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This is my 3rd Order IVP. D[y[x], {x, 3}] == (m/2)*((D[y[x], x])^2 - 1) - ((m + 1)/2)* y*(D[y[x], {x, 2}])^2. Initial conditions given are as follows y[0] = 0; y'[0] = 0; y''[0] = 1; (* This is the First assumed value of y''[0] for initializing shooting method using RK 4rth order method *)

For this problem y'[infinity] is equal to 1, I want to loop the shooting method using Runge Kutta 4 th order method in such a way that after calculations, it will check the y'[] value at sufficiently large value of x and from experimental results, for x ~ 7-8, f'[] is 1.

To Use RK method I redefined the equation like

D[l[x], {x, 2}] ==  z; D[y[x], x] ==  l; D[z[x], x] == (m/2)*(l^2 - 1) - ((m + 1)/2)*z*y;

Now I could write one loop for RK method but this is proving out a bit tough for me because I have to generate a list for different values of f[0] { 0,1} and m is also a variable with range (0-1) step size 0.01 & then I have to deal with 3 equations simultaneously. I am solving this over C programming but its not that efficient. I dont know how to go ahead. If anyone could throw some light.....

Michael E2
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user11948
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  • Where's the definition of m? Then your equation involves simple mistake, y*(D[y[x], {x, 2}])^2 should be y[x] (D[y[x], {x, 2}])^2. Finally, I tried to set m = 1, and this equation is easily solved by NDSolve. – xzczd Mar 03 '14 at 02:17
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    Have there been some recent homework assignments on the shooting method? Seems to have been a few related questions recently. Is this a homework problem? – Mike Honeychurch Mar 03 '14 at 04:17
  • @xzczd- How could you solve this ODE with NDsolve. What you assumed the value of f''[0]. Since we only know f[0] and f'[0]. We do know f'[infinity]= 1 but we cant use this condition in NDSolve, that's why we have to use shooting method I guess. m is a list you can assume - {0.1,0.2,0.3...........1.0}. f[0] is same list. I wish to make a Module function which takes these euqation, perform RK method over them and solved and plot the results for different m and f[0]. – user11948 Mar 03 '14 at 06:02
  • @MikeHoneychurch- I dont know about any homework but this a problem related to Boundary layer flow with Wedge effect and Blowing and suction together. – user11948 Mar 03 '14 at 06:03
  • @MikeHoneychurch this also is applied as an approximation method in quantum chemistry. It converges slowly. – olliepower Mar 03 '14 at 06:30
  • ** I used NDSolve but the results I am getting are totally wrong, dont know why** 
    s = NDSolve[{2*y'''[x] + ((1/3) + 1) y[x]*y''[x] + 
 2*(1/3)* (1 - (y'[x])^2) == 
0, \[Theta]''[x] + 0.5*0.72*(1/3 + 1)*y[x]*\[Theta]'[x] == 0,

    y[0] == 0, y'[0] == 0, y'[6.6] == 1, [Theta][0] == 0, [Theta][6.4] == 1}, {y, y', y'', [Theta]}, {x, 0, 20}]

     – user11948 Mar 03 '14 at 07:03
  • Using NDSolve for this problem is giving incorrect answers Reason might be that since f'[infinity]=1 but I cant use this in my NDSolve argument The function increases rapidly and f'[0] is practically one at x ~ 6.6 If I use this condition, the values I am getting are negative means if f[2]= x then what I am getting is f[2]= -x. Thats why I said we need to use shooting technique – user11948 Mar 03 '14 at 07:49
  • s = NDSolve[{2*y'''[x] + ((1/3) + 1) y[x]*y''[x] + 2*(1/3)* (1 - (y'[x])^2) == 0, \[Theta]''[x] + 0.5*0.72*(1/3 + 1)*y[x]*\[Theta]'[x] == 0, y[0] == 0, y'[0] == 0, y'[7] == 1, \[Theta][0] == 0, \[Theta][6.4] == 1}, {y, y', y'', \[Theta]}, {x, 0, 20}]

    Plot[Evaluate[{y[x], y'[x], y''[x], \[Theta][x]} /. s], {x, 0, 5}]

     – user11948 Mar 03 '14 at 07:51
  • Seems that since you add a - after my name, I didn't receive the message for your comment. I chose the initial condition y''[0] == 1 for I didn't notice it's just assumed and the actually boundary is f'[Infinity] == 1, then, if you want to apply shooting method, you may be interested in this and this post. – xzczd Mar 11 '14 at 13:43

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