# How do we solve third order nonlinear differential equation f’’’+ff’-f’2-Re2 f’=0

How do we solve third order nonlinear differential equation f’’’+ff’-f’2-Re2 f’=0 f(0)=0,f'(0)=1,f(Infinity)=0

From the OP's comment:

ic1 = 0; ic2 = 0; ic3 = 1 a = 0.0;
sol = NDSolve[{y'''[x] + y[x] y'[x] - y'[x]^2 - a^2 y'[x] == 0,
y[0] == ic1, y'[Infinity] == ic3, y'[0] == ic2}, y[x], PlotRange];
Plot[y[x] /. sol, {x, 0, finish}].


I have used this code to solve using mathematica but i didnt get it

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Please show what have you tried –  belisarius Mar 25 '14 at 6:54
You can start with DSolve. If that fails, then try NDSolve. –  Nasser Mar 25 '14 at 7:23
Is this a question about Mathematica (the software) or math in general? –  Yves Klett Mar 25 '14 at 8:58
ic1 = 0; ic2 = 0; ic3 = 1 a = 0.0; sol = NDSolve[{y'''[x] + y[x] y'[x] - y'[x]^2 - a^2 y'[x] == 0, y[0] == ic1, y'[Infinity] == ic3, y'[0] == ic2}, y[x], PlotRange]; Plot[y[x] /. sol, {x, 0, finish}]. I have used this code to solve using mathematica but i didnt get it –  Priyadharshini Mar 25 '14 at 9:28

It turns out the issue is to do with mathematics and not Mathematica.

The OP's code yields an error:

ic1 = 0; ic2 = 0; ic3 = 1; a = 0;
sol = NDSolve[
{y'''[x] + y[x] y'[x] - y'[x]^2 - a^2 y'[x] == 0,
y[0] == ic1, y'[0] == ic2, y'[Infinity] == ic3},
y[x], {x, 0, 1000}];


NDSolve::ndsv: Cannot find starting value for the variable y'. >>

Translated into mathematics, it means this.

Suppose $\lim_{x \rightarrow \infty} y'(x) = 1$. Therefore $y \rightarrow \infty$. According to the differential equation $y'''=-y\,y'+(y')^2 \approx 1-y$ as $x \rightarrow \infty$. Hence $y''' \rightarrow -\infty$, which contradicts $y' \rightarrow 1$.

Therefore there is no such solution.

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In addition to the analysis of Michael E2 there are errors in Mathematica code even if to put the boundary condition not at infinity, but at some constant value. In fact they are corrected in the Michael E2's answer above. I only slightly change them to put the boundary at a finite distance, say, at xmax=100, and with that to plot the result:

    ic1 = 0;
ic2 = 0;
ic3 = 1 ;
a = 0.;
xmax = 100;
sol = NDSolve[{y'''[x] + y[x] y'[x] - y'[x]^2 - a^2 y'[x] == 0,
y[0] == ic1, y'[xmax] == ic3, y'[0] == ic2}, y[x], {x, 0, xmax}]


yielding the result in the form of the interpolation function. One can plot it:

Plot[y[x] /. sol, {x, 0, xmax}]


It should look like the following that supports the analytic result of Michael E2.

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THANK YOU VERY MUCH ALEXEI.how do we find y''(0) by using previous one –  Priyadharshini Mar 25 '14 at 14:49
It is perhaps worth noting that the solution to the BVP is not unique. –  Michael E2 Mar 25 '14 at 14:52
How can i proceed to get y''(0) –  Priyadharshini Mar 25 '14 at 15:02
Kindly help me Michael –  Priyadharshini Mar 25 '14 at 16:24
@Priyadharshini You need to put @ in front of my name for me to get notified of your comment. You can get y''[0] with D[y[x] /. First[sol], {x, 2}] /. x -> 0, if you use Alexei's sol. –  Michael E2 Mar 25 '14 at 20:30