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I am working through a paper before continuing on with my research but have become stuck with combining these two differentials.

The differentials I have is:

$ \frac{\partial \rho v}{\partial t} + \frac{\partial U\rho v}{\partial s} = \frac{B}{\mu} \frac{\partial b}{\partial s} $ (1)

$ \frac{\partial{b}}{\partial t} +\frac{\partial U b}{\partial s} =B \frac{\partial v}{\partial s} $ (2)

Noting that U=U(s,t) and v(s,$\theta$), b(s, $\theta$) and $\rho=\rho(t)$. I am trying to get to the solution (6) in the paper but cannot get the first two terms or the last term! To get the middle terms I have divided by B in the second equation and used parts to expand the second term in the first.

(6)

$ \frac{\partial^2 b}{\partial t^2}+\frac{\partial^2(Ub)}{\partial t \partial s} +\frac{\partial}{\partial s}\left(U(\frac{\partial b}{\partial t} + \frac{\partial Ub}{\partial s})\right) - \frac{B^2}{\rho \mu}\frac{\partial^2b}{\partial s^2}=0$

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We have the following equations

$$ \frac{\partial \rho v}{\partial t} + \frac{\partial U\rho v}{\partial s} = \frac{B}{\mu} \frac{\partial b}{\partial s} \qquad\text{and} $$

$$ \frac{\partial{b}}{\partial t} +\frac{\partial U b}{\partial s} =B \frac{\partial v}{\partial s}. $$

Differentiate the second equation with respect to $t$ to obtain:

$$ \frac{\partial^2{b}}{\partial t\partial t} +\frac{\partial^2 U b}{\partial t\partial s} =B \frac{\partial^2 v}{\partial t\partial s}=B \frac{\partial}{\partial s}\left[\dfrac{\partial v}{\partial t}\right]. (1)$$

If we rearrange the first equation

$$ \frac{\partial \rho }{\partial t}v +\rho \frac{\partial v}{\partial t}+ \rho\frac{\partial Uv}{\partial s} = \frac{B}{\mu} \frac{\partial b}{\partial s} $$

and solve for $\partial v/\partial t$ we obtain:

$$ \frac{\partial v}{\partial t} =\frac{B}{\mu\rho} \frac{\partial b}{\partial s}- \frac{\partial Uv}{\partial s}-\dfrac{1}{\rho}\frac{\partial \rho }{\partial t}v. $$

Using this last expression for the equation $(1)$ we obtain: $$\frac{\partial^2{b}}{\partial t\partial t} +\frac{\partial^2 U b}{\partial t\partial s} =B \frac{\partial}{\partial s}\left[\frac{B}{\mu\rho} \frac{\partial b}{\partial s}- \frac{\partial Uv}{\partial s}-\dfrac{1}{\rho}\frac{\partial \rho }{\partial t}v\right]$$

$$\frac{\partial^2{b}}{\partial t^2} +\frac{\partial^2 U b}{\partial t\partial s} =\frac{B^2 }{\mu\rho}\frac{\partial^2 b}{\partial s^2}-B\dfrac{\partial^2Uv}{\partial s^2}-\dfrac{B}{\rho} \dfrac{\partial^2\rho}{\partial s\partial t}v-\dfrac{B}{\rho}\dfrac{\partial \rho}{\partial t}\dfrac{\partial v}{\partial s}.$$

In the last expression, we can drop the partial derivative of $\rho$ with respect to $t$ and $s$

$$\frac{\partial^2{b}}{\partial t^2} +\frac{\partial^2 U b}{\partial t\partial s} =\frac{B^2 }{\mu\rho}\frac{\partial^2 b}{\partial s^2}-B\dfrac{\partial^2Uv}{\partial s^2}-\dfrac{B}{\rho}\dfrac{\partial \rho}{\partial t}\dfrac{\partial v}{\partial s}.$$

I think there is something missing in your problem description because I don't see an obvious way to rewrite the last expression into the expression in your question.

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