Quasilinear PDE

Quasilinear PDE
A first-order quasilinear PDE with two independent variables has the general form
f(x,y,u) u_x+g(x,y,u) u_y=h(x,y,u)
The system of ordinary differential equations are
dx/f(x,y,u) =dy/g(x,y,u) =du/h(x,y,u)
The following examples explain how to find the solution u(x,t).
Example 1: Solve the initial value problem u_t+xu_x=u with u(x,0)=x^2
Solution :
dt/1=dx/x=du/u
dt/1=dx/x ? t+C_1= ln?x
x=Ce^t ?C=xe^(-t)
dt/1=du/u ? t+C_2= ln?u
u=Ke^t
u(x,t)=g(xe^(-t) ) e^t
u(x,0)=x^2 ? g(x)=x^2
u(x,t)=(xe^(-t) )^2 e^t=x^2 e^(-2t) e^t=x^2 e^(-t)

Example 2: Solve the initial value problem u_t+xu_x=-u^2 with u(x,0)=sin?x

Solution :
dt/1=dx/x=du/(-u^2 )
dt/1=dx/x ? t+C_1= ln?x
x=Ce^t ?C=xe^(-t)
And dt/1=du/?-u?^2
t+K=( 1 )/( u )
u=1/(t+K)
u(x,t)=1/(t+g(xe^(-t) ) )
u(x,0)=sin?x ? 1/g(x) =sin?x ? g(x)=csc?x
u(x,t)=1/(t+csc?(xe^(-t) ) )

Example 3: Solve the initial value problem xu_t-2xtu_x=2tu with u(x,0)=x^3
Solution :
dt/x=dx/(-2xt)=du/2tu
dt/x=dx/(-2xt) ? 2tdt=-dx

t^2+x=C
And dx/(-2xt)=du/2tu ? dx/(-x)=du/u
-ln?x+ln?K=ln?u ? u=( K )/x
Then u(x,t)=(g(t^2+x))/x
u(x,0)=x^3 ? (g(x))/x=x^3 ? g(x)=x^4
u(x,t)=?(t^2+x)?^4/x






Example 4: Solve the initial value problem u_t+4u_x=u^2 with u(x,0)=( 1 )/(1+x^2 )
Solution :
dt/1=dx/4=du/u^2
dt/1=dx/4 ? dx= 4dt
x=4t+C ? x-4t=C
And dt/1=du/u^2 ? t+K=-( 1 )/( u )
u=(-1 )/( t+K)
Then u(x,t)=(-1 )/(t+g(x-4t) )
u(x,0)=( 1 )/(1+x^2 ) ? (-1)/g(x) =( 1 )/(1+x^2 )
g(x)=-(x^2+1)
u(x,t)=(-1 )/(t-(x-4t)^2-1)


H.W: Solve the initial value problems
1. u_t-2u_x=u^2 u(x,0)=sin?x Ans: u(x,t)=(-1 )/(t-csc?(x+2t) )
2. u_t+xu_x=( x )/2u with u(x,0)=x Ans: u(x,t)=?(x+ x^2 e^(-2t)-xe^(-t) )