I have this equation:

$$\Omega \frac{d}{dt}v_x - \frac{d^2}{dz^2}v_x = 0$$ And I am finding a solution in a form of:

$$v_x = A(x,y,z) e^{it} $$ where $A(x,y,z$ is a complex amplitude. After simple substitution I am getting:

$$ A''-i\Omega A. $$ And here I stumbled. I know that oscillator equation has the form of: $A''+ \omega_0^2A=0$, but in that case $\omega_0$ is a real number, whereas in my case if I, for example, substitute $\omega_0^2=-i\Omega$ I will get the same equation where $\omega_0$ is a complex number.. So if I do this than I can write a solution in the following form:

$$A = C_1e^{i\omega_0z} + C_2e^{-i\omega_0z}$$ is it eligible or not?

I know from physical view that $v_x$ has to have a periodic solution (I know it from physical statement of my problem) so it correlates with physical sence. Exponent is responsible for periodic time in the system while Amplitude also has periodic behaviour.

But I am confused: $C_1 $ and $C_2 $ are the complex amplitude? Because if I try to reformulate the problem to its initial values like: $\omega_0 = \sqrt{-i\Omega}$ than after proccessing the result will be: $\omega_0 = \sqrt{\Omega}\sqrt{-1}\sqrt{i} = - 0.7 \sqrt{\Omega} + 0.7 \sqrt{\Omega} i$ and after substituting this to :

$$A = C_3e^{-z}e^{-0.7i\sqrt{\Omega}z} + C_4e^{z}e^{0.7i\sqrt{\Omega}z}$$ which is not periodical solution anymore.. I am so confused here.. How should I tackle this problem..

Thank you in advance!


Of course, since $v_x(x, y, z, t)$, but one has to differentiate over $z$ and $t$ . It gets dependency from $z$ in explicit form and dependency from $x, y$ goes to 'constants', dependency from $t$ vanishes since we find solution that already includes $t$ in exponent and

$$A = C_1(x, y)e^{i\omega_0z} + C_2(x, y)e^{-i\omega_0z}$$

  • 1
    $\begingroup$ Are those supposed to be partial derivatives? $\endgroup$ – K.defaoite Jun 11 '20 at 1:14
  • $\begingroup$ @K.defaoite Please , see Edit One $\endgroup$ – John Jun 11 '20 at 5:13
  • $\begingroup$ I looked at your edit but I still don’t understand why the derivatives aren’t written as partial derivatives. $\endgroup$ – Radial Arm Saw Jun 12 '20 at 0:49

I suppose that $x,y,z$ are NOT function of $t$ because this is not specified in the wording of the question.

$$\Omega \frac{\partial v_x}{\partial t} - \frac{\partial^2 v_x}{\partial z^2} = 0$$ As far as I can understand you are looking for a solution on the form :

$$v_x = A(x,y,z) e^{it}\quad\implies\quad \begin{cases} \frac{\partial v_x}{\partial t}=i A(x,y,z) e^{it} \\ \frac{\partial^2 v_x}{\partial z^2}=\frac{\partial^2 A(x,y,z)}{\partial z^2}e^{it} \end{cases}$$ $$\Omega i A(x,y,z) e^{it} - \frac{\partial^2 A(x,y,z)}{\partial z^2}e^{it} = 0$$ $$i\Omega A(x,y,z) - \frac{\partial^2 A(x,y,z)}{\partial z^2} = 0$$ Since there is no partial derivatives with regards to $x$ and $y$ the PDE is reduced to an ODE with regard to the variable $z$ only. $x,y$ must be considered as parameters. $$i\Omega A-A''=0$$ $$A(x,y,z)=c_1(x,y)\exp\left(\frac{1-i}{\sqrt{2\Omega}}z\right)+c_2(x,y)\exp\left(-\frac{1-i}{\sqrt{2\Omega}}z\right)$$ Of course $c_1(x,y)$ and $c_2(x,y)$ can be complex.

One can separate the sinusoidal terms as : $$\exp\left(\frac{1-i}{\sqrt{2\Omega}}z\right)=\exp\left(\frac{1}{\sqrt{2\Omega}}z\right)\cos\left(\frac{1}{\sqrt{2\Omega}}z\right) -i\exp\left(\frac{1}{\sqrt{2\Omega}}z\right)\sin\left(\frac{1}{\sqrt{2\Omega}}z\right) $$

Finally $v_x(x,y,z,t)$ is obtained with arbitrary functions $c_1(x,y)$ and $c_2(x,y)$ into it.

The integration with respect to $x$ leads to $v(x,y,z,t)$ with $c_1$ replaced by $C_1(x,y)=\int c_1(x,y)dx+f_1(y)$ and $c_2$ replaced by $C_2(x,y)=\int c_2(x,y)dx+f_2(y)$. All are arbitrary complex and/or real functions.


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