From: "David Zachmann"
Subject: The Solution
Date: Wed, 8 Dec 1999 06:27:35 -0600
Newsgroups: sci.math
Keywords: example -- solving a simple linear PDE
(1) The problem is Ut + U*Ux = C*Uxx, subject to the boundary values
U(x, 0) = F(x), with x a real number with t>0
(2) Under these restrictions the problems can be solved. But first we must
come up with a transformation that reduces this nonlinear equation, if
possible, to a linear diffusion equation.
(3) To this end, first, we write (1) in the form analogous to a
conservation law
Ut + Dx[(1/2)U^2 - C*Ux] = 0
(4) Now (3) can be viewed as the compatibility condition for a function G to
exist such that
(i) U = Gx, and
(ii) C*Ux - (1/2)U^2 = Gt.
(5) I now substitute the value of U from (4)(i) in (4)(ii) to obtain
C*Gxx - (1/2)*(Gx)^2 = Gt
(6) Now I introduce the change of variables G = -2*C*ln(Q),
So that
(i) Gt = -2*C*Qt/Q
(ii) U = Gx = -2C*Qx/Q
(iii) Gxx = -2C*(Qxx*Q - (Qx)^2)/Q^2
(7) Substituting (7)(i) and (7)(ii) and (7)(iii) into (5), I obtain,
-2C^2 * (Qxx*Q - (Qx)^2)/Q^2 - (1/2)[-2C*Qx/Q]^2 = (-2/C)*Qt/Q
OR
(Qxx*Q - (Qx)^2)/Q^2 + (Qx/Q)^2 = C*Qt/Q
OR
Qxx/Q - (Qx)^2/Q^2 + (Qx)^2/Q^2 = C*Qt/Q
OR
Qxx/Q = C*Qt/Q
OR
Qxx = C*Qt
(8) I now solve equation (7), subject to the initial condition
Q(x,0) = Q*(x), x in the reals.
(9) This can be written in terms of our original initial condition, viz.,
U(x, 0) = F(x) by using the transformation (6)(ii), viz,
U = Gx = -2C*Qx/Q. This results in
F(x) = U(x,0) = -2C*Qx(x,0)/Q(x,0)
(10) Integrating this result gives us
Q(x,0) = Q*(x) = exp{(-1/(2C)*DefIntg[F(a*)da*, a*=0, a*=x]
(11) The Fourier transform can be used to solve the linear initial value
problem
(i) Qxx = C*Qt
(ii) Q(x,0) = Q*(x), x in the reals.
Utilizing The Fourier Transform (11) can be solved in the standard way.
Doing so, I get
Q(x,t) = (1/(2*Sqrt(pi*C*t))*DefIntg{Q(Z)*exp[-(x - Z)^2/(4*C*t)dZ,
z= -infinity, z=+infinity},
Where Q(Z) is given by (10).
(12) I then substitute the value of Q(Z) to rewrite Q(x,t) in (11) in the
form
Q(x,t) = (1/(2*Sqrt(pi*C*t))*DefIntg{exp[-f /(2*C)]dZ, Z = -infinity, Z = +
infinity]
Where
f(Z,x,t) = DefIntg[F(a*)da*, a*=0, a*=Z] + [(x - Z)^2]/(2t)
(13) Thus
Qx =
- (1/(4*C*Sqrt(pi*C*t))*DefIntg{[(x - Z)/t * exp(-f/(2*C))dZ, Z = -infinity,
Z= + infinity}
(14) Therefore, the exact solution of our initial value problem is
U(x,t) = f(x,t)/g(x,t)
Where
(i) f(x,t) = DefIntg{[(x - Z)/t * exp(-f/(2*C))dZ, Z = -infinity, Z= +
infinity}
and
(ii) g(x,t) = DefIntg{ exp[-f/(2*C)]dZ, Z = -infinity, Z = + infinity]
FINIS
David Zachmann
------------------------
Maksim Shabrov wrote in message
news:384CB1C7.6A595E9B@brown.edu...
> Hi,
> does anyone have any ideas on how to solve the following:
>
> Ut+U*Ux=C*Uxx (C=const) ?
>
> make Ux=U1 and solve the system:
>
> Ut-C*U1x+U*U1=0
> Ux-U1=0
> ??
> Thanks.
> Maxim.