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True Cartesian curvature equation, trying to solve it

  1. May 8, 2015 #1
    1. The problem statement, all variables and given/known data
    Solve the following equation:

    v is the dependent variable, x is the independent variable

    2. Relevant equations

    [itex]\frac{d^2v/dx^2}{(1+\frac{dv}{dx}^2)^{3/2}}=1[/itex]

    3. The attempt at a solution

    Hi,

    I am trying to solve the following equation:

    [itex]\frac{d^2v/dx^2}{(1+\frac{dv}{dx}^2)^{3/2}}=1[/itex]

    I used separation of variables as follows:

    Let [itex] u = \frac{dv}{dx} [/itex]

    [itex]\frac{du/dx}{(1+u^2)^{3/2}}=1[/itex]

    Separate the variables and integrate:

    [itex]\frac{du}{(1+u^2)^{3/2}}=dx[/itex]

    [itex]\frac{u}{(1+u^2)^{1/2}}=x + C[/itex]

    [itex]u= \sqrt{\frac{(x+C)^2}{1-(x+C)^2}} or -\sqrt{\frac{(x+C)^2}{1-(x+C)^2}} [/itex]

    Why is this not a valid solution when substituting back into the above equation for u and du/dx?
     
  2. jcsd
  3. May 8, 2015 #2

    Zondrina

    User Avatar
    Homework Helper

    You have made a mistake somewhere in the final step. Here is what I get:

    Screen Shot 2015-05-08 at 8.34.50 AM.png
     
  4. May 8, 2015 #3

    Mark44

    Staff: Mentor

    If a fraction is equal to 1, then the numerator must be equal to the denominator.

    So this equation: [itex]\frac{d^2v/dx^2}{(1+\frac{dv}{dx}^2)^{3/2}}=1[/itex]
    could be written as [itex]\frac{d^2v}{dx^2} = {(1+\frac{dv}{dx}^2)^{3/2}}[/itex]
    or, since the Leibniz notation is a bit clumsy here, as this:
    v'' = (1 + (v')2)3/2
    The first equation and either of the next two equations say exactly the same thing.
     
    Last edited: May 8, 2015
  5. May 8, 2015 #4
    Thanks for your replies.

    Zondrina: I believe that's the same answer, just in a different form as 1 = [1 - (x + C)^2] / [ 1 - (x+C)^2 ] ?

    Mark44: I evaluated the integral in mathematica and matlab, and the same answer is given. I am integrating 1/(1+u^2)^(3/2) over u not x?
     
  6. May 8, 2015 #5

    Mark44

    Staff: Mentor

    Edit: I checked your work and it's fine. Disregard what I said. I have edited my previous post.
     
    Last edited: May 8, 2015
  7. May 8, 2015 #6
    Thanks for checking. It strikes me as odd that Matlab cannot solve this ODE and that the above procedure I used does not turn out to be a solution when back substituting for u and du/dx:

    >> syms u(x)
    >> dsolve(diff(u,x)/(1+u^2)^(3/2)==1)
    Warning: Explicit solution could not be found.
    > In dsolve at 194
    ans =
    [ empty sym ]
     
  8. May 9, 2015 #7

    Zondrina

    User Avatar
    Homework Helper

    Solving the equation in post #2 yields the solution:

    $$v = \pm \left[- \frac{(x+C)}{\sqrt{\frac{1}{1 - (x+C)^2} - 1}} + K\right] = \mp \left[\frac{(x+C)}{\sqrt{\frac{1}{1 - (x+C)^2} - 1}} - K\right] = \mp \left[\sqrt{1 - (x+C)^2} - K \right]$$

    If there were initial conditions, you could even solve for the constants ##C## and ##K##.
     
  9. May 9, 2015 #8
    Matlab's simplifier was the problem, I verified the solution by hand, thanks for the help.
     
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