1st Order PDE with Variable coefficients

1st Order PDE with Variable coefficients
Let’s consider the linear first order variable coefficients partial differential equation a(x,t)u_t+ b(x,t)u_x = f(x,t) with an initial condition u(x,0)=g(x) where a(x,t) ,b(x,t), f(x,t) and g(x) are given functions.
The system of ordinary differential equations are
dt/(a(x,t) )=dx/b(x,t) =du/(f(x,t))
The following examples explain how to find the solution u(x,t).
Example 1: Solve the initial value problem u_t+tu_x=0 with u(x,0)=cos?2x
Solution :
dt/(1 )=dx/t? tdt=dx
( t^2)/2+C=x ? C=x-( t^2)/2
Therefore u(x,t)=g(x-( t^2)/2) is a general solution
u(x,0)=cos?2x ? u(x,0)=g(x)=cos?2x
u(x,t)=cos??2(x-( t^2)/2)=? cos?(2x-t^2 )
Example 2: Solve the initial value problem tu_t+xu_x=0 with u(x,1)=cos?x
Solution :
dt/t=dx/x ? ln?t= ln?x+ln?C
( t )/( x )=C ? ( x )/( t )=C_1
u(x,t)=g(( x )/( t ))
u(x,1)=cos?x ? g(x)= cos?x
u(x,t)=cos??(( x )/( t ))?


Example 3: Solve the initial value problem u_t+xu_x=xt with u(x,0)=sin?x
Solution :
dt/(1 )=dx/x=du/xt
dt/(1 )=dx/x ? t+C_1=ln?x ? x=Ce^t
C=xe^(-t)
dt/(1 )=du/xt ? du=xtdt
du=Cte^t dt
u(x,t)=C(t-1) e^t+K
u(x,t)=x(t-1)+g(xe^(-t) )
u(x,0)=sin?x ? -x+g(x)=sin?x
g(x)=x+sin?x
u(x,t)=x(t-1)+xe^(-t)+sin?(xe^(-t) )


H.W: Solve the initial value problems
1. tu_t-2xu_x=0 with u(x,1)=tan^(-1)?2x Ans: u(x,t)=tan^(-1)?(2xt^2 )

2. u_t+u_x=2x with u(x,0)=e^x Ans: u(x,t)=e^(x-t)+2xt-t^2

3. u_t+xtu_x=0 with u(x,0)=sin?x Ans: u(x,t)=sin??(xe^(-t^2?2))?