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September 13th, 2015, 10:01 AM  #1 
Member Joined: Sep 2013 Posts: 31 Thanks: 0  Transport eq. with two initial conditions?
Given the usual homogenous transport equation: $\displaystyle \partial_t \rho+ v \partial_x \rho=0$ with initial conditions: $\displaystyle \rho(0,t)=g(t)$ and $\displaystyle \rho(x,0)=0$. How do you find $\displaystyle \rho(x,t)$? 
September 24th, 2015, 06:11 AM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 
Since this is a partial differential equation with two independent variables, of course there are two conditions, one for each variable. (Technically, the condition for x is a "boundary condition", not an "initial condition". Your differential equation is $\displaystyle \frac{\partial \rho}{\partial x}+ v\frac{\partial \rho}{\partial t}= 0$ (are we to assume that v is a constant?) with initial conditions $\displaystyle \rho(0, t)= g(t)$ and $\displaystyle \rho(x, 0)= 0$. It should be easy to see that $\displaystyle \rho(x, t)= F(x vt)$ is a solution for F any differentiable function. It should also be easy to show that no such function satisfies both conditions. Last edited by Country Boy; September 24th, 2015 at 06:15 AM. 
September 25th, 2015, 07:54 PM  #3 
Global Moderator Joined: Dec 2006 Posts: 20,919 Thanks: 2201 
The equation given is $\dfrac{\partial \rho}{\partial t}+ v\dfrac{\partial \rho}{\partial x}= 0$, not $\dfrac{\partial \rho}{\partial x}+ v\dfrac{\partial \rho}{\partial t}= 0$. It is satisfied by $\rho(x, t) = F(x  vt)$, where $F$ is differentiable. The condition $\rho(x, 0) = 0 \implies F(x) = 0 \implies \rho(x, t) = 0$, so $\rho(0, t) = g(t)$ can't also be satisfied unless $g(t) = 0$. Maybe the problem was mistyped. 

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conditions, initial, transport 
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