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Mar 31, 2019 at 15:42 history edited Michael E2
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Sep 22, 2017 at 5:57 history edited xzczd
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Mar 11, 2014 at 13:43 comment added xzczd 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.
Mar 3, 2014 at 7:51 comment added user11948 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}]
Mar 3, 2014 at 7:49 comment added user11948 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
Mar 3, 2014 at 7:03 comment added user11948 ** 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}]
Mar 3, 2014 at 6:30 comment added olliepower @MikeHoneychurch this also is applied as an approximation method in quantum chemistry. It converges slowly.
Mar 3, 2014 at 6:03 comment added user11948 @MikeHoneychurch- I dont know about any homework but this a problem related to Boundary layer flow with Wedge effect and Blowing and suction together.
Mar 3, 2014 at 6:02 comment added user11948 @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].
Mar 3, 2014 at 4:17 comment added Mike Honeychurch 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?
Mar 3, 2014 at 2:17 comment added xzczd 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.
Mar 3, 2014 at 2:02 history asked user11948 CC BY-SA 3.0