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Initial value problem

  1. May 14, 2014 #1

    jjr

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    1. The problem statement, all variables and given/known data

    The problem is from Walter Gautschi - Numerical Analysis, exercise 5.1.

    Consider the initial value problem

    [itex]\frac{dy}{dx}=\kappa(y+y^3), 0\leq x\leq1; y(0)=s[/itex]

    where [itex]\kappa > 0[/itex] (in fact, [itex]\kappa >> 1[/itex]) and s > 0. Under what conditions on s does the solution [itex]y(x) = y(x;s)[/itex] exist on the whole interval [0,1]? {Hint: find y explicitly.}

    3. The attempt at a solution

    Following the hint, I tried solving it as a separable differential equation:

    [itex]\frac{dy}{y+y^3} = \kappa dx[/itex]

    (Using wolframalpha here)

    [itex]log(y) - \frac{1}{2}log(y^2+1) = \kappa x[/itex]

    [itex]10^{log(y)-\frac{1}{2}log(y^2+1)} = 10^{\kappa x}[/itex]

    Ending up with

    [itex]\frac{y}{2} + \frac{1}{2y} = 10^{-\kappa x}[/itex]

    Not sure how to solve this, so used wolframalpha again and got:

    [itex]y = 10^{-\kappa x} \pm \sqrt{10^{-2 \kappa x} - 1} [/itex]

    Evidently [itex]y(0) = 1[/itex], which means that s has to be equal to 1.

    This doesn't seem right to me. Especially considering the work needed to find y explicitly, seeing as how this is not a course in differential equations. So what I am missing here?
     
  2. jcsd
  3. May 14, 2014 #2

    pasmith

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    I see no constant of integration; the right hand side should be [itex]\kappa x + C[/itex].
     
  4. May 16, 2014 #3

    jjr

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    Thank you, and right you are; that was quite sloppy of me. So starting then at

    [itex]\frac{y}{2}+\frac{1}{2y}=C10^{\kappa x}[/itex]

    yielding

    [itex]y = C10^{-\kappa x} \pm \sqrt{ C^{2}10^{-2 \kappa x} - 1 }[/itex]

    so

    [itex] y(0) = s = C \pm \sqrt{C^2-1}[/itex]

    Which imposes a constraint on C: [itex] |C| \geq 1 [/itex]

    But they are asking for what conditions on s the solution [itex]y(x) = y(x;s)[/itex] exists on the interval [0,1]? I am having some trouble seeing the connection here, especially in the context of the chapter "Initial Value Problems for ODEs: One-Step Methods". Any thoughts?
     
  5. May 16, 2014 #4

    pasmith

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    [tex]\log(y) - \frac12 \log(y^2 + 1) = \log\left( \frac{y}{\sqrt{1 + y^2}}\right).[/tex] This is not in general equal to [itex]\log(y/2 + 1/(2y))[/itex], which is what your expression implies (and I must apologise for not pointing this out in my first reply).

    Also: in calculus, "log" means "natural logarithm", so if [itex]a = \log b[/itex] then [itex]b = e^a[/itex].
     
  6. May 16, 2014 #5

    jjr

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    Ah, that explains the seemingly frivolous interchangeable use of log and ln I've sometimes seen. And yes, I see that I've made en error when dealing with the [itex] \frac{1}{2} [/itex] factor. Well, being more explicit this time and starting from (I will for my own sake switch to using ln now):

    [itex] ln(y) - \frac{1}{2}ln(y^2+1) = \kappa x + C_1 [/itex]

    [itex] ln(y) - ln( \sqrt{y^2+1}) = \kappa x + C_1 [/itex]

    [itex] e^{ln(y) - ln(\sqrt{y^2+1})} = C_2 e^{\kappa x} [/itex]

    [itex] \frac{e^{ln(y)}}{e^{ln(\sqrt{y^2+1})}} = C_2 e^{\kappa x} [/itex]

    [itex] \frac{y}{\sqrt{(y^2+1)}} = C_2 e^{\kappa x} [/itex]

    [itex] \frac{y^2}{y^2+1} = C_3 e^{2\kappa x} [/itex]

    [itex] \frac{y^2+1}{y^2} = C_4 e^{-2\kappa x} [/itex]

    [itex] 1 + y^{-2} = C_4 e^{-2\kappa x} [/itex]

    [itex] y^{-2} = C_4 e^{-2\kappa x} - 1 [/itex]

    [itex] y^2 = \frac{1}{C_4 e^{-2\kappa x} -1} [/itex]

    [itex] y = \pm \frac{1}{\sqrt{C_4 e^{-2\kappa x} -1}} [/itex]

    So

    [itex] y(0) = s = \pm \frac{1}{\sqrt{C_4 - 1}} [/itex]

    and thus [itex] C_4 > 1 [/itex].

    Hopefully the math checks out this time. But I am still not sure how to use this in the context of the exercise. Any ideas?
     
  7. May 16, 2014 #6

    ehild

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    It was given that s>0. Find the integration constant C4 in terms of s. Than write y(x) using s, and see what should be true for s, so y(x,s) is defined for the whole interval [0,1].

    ehild
     
  8. May 16, 2014 #7

    jjr

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    Perfect, thanks!
     
  9. May 16, 2014 #8

    ehild

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    You are welcome:smile:

    ehild
     
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