# The energy in the linear stability analysis

1. Nov 12, 2014

### jollage

Hi,

I'm considering the energy evolution of variables of different orders in a partial differential equation.

The PDE is nonlinear, which can be written as

$\frac{\partial u}{\partial t} = \mathcal{N}u$

where $\mathcal{N}$ is a nonlinear operator in space and time. Now I want to check the linear stability analysis of this PDE and I assume

$u=u_0(x) + \epsilon u_1(x,t) + \epsilon^2 u_2(x,t) + H.O.T$

where $u_0(x)$ is only a function of space, not of time (so, it's a time mean of the variable) and the other orders are modulated by a small quantity $\epsilon$. Substituting this decomposition into the nonlinear PDE, neglecting the high order terms and collecting the terms of same order of $\epsilon$, we will get the evolution equation for $u_1$ and $u_2$, which are

$\frac{\partial u_1}{\partial t} = \mathcal{L} u_1$

$\frac{\partial u_2}{\partial t} = \mathcal{L} u_2 + \mathcal{N}_1$

where $\mathcal{L}=\mathcal{L}(x)$ is the linear operator and $\mathcal{N}_1=\mathcal{N}_1(x,t)$ is the nonlinear terms of the first order variables $u_1$ acting on the second order variable $u_2$.

This derivation is trivial. Now I check the energy evolution of the variables $\frac{1}{2}u_1^*u_1$ and $\frac{1}{2}u_2^*u_2$, and the two equations will become

$\frac{\partial}{\partial t} (\frac{1}{2}u_1^*u_1) = u_1^* \frac{1}{2}(\mathcal{L} + \mathcal{L}^*) u_1$

$\frac{\partial}{\partial t} (\frac{1}{2}u_2^*u_2) = u_2^* \frac{1}{2}(\mathcal{L} + \mathcal{L}^*) u_2 + \frac{1}{2}(u_2^*\mathcal{N}_1 + \mathcal{N}_1^* u_2)$

My question is regarding the above equations.

(1) Since $\epsilon$ is small, we expect that $\epsilon u_1(x,t)$ is much smaller than $\epsilon^2 u_2(x,t)$ provided that $u_1(x,t)$ is of the same order of $u_2(x,t)$. In this case, is it meaningful to discuss the energy $\frac{1}{2}u_1^*u_1$ and $\frac{1}{2}u_2^*u_2$ in the same context? Though $\frac{1}{2}u_1^*u_1$ and $\frac{1}{2}u_2^*u_2$ is of the same order, but $\epsilon^2\frac{1}{2}u_1^*u_1$ and $\epsilon^4\frac{1}{2}u_2^*u_2$ is two orders of $\epsilon$ apart.

To be more specific, let me consider the following example. If the eigenvalue of $\mathcal{L}$ is negative, it means that the energy $u_1$ decays monotonically. If the nonlinear part $\frac{1}{2}(u_2^*\mathcal{N}_1 + \mathcal{N}_1^* u_2)$ causes a large energy growth of $u_2$, will this lead to energy growth of $u$?

(2) The two equations are only one-way coupled, so we can solve the first equation independently. I'm thinking a dynamical system. With different initial conditions for $u_1$, the energy evolution of $\frac{1}{2}u_1^*u_1$ could be different. Assume that we manage to pick up the optimal initial condition of $u_1$(to avoid ambiguity, its norm is 1) which gives rise to the optimal energy growth over all the initial conditions, then we use this path of energy evolution for $\frac{1}{2}u_1^*u_1$ in the second equation, my question is that is this the optimal energy growth path for $\frac{1}{2}u_2^*u_2$? Or there exists some interaction of $u_2$ and $\mathcal{N}_1$ such that it can bypass this optimal path?

Thanks.

2. Nov 12, 2014

### the_wolfman

1) By considering terms of order $\epsilon^2$ you've move beyond a "linear" analysis and are doing a full perturbative analysis.

2)Perturbation theory is very powerful, but its also involves a lot of algebra. It is very easy to miss terms, or include terms at the wrong order. In the above analysis you do both. Before trying to generalize to generic "operators" I recommend expanding a few simple nonlinear equations. For instance you could try $\frac{d^2 u}{d^2 x}= u^2$

3) In general the zero order term is not the "time mean.

4) You are missing the first order and third order energy equations (see comment 2). The first order equation will be of the form $u_1 u_0$ and the third order will be of the form $u_3 u_0 + u_1 u_2$

3. Nov 12, 2014

### jollage

Hi the_wolfman,

(1) Yes, you are right, in the title I'd better remove "linear"

(2) & (4) It's my fault not being clear. But the operator $\mathcal{L}$ actually includes $u_0 u_1$ you've pointed out (because I assume all the linear part of $u_1$ is absorbed in $\mathcal{L}$ in the first equation). By the way, the operator $\mathcal{N}_1$ includes $u_1 u_1$ in the second equation. When you do perturbation analysis, the process depends on which part you throw away. The term $u_3 u_0$ and $u_1 u_2$ should appear in the dynamic equation for $u_3$ (if I carry on my decomposition to $\epsilon^3$). But I don't want to do that, the second order $u_2$ is what I consider now. In the perturbation analysis, you have to truncate the series at some point.

(3) You are right. But as in my thread, I force it to be time-independent.

(4) In the energy equation, again, it depends on which part you throw away. In $\mathcal{N}_1 u_2$ for example, I have the terms like $u_1 u_1 u_2$. In my derivation, I got the dynamic equation for $u_2$, I then multiply this equation by $u_2^*$ to get the energy equation. By doing so, I didn't consider the energy from $u_3 u_0$ (otherwise I should have a governing equation for $u_3$).

(*) The original governing equation in the first thread should be $\frac{\partial u}{\partial t} = \mathcal{N}(u)$

4. Nov 14, 2014

### the_wolfman

Lets think carefully about the energy of you system.

The energy evolution equation for the original system is
$u^* \frac{\partial u}{\partial t}= u^* N\left(u\right)$

Use this to calculate the energy of your model where you've truncated at $u_3$:
$u^* \frac{\partial u}{\partial t}=\left(u_0^* + \epsilon u_1^* + \epsilon^2 u_2^*\right)\frac{\partial \left(u_0 + \epsilon u_1 + \epsilon^2 u_2\right) }{\partial t}.$

You've correctly identified the terms $u_1^* \frac{\partial u_1}{\partial t}$ and $u_2^* \frac{\partial u_2}{\partial t}$ but your missing cross terms like $u_1^* \frac{\partial u_2}{\partial t}$.

It dose not make since to talk about the energy of your system if you only keep $u_1^* u_1$ and $u_2^* u_2$ but neglect the cross terms!

If you keep higher order terms and repeat the calculations of the total energy you will see that there are cross terms like $u_1^* \frac{\partial u_3}{\partial t}$ and $u_0^* \frac{\partial u_4}{\partial t}$ that contribute to the 4-th order energy equation.

Now I agree that you have to truncate your expansion. But you have to be careful when you start considering high-order energy terms. For instance the energy term $u_2^2$ is order $\epsilon^4$ small. If you want to argue that this term is significant to a particular problem, then be consistent and include all energy terms that are accurate to $\epsilon^4$. You can't ignore the evolution $u_3$ and $u_4$!

Maybe you don't care about $\epsilon^4$ accuracy, but instead you might want to know if your truncated model conserves energy. Its a desirable property for models to have. In this case the neglect of $u_3$ and $u_4$ in your energy analysis is justified. These terms don't exist in the model. However be careful not to assign physical significance to high-order energy terms. They may be necessary to conserve energy, but the representation of the dynamics at these high orders is incomplete.

5. Dec 1, 2014

### jollage

What you wrote there brings me to think whether the energy analyses we are talking about are the same thing. I am studying turbulence and the transition to turbulence. There, the hot topic is the flow stability or instability. Usually the stability analysis starts with the Reynolds' decomposition $u(x,t)=U(x)+u'(x,t)$ and then the linearization is followed. The energy of fluctuation, for me, is actually meaning the variance of the fluctuation, i.e., $u'(x,t)^*u'(x,t)$. So I am more familiar to just considering the energy of the fluctuation at that order.

The problem of energy at different orders as you pointed out is interesting to me. Could you please tell me any field or subject employing such energy analysis (i.e., the energy energy in the flavor of considering different orders)? Thank you very much.

6. Dec 3, 2014

### the_wolfman

While similar, the turbulent decomposition and linearization/perturbation theory are two different techniques. In turbulance, we are often interested in average over bulk quantities. When you consider the bulk energy, the average of the first order energy $\left<U\left(x\right)\tilde u\left(x,t\right)\right>$ is zero exactly. (We assume that bulk flow changes so slowly, that we can treat it as a constant when averaging. The perturbed flow averages to zero by definition). The bulk energy comes from the bulk flow and the quadratic term $\left<\tilde u\left(x,t\right)\tilde u\left(x,t\right)\right>$.

There are cases in turbulance where you want to know what the instantaneous energy is doing. In this case, you don't perform the averaging and the first order terms are important.

I study plasma physics. Both turbulence and stability are hot issues, and we employ both techniques. These techniques are commonly used in many other fields that study nonlinear dynamics.

7. Dec 3, 2014

### jollage

Thanks. I see. Could tell me any references on the energy evolution of the lower orders? I want to see how people deal with the evolution of the $\epsilon^3$ or $\epsilon^4$ order in the following

$u^2 = U^2 + 2\epsilon U u_1 + \epsilon^2(u_1^2+2U u_2) + 2\epsilon^3 u_1 u_2 + \epsilon^4 u_2^2$ under the decomposition $u(x,t)=U(x) + \epsilon u_1(x,t)+\epsilon^2 u_2(x,t)$.

If not, the reference on the general energy analysis would also help. Thanks.