Re: mixed partial derivatives
- To: mathgroup at smc.vnet.net
- Subject: [mg91270] Re: mixed partial derivatives
- From: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>
- Date: Wed, 13 Aug 2008 04:38:15 -0400 (EDT)
- Organization: The Open University, Milton Keynes, UK
- References: <g7rij3$i8v$1@smc.vnet.net>
Narasimham wrote:
> When mixed derivatives are not allowed, what is the fix? TIA
>
> p=D[X[u,v],u]; q=D[X[u,v],v]; r =D[X[u,v],u,u]; s=D[X[u,v],u,v];
> t=D[X[u,v],v,v];
> GC[u_,v_] = (r t - s^2)/(1+p^2 + q^2)^2
> NDSolve[{GC[u,v]== 1, X[u, 0] == Cosh[u], Derivative[0,1][X][u, 0] ==
> 0,
> X[-3, v] == X[3, v]},X[u,v], {u,0,3},{v,0,3}]
Is your question related to the following messages I got on my system or
it is something completely different?
In[1]:=
p = D[X[u, v], u];
q = D[X[u, v], v];
r = D[X[u, v], u, u];
s = D[X[u, v], u, v];
t = D[X[u, v], v, v];
GC[u_, v_] = (r t - s^2)/(1 + p^2 + q^2)^2
NDSolve[
{GC[u, v] == 1,
X[u, 0] == Cosh[u],
Derivative[0, 1][X][u, 0] == 0,
X[-3, v] == X[3, v]}, X[u, v], {u, 0, 3}, {v, 0, 3}]
Out[3]=
(1,1) 2 (0,2) (2,0)
-X [u, v] + X [u, v] X [u, v]
------------------------------------------
(0,1) 2 (1,0) 2 2
(1 + X [u, v] + X [u, v] )
During evaluation of In[1]:= NDSolve`FiniteDifferenceDerivative::ordred:
There are insufficient points in dimension 1 to achieve the requested
approximation order. Order will be reduced to 1.
During evaluation of In[1]:= NDSolve`FiniteDifferenceDerivative::conw:
There are insufficient points in dimension 1 to generate consistent
finite different weights.
Out[4]=
(1,1) 2 (0,2) (2,0)
-X [u, v] + X [u, v] X [u, v]
NDSolve[{------------------------------------------ == 1,
(0,1) 2 (1,0) 2 2
(1 + X [u, v] + X [u, v] )
(0,1)
X[u, 0] == Cosh[u], X [u, 0] == 0, X[-3, v] == X[3, v]}, X[u, v],
{u, 0, 3}, {v, 0, 3}]
Regards,
-- Jean-Marc