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I have the equation $$ \frac {\partial v}{\partial t}= \gamma \left(1-\frac{v}{v_0}\right)+\alpha \left(1-\frac{v}{v_0}\right)\rho-\beta (\rho-\rho_0) $$ and the mass conservation equation $$ \frac{\partial\rho}{\partial t}+\frac{\partial(v\rho)}{\partial x}=0 $$ where $v \equiv v(x,t)$ and $\rho \equiv \rho(x,t)$.

I'd like to combine these two to obtain an equation for $\rho$, eliminating $v$, but I can not see how.

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The most straightforward, but tedious, way is to generate six equations by computing derivatives of the original equations wrt $x$ and $t$. You can then express them in matrix form $$ \begin{bmatrix} A_{11} & A_{12} & A_{13} & A_{14} & A_{15} & A_{16} \\ A_{21} & A_{22} & A_{23} & A_{24} & A_{25} & A_{26} \\ \dots \\ A_{61} & A_{62} & A_{63} & A_{64} & A_{65} & A_{66} \end{bmatrix} \begin{bmatrix} v \\ \frac{\partial v}{\partial x} \\ \frac{\partial v}{\partial t}\\ \frac{\partial^2 v}{\partial x^2} \\ \frac{\partial^2 v}{\partial x\partial t} \\\frac{\partial^2 v}{\partial t^2} \end{bmatrix} = \begin{bmatrix} C_1 \\ C_2 \\ C_3 \\ C_4\\ C_5 \\ C_6 \end{bmatrix} $$ Solving for the unknowns will give you expressions that contain only $\rho$ and its derivatives that can be used to eliminate $v$ from your equations.

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  • $\begingroup$ well... that worked, but I got an equation 12 lines long. $\endgroup$ Apr 11, 2014 at 3:06
  • $\begingroup$ Now all you have to do is simplify the equation :) $\endgroup$ Apr 11, 2014 at 3:12
  • $\begingroup$ hahaha are you kidding?? that impossible Biswajit Banerjee! thanks I had not seen this method before anyway $\endgroup$ Apr 11, 2014 at 3:23
  • $\begingroup$ You can simplify things slightly if you assume that $\rho \equiv \rho(x)$, then all the $\partial\rho/\partial t$ terms become zero. $\endgroup$ Apr 11, 2014 at 3:33
  • $\begingroup$ Alternatively, you can try to repeat the process after multiplying the first equation by $\rho$ on both sides and see if terms drop out. $\endgroup$ Apr 11, 2014 at 3:43

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