Re: how to find complete integral of pde
- To: mathgroup at smc.vnet.net
- Subject: [mg74436] Re: how to find complete integral of pde
- From: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>
- Date: Wed, 21 Mar 2007 02:51:15 -0500 (EST)
- Organization: The Open University, Milton Keynes, UK
- References: <etlqcn$rap$1@smc.vnet.net><etnif8$dl7$1@smc.vnet.net> <etof1d$i2v$1@smc.vnet.net>
bhargavi wrote:
> hi, tx alot for ur reply.
> i copied input from mathematica5.2, n pasting here.
> \!\(CompleteIntegral[D[u[x, y], x]*1.5*\((
> 1 - 4\ y\^2)\) == \ Br\ 144\ y\^2\ + D[u[x, y], y]^2, u[x, y],
> {x, y}]\)
[snip]
The function CompleteIntegral does not reside in Mathematica kernel;
therefore, you must load the package DSolveIntegrals before you can use
CompleteIntegral. For instance
In[1]:=
Remove[CompleteIntegral]
<< "Calculus`DSolveIntegrals`"
CompleteIntegral[D[u[x, y], x]*1.5*(1 - 4*y^2) ==
Br*144*y^2 + D[u[x, y], y]^2, u[x, y], {x, y}]
Out[3]=
{{u[x, y] ->
2
x B[1] - 6. y Sqrt[-1. Br y + 0.0104167 B[1.] -
2
0.0416667 y B[1.]] + B[2] -
((0. + 0.0625 I) B[1.]
Log[(0. - 2. I) y
Sqrt[1. Br + 0.0416667 B[1.]] +
2
2. Sqrt[-1. Br y + 0.0104167 B[1.] -
2
0.0416667 y B[1.]]]) /
Sqrt[1. Br + 0.0416667 B[1.]]},
{u[x, y] ->
2
x B[1] + 6. y Sqrt[-1. Br y + 0.0104167 B[1.] -
2
0.0416667 y B[1.]] + B[2] +
((0. + 0.0625 I) B[1.]
Log[(0. - 2. I) y
Sqrt[1. Br + 0.0416667 B[1.]] +
2
2. Sqrt[-1. Br y + 0.0104167 B[1.] -
2
0.0416667 y B[1.]]]) /
Sqrt[1. Br + 0.0416667 B[1.]]}}
Regards,
Jean-Marc