Solving the one-dimensional incompressible Navier-Stokes Equations I am interested in solving the PDE system
$$\frac{\partial\rho}{\partial t}(x(t),t)+u(t)\frac{\partial \rho}{\partial x}(x(t),t)=0,  \qquad (\text{EQ} \ 1)$$
$$\rho(x(t),t) u'(t)=-\frac{\partial p}{\partial x}(x(t),t)+\rho(x(t),t)g(x(t),t) \qquad (\text{EQ} \ 2)$$
where $p(x(t),t)$, $g(x,t)$, $u(0)=u_0$, and $f(x)=\rho(x,0)$ are known.
I attempted to solve this PDE with the method of characteristics.
For an arbitrary variable $s$ I defined $h(s)=\rho(x(t(s)),t(s))$.
By the multivariable chain rule,
$$h'(s)=\frac{d}{ds}\rho(x(t(s)),t(s))=\frac{\partial\rho}{\partial x}\frac{dx}{ds}+\frac{\partial \rho}{\partial t}\frac{dt}{ds}.$$
Setting $$\frac{dt}{ds}=1, \qquad \text{(EQ 3)}$$
$$\ \frac{dx}{ds}=u(t(s)), \qquad \text{(EQ 4)}$$
$$h'(s)=\frac{\partial\rho}{\partial x}u(t(s))+\frac{\partial \rho}{\partial t}(1)=0$$ by (EQ 1).
Integrating EQ 3 with respect to $t$, I get
$$t=s+C$$
for some unknown constant $C$.
Setting $t(0)=0$, I get $t(s)=s$.
So EQ 4 becomes
$$\frac{dx}{dt}=u(t).$$
Integrating with respect to $t$, I obtain
$$x(t)-x(0)=\int_0^t u(s) ds.$$
Hence
$$x(0)=x(t)-\int_0^t u(s) ds$$
Since $h'(s)=0$, h is constant and $h(t)=h(0)$.
By the definition of $h$,
$$\rho(x(t),t)=h(t)=h(0)=\rho(x(0),0)=f(x(0))=f\left(x(t)-\int_0^t u(s)\right)\qquad \text{(EQ 5)}$$.
Substituting EQ 5 into EQ 2, I get
$$f\left(x(t)-\int_0^t u(s)\right)u'(t)=-\frac{\partial p}{\partial x}(x(t),t)+f\left(x(t)-\int_0^t u(s)\right)$$
To remove the integral, I defined an auxiliary function $U(t)=\int_0^t u(s)ds$.
Then, by the Fundamental Theorem of Calculus, $U'(t)=u(t)$ and $u'(t)=U''(t)$.
Hence 
$$f\left(x(t)-U(t)\right)U''(t)=-\frac{\partial p}{\partial x}(x(t),t)+f\left(x(t)-U(t)\right)g(x(t),t)$$
$$U''(t)=-\left(f\left(x(t)-U(t)\right)\right)^{-1}\frac{\partial p}{\partial x}(x(t),t)+g(x(t),t).$$
Since $x(t)$, it makes to either set $x(t)=0$ or $x(t)=U(t)$.
Setting $x(t)=0$, I obtain
$$U''(t)=-\left(f\left(-U(t)\right)\right)^{-1}\frac{\partial p}{\partial x}(0,t)+g(0,t).$$
Since the functions $p$, $g$, and $f$ are known but arbitrary this is equivalent to solving the problem
$$U''(t)=F(U(t),t)\qquad \text{(EQ 6)}$$
for some known but arbitrary function $F$.
The initial conditions for this problem would be $U(0)=\int_0^0 u(s) ds=0$ and $U'(0)=u(0)=u_0$.
EQ 6 is generally a second-order nonlinear ODE which contains both the independent variable $t$ and the dependent variable $U$ which makes it difficult if not impossible to solve.
If I were able to solve for $u$, then I know that $\rho(x(t),t)=f\left(x(t)-\int_0^t u(s)\right)$, so I would have a closed form solution for $\rho$.
I think that the method of characteristics might not be the way I should be approaching this problem. The other way I think of solving PDEs is with separation of variables since (EQ 1) and (EQ 2) are linear in $\rho$, but I have no boundary condition on $\rho$. Additionally separation of variables usually relies on a trigonometric function being derived from a second-order spartial derivative but (EQ 1)-(EQ 2) is a first-order system.
The question asks for "implicit solutions" for $\rho$ and $u$. I interpret this as meaning I need to derive an equation involves $u$, the functions that you would accept in a closed-form solution (trigonometric functions, logarithms, etc), and possibly $\int_0^t u(s)ds$? If an "implicit solution" can involve a second-order ODE then I would be done but I doubt that this is the case. Because (EQ 5) gives a closed-form solution for $\rho$ I am only concerned with finding $u$.
I put the linear PDE and the non-linear equations tags because the system is nonlinear when $\rho$ and $u$ are considered simultaneously, but it is linear in $\rho$ and in $u$. Correct me if I am wrong.
EDIT: The question did not make it clear whether $p$ and $g$ had any dependence on $x$ or $t$ (it did not give arguments for $g$ and $\frac{\partial p}{\partial x}$ like it did for $\rho$ and $u$). I don't know much about fluid dynamics so I assumed that they both depended on both $x$ and $t$. However, if $p$ and $g$ are both independent of $t$, then I know exactly how to solve this problem. Reading elsewhere I found that $g$ is generally independent of both $x$ and $t$. If it is reasonable to assume that pressure is independent of time then that would explain why the problem shouldn't have an unsolvable nonlinear second order ODE in it.
Thanks, Andrew Murdza
 A: I figured out the answer to the question a while ago but I hadn't made the time to post it here. In case anyone has the same question the rest of proof is as follows:
Ignore everything after the line
Substituting EQ 5 into EQ 2, we get
$$f\left(x(t)-\int_0^t u(s)\right)u'(t)=\frac{\partial p}{\partial x}(x(t),t)+f\left(x(t)-\int_0^t u(s)\right)g(x(t),t).$$
Isolating $u'(t)$, we find that
$$u'(t)=\frac{1}{f\left(x(t)-\int_0^t u(s)\right)}\frac{\partial p}{\partial x}(x(t),t)+g(x(t),t).$$
Since $x(t)$ is arbitrary, we may set $x(t)=\int_0^t u(s)$ to get
$$u'(t)=\frac{1}{f\left(0\right)}\frac{\partial p}{\partial x}\left(\int_0^t u(s),t\right)+g\left(\int_0^t u(s),t\right).$$
By the Fundamental Theorem of Calculus we have
$$u(T)-u(0)=\int_0^T u'(t)=\int_0^T \left(\frac{1}{f(0)}\frac{\partial p}{\partial x}\left(\int_0^t u(s),t\right)+g\left(\int_0^t u(s),t\right)\right).\qquad \text{(EQ 7)}\qquad \text{(EQ 8)}$$
Rewriting EQ 7, we conclude that
$$u(T)=u(0)+\frac{1}{f(0)}\int_0^T \frac{\partial p}{\partial x}\left(\int_0^t u(s),t\right)+\int_0^T g\left(\int_0^t u(s),t\right)$$
The problem also assumes $g$ is constant (this is not explicitly stated but $g$ is constant throughout the book that the question is from), so EQ 8 simplifies to
$$u(T)=u(0)+\frac{1}{f(0)}\int_0^T \frac{\partial p}{\partial x}\left(\int_0^t u(s),t\right)+\int_0^T g=u(0)+\frac{1}{f(0)}\int_0^T \frac{\partial p}{\partial x}\left(\int_0^t u(s),t\right)+gT$$ 
What made this problem difficult for me is that I did not realize that an implicit solution could be an integral equation in the unknown (to me differential equations seem easier to work with than integral equations but others must disagree)
