I have the following differential equations.
$$ \tau V\frac{\partial r}{\partial x} + W\frac{\partial ^2r}{\partial x^2} + [r - Cu(1-r^2)][1-r^2]=0,$$
$$ V\frac{\partial u}{\partial x} + D\frac{\partial^2 u}{\partial x^2} - \frac{V}{2}\frac{\partial r}{\partial x}=0,$$ with far field boundary condition as $u_{x=\infty} = 0.6.$ and $u_{x=0} = 0$. Here, $\tau, V, W, D, C$ are dimensionless constants.
Now, I would like to solve the second equation subject to boundary conditions. But the last term in the second equation confuses me on how to proceed. Can anyone help me out with this ?
thanks
Disclaimer: I've just realized you can't actually solve for $r(x)$ in the first equation, which makes my answer kind of useless. I'll leave it up anyway, to learn from my mistakes.
Suppose you've already solved the first equation, and $r(x)$ is known, then the second equation is linear and non-homogeneous, and can easily be solved using variations of parameter
Let $a = \frac{V}{D}$
$$ \frac{d^2u}{dx^2} + a\frac{du}{dx} = \frac{a}{2}\frac{dr}{dx} $$
The homogeneous equation has solution $u_h = c_1 + c_2e^{-ax}$, so guess a particular solution of the form
$$ u_p(x) = p(x) + q(x)e^{-ax} $$
Then
\begin{align} p'(x) + q'(x)e^{-ax} &= 0 \\ -aq'(x)e^{-ax} &= \frac{a}{2}r'(x) \end{align}
or $p'(x) = \frac{1}{2}r'(x)$ and $q'(x) = -\frac{1}{2}r'(x)e^{-ax}$
You can integrate to find
\begin{align} p(x) &= \frac{1}{2}r(x) \\ q(x) &=-\frac12\int r'(x) e^{ax} = -\frac12 r(x)e^{ax}\ dx + \frac{a}{2}\int r(x)e^{ax}\ dx \end{align}
Therefore
$$ u_p(x) = \frac{a}{2}e^{-ax}\int_0^x r(t)e^{at}\ dt $$
Matching the boundary conditions, we find
$$ u(x) = u_{\infty}(1-e^{-ax}) + \frac{a}{2}e^{-ax}\int_0^x r(t)e^{at}\ dt $$
where $u_\infty$ is some constant depending on the behavior of $r(x)$ at ${x\to\infty}$