hoomanya said:
\partial(yv)/\partialt=f(y,v,t)
where v is velocity = dx/dt, and y is a material property.
In addition:
\partialv/\partialx = constant
So, let me make sure I understand this. In your first equation, you're differentiating the product of y and v with respect to time:
<br />
\begin{eqnarray*}<br />
f(y,v,t) & = & \frac{\partial yv}{\partial t} \\<br />
& = & y\frac{\partial v}{\partial t}+v\frac{\partial y}{\partial t} \\<br />
& = & y\frac{\partial^2 x}{\partial t^2}+\frac{\partial x}{\partial t}\frac{\partial y}{\partial t} \\<br />
\end{eqnarray*}<br />
and your second equation is
<br />
\frac{\partial v}{\partial x} = C<br />
I'm having some trouble understanding that second equation. It's not clear to me how we regard v as a function of x so as to take the derivative. But here's my attempt. I'm picturing this as a particle moving in the +x direction under the influence of some sort of force. Let's assume it starts off at x=0, and that it has velocity v0. (It has to have some velocity to start with, or it'll never move away from its starting point.) Then it's going to move in the positive direction with increasing velocity, the velocity at time t always equal to v0 + C x(t). to get from x to x(t)+dx will take time dt=\frac{dx}{v_0+C x(t)}, giving the following DE for x(t):
<br />
\frac{dx}{dt}=v_0+C x(t)<br />
with initial condition x(0) = 0.
This has solution
<br />
\begin{eqnarray*}<br />
x(t) & = & \frac{v_0}{C}\left(e^{Ct}-1\right) \\<br />
v(t) & = & v_0 e^{Ct}<br />
\end{eqnarray*}<br />
I'll pause there to ask if this looks right. Also, could you explain what the Hell this is? It would really help to have an idea where the equations come from.
