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Can an integral that is a variable of itself be solved analytically?

  1. Dec 1, 2011 #1
    Almost two months ago I posted the following question:
    I probably wasn't specific enough in my question to really get the answer I wanted. So I now ask, "Can an integral that is a variable of itself be solved analytically?"
     
    Last edited: Dec 1, 2011
  2. jcsd
  3. Dec 1, 2011 #2

    Simon Bridge

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    If:
    [tex]I=\int f(I)dt[/tex]
    (where I is a function of t)

    Then:
    [tex]\frac{dI}{dt}=f(I)[/tex]

    ... will have the same solution for I.

    Which is 1st order non-linear.
    So you will have analytic solutions under the same conditions.

    (note: differentiating both sides was suggested in your original post.)
     
  4. Dec 3, 2011 #3
    I guess what I want is an algebraic solution. Let's try this now:

    [tex]\frac{dI}{dt} = \frac{V_f \left(1-e^{-\frac{t}{RC}}\right) - RI}{L}[/tex]

    [tex]L\frac{dI}{dt} = V_f\left(1-e^{-\frac{t}{RC}}\right) - RI[/tex]
    [tex]RI = V_f\left(1-e^{-\frac{t}{RC}}\right)-L\frac{dI}{dt}[/tex]
    [tex]I = \frac{V_f\left(1-e^{-\frac{t}{RC}}\right)-L\frac{dI}{dt}}{R}[/tex]

    Now the problem is, "What is [itex]\frac{dI}{dt}[/itex]?" I already had defined it in terms of [itex]I[/itex]. Now I am simply back where I started. I just swapped the terms, and I still don't see a closed-form solution.
     
  5. Dec 3, 2011 #4

    Mute

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    Your notation could be confusing. Without the limits someone might think you are differentiating with respect to the dummy integration variable, which would not be good. You should write

    If

    [tex]I(t) = \int_0^t dt' f(I(t'),t'),[/tex]

    then

    [tex]\frac{dI(t)}{dt} = f(I(t),t).[/tex]

    (I added an extra argument for t on its own, since it depends on t separately too).

    Do you know what a differential equation is? Do you know how to solve one? Your equation is relatively simple and can be solved with an integrating factor. See here.
     
  6. Dec 3, 2011 #5
    I suggest you look at easier ones first. See, "A First Course in Integral Equations" by Abdul-Majid Wazwas. No, I'm not kidding. That is his name. Don't make fun. Start with this one:

    [tex]I(t)=\int_0^t k\left(1-e^{-x/c}\right) I(x)dx[/tex]

    Now differentiate:

    [tex]\frac{dI}{dt}=k\left(1-e^{-t/c}\right)I(t)[/tex]

    separate variables:

    [tex]\frac{dI}{I}=k\left(1-e^{-t/c}\right) dt[/tex]

    integrate:

    [tex]\int_{I_0}^I \frac{dI}{I}=k\int_{t_0}^{t} \left(1-e^{-t/c}\right)dt[/tex]

    [tex]\ln(I)-\ln(I_0)=k\left(t+ce^{-t/c}\right)_{t_0}^t[/tex]

    Do all that and I get:

    [tex]I(t)=I_0\exp\left\{k[(t+ce^{-t/c})-(t_0+ce^{-t_0/c})]\right\}[/tex]
     
  7. Dec 3, 2011 #6

    Simon Bridge

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    The differential equation does yield an algebraic solution. It's the first step.
    Especially as in your case f(I)=RI/L

    @Mute: Yes I noticed that and was concerned - now you've pointed it out I can relax :) OP seems not to understand the differential equation.

    Semi walk-through:

    OPs integral is of the form

    [tex]y(x)=\int f(x).dx - \int ay(x).dx[/tex]
    ... and only the second term is causing trouble.
    Focussing on the problematic term - differentiate both sides:

    [tex]\frac{dy}{dx}=ay[/tex]

    ...rearrange and integrate both sides:

    [tex]\int \frac{dy}{y}=a\int dx[/tex]

    which yields:

    [tex]\ln{|y|}=ax+c[/tex]
    ... where c is the constant of integration - the actual problem has definite integrals so apply limits. Anyway - need to make y the subject so we take the natural exponent of both sides.

    [tex]y=Ce^{ax}[/tex]
    ... where C=ec

    ... so what did I miss?

    Of course, it may be more like:

    [tex]I=\frac{R}{L}\int_0^T i(t)dt[/tex]
    ... which is to say the integral is expected to turn out a single number, rather than:

    [tex]I=\frac{R}{L}\int_0^t i(t^\prime)dt^\prime[/tex]
    ... but I don't want to make it too easy :)
    [between this and #5, it should be easy.]
     
  8. Dec 4, 2011 #7
    It appears that one can go from:

    [tex]\ln(I)-\ln(I_0)=k\left(t+ce^{-t/c}\right)_{t_0}^t[/tex]

    to

    [tex]\ln(I)=\ln(I_0)+k\left(t+ce^{-t/c}\right)_{t_0}^t[/tex]

    to

    [tex]I(t)=I_0\exp\left\{k[(t+ce^{-t/c})-(t_0+ce^{-t_0/c})]\right\}[/tex]

    Now I see why that is (Answer: Logarithm of the product gives us the sum of the logarithm of the factors). I just didn't know how to recognize starting with the sum of the logarithms first (i.e. Sum of the logarithm of the factors gives the logarithm of the product).

     
    Last edited: Dec 4, 2011
  9. Dec 4, 2011 #8
    That's not correct. You'd have to take log of both sides first as in:

    [tex]\log(I(t))=\log(I_0)+k[(t+ce^{-t/c})-(t_0+ce^{-t_0/c})[/tex]
     
  10. Dec 4, 2011 #9
    http://en.wikipedia.org/wiki/Integrating_factor

    The integrating factor cannot allow [itex]Q(x)[/itex] to be zero for all [itex]x[/itex], otherwise the quantity:

    [tex]y = \frac{\int Q(x) M(x)\, dx}{M(x)}[/tex]

    ....would also be zero. This would not help me.

    So I have to set [tex]Q(x)[/tex] to the value of the first term. Therefore, given an ordinary differential equation of the form:

    [tex]y'+P(x)y = Q(x)[/tex]

    [tex]I' + \frac{R}{L}I = \frac{V}{L}\left(1-e^{-t\frac{1}{RC}}\right)[/tex]
    [tex]y' = I'[/tex]
    [tex]P = \frac{R}{L}[/tex]
    [tex]y = I[/tex]
    [tex]Q = \frac{V}{L}\left(1-e^{-t\frac{1}{RC}}\right)[/tex]

    The integrating factor:
    [tex]M(x)=e^{\int P(x)\,dx}[/tex]

    ....is therefore:

    [tex]M(x)=e^{t\frac{R}{L}}[/tex]

    Therefore, [itex]y[/itex] equals:

    [tex]y = \frac{\int \left(\frac{V}{L}\left(1-e^{-t\frac{1}{RC}}\right)\right) \left(e^{t\frac{R}{L}}\right)\, dt}{e^{t\frac{R}{L}}}[/tex]

    To solve integral of the numerator, I used:

    http://www.quickmath.com/webMathema...1=(V/L)*(1-e^(-t/a))*(e^(t/b))&v2=t&v3=0&v4=T

    If my math is correct, then the solution is:

    [tex]y=\frac{\frac{V}{R}\left(\frac{(((L/R)^2 - LC)e^{T/(RC)} + LC)e^{\frac{TR}{L}-\frac{T}{RC}} - \left(\frac{L}{R}\right)^2}{\frac{L}{R} - RC}\right)}{e^{\frac{TR}{L}}}[/tex]

    However, when [itex]T[/itex] becomes modestly large, my MS Excel spreadsheet doesn't seem to be able to handle it. So I can't verify the correctness of this using MS Excel.

    At least I can plot it on the QuickMath site. It is also in the expected shape:

    http://www.quickmath.com/webMathema...=T&v8=y&v9=0&v10=20&v11=0&v12=1&v16=light-red

    L/R=10
    RC=1
    V=100
    L=1000
    R=100
    V/R=1
     
    Last edited by a moderator: May 5, 2017
  11. Dec 5, 2011 #10

    Simon Bridge

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    That looks over-complicated.
    However, you have seen that your original question has been answered.

    In the following I will use lower case for the time varying current i(t) and upper case I to denote fixed values - I find that easier to read. I'll repeat the first part of what you did for clarification. As always, check my working[1]: this is supposed to be illustrative, not correct.

    In standard form, the DE is:

    [tex] L\frac{d}{dt}i + Ri = V_f \left ( 1- e^{-t/RC} \right )[/tex]

    So you need an integrating factor of [itex] e^{Rt/L}[/itex]:

    [tex] i(t)=\frac{V_f}{L}e^{-Rt/L}\int e^{Rt/L}(1-e^{-t/RC})dt[/tex]
    ... expand the integrand:


    [tex] i(t)=\frac{V_f}{L}e^{-Rt/L}\left ( i_1 - i_2\right )[/tex]
    ... where:
    [tex]\begin{align}
    i_1 & =\int e^{Rt/L}dt\\
    i_2 & =\int \exp{\left [(\frac{R}{L}-\frac{1}{RC})t \right ]}dt
    \end{align}[/tex]

    ... using [itex]a\int e^{at}dt = e^{at}[/itex], divide through by L, gives:

    [tex] i= V_f\left [
    \frac{1}{R}\exp{\left [ ( \frac{R}{L} )t \right ] }
    - \frac{RC}{R^2C-L}\exp{\left [(\frac{R}{L}-\frac{1}{RC})t \right ]}
    \right ] e^{-Rt/L} +c [/tex]
    ... where c is the constant of integration, determined from initial condition (i.e. [itex]i(0)=I_0[/itex])
    [tex]I_0 = \frac{V_f}{R}- \frac{RCV_f}{R^2C-L} + c[/tex]

    If I expand the brackets and then collect like terms I get:

    [tex]
    i= V_f\left [
    \frac{1}{R}
    - \frac{RC}{R^2C-L}\exp{\left [-(\frac{1}{RC})t \right ]}
    \right ] +c
    [/tex]

    Which should give an idea if how it behaves.
    When [itex]t \gg RC[/itex] the exponential term vanishes, leaving:

    [tex] i(t \gg RC) \rightarrow \frac{V_f}{R} + c =I_f [/tex]

    Putting I0=0 and (given figures) RC=1, V=100, L=1000, R=100; the equation becomes:

    [tex]i(t)= \frac{1}{9}\left ( e^{-t} - 1 \right )
    [/tex]

    The current starts at 0 and decays exponentially to -(1/9) units with a mean-time of 1s.

    attachment.php?attachmentid=41578&stc=1&d=1323072129.png


    ----------------------
    [1] I am not going to guarantee any of these calculations are 100% correct or correctly performed. Finding my mistakes is left as an exercise for the student xD
     

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  12. Dec 5, 2011 #11
    Whooops! A bit of typo there. It's actually supposed to be:

    [tex]y=\frac{\frac{V}{L}\left(\frac{(((L/R)^2 - LC)e^{T/(RC)} + LC)e^{\frac{TR}{L}-\frac{T}{RC}} - \left(\frac{L}{R}\right)^2}{\frac{L}{R} - RC}\right)}{e^{\frac{TR}{L}}}[/tex]

    The graphs I showed reflect this equation (not the typo version).

    By the way: The graph at http://www.quickmath.com/webMathema...=T&v8=y&v9=0&v10=20&v11=0&v12=1&v16=light-red only works if you clear the second and third equations, [itex]y=sin(x)[/itex] and [itex]x^2+y^2=1[/itex] as they are not needed in the problem, otherwise one gets the message "When defining an equation to graph please use variables T and y only. You have: x in `y = sin(x)`". After those two are deleted, leaving behind the line where

    y=(((10^2-10)*%e^(T)+10)*%e^(T/10-T)/(10-1)-10^2/(10-1))*100/1000/(%e^(T/10))

    is written, then the graph can be successfully plotted by clicking on the plot button.

    To avoid that issue, it is simpler just to click this link http://www.quickmath.com/webMathema...1))*100/1000/(%e^(x/10))&v2=0&v3=20&v4=0&v5=1

     
    Last edited by a moderator: May 5, 2017
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