Deriving the 1-D Linear Convection Equation

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In the discussion on deriving the 1-D linear convection equation, participants clarify the transition from the 1-D nonlinear convection equation under the assumptions of inviscid flow, no pressure gradient, and no body forces. The key question raised is why only one instance of the velocity term 'u' is replaced with a constant 'c' during the derivation, while others remain unchanged. The correct approach involves linearizing the equation by expressing 'u' as a sum of a constant velocity 'u0' and a small perturbation 'u''. This leads to the linearized equation that describes wave propagation. The conversation concludes with an acknowledgment of the need for further exploration into linearization techniques.
Mr_Acceleration
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With the assumptions of Inviscid flow, no pressure gradient and no body force terms in 1-D Navier Stokes becomes 1-D nonlinear convection equation;
sadasd.png

And if we assume velocity of wave propagation is constant value c, equation becomes 1-D linear convection equation;
sadasd.png

This is online derivation and my question is why only the u value outside of partial derivatives is replaced with c? we have 3 u term there. I was thinking they were all same term. So why do we only replace one of them?
 
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I don't think this is correct. Can you please provide the full derivation ?
 
dRic2 said:
I don't think this is correct. Can you please provide the full derivation ?
123.png

Here is the derivation.
 
This is correct so far. The problem is this:

Mr_Acceleration said:
And if we assume velocity of wave propagation is constant value c, equation becomes 1-D linear convection equation;

This is not. That's why I'd like a full derivation.

Anyway the reasoning is the following:
$$ \frac {\partial u} {\partial t} + u \frac {\partial u} {\partial x} = 0$$
is NON-linear. What you usually do is linearize it, by writing ##u = u_0 + u'## where ##u_0## is an unperturbed constant velocity and ##u'## is a small perturbation. If you substitute into the diff equation ##\frac {\partial u_0} {\partial t} = \frac {\partial u_0} {\partial x} = 0## because ##u_0## is a constant. You finally end up with
$$ \frac {\partial u'} {\partial t} + u_0 \frac {\partial u'} {\partial x} = 0$$
 
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Thank you for your answer. I didn't know about this. Now i will search linearization with pertubation.
 
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