# What's exactly the meaning of Linearization?

1. Sep 6, 2013

### jollage

Hi all,

I want to discuss about the assumptions in the linearization. By linearization, I mean the following classic procedure.
(1) Original nonlinear governing equation is $\frac{\partial u}{\partial t}+u\frac{\partial u}{\partial x}=\mathcal{L}(u)$, RHS is a linear operator
(2) introduce the decomposition u=U+u'
(3) expand the nonlinear solution to $\frac{\partial (U+u')}{\partial t}+U\frac{\partial U}{\partial x}+U\frac{\partial u'}{\partial x}+u'\frac{\partial U}{\partial x}+u'\frac{\partial u'}{\partial x}=\mathcal{L}(u')$
(4) ASSUME THE BASE FLOW U SATISFYING THE NONLINEAR EQUATION, by subtraction, we have $\frac{\partial u'}{\partial t}+U\frac{\partial u'}{\partial x}+u'\frac{\partial U}{\partial x}+u'\frac{\partial u'}{\partial x}=\mathcal{L}(u')$
(5) Neglect the quadratic term, we can have the linear governing equation.

My question is about the assumption in step 4. How to understand the assumption that U satisfy the nonlinear equation while at the same time U+u' is the solution to the nonlinear equation?

What I can think of is (1) Let's say the base flow U is independent on x, equation in (4) becomes $\frac{\partial u'}{\partial t}+U\frac{\partial u'}{\partial x}+u'\frac{\partial u'}{\partial x}=\mathcal{L}(u')$, this is also $\frac{du'}{dt}+u'\frac{\partial u'}{\partial x}=\mathcal{L}(u')$, so in this case, it's just a problem of changing the reference coordinate, i.e., if we are moving along with the base flow U, we see the same solution as the original nonlinear one in its reference coordinate (let's say the one sticking to the ground.)

(2) If the base flow U is not independent on x, then the third term in the equation of (4) will modulate the amplitude of the solution u' according to the x-dependence of U. This is the influence of the base flow U exerting on u', a price u' has to pay living on U.

Could anyone say something about the change of phase between the nonlinear equations in (1) and (4)?

Thanks.

Jo

2. Sep 6, 2013

$$\dfrac{\partial u}{\partial t} + u\dfrac{\partial u}{\partial x} = \mathcal{L}(u),$$

where $\mathcal{L}$ is a linear operator. Then when you introduce disturbance quantities $u = U + u^{\prime}$, you get the following:
$$\dfrac{\partial (U + u^{\prime})}{\partial t} + (U + u^{\prime})\dfrac{\partial (U + u^{\prime})}{\partial x} = \mathcal{L}(U + u^{\prime}).$$

This expands to
$$\dfrac{\partial U}{\partial t} + \dfrac{\partial u^{\prime}}{\partial t} + U \dfrac{\partial U}{\partial x} + U \dfrac{\partial u^{\prime}}{\partial x} + u^{\prime} \dfrac{\partial U}{\partial x} + u^{\prime} \dfrac{\partial u^{\prime}}{\partial x}= \mathcal{L}(U + u^{\prime}).$$

Now here is the error I see in your original post. Because $\mathcal{L}$ is linear, then by definition, $\mathcal{L}(U + u^{\prime}) = \mathcal{L}(U) + \mathcal{L}(u^{\prime})$. Then, that leads us to
$$\dfrac{\partial U}{\partial t} + \dfrac{\partial u^{\prime}}{\partial t} + U \dfrac{\partial U}{\partial x} + U \dfrac{\partial u^{\prime}}{\partial x} + u^{\prime} \dfrac{\partial U}{\partial x} + u^{\prime} \dfrac{\partial u^{\prime}}{\partial x}= \mathcal{L}(U) + \mathcal{L}(u^{\prime}).$$

To illustrate your step 4, it helpful to rearrange some terms. It should be more clear if we rewrite the previous equation as follows:

$$\left\{ \dfrac{\partial U}{\partial t} + U \dfrac{\partial U}{\partial x} \right\} + \dfrac{\partial u^{\prime}}{\partial t} + U \dfrac{\partial u^{\prime}}{\partial x} + u^{\prime} \dfrac{\partial U}{\partial x} + u^{\prime} \dfrac{\partial u^{\prime}}{\partial x}= \left\{\mathcal{L}(U)\right\} + \mathcal{L}(u^{\prime}).$$
Take a look at the sections I highlighted in curly braces. These are the exact terms from the original equation solved for the base flow. In other words, if you substituted just the undisturbed flow into the original equation, you would get
$$\dfrac{\partial U}{\partial t} + U\dfrac{\partial U}{\partial x} = \mathcal{L}(U),$$
and if you subtract that equation from the expanded equation above it, the terms in the curly braces drop out, leaving
$$\dfrac{\partial u^{\prime}}{\partial t} + U \dfrac{\partial u^{\prime}}{\partial x} + u^{\prime} \dfrac{\partial U}{\partial x} + u^{\prime} \dfrac{\partial u^{\prime}}{\partial x} = \mathcal{L}(u^{\prime}).$$

So far, this isn't the meaning of linearization. The equation is still nonlinear. You linearize it by making the assumption that $u^{\prime}$ is a small quantity, so any term of $\mathcal{O}(u^{\prime 2})$ is a very small and can be neglected in relation to the other terms. That is why you then drop the last term on the LHS to get the linearized equation,
$$\dfrac{\partial u^{\prime}}{\partial t} + U \dfrac{\partial u^{\prime}}{\partial x} + u^{\prime} \dfrac{\partial U}{\partial x} = \mathcal{L}(u^{\prime}).$$

There shouldn't be much if any phase shift between the nonlinear equation you started with and the linearized equation so long as $u^{\prime}$ remains small. As that parameter becomes increasingly large, the phase and amplitude will (likely) start to stray from the true, nonlinear solution.