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