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Order of variables: p[x,t] or p[t,x]

  • To: mathgroup at smc.vnet.net
  • Subject: [mg6565] Order of variables: p[x,t] or p[t,x]
  • From: "w.meeussen" <w.meeussen.vdmcc at vandemoortele.be>
  • Date: Tue, 1 Apr 1997 19:01:01 -0500 (EST)
  • Sender: owner-wri-mathgroup at wolfram.com

(* to be read in Courier font *)
hi,

I stumbled onto some weird stuff & need some advice :
in solving the DE:
eqn1=D[p[x,t],t]==- v D[p[x,t],x]
representing the evolution of a population p[x,t] of droplets with curvature
x and time t,


DSolve[eqn1,p[x,t],{x,t}]//Simplify
 
                      x
{{p[x, t] -> C[1][t - -]}}
                      v

this gives me some unspecified function C[1] of time and x.
Ok, what I realy want is p[x-t v,0] or C[1][t-x/v] with
C[1][-x/v] 'equivalent to' p[x,0] somehow.

try reversing the order of variables:
 
eqn3=D[p[t,x],t]==- v D[p[t,x],x]
gives
 (1,0)               (0,1)
p     [t, x] == -(v p     [t, x])
with:
DSolve[eqn3,p[t,x],{t,x}]
resulting in
{{p[t, x] -> C[1][-(t v) + x]}}

this time, the dummy function C[1] is given in function of x.

So, now I hoped for a direct 'equivalence' between p[0,x] and C[1][x] and
tried to add p[0,x] to the eqn3:

like:
DSolve[{eqn3,p[0,x]==q[x]},p[t,x],{t,x}]
BUT IT's NO GOOD:
         (1,0)               (0,1)
DSolve[{p     [t, x] == -(v p     [t, x]), 
 
   p[0, x] == q[x]}, p[t, x], {t, x}]

**************************************************************
the order of variables seems to define the way the solution is
represented: in terms of x (curvature, dimensions 1/meters) or t (time,
dimensions seconds).
**************************************************************
The real problem I'm faced with is somewhat more complex:

eqn2=D[p[x,t],t]==- v E^(x a) D[p[x,t],x]

                         1
{{p[x, t] -> C[1][t + --------]}}
                       x a
                      E    v a

this one will not react favorably to changing the order of variables from
x,t to t,x.

The 'ultimate unabridged' problem is:

flux = (4 (ceq - c E^(a x)) Pi (1 + x d))/(x^2 d)

eqn5=D[p[x,t],t]==- (flux v) D[p[x,t],x]

It works out like:
DSolve[eqn5,p[x,t],{x,t}]

{{p[x, t] -> 
 
    C[1][-(-4 Pi t v + 
 
                    2
         (d DSolve`t  C[1][DSolve`t, 0, x]) / 
 
                     a DSolve`t
          ((ceq - c E          ) (1 + d DSolve`t))) / 
 
      (4 Pi v)]}}

but now I lost it completely: 
its still of the general form C1[ stuff in x & t ]
but there is a second occurence of C1[dummy,0,x] in there !
WITH THREE ARGUMENTS !?!
I don't understand its meaning! does it have any?
how is it linked to p[x,0] ?
********************************************************

can somebody 
             get me
                    on the tracks
                                  again, please ?

********************************************************

A propos,

introducing p[x,0]==q[x] still doesn't work,
and reversing the variables this time ruins it :

In[20]:=
eqn6=D[p[t,x],t]==- (flux v) D[p[t,x],x]
DSolve[eqn6,p[t,x],{t,x}]
Solve::tdep: 
   The equations appear to involve transcendental
    functions of the variables in an essentially
    non-algebraic way.

and bounces the input back.

More A Propos,
yes, I tried LaplaceTransform following the recepy in TheBook&Help
under example 'One Dimensional Heat Conduction'.
but that didn't even solve eqn1.


greetings from Flanders,

wouter.
(A Primitive flemishing Flemish)





Dr. Wouter L. J. MEEUSSEN
eu000949 at pophost.eunet.be
w.meeussen.vdmcc at vandemoortele.be



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