       Re: mixed partial derivatives

• To: mathgroup at smc.vnet.net
• Subject: [mg91450] Re: mixed partial derivatives
• From: Narasimham <mathma18 at hotmail.com>
• Date: Fri, 22 Aug 2008 03:12:15 -0400 (EDT)
• References: <g7rij3\$i8v\$1@smc.vnet.net> <g7uo04\$5tc\$1@smc.vnet.net>

```On Aug 13, 6:36 pm, Jens-Peer Kuska <ku... at informatik.uni-leipzig.de>
wrote:
> Hi,
>
> X[-3, v] == X[3, v]
>
> will not work with mixed derivatives aspecial on the
> interval u in [0,3] and not u in [-3,3]
> and your equation is singular.
>
> Regards
>    Jens
>
> 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}]
>
> > Narasimham

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], X[0, v] == Cos[v],
Derivative[0, 1][X][0, v] == 0, Derivative[1, 0][X][u, 0] == 0},  X[u,
v], {u, 0, 3}, {v, 0, 3}]

Here also output says mixed partials are not allowed.

Narasimham

```

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