dt, you are integrating a piecewise continuous function. Is the result of that necessarily differentiable? Try to think of a counterexample.
     
  7. Jan 18, 2013 #6
    Re: Re: Is This Line Integral Differentiable?

    I don't see why my integrand is piecewise continuous.. U, V are everywhere continuous..

    My problem is ψ..
     
    Last edited: Jan 18, 2013
  8. Jan 18, 2013 #7

    Dick

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    [itex]d \vec r(t)=\vec r'(t) dt[/itex]. If you are only given that r(t) is piecewise smooth then r'(t) is only piecewise continuous. Think of the one dimensional example r(t)=|t|. I'd call that piecewise smooth.
     
  9. Jan 18, 2013 #8
    I understand that bit now.
    Here's my attempt at a solution.
    Hopefully you will be able to understand where my question comes from and help me.
     

    Attached Files:

  10. Jan 18, 2013 #9

    Dick

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    Yeah, that helps. I was picturing how you were differentiating the psi wrong. But I think the whole 't' thing is leading you astray. H1 and H2 aren't functions of t. t is a dummy variable. When you've got something like [itex]\int_0^1 f(xt,y) d(xt)[/itex] change the variable to x'=xt. Now you've got [itex]\int_0^x f(x',y) d(x')[/itex]. What's the x derivative of that?
     
  11. Jan 18, 2013 #10
    I don't understand a few things:
    First, I don't see d(xt) anywhere, only dt. Where did the integral with d(xt) come from?
    Second, can you please explain how is it that I obtain an integral with a variable limit from an integral with a constant limit? Also the bit with x and x', what are they exactly?

    As far as what you asked, the derivative of that integral is f(x', t) because of the fundamental theorem.
     
  12. Jan 18, 2013 #11

    Dick

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    Nope, the derivative can't be f(x',t). x' is a dummy variable of integration. So is t. It's f(x,y). In my example x is the constant. x'=xt. x' is just a change of variables from t. In your formula for H2(t) you have x0*v(x0*t,y) dt. That's the same as v(x0*t,y) d(x0*t) since x0 is constant. You really want to change variables to x'=x0*t.
     
  13. Jan 18, 2013 #12
    Okay, gotcha.
    So your x is my x₀, and I understand the why the derivative is f(x,y).
    I'm still unsure of where the d(xt) came from. I have xU(xt,y)d(t) in my integral. Can you move the x inside the parenthesis because it's a constant? "d" is like a linear operator?
     
    Last edited: Jan 18, 2013
  14. Jan 18, 2013 #13

    Dick

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    Exactly. You can move x into the d part because it's a constant. d(xt)=xdt. Suggests you do the change of variables x'=xt. So dx'=xdt.
     
  15. Jan 18, 2013 #14
    You said that in your example, x is a constant.
    Doesn't that mean that [itex]\int _{ 0 }^{ x }{ f\left( x',y \right) dx' }[/itex] is just a number? Why can we suddenly take the derivative with respect to x which is a constant?

    I think I have a fundamental problem of understanding here..
     
  16. Jan 18, 2013 #15

    Dick

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    x is a constant with respect to the dt or dx' integration. That's all that I mean by 'constant'.
     
  17. Jan 18, 2013 #16
    I think i'm not being clear enough, so I will write down on paper what's bothering me in each step of the problem. Right now I have a mess in my head.
    Thanks alot for bearing with me!
     

    Attached Files:

  18. Jan 18, 2013 #17
    Figured everything out. Thanks alot for all your help!
     
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