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.