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I want to discuss about the assumptions in the linearization. By linearization, I mean the following classic procedure.

(1) Original nonlinear governing equation is [itex]\frac{\partial u}{\partial t}+u\frac{\partial u}{\partial x}=\mathcal{L}(u)[/itex], RHS is a linear operator

(2) introduce the decomposition u=U+u'

(3) expand the nonlinear solution to [itex]\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')[/itex]

(4) ASSUME THE BASE FLOW U SATISFYING THE NONLINEAR EQUATION, by subtraction, we have [itex]\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')[/itex]

(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 [itex]\frac{\partial u'}{\partial t}+U\frac{\partial u'}{\partial x}+u'\frac{\partial u'}{\partial x}=\mathcal{L}(u')[/itex], this is also [itex]\frac{du'}{dt}+u'\frac{\partial u'}{\partial x}=\mathcal{L}(u')[/itex], 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