1st Order PDE with Constant Coefficients

Partial Differential Equation PDE

A differential equation involving partial derivatives of a dependent variable(one or more) with more than one independent variable is called a partial differential equation, hereafter denoted as PDE. For example the following equations are PDEs:

a ?u/?t+b ?u/?x=f(x,t) (au_t+bu_x=f(x,t))
(?^2 u)/(?t^2 )-c^2 (?^2 u)/(?x^2 )=0 ( ?u_tt-c^2 u_xx=0?^ ) Wave Equation
?u/?t-T (?^2 u)/(?x^2 )=0 ( ? u_t-Tu_xx=0?^ ) Heat Equation
(?^2 u)/(?x^2 )+(?^2 u)/(?y^2 )=0 (u_xx+u_yy=0) Laplace Equation

1st Order PDE with Constant Coefficients
Let’s consider the linear first order constant coefficient partial differential equation au_t+ bu_x = f(x,t) with an initial condition u(x,0)=g(x) where a?0 and b are constants, and f(x,t) and g(x) are given functions.
This is a typical initial value problem for PDEs.
Our objective is to find the solution u(x,t).

?(Solution Formula u(x,t)=g(x-( b )/( a ) t)+( 1 )/( a ) ?_0^t?f(x-( b )/( a ) t+( b )/( a ) ?,?) d?)
Example 1: Solve the initial value problem
u_t+4u_x=0 with u(x,0)=( 1 )/(1+x^2 )
Solution? u(x,t)=g(x-4t)
u(x,0)=( 1 )/(1+x^2 ) ? g(x)=( 1 )/(1+x^2 )
u(x,t)=( 1 )/(1+?(x-4t)?^2 )
Example 2: Solve the initial value problem
2u_t+ 3u_x =0 with u(x,0)=(1+e^x)/(1+e^4x )
Solution:
f(x,t)=0 and g(x)=(1+e^x)/(1+e^4x ) (a=2,b=3)
u(x,t)=g(x-( b )/( a ) t)+( 1 )/( a ) ?_0^t?f(x-( b )/( a ) t+( b )/( a ) ?,?) d?
u(x,t)=g(x-( 3 )/( 2 ) t)+( 1 )/( 2 ) ?_0^t?0 d?=(1+e^(x-3t?2))/(1+e^4(x-3t?2) )
Example 3: Solve the initial value problem
2u_t+ 5u_x =( 1 )/( 4 ) cos?x+( 1 )/( 6 ) sin?4t with u(x,0)=e^(-x^2?8)
Solution:
f(x,t)=( 1 )/( 4 ) cos?x+( 1 )/( 6 ) sin?4t and g(x)=e^(-x^2?8) (a=2,b=5)
u(x,t)=g(x-( b )/( a ) t)+( 1 )/( a ) ?_0^t?f(x-( b )/( a ) t+( b )/( a ) ?,?) d?
u(x,t)=g(x-( 5 )/( 2 ) t)+( 1 )/( 2 ) ?_0^t?f(x-( 5 )/( 2 ) t+( 5 )/( 2 ) ?,?) d?
=e^(-(x-5t?2)^2?8)+( 1 )/( 2 ) ?_0^t?(( 1 )/( 4 ) cos?(x-( 5 )/( 2 ) t+( 5 )/( 2 ) ?)+( 1 )/( 6 ) sin?(4?) ) d?
=e^(-(x-5t?2)^2?8)+( 1 )/( 2 ) ? [( 1 )/( 10 ) sin??(x-( 5 )/( 2 ) t+( 5 )/( 2 ) ?)-( 1 )/( 24 ) cos?(4?) ? ]?|_0^t
=e^(-(x-5t?2)^2?8)+( 1 )/( 20) sin??(x)-( 1 )/( 48 ) cos?(4t) ?-1/20 sin??(x-( 5 )/( 2 ) t)+( 1 )/48?


Example 4: Solve the initial value problem u_t+ 2u_x =6te^(-2x) with u(x,0)=x^2
Solution:
f(x,t)=6te^(-2x) and g(x)=x^2 (a=1,b=2)
u(x,t)=g(x-( b )/( a ) t)+( 1 )/( a ) ?_0^t?f(x-( b )/( a ) t+( b )/( a ) ?,?) d?
u(x,t)=g(x-2t)+?_0^t?f(x-2t+2?,?) d?
u(x,t)=(x-2t)^2+?_0^t??6?e^(-2(x-2t+2?) ) ? d?
6? e^(-2x+4t-4?)
6 ((-1)?4) e^(-2x+4t-4?)
0 (1?16) e^(-2x+4t-4?)
u(x,t)=(x-2t)^2+[-3/2 ?e^(-2x+4t-4?)-3/8 e^(-2x+4t-4?) ]_0^t
u(x,t)=(x-2t)^2-3/2 te^(-2x)-3/8 e^(-2x)+3/8 e^(e^(-2x+4t) )

H.W(221) : Solve the initial value problem
1. 2u_t+ 5u_x =0 with u(x,0)=1?((1+e^x ) ). Ans: u(x,t)=1?((1+e^(x-5t/2) ) )
2. 2u_t+ 3u_x =( 1 )/( 4 ) sin?x with u(x,0)=(1+e^x)/(1+e^4x )
Ans: u(x,t)=(1+e^(x-3t?2))/(1+e^4(x-3t?2) )-( 1 )/( 12 ) cos?x+( 1 )/( 12 ) cos?(x-3t/2)
3. 3u_t+ 2u_x =cos?t with u(x,0)=sin?x
Ans: u(x,t)=( 1 )/( 3 ) sin?t+sin?(x-( 2 )/( 3 ) t)