Differential equations uh oh

Discussion in 'Physics & Math' started by Fudge Muffin, Jun 3, 2013.

  1. Fudge Muffin Fudge Muffin Registered Senior Member

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    148
    I was thinking there, if I was really pumped up for my physics exam, we could say the rate at which I write words is proportional to the number I have already written.

    Then, I made a differential equation as normal... \( dW/dt = k dW \)

    but then I got..

    \( lnW = kt + c \)

    which was all fine until i considered... hang on, at the start of the exam I have written no words. How could I take \( ln0 \) ????

    I can progress no further!
     
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  3. Fraggle Rocker Staff Member

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    Obviously you can't start at zero because that's a singularity: before t=0 your speed was a constant 0. (If "singularity" is the right word--last time I studied this stuff was in 1963.) You have to take your first data point at, say, one minute into the exam.

    BTW, the logarithm of zero is minus infinity. You might actually be able to plug that into your formula, but why bother?
     
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  5. Tach Banned Banned

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    Think about this: \(W(t)=e^{kt} e^C\)

    If you think that at \(t=0\) you have written your first word, you get the boundary condition \(1=W(0)=e^C\) , meaning \(C=0\). So, \(W(t)=e^{kt} \).

    You can get next \(k=\frac{dW}{dt}|_{t=0}\)
     
    Last edited: Jun 4, 2013
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  7. eram Sciengineer Valued Senior Member

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    You could use dW/dt = kW, which would yield an exponential function.
     
  8. Tach Banned Banned

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    5,265
    He did, the \(dW\) in the RHS of his ODE is just a typo.
     
  9. eram Sciengineer Valued Senior Member

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    1,877
    Oh yeah.

    @Fudge Muffin At t = 0, the exponential function already has a certain value. So perhaps you cheated and wrote some stuff before you were allowed to.

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  10. ash64449 Registered Senior Member

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    What is typo?
     
  11. rpenner Fully Wired Valued Senior Member

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    \( W' = k W\) is one of the classic differential equations. Here is how you solve it:
    \(\begin{eqnarray} W' & = & k W & & & \quad \quad \quad & \frac{W'}{W} & = & k & & & & \textrm{Divide both sides by W, a step that assumes W is nowhere 0 so solutions that pass through 0 must be checked} \int \frac{W'}{W} dt & = & \int k \, dt & & & & \textrm{Setup both sides to be integrated by t} \ln W + C_1 & = & k t + C_2 & & & & \textrm{Carry out the integration, with u-substitution} \ln W & = & k t + C & & & & \textrm{Accumulate the integration constants into a single constant} W & = & W_0 e^{k t} & & & & \textrm{Exponentiate both sides} W' & = & k W_0 w^{k t} & = & k W & & \textrm{Differentiate both sides to check out equation holds when W \leq 0} \end{eqnarray}\)
    Even when \(W_0 \leq 0\) this equation satisfies the differential equation.
    Since \(e^x\) is for no value of x equal to zero, the equation \(W = W_0 e^{k t} \) can only be solved by setting \(W_0\) to zero in which case \(W(t) = 0\) for all t. If this equation holds, your exam experience is going to be unpleasant.

    However, your assumption that \(W' = k W\) was baseless. It is a model that relies on no scientific theory and must be a stand-alone hypothesis. So either you will write no words on the physics exam and tend to support your hypothesis or you will write at least one word and reject it.

    But words are discrete and not continuous, so your model in continuous variables doesn't make a lot of sense. By phrasing a similar model with discrete words, we get an asymptotic form for large numbers of words without requiring that the number of words which starts at zero remain at zero.

    If for large W, \(W(t) \approx W_0 e^{kt}\) approximately holds then \(t(W) \approx \frac{1}{k} \ln \frac{W}{W_0}\) approximately holds and \(t'(W) \approx \frac{1}{k W_0 W}\). So the time between successive words is approximately inversely proportional to the number of words. (Another reason to expect the continuous function to be flat when it passes through a point with zero words.) Now if we construct a discrete model where each word is completed in additional time inversely proportional to some linear function of the number of words we have a sequence of word arrival times:
    \(\Delta t_n = \frac{1}{a n + b}\) and thus \(t(W) = \sum_{k=0}^{W-1} \frac{1}{a k + b} = \frac{1}{a} \left[ \Phi(W + \frac{b}{a}) - \Phi(\frac{b}{a}) \right] \) where \(\Phi(x) = \frac{d \quad}{d x} \ln (x-1)! \approx \ln \left(x - \frac{1}{2} \right) + \frac{1 + x}{24 x^3}\).

    Thus for large W, \(t(W) \approx \frac{1}{a} \ln \frac{2 a W + 2 b - a}{2 b - a} - \frac{a^2 + a b}{24 b^3}\) with inverse:
    \(W \approx \frac{2 b - a}{2 a} \left[ e^{\frac{a^2 + a b}{24 b^3}} e^{ a t} - 1 \right] \)
     

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