0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ I would rather say that the last term in the first equation is the most difficult to deal with. Is $x$ the only independent variable? Why are those partial differential equations instead of ordinary equations? There are a handful numerical methods that would help you to solve these equations if they are ordinary. $\endgroup$ – rafa11111 Dec 18 '18 at 12:51
  • $\begingroup$ @rafa11111 Ok. let us say these are ordinary equations. I would like to solve it analytically. $\endgroup$ – newstudent Dec 18 '18 at 12:52
  • $\begingroup$ If you are sure these are ordinary equations, please edit the question to address so (change the equation, the tag, etc.). I don't think there is an analytical solution to these equations, mainly due to the nonlinear term in the first equation. If $C$ is a small number, you can use a perturbation method, that provides an approximate analytical solution. $\endgroup$ – rafa11111 Dec 18 '18 at 13:01
  • $\begingroup$ You can integrate the second equation once then substitute for $r(x)$ in the first to give an equation in $u(x)$ alone. I think your journey will end there! $\endgroup$ – user121049 Dec 19 '18 at 8:00
  • $\begingroup$ You need some more boundary conditions on $r$. The only hope I see is to assume $r$ stays small so you can set $1-r^2\approx 1$ and linearize the first equation. The two equations are then easy to solve. Then check $r$ really stays small. $\endgroup$ – user121049 Dec 20 '18 at 9:00
1
$\begingroup$

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 ${x\to\infty}$

$\endgroup$
  • $\begingroup$ There is a $u(x)$ in the first equation so how do you solve it for $r(x)$? $\endgroup$ – user121049 Dec 19 '18 at 20:37
  • $\begingroup$ @user121049 Wow I didn't even notice. Thanks for pointing it out. $\endgroup$ – Dylan Dec 20 '18 at 3:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.