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September 13th, 2015, 10:01 AM   #1
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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 $\displaystyle \rho(x,t)$?
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September 24th, 2015, 06:11 AM   #2
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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.
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September 25th, 2015, 07:54 PM   #3
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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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