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Coupled non-linear boundary value ODEs

Posted 10 years ago
How can you solve these 4 coupled non-linear boundary value ODEs ("Von Kármán Swirling flow" problem) in MATHEMATICA?
 my equations are:
F" = F^2 + HF' - G^2
G" = 2FG + HG'
H' = -2F
P' = 2FH - 2F'
Boundary Condition: F(0)=0 F(inf)=0 G(0)=1 G(inf)=0 H(0)=0 P(0)=0
Thanks
POSTED BY: vahid khalili
3 Replies
Could you write this in proper Mathematica syntax?  A function, say F, should be written with its arguments as F. Also, by HF, od you mean H times F or do you mean the single functin called HF?  And os on.  By tdoing that you will bothe learn Mathematica syntax so you can follow the discussion and you will let us know what the actual formulation of your problem is.  
POSTED BY: David Reiss
Posted 10 years ago
my Functions is F(t), G(t), H(t) and P(t)
and equations are :
F"[t] = (F[t])^2 + H[t]F'[t] - (G[t])^2
G"[t] = 2F[t]G[t] + H[t]G'[t]
H'[t] = -2F[t]
P'[t] = 2F[t]H[t] - 2F'[t]

Boundary Condition: F(0)=0 F(inf)=0 G(0)=1 G(inf)=0 H(0)=0 P(0)=0
POSTED BY: vahid khalili
Note that you need to use == for equality and that it's usually a good idea to use lower case variables as Mathematica variables are upper case. However, it appears that your problem cannot be solved symbolically.
 In[2]:= DSolve[{f''[t] ==  (f[t])^2 + h[t] f'[t] - (g[t])^2,
   g''[t] == 2 f[t] g[t] + h[t] g'[t],
   h'[t] == -2 f[t],
   p'[t] == 2 f[t] h[t] - 2 f'[t]}, {f[t], h[t], g[t], p[t]}, t]
 
 Out[2]= DSolve[{(f^\[Prime]\[Prime])[t] ==
    f[t]^2 - g[t]^2 + h[t] Derivative[1][f][t], (g^\[Prime]\[Prime])[
     t] == 2 f[t] g[t] + h[t] Derivative[1][g][t],
   Derivative[1][h][t] == -2 f[t],
  Derivative[1][p][t] == 2 f[t] h[t] - 2 Derivative[1][f][t]}, {f[t],
  h[t], g[t], p[t]}, t]
POSTED BY: Frank Kampas
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