Fluid Mechanics question,velocity potential

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SUMMARY

The discussion focuses on the theoretical aspects of velocity and force potential in fluid mechanics, specifically addressing the relationship between acceleration and the velocity potential function, denoted as Ω. The user expresses confusion regarding the derivation of the acceleration expression in terms of partial derivatives. The response clarifies that the acceleration in the x-direction can be expressed as u' = u(∂u/∂x) + v(∂u/∂y) + w(∂u/∂z) + (∂u/∂t), where u, v, and w are the velocity components derived from the potential function Ω. This clarification emphasizes the importance of correctly applying partial derivatives in fluid dynamics.

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I am new in this place, am studying civil engineering in Spain,Madrid, and is something I do not understand in a theoretical exposition of the velocity/force potential.

They suppose that the external force that it acts in each point of the fluid and the speed, derive from scalar , so they admit a potential :

\frac{\vec{F}}{m}= - \vec{\nabla} U

\vec{V}= \vec{\nabla} \Omega


If the acceleration depends on the coordinates of the point and the time \vec{a} ( u',v',w') = f(x,y,z,t) :

u'= \frac{du}{dt}= \frac{\partial u}{\partial x} \frac{\partial x}{\partial t} +\frac{\partial u}{\partial y } \frac{\partial y}{\partial t} + \frac{\partial u}{\partial z}\frac{\partial z}{\partial t} + \frac{\partial u}{\partial t} and thus with the other coordinates of the acceleration


And here my doubt comes, I do not understand as they obtain to this expression:

u'= \frac{\partial^2 \Omega}{\partial x^2} \frac{\partial \Omega}{\partial x} +\frac{\partial^2 \Omega}{\partial x \partial y} \frac{\partial \Omega}{\partial y} + \frac{\partial^2 \Omega}{\partial x \partial z}\frac{\partial \Omega}{\partial z} + \frac{\partial^2 \Omega}{\partial x \partial t}




if somebody can help to understand it me, would be thanked for. Thank you very much
 
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First of all, don't use partials where they don't belong!

Now, we have
\frac{dx}{dt}=u,\frac{dy}{dt}=v,\frac{dz}{dt}=w
Thus, we may write the expression for the acceleration in the x-direction, i.e, u' as:
u'=u\frac{\partial{u}}{\partial{x}}+v\frac{\partial{u}}{\partial{y}}+w\frac{\partial{u}}{\partial{z}}+\frac{\partial{u}}{\partial{t}}

Now, insert:
u=\frac{\partial\Omega}{\partial{x}},v=\frac{\partial\Omega}{\partial{y}},w=\frac{\partial\Omega}{\partial{z}}

See if you get it right now!
 
Compound functions always was my nightmare\frac{\partial}{\partial t} \left(\frac{\partial \Omega}{\partial x}\right)=\frac{\partial}{\partial x} \left(\frac{\partial \Omega}{\partial x}\right) \underbrace{\frac{dx}{dt}}_{u}+\frac{\partial}{\partial y} \left(\frac{\partial \Omega}{\partial x}\right) \underbrace{\frac{dy}{dt}}_{v} + \frac{\partial}{\partial z} \left(\frac{\partial \Omega}{\partial x}\right) \underbrace{ \frac{dz}{dt}}_{w}+ \frac{\partial}{\partial t} \left(\frac{\partial \Omega}{\partial x}\right) \frac{dt}{dt}


that is what it did not see, thank you very much to solve the doubt to me
 

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