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Check my separable diff EQ work?

  1. Jun 4, 2012 #1

    ElijahRockers

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    Gold Member

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

    A model for the shape of a tsunami is given by

    [itex]\frac{dW}{dx} = W\sqrt{4-2W}[/itex]

    where W(x) > 0 is the height of the wave expressed as a function of its position relative to a point off-shore.

    Find the equilibrium solutions, and find the general form of the equation. Use graphing software to graph the direction field, and sketch all solutions that satisfy the initial condition W(0) = 2.

    2. Relevant equations

    [itex]\int \frac{dy}{y\sqrt{4-2y}} = -tanh(\frac{1}{2}\sqrt{4-2y})[/itex]

    3. The attempt at a solution

    i'm pretty sure the equilibrium solutions are w = 0,2

    but i have never seen or used hyperbolic trig functions, so I guess I was just wondering if they work the same way as regular trig functions.

    It doesn't seem hard, I guess I would just like someone to verify my answer for the general form:

    [itex]W(x) = 2-2arctanh^2(-x+C)[/itex]

    If anyone gets anything different let me know and I can show my work, thanks.

    As for the sketching, as far as I can tell W(any x)=2 is a horizontal straight line, which seems pretty boring to sketch...
     
    Last edited: Jun 4, 2012
  2. jcsd
  3. Jun 4, 2012 #2

    HallsofIvy

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    Yes, the equilibrium solutions are W= 0 and W= 2. And, yes, W= 2 is a pretty boring graph! As is W= 0. But what happens for other values of W? The problem suggests you use graphing software but a rough sketch of the direction fields is not difficult.

    If W< 0, [itex]\sqrt{4- 2W}[itex] is positive so the product, [itex]W\sqrt{4- 2W}[/itex] is negative. If 0< W< 2, W is positive, [itex]\sqrt{4- 2W}[/itex] is still positive so [itex]W\sqrt{4- 2W}[/itex] is positive. If W> 2, 4- 2W< 0 so [itex]\sqrt{4- 2W}[/itex] is not a real number.
     
  4. Jun 4, 2012 #3

    ElijahRockers

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    Interesting, I would've been confused if I didn't have the software, but your method makes a lot of sense. So then, W=2 is locally asymptotically stable, and W=0 would be unstable.
     
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