Solve third order, non-linear, ODE with indefinite BC's?

Topher925
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I have a third order, non-linear, homogeneous, constant coefficient ODE that I need to solve but have no idea how to do it. To make matters worse, one of the boundary conditions are indefinite. Here's the equation,

y''' + y*y'' - y'^2 + 1 = 0

and the BC's
y(0) = 0
y'(0) = 0
y'(\infty) = 1

Does this equation have a general or analytical (non numerical) solution? I have access to Maple, just not at the moment, and would like to try and solve it analytically without computer first. Any suggestions on how to solve it? Can it be solved?
 
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Topher925 said:
I have a third order, non-linear, homogeneous, constant coefficient ODE that I need to solve but have no idea how to do it. To make matters worse, one of the boundary conditions are indefinite. Here's the equation,

y''' + y*y'' - y'^2 + 1 = 0

and the BC's
y(0) = 0
y'(0) = 0
y'(\infty) = 1

Does this equation have a general or analytical (non numerical) solution? I have access to Maple, just not at the moment, and would like to try and solve it analytically without computer first. Any suggestions on how to solve it? Can it be solved?

It's called the Falkner-Skan equation but maybe you know that already. Maybe a Google can help you. Wikipedia says it can't be solve analytically. Here's my attempt to solve it numerically in a limited range, say (0,10) in Mathematica:

Code:
myend = 10;
sols = Map[
   First[NDSolve[{y'''[t] + y[t] y''[t] - (y'[t])^2 + 1 == 0, 
       y[0] == y'[0] == 0, y'[myend] == 1}, y, t, 
      Method -> {"Shooting", 
        "StartingInitialConditions" -> {y[0] == 0, y'[0] == 0, 
          y''[0] == #}}]] &, {1.7, 1.75, 1.8}];
Plot[Evaluate[y[t] /. sols], {t, 0, myend}, 
 PlotStyle -> {Black, Blue, Green}, AspectRatio -> 4/3]

And that code returns y''(0)\approx 1.233
 
Last edited:
see attached... it solves falkner skan in maple..

any questions, fire away
 

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Thread 'Direction Fields and Isoclines'
I sketched the isoclines for $$ m=-1,0,1,2 $$. Since both $$ \frac{dy}{dx} $$ and $$ D_{y} \frac{dy}{dx} $$ are continuous on the square region R defined by $$ -4\leq x \leq 4, -4 \leq y \leq 4 $$ the existence and uniqueness theorem guarantees that if we pick a point in the interior that lies on an isocline there will be a unique differentiable function (solution) passing through that point. I understand that a solution exists but I unsure how to actually sketch it. For example, consider a...
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