1
$\begingroup$

I have a set of two PDEs: $$\partial_{\tau}\theta+\partial_{\eta}\psi=0$$ $$\partial_{\tau}\psi=-\partial_{\eta}\theta+\alpha\partial_{\eta}^{2}\psi$$

These can be combined into a wave equation of the form: $$\partial_{\tau}^{2}\theta=\partial_{\eta}^{2}\left[\theta+\alpha\partial_{\tau}\theta\right] $$

which with the ansatz: $$\theta=\sin\left(\eta\right)\exp\left(-\omega\tau\right)$$

gives the dispersion relation: $$\omega^2-\alpha\omega+1=0$$

The solution to the dispersion relation is $\omega=\omega_r+i\omega_i$ with $\omega_r=\frac{1}{2}\alpha$ and $\omega_i=\pm\sqrt{1-\frac{1}{4}\alpha^2}$.

Now i would like to determine $\psi\left(\eta,\tau\right)$, I attempted to do this by using the original equations in a rewritten form:

  1. $$\partial_{\eta}\psi = -\partial_{\tau}\theta \rightarrow \psi\left(\eta,\tau\right) = -\omega\cos\left(\eta\right)\exp\left(-\omega\tau\right)+K_1$$

  2. $$\partial_{\tau}\psi = -\partial_{\eta}\theta-\alpha\partial_{\eta}\partial_{\tau}\theta \rightarrow \psi\left(\eta,\tau\right) = \left(\frac{1}{\omega}-\alpha\right)\cos\left(\eta\right)\exp\left(-\omega\tau\right) + K_2$$

but as far as i can see these two partial solutions cannot be combined to satisfy both original equations simulateously. How do i go about getting a solution for $\psi$?

$\endgroup$
  • $\begingroup$ How is the solution you derived a solution? since you end up with $\omega^2$ on the l.h.s and something linear on the r.h.s? unless its a dispersion relation that you obtain? or I am being silly. $\endgroup$ – Chinny84 Feb 8 '16 at 11:06
  • $\begingroup$ @Chinny84 - I indeed obtain a dispersion relation: $\omega^2 - \alpha\omega + 1=0$, from there $\omega=\omega_r+i\omega_i$ with $\omega_r=\frac{1}{2}\alpha$ and $\omega_i=\pm\sqrt{1-\frac{1}{4}\alpha^2}$. I didn't include it in the question because i wanted to keep it short... $\endgroup$ – nluigi Feb 8 '16 at 11:15
  • $\begingroup$ @Chinny84 - do you mean the wave equation or the continuity/navier-stokes equations? I think i have gone over the derivations about a hundred times now and i think i atleast got the pde's correct. Please correct me if i am wrong... $\endgroup$ – nluigi Feb 8 '16 at 11:22
  • $\begingroup$ no i was wrong! Sorry about that haste remark. I will work on this. I used both in way, since both form foundations to Magneto-Hydrodynamics. $\endgroup$ – Chinny84 Feb 8 '16 at 11:24
  • $\begingroup$ @Chinny84 - I actually think i found a solution and am writing an answer as we speak... perhaps you can comment on it in a few minutes? $\endgroup$ – nluigi Feb 8 '16 at 11:26
0
$\begingroup$

As it turns out the two results for $\psi$ are actually equivalent as by the dispersion relation:

$$\frac{1}{\omega}-\alpha=\frac{1}{\omega}\left(1-\alpha\omega\right)=-\omega$$

Alternative method: First, redefine $\theta$ and $\psi$ using a scalar $\phi$: $$\theta=\partial_{\eta}\phi\quad\psi=-\partial_{\tau}\phi$$

This ensures that: $$\partial_{\tau}\theta+\partial_{\eta}\psi=0$$ is exactly satisfied and yields a wave equation for $\phi$ from the other equation: $$\partial_{\tau}^{2}\phi=\partial_{\eta}^{2}\left[\phi+\alpha\partial_{\tau}\phi\right]$$

The solution for $\phi$ has a similar form as previously found for $\theta$: $$\phi\left(\eta,\tau\right)=\left[A\sin\left(\eta\right)+B\cos\left(\eta\right)\right]\exp\left(-\omega\tau\right) $$ where the constants $A$ and $B$ are yet to be determined.

From the definitions for $\theta$ and $\psi$ I find: $$\theta=\partial_{\eta}\phi=\left[A\cos\left(\eta\right)-B\sin\left(\eta\right)\right]\exp\left(-\omega\tau\right)$$

$$\psi=-\partial_{\tau}\phi=\omega\left[A\sin\left(\eta\right)+B\cos\left(\eta\right)\right]\exp\left(-\omega\tau\right) $$

Now if I am looking for a solution of $\psi$ which corresponds to a $\theta$ distribution: $$\theta=\sin\left(\eta\right)\exp\left(-\omega\tau\right) $$ I require $A=0$ and $B=-1$ such that: $$\psi=-\omega\cos\left(\eta\right)\exp\left(-\omega\tau\right) $$

$\endgroup$

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.