Re: Help in solving PDF equations
- To: mathgroup at smc.vnet.net
- Subject: [mg52245] Re: Help in solving PDF equations
- From: Paul Abbott <paul at physics.uwa.edu.au>
- Date: Wed, 17 Nov 2004 02:20:13 -0500 (EST)
- Organization: The University of Western Australia
- References: <cnbn3o$9va$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
In article <cnbn3o$9va$1 at smc.vnet.net>, Wei Wang <weiwang at baosteel.com>
wrote:
> Could anybody please help me in solving the following PDF equation?
>
> (p + r/rou)*(1- (Derivative[1, 0][f][x, t])^2 - k * Derivative[0, 1][f][x,
> t])/Sqrt[1 + (Derivative[1, 0][f][x, t])^2] ==0
Clearly, the left-hand side of this expression vanishes if its numerator
does:
de = (1- Derivative[1, 0][f][x, t]^2 - k Derivative[0, 1][f][x, t]) == 0
and Mathematica can solve this partial differential equation in closed
form:
sol = DSolve[de, f, {x, t}]
> with initial or boundary conditions as
>
> f[x, 0] == -1
Applying this initial condition
f[x, 0] == -1 /. sol
one can find the two undetermined coefficients:
coefs = SolveAlways[#,x]& /@ % // Union
The solution then reads
f[x, t] /. sol /. coefs // Flatten // Union
{t/k - 1}
which is independent of x.
Note that the denominator of the original equation reduces to unity as
can be seen by evaluating
1 + Derivative[1, 0][f][x, t]^2 /. sol /. coefs
> Derivative[0, 1][f][x, 0] == p/k
This condition can only be satisfied if p == 1 as can be seen from
Derivative[0, 1][f][x, 0] == p/k /. sol /. coefs
> Derivative[1, 0][f][R0, t] == 0
This condition is satisfied since f is independent of x,
Derivative[1, 0][f][R0, t] == 0 /. sol /. coefs
> f[x, t] == If[0<= x <=1, Sqrt[1 - x^2]]
This does not make sense to me. The right-hand side does not depend on
t. If you mean this to hold for a fixed t then it is incompatible with
the general solution determined above.
> where, p, k, R0 and rou are constants.
The supplied conditions determine p but k, R0, and rou are not
determined.
Cheers,
Paul
--
Paul Abbott Phone: +61 8 6488 2734
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