I'm working on global seismology and I'm currently facing troubles understanding how an equation is obtained. The equation concerned is the following one : $$ \rho^{E1} = -\nabla \cdot (\rho^0\mathbf{s}) $$ (From the book Theoretical Global Seismology, Princeton Press : Google Book Link equation (3.46) p.65) Where $\rho^{E1}$ is the first order perturbation of the density and $\rho^0$ the density in the reference state. $\mathbf{s}$ is the displacement field.

It comes from the mass conservation equation which has been integrated in time : $$ \partial_t\rho^E + \nabla \cdot (\rho^E \mathbf{u}^E) = 0 $$

and the following decomposition : $$ \rho^E(\mathbf{x}, t) = \rho^0(\mathbf{x}) + \rho^{E1}(\mathbf{x}, t) $$

Where $\rho^0$ is independent of time and $\mathbf{u}$ is the velocity field. How do you obtain the first equation from the second one ? If someone could help me out on this one it's be really nice ! Cheers

  • $\begingroup$ Is $\rho^0$ independent of time? Any more properties that could be useful? $\endgroup$
    – Danu
    Feb 25 '14 at 16:42
  • $\begingroup$ It is, I edited the post $\endgroup$
    – Amzocks
    Feb 26 '14 at 8:37

If you substitute the decomposition in, you get:

$$ \partial_t \rho^0 + \partial_t \rho^{E1} + \nabla \cdot (\rho^0 \mathbf{u}^E) + \nabla \cdot (\rho^{E1} \mathbf{u}^E) = 0 $$

Typically the decomposition used assumes that $\rho^0$ is constant in time and that $\rho^{E1}$ is random in time, such that it's mean value is 0. Therefore, $\partial_t \rho^0 = 0$.

So when you integrate over all time, the value of $\overline{\rho}^{E1} = 0$. Which means $\int_t \nabla \cdot (\rho^{E1} \mathbf{u}^E) = 0$ and $\int_t \nabla \cdot (\rho^0 \mathbf{u}^{E}) = \nabla \cdot (\rho^0 \int_t \mathbf{u}^E)$ and of course the integral of the velocity is just the displacement field. So the final expression is:

$$ \rho^{E1} = -\nabla \cdot (\rho^0 \mathbf{s})$$


Based on the comments, there are two things to clarify. First is the decomposition itself. Typically when a term is split as done here, it is split into a mean and a fluctuating component. The fluctuating component has a zero mean. This has to be true because $\rho^0$ is defined as the mean component (brackets indicate taking the mean):

$$ \left< \rho^E \right> = \left< \rho^0 + \rho^{E1} \right> = \rho^0 $$

which means $\left< \rho^{E1} \right> = 0$ by definition. Otherwise $\rho^0$ wouldn't be the mean component.

For why the term $\int_t \nabla \cdot (\rho^{E1}\mathbf{u}^E)$... Perform the same decomposition on $\mathbf{u}^E$, $\mathbf{u}^E = \mathbf{u}^0 + \mathbf{u}^{E1}$. When you substitute that in, you get:

$$\int_t \nabla \cdot (\rho^0 \mathbf{u}^0 + \rho^0 \mathbf{u}^{E1} + \rho^{E1} \mathbf{u}^0 + \rho^{E1} \mathbf{u}^{E1})$$

So now, we transpose the operators: $\int_t \nabla \cdot () = \nabla \cdot \int_t ()$ because we consider them to be linear operators. If we consider those terms and group the first two back together into $\rho^0 \mathbf{u}^E$, then we see $\rho^0$ is independent of time and moves outside the integral while $\int_t \mathbf{u}^E = \mathbf{s}$ and you get that term.

Now you're left with two terms: $\nabla \cdot \int_t \rho^{E1} \mathbf{u}^0$ and $\nabla \cdot \int_t \rho^{E1} \mathbf{u}^{E1})$. Since we are interested in linear effects only, that second term is considered a higher order term and is neglected as insignificant. If it is significant, than our linear assumption breaks down but we're assuming things are linear.

So then the term $\nabla \cdot \int_t \rho^{E1} \mathbf{u}^0$ is easy to see that it's zero. The $\mathbf{u}^0$ moves outside the integral, and the time integral of $\rho^{E1}$ is zero based on the decomposition given above.

All of that said, this is very standard procedure for linearizing equations. Any textbook in engineering that talks about linear assumptions or stability of solutions will cover this procedure in detail, whether it's structural mechanics, turbulence, aerodynamics... Learn the procedure, practice it, because it's an essential part of the analysis of differential equations.

  • $\begingroup$ The integration of $\int_t \nabla \cdot (\rho^{E1} \mathbf{u}^E) = 0$ is less hand-wavy if you also decompose $\mathbf{u}^E = \mathbf{u}^0 + \mathbf{u}^{E1}$ and neglect the product of fluctuating quantities. It should give you the same answer, but it's a more explicit way to show why the integral is 0. $\endgroup$
    – tpg2114
    Feb 25 '14 at 17:23
  • $\begingroup$ Thanks a lot man, but I still don't get why first, $\overline{\rho}^{E1} = 0$ and also $\int_t \nabla.(\rho^{E1}\mathbf{u}^E) = 0$. How can we state these equalities ? $\endgroup$
    – Amzocks
    Feb 26 '14 at 7:50
  • 1
    $\begingroup$ @Amzocks I've updated it, it's more detailed than I wanted it to be because I was hoping to leave the details to you to prove. The procedures here are really important to understand because they are everywhere in physics and engineering, so please work through the math on your own using what's here to verify. And practice, practice, practice! This won't be the only time you need to linearize equations! $\endgroup$
    – tpg2114
    Feb 26 '14 at 15:13

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