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Consider $$ \frac{\partial u}{\partial t}=fv-g\frac{\partial\eta}{\partial x}\\ \frac{\partial v}{\partial t}=-fu-g\frac{\partial\eta}{\partial y}\\ \frac{\partial\eta}{\partial t}=-H_0\frac{\partial u}{\partial x}-H_0\frac{\partial v}{\partial y}. $$ Here $u,v$ and $\eta$ are functions of $(x,y,t)$, $f\in\mathbb{R}$ is a parameter, $g$ is the gravitational acceleration and $H_0$ is a constant.

Now make the ansatz $$ (u,v,\eta)^T=e^{i(kx+ly)+\lambda t}\cdot (u_0,v_0,\eta_0)^T. $$

Now it is said that inserting this ansatz into the equation, we get $$ 0=\lambda(\lambda^2+gH_0(k^2+l^2)+f^2). $$

How to get this?

If I insert the ansatz in the system, I only get three equations for $\lambda$, namely $$ \lambda=u_0^{-1}(fv_0-ikg),\\ \lambda=v_0^{-1}(-fu_0-ilg\eta_0),\\ \lambda=\eta_0^{-1}(H_0u_0ik-H_0v_0il) $$

--- By the way: Where does tis ansatz come from?

Is it a product ansatz $$ (u(x,y,t),v(x,y,t),\eta(x,y,t))=(u(t),v(t),\eta(t))\cdot (u(x,y),v(x,y),\eta(x,y)) $$ and choosing the space-part $(u(x,y),v(x,y),\eta(x,y))$ to be eigenmodes/ Fouriermodes oscillating with same spatialbehaviour? I.e. is the idea maybe to write solutions in fouriermode basis and tehrefore to start with fouriermodes as solutions (for the space-dependent part)?

mathfemi
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We have a system of three equations:

\begin{equation} \begin{cases} u_0\lambda=fv_0-g\eta_0ik\\v_0\lambda=-fu_0-g\eta_0il\\ \eta_0\lambda=-H_0u_0ik-H_0v_0il \end{cases} \begin{cases} u_0={fv_0-g\eta_0ik\over\lambda}\\v_0\lambda=-fu_0-g\eta_0il\\ \eta_0\lambda=-H_0u_0ik-H_0v_0il \end{cases} \end{equation}

we get $u_0$ from the first equation and we replace it in the other ones, so:

\begin{equation} \begin{cases} u_0={fv_0-g\eta_0ik\over\lambda}\\v_0\lambda={-f^2v_0+fg\eta_0ik\over\lambda}-g\eta_0il\\ \eta_0\lambda=-H_0{fv_0-g\eta_0ik\over\lambda}ik-H_0v_0il \end{cases} \end{equation}

Now we make some manipulation with the second and third equation:

\begin{equation} \begin{cases} v_0(\lambda^2+f^2)=\eta_0(fgik-\lambda gil)\\ \eta_0(\lambda^2+H_0k^2g)=v_0(-H_0ikf-\lambda H_0il) \end{cases} \begin{cases} {v_0\over\eta_0}={(fgik-\lambda gil)\over(\lambda^2+f^2)}\\ {\eta_0\over v_0}={(-H_0ikf-\lambda H_0il)\over(\lambda^2+H_0k^2g)} \end{cases} \begin{cases} {v_0\over\eta_0}={gi(fk-\lambda l)\over(\lambda^2+f^2)}\\ {\eta_0\over v_0}={-H_0i(kf+\lambda l)\over(\lambda^2+H_0k^2g)} \end{cases} \end{equation}

Now we multiply these equations:

\begin{equation} {v_0\over\eta_0}{\eta_0\over v_0}={gi(fk-\lambda l)\over(\lambda^2+f^2)}{-H_0i(kf+\lambda l)\over(\lambda^2+H_0k^2g)}\longrightarrow 1=gH_0{(f^2k^2-\lambda^2l^2)\over(\lambda^2+f^2)(\lambda^2+H_0k^2g)} \end{equation}

Hence we get:

$$\lambda^4+\lambda^2H_0k^2g+f^2\lambda^2+f^2H_0k^2g=gH_0f^2k^2-gH_0l^2\lambda^2$$ $$\lambda^4+\lambda^2H_0k^2g+f^2\lambda^2+gH_0l^2\lambda^2=0$$ $$\lambda^2[\lambda^2+gH_0(k^2+l^2)+f^2]=0$$

MattG88
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  • Thank you. And for the other part of my question: Do you have an idea whats the idea behind the ansatz? Its always a bit strange when an ansatz is given without justifying it. – mathfemi Dec 24 '16 at 15:39
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    It is just a shortcut...you can also get $v_0$ from the first equation and replace it in the second one; $\eta_0$ will be symplified and you'll get the same equation with $\lambda$. – MattG88 Dec 24 '16 at 15:39
  • I'm not expert in pde, but your answer might work: we just try taking as ansats a function that is factored in time and spatial part and they are independent.. – MattG88 Dec 24 '16 at 15:47
  • I just do not understand where the ansatz $\exp(i(kx+ly))(u_0,v_0,\eta_0)^T$ for the spatial part comes from. – mathfemi Dec 24 '16 at 15:50
  • Is this an equation for cosmological model? If so...which model? – MattG88 Dec 24 '16 at 15:52