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I am trying to initialize a traveling wave for a 1d simulation as one can see from the attached figure. enter image description here

Such that it will be traveling to the right. However, I cannot initialize the right velocity profile, and this makes the initial pressure distribution tends to be more uniform to reach the same pressure of the surrounding fluid !

Can any one provide some support?

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can show a diagram of how it breaks into two rarefaction waves? – udiboy1209 Nov 14 '13 at 14:38
@udiboy, sorry, I missexplained myself. see the Update – user2536125 Nov 14 '13 at 14:51
What is the medium in which the wave is propagating? And, more specifically, which model are you using for your simulation? – user23660 Nov 14 '13 at 15:08
@user23660 I am using the 1d euler equation, and the gas is simply air (ideal ). – user2536125 Nov 14 '13 at 15:10
Would be a better home for this question? – Qmechanic Nov 14 '13 at 17:50

We are trying to get the simple wave solution, so one can assume the dependence of the functions defining the solution (namely $u$, $p$ and $\rho$) only on a single combination of variables $x$ and $t$. In case of weak sound wave this combination would be $x - c t$, but nonlinear effects would makes this more complicated. Nevertheless, we still can choose one of the functions, for example $\rho$, as an independent variable on which the other two would depend and write $$ \rho = \rho(x,t),\quad p= p(\rho), \quad u = u(\rho) . $$

We can than substitute these into continuity equation and Euler equation: $$ \dot{\rho}+\rho' u + \rho \frac{d u}{d\rho} \rho ' = 0, \tag{1} $$ $$ \frac{du}{d\rho}\dot{\rho}+ u \frac{du}{d\rho} \rho' + \frac{c^2(\rho)}{\rho}\rho ' =0,\tag{2} $$ where $\rho'= \dfrac{\partial \rho}{\partial x}$, $\dot{\rho}= \dfrac{\partial \rho}{\partial t}$. The local speed of sound is defined by $c^2(\rho)= \dfrac{d p}{d \rho}$ and could be found using adiabatic equation for an ideal gas.

For the initial conditions on the velocity $u$ we could solve (1) and (2) for $\frac{du}{d\rho}$ (also eliminating $\dot{\rho}$): $$ \dfrac {du}{d\rho} = \pm \frac{c(\rho)}{\rho}, $$ where two sign choices correspond to simple waves traveling to the right (+) and left (-). Integrating we obtain: $$ u = \pm \int \frac{c}{\rho} d\rho = \pm \int \frac{dp}{\rho c}. $$ The final explicit result could be obtained by using the adiabatic process equation: $p \rho^{-\gamma}= p_0 \rho_0^{-\gamma}$.

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Thanks for the answer, however, I think I did not explaine myself well. By 1D Euler I meant, the continuity, energy and momentum equations without source terms (no friction or heat transfer). so I can't see how your development can make a pure right(or left) traveling wave! – user2536125 Nov 14 '13 at 17:16
That's what I used. Only instead of energy equation I used (equivalent for our purposes) adiabatic condition. The fact that the wave is traveling in one direction is expressed in dependence $u(\rho)$ and $p(\rho)$. – user23660 Nov 14 '13 at 17:30
What do you mean by "pure" traveling wave? For a single fluid, the Euler equations will not give a simple traveling wave solution. Only two options exist: expansion wave or compression wave. Expansion waves are described in the above answer and involve characteristics spreading out. Compression waves involve characteristics converging to form a shock wave. – SimpleLikeAnEgg Nov 14 '13 at 17:50
@SimpleLikeAnEgg , by a pure traveling wave, it's like if one had a disturbance source at the left end of the tube, this source will generate a pulse which will automatically travel to the right direction – user2536125 Nov 15 '13 at 6:50
@user2536125 Disturbances tend to spread out in all directions in fluids. With periodic boundary conditions you could potentially initialize a traveling sound wave. A sound wave is a very weak wave, but with the Euler equations it would eventually form shocks. Otherwise, if you make the left boundary condition be that of higher pressure and density you will create a rightward traveling shock. Good luck. – SimpleLikeAnEgg Nov 15 '13 at 19:28

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