# Combining two results from partial integration

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 $\psi$?

• 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. – Chinny84 Feb 8 '16 at 11:06
• @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... – nluigi Feb 8 '16 at 11:15
• @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... – nluigi Feb 8 '16 at 11:22
• 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. – Chinny84 Feb 8 '16 at 11:24
• @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? – nluigi Feb 8 '16 at 11:26

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: $$\psi=-\omega\cos\left(\eta\right)\exp\left(-\omega\tau\right)$$