A one-dimensional compressible fluid blob starts at \(t = 0\) with uniform density \(\rho \equiv 1\) on \(1 \leq x \leq 2\).
It obeys the conservation of mass equation \[\frac{\partial \rho}{\partial t} + \frac{\partial (\rho \mathcal{v})}{\partial x} = 0\] with the (Eulerian) velocity field given as \(\mathcal{v}(x, t) = x^2e^{-3t}\).
- Find the density of the blob for \(t \geq 0\) as a function of position and time, \(\rho = \rho(x, t)\).
- Find the positions of the free-boundaries, \(x_1(t), x_2(t)\) (the left and right edges of the blob).
- Use your results from the above parts to directly evaluate the integral \[\int_{x_1(t)}^{x_2(t)} \rho(x, t) \:dx\] and show that this is consistent with the Reynolds transport theorem.
A)
We start with the conservation of mass equation: \[\frac{\partial \rho}{\partial t} + \frac{\partial (\rho \mathcal{v})}{\partial x} = 0\]
We use the Chain Rule, plug in what was given, and rearrange: \[\frac{\partial \rho}{\partial t} + \frac{\partial \rho}{\partial x}\mathcal{v} + \frac{\partial \mathcal{v}}{\partial x}\rho = 0\]
\[\frac{\partial \rho}{\partial t} + \underbrace{(x^2e^{-3t})}_{c(x)}\frac{\partial \rho}{\partial x} = \underbrace{-2xe^{-3t}\rho}_{r(x,\rho)}\]
We are now ready to use the method of characteristics!
First, we will construct the ODE describing the path:
\[\frac{dy}{d\tau} = c(y(\tau)) = y^2e^{-3\tau}\] \[y(t) = x\] \[y(0) = A\]
We will solve the above, and then solve for \(A\):
\[y(\tau) = \left(\frac{1}{A}+\frac{e^{-3\tau}}{3}-\frac{1}{3}\right)^{-1}\] \[y(t) = x = \left(\frac{1}{A}+\frac{e^{-3t}}{3}-\frac{1}{3}\right)^{-1}\] \[\Rightarrow A = \left(\frac{1}{x}-\frac{e^{-3t}}{3}+\frac{1}{3}\right)^{-1}\]
Now, we have:
\[\frac{d\rho}{dt} = r(y(t),\rho(t)) = -2y(t)e^{-3t}\rho\] \[\Rightarrow \frac{1}{\rho}d\rho = -2e^{-3t}y(t)\:dt\] \[\Rightarrow \int_{0}^{t} \frac{1}{\rho}d\rho = \int_{0}^{t} -2e^{-3\tau}y(\tau)\:d\tau\]
by separation of variables. We will plug in \(y(\tau)\) and simplify:
\[\int_{0}^{t} \frac{1}{\rho}d\rho = \int_{0}^{t} -2e^{-3\tau}\left(\frac{1}{A}+\frac{e^{-3\tau}}{3}-\frac{1}{3}\right)^{-1}\:d\tau\]
\[\Rightarrow \ln(\rho(t))-\underbrace{\ln(\rho(0))}_{=1} = \int_{0}^{t} \frac{-6Ae^{-3\tau}}{3+Ae^{-3\tau}-A}\:d\tau\] \[\Rightarrow \ln(\rho(t)) = 2\ln(Ae^{-3\tau}-A+3)\Biggr|_{0}^{t}\] \[\Rightarrow \ln(\rho(t)) = 2\ln(Ae^{-3t}-A+3)-2\ln(3)\] \[\Rightarrow \rho(t) = \frac{(Ae^{-3t}-A+3)^2}{9}\]
From this point, we plug in \(A\) that we previously solved for:
\[\boxed{\rho(x,t) = \left(\frac{x}{3}(e^{-3t}-1)-1\right)^{-2}}\]
B)
We have that \(x\) starts off bounded (\(1 \leq x \leq 2\)) so…
\[x_1(0) = 1 \Rightarrow A_1 = 1\] \[x_2(0) = 2 \Rightarrow A_2 = 2\]
from the setup in Part A. We also had:
\[x := \left(\frac{1}{A}+\frac{e^{-3t}}{3}-\frac{1}{3}\right)^{-1}\]
This gives us:
\[x_1(t) = \left(\frac{1}{A_1}+\frac{e^{-3t}}{3}-\frac{1}{3}\right)^{-1} \Rightarrow \left(\frac{e^{-3t}}{3}+\frac{2}{3}\right)^{-1} = \boxed{\frac{3}{2+e^{-3t}}}\] \[x_2(t) = \left(\frac{1}{A_2}+\frac{e^{-3t}}{3}-\frac{1}{3}\right)^{-1} \Rightarrow \left(\frac{e^{-3t}}{3}+\frac{1}{6}\right)^{-1} = \boxed{\frac{6}{1+2e^{-3t}}}\]
C)
Combining Parts A & B:
\[\int_{x_1(t)}^{x_2(t)} \rho(x, t) dx = \int_{\frac{3}{2+e^{-3t}}}^{\frac{6}{1+2e^{-3t}}} \left(\frac{x}{3}(e^{-3t}-1)-1\right)^{-2} dx\]
We will use u-substitution:
\[u = \frac{x}{3}(e^{-3t}-1)-1, du = \frac{e^{-3t}-1}{3}\]
\[\Rightarrow \frac{3}{e^{-3t}-1}\int u^{-2} du\] \[ = -\frac{3}{\left(e^{-3t}-1\right)u}\]
Now, we will integrate with bounds:
\[ = -\frac{3}{\left(e^{-3t}-1\right)\left(\frac{x}{3}(e^{-3t}-1)-1\right)}\Biggr|_\frac{3}{2+e^{-3t}}^\frac{6}{1+2e^{-3t}}\]
The process of simplifying is omitted (try it out yourself!), but in the end, we arrive at:
\[\frac{2e^{-3t}+1}{e^{-3t}-1} - \frac{e^{-3t}+2}{e^{-3t}-1} = \frac{e^{-3t}-1}{e^{-3t}-1} = \boxed{1}\]
The Reynolds transport theorem gives us the following formula:
\[\frac{d}{dt}\left(\int_{a(t)}^{b(t)}\rho(x,t)\:dx\right) = \int_{a(t)}^{b(t)}\left[\frac{\partial}{\partial t}\rho(x,t)+\frac{\partial}{\partial x}\left(\rho(x,t)\mathcal{v}(x,t)\right)\right] dx\]
We just solved for the integral on the left hand side and the integrand on the right hand side was given in the problem:
\[ \text{LHS} = \frac{d}{dt}\:1 = 0\] \[ \text{RHS} = \int_{a(t)}^{b(t)}0\:dx = 0\]
So, the results are consistent with the transport theorem.