rotational and inviscid

LuckyTran · November 3, 2023, 16:33

So now you have the initial value problem.

If the initial value is 0. The change when the initial value is 0 is 0. Hence, when it is 0, it stays 0.

There are numerous analogues. The simple harmonic oscillator for example is y''=y. If you put y(t=0)=0, then y''=y'=y=0. A simple harmonic oscillator doesn't oscillate if there is no initial oscillation even though the system is capable of oscillating.

FMDenaro · November 3, 2023, 17:15

Quote:

Originally Posted by ichamail

Let me use some different notation just in case we implie the same thing but we "lost in translation"

$\frac{d \vec{\zeta} }{dt}(t) = \left( \vec{\zeta}(t) \cdot \nabla \right) \vec{V}(t)$

If we know that $\vec{\zeta}(t_0) = 0$ then, all I can say is

$\frac{d \vec{\zeta} }{dt}(t_0) = \left( \vec{\zeta}(t_0) \cdot \nabla \right) \vec{V}(t_0) = 0$

This is all I can understand. If you are telling me that there is something more than that, I'm sorry but maybe I need to go back to the basics, and the come back again to ask for help

Ok, let use the equation and integrate exactly in time

z(t)=z(t0) + Int[t0,t] z(tau).Grad v(tau) dtau

note that we integrate along the path-line dx/dt=v[x(t),t].

Thus, z(t) is a non-vanishing function if along the path-line there is a non-zero contribution of the integral.

ichamail · November 3, 2023, 17:15

Quote:

Originally Posted by LuckyTran

So now you have the initial value problem.

If the initial value is 0. The change when the initial value is 0 is 0. Hence, when it is 0, it stays 0.

There are numerous analogues. The simple harmonic oscillator for example is y''=y. If you put y(t=0)=0, then y''=y'=y=0. A simple harmonic oscillator doesn't oscillate if there is no initial oscillation even though the system is capable of oscillating.

I have to disagree on that. For the case of simple harmonic oscillator you need two initial conditions. You can have y(0)=0 and y'(0) = 1, where 1 would be the maximum speed. On the other hand y=0 is definately a solution of this ode and it means that when you are at rest you stay at rest. But a solution must satisfy both the equation and and the initial values.

But for the case of vorticity that you need only one initial vvalue I agree

LuckyTran · November 3, 2023, 18:36

So now it is suddenly obvious how y(t=0)=0 does not necessarily imply y'(t=0)=0?

Interesting.....

So do you get it now?

ichamail · November 4, 2023, 13:53

Quote:

Originally Posted by LuckyTran

So now it is suddenly obvious how y(t=0)=0 does not necessarily imply y'(t=0)=0?

Interesting.....

So do you get it now?

I never said $y(t=0)= 0 \implies y'(t=0) = 0$ . I said that $y = 0 \implies y' = 0$ . These things are different.

is the same thing as

that implies

. Just like

, which is not an independent variable here but a function of time (

), vorticity $\vec{\zeta}$ is a function of time and space ( $\vec{\zeta} =\vec{\zeta}(x,y,z,t)$ ), and not an independent variable. So setting $\vec{\zeta} = 0$ , is like saying that $\vec{\zeta}(x,y,z,t) = 0$ which definately implies that $\frac{d \vec{\zeta}(t)}{dt} = 0$ . One the other hand setting $\vec{\zeta}(t=0,x,y,z) = 0$ does not implies that $\frac{d\zeta(t)}{dt} = 0$ , or $\frac{d\zeta}{dt}(t=0) = 0$ (assuming of course that you don't know the pde)

The point is that now I understand what you say, about knowing the initial value $\vec{\zeta}(0, x,y,z) = 0$ , and then calculating from the equation of vorticity the time derivative at t=0 $\frac{d \vec{\zeta} }{dt}(t_0) = \left( \vec{\zeta}(t_0) \cdot \nabla \right) \vec{V}(t_0) = 0$ can lead as to the colclusion that $\frac{d \vec{\zeta} }{dt}(t) = 0$

I just think, the way I would explain it now is this:

we have this pde
$\frac{d \vec{\zeta} }{dt}(t) = \left( \vec{\zeta}(t) \cdot \nabla \right) \vec{V}(t)$

$\vec{\zeta}(t) = \vec{0}$ is a solution to this pde, because it satisfies the equation. $\frac{d \vec{\zeta} }{dt}(t) - \left( \vec{\zeta}(t) \cdot \nabla \right) \vec{V}(t) = \frac{d \vec{0} }{dt} - \left( \vec{0} \cdot \nabla \right) \vec{V}(t) = \vec{0} - \vec{0} = \vec{0}$
THOUGH, It doesn't satisfies EVERY initial condition.

In case that the initial condition is $\vec{\zeta}(t_0) = 0$ , then this solution satisfies both the pde and the initial condition.
So we can say $\vec{\zeta}(t) = \vec{0} \implies \frac{d \vec{\zeta} }{dt}(t) = 0$

LuckyTran · November 4, 2023, 14:11

That point is that you don't know y'(t=0) is anything until you apply it to the PDE regardless of what y(t=0). You are saying it... I really don't get why you don't understand your own words and why you're confusing yourself.

Note that these two statements are not equivalent:
$\vec{\zeta}(t) = \vec{0} \implies \frac{d \vec{\zeta} }{dt}(t) = 0$ is trivial because as you write $\vec{\zeta}(t) = \vec{0} \implies \frac{d \vec{0} }{dt}(t) = 0$ . It's trivial because this is true , everywhere, all the time and has nothing to do with PDE's. The derivative of a constant function is 0.

What you want to write is:
$\vec{\zeta}(t=t_0) = \vec{0} \implies \frac{d \vec{\zeta} }{dt}(t) = 0$ for any/all real t. This statement is only true due to the vorticity transport equation for this particular case of inviscid irrotational flow and under the condition that the vorticity was zero at least once.

I know I didn't fully write out in detailed explicit notation what was meant by my words, but you cannot fancy it up and improperly insert notation because you can end up writing something inconsistent.

FMDenaro · November 4, 2023, 14:14

Quote:

Originally Posted by ichamail

I never said $y(t=0)= 0 \implies y'(t=0) = 0$ . I said that $y = 0 \implies y' = 0$ . These things are different.

is the same thing as

that implies

. Just like

, which is not an independent variable here but a function of time (

), vorticity $\vec{\zeta}$ is a function of time and space ( $\vec{\zeta} =\vec{\zeta}(x,y,z,t)$ ), and not an independent variable. So setting $\vec{\zeta} = 0$ , is like saying that $\vec{\zeta}(x,y,z,t) = 0$ which definately implies that $\frac{d \vec{\zeta}(t)}{dt} = 0$ . One the other hand setting $\vec{\zeta}(t=0,x,y,z) = 0$ does not implies that $\frac{d\zeta(t)}{dt} = 0$ , or $\frac{d\zeta}{dt}(t=0) = 0$ (assuming of course that you don't know the pde)

The point is that now I understand what you say, about knowing the initial value $\vec{\zeta}(0, x,y,z) = 0$ , and then calculating from the equation of vorticity the time derivative at t=0 $\frac{d \vec{\zeta} }{dt}(t_0) = \left( \vec{\zeta}(t_0) \cdot \nabla \right) \vec{V}(t_0) = 0$ can lead as to the colclusion that $\frac{d \vec{\zeta} }{dt}(t) = 0$

I just think, the way I would explain it now is this:

we have this pde
$\frac{d \vec{\zeta} }{dt}(t) = \left( \vec{\zeta}(t) \cdot \nabla \right) \vec{V}(t)$

$\vec{\zeta}(t) = \vec{0}$ is a solution to this pde, because it satisfies the equation. $\frac{d \vec{\zeta} }{dt}(t) - \left( \vec{\zeta}(t) \cdot \nabla \right) \vec{V}(t) = \frac{d \vec{0} }{dt} - \left( \vec{0} \cdot \nabla \right) \vec{V}(t) = \vec{0} - \vec{0} = \vec{0}$
THOUGH, It doesn't satisfies EVERY initial condition.

In case that the initial condition is $\vec{\zeta}(t_0) = 0$ , then this solution satisfies both the pde and the initial condition.
So we can say $\vec{\zeta}(t) = \vec{0} \implies \frac{d \vec{\zeta} }{dt}(t) = 0$

Why? The derivative is still a function of time, you cannot assess in general this conclusion.

ichamail · November 4, 2023, 14:53

Quote:

Originally Posted by FMDenaro

Why? The derivative is still a function of time, you cannot assess in general this conclusion.

It is not a general conclusion. I prove it down below

FMDenaro · November 4, 2023, 14:59

Quote:

Originally Posted by ichamail

It is not a general conclusion. I prove it down below

I don't agree with the part highlighted in red. You can easily see that

z(t)=z(t0)+(t-t0)*dz/dt|t0 + (t-t0)^2/2 *d^2z/dt^2|t0 + ...

Therefore you have to assess that all derivatives are zero, not only the first derivative.
In your case, dz/dt|t0=0 does not imply that dz/dt(t) is zero for all time.

ichamail · November 4, 2023, 15:10

Quote:

Originally Posted by FMDenaro

I don't agree with the part highlighted in red. You can easily see that

z(t)=z(t0)+(t-t0)*dz/dt|t0 + (t-t0)^2/2 *d^2z/dt^2|t0 + ...

Therefore you have to assess that all derivatives are zero, not only the first derivative.
In your case, dz/dt|t0=0 does not imply that dz/dt(t) is zero for all time.

Do you disagree with the proof I wrote under the highlighted part?

In the red highlighted part l didn't mean tha l came to this conclusion without the part bellow.

FMDenaro · November 4, 2023, 15:20

You wrote:

Quote:

Originally Posted by ichamail

The point is that now I understand what you say, about knowing the initial value $\vec{\zeta}(0, x,y,z) = 0$ , and then calculating from the equation of vorticity the time derivative at t=0 $\frac{d \vec{\zeta} }{dt}(t_0) = \left( \vec{\zeta}(t_0) \cdot \nabla \right) \vec{V}(t_0) = 0$ can lead as to the colclusion that $\frac{d \vec{\zeta} }{dt}(t) = 0$

I just think, the way I would explain it now is this:

we have this pde
$\frac{d \vec{\zeta} }{dt}(t) = \left( \vec{\zeta}(t) \cdot \nabla \right) \vec{V}(t)$

$\vec{\zeta}(t) = \vec{0}$ is a solution to this pde, because it satisfies the equation. $\frac{d \vec{\zeta} }{dt}(t) - \left( \vec{\zeta}(t) \cdot \nabla \right) \vec{V}(t) = \frac{d \vec{0} }{dt} - \left( \vec{0} \cdot \nabla \right) \vec{V}(t) = \vec{0} - \vec{0} = \vec{0}$
THOUGH, It doesn't satisfies EVERY initial condition.

In case that the initial condition is $\vec{\zeta}(t_0) = 0$ , then this solution satisfies both the pde and the initial condition.
So we can say $\vec{\zeta}(t) = \vec{0} \implies \frac{d \vec{\zeta} }{dt}(t) = 0$

I wrote the Taylor expansion in time that shows you can have a non-vanishing functions z(t) and z'(t) even if z(t0)=0 and z'(t0)=0.

FMDenaro · November 4, 2023, 15:25

The key is that you want to start assuming z(t)=0. Actually, we have only the equation and the initial condition at t0, no more conditions are assumed.

ichamail · November 4, 2023, 15:36

Quote:

Originally Posted by LuckyTran

That point is that you don't know y'(t=0) is anything until you apply it to the PDE regardless of what y(t=0). You are saying it... I really don't get why you don't understand your own words and why you're confusing yourself.

Note that these two statements are not equivalent:
$\vec{\zeta}(t) = \vec{0} \implies \frac{d \vec{\zeta} }{dt}(t) = 0$ is trivial because as you write $\vec{\zeta}(t) = \vec{0} \implies \frac{d \vec{0} }{dt}(t) = 0$ . It's trivial because this is true , everywhere, all the time and has nothing to do with PDE's. The derivative of a constant function is 0.

What you want to write is:
$\vec{\zeta}(t=t_0) = \vec{0} \implies \frac{d \vec{\zeta} }{dt}(t) = 0$ for any/all real t. This statement is only true due to the vorticity transport equation for this particular case of inviscid irrotational flow and under the condition that the vorticity was zero at least once.

I know I didn't fully write out in detailed explicit notation what was meant by my words, but you cannot fancy it up and improperly insert notation because you can end up writing something inconsistent.

I never said that.

I know saying $\vec{\zeta}(t) = \vec{0}$ is a trivial solution. But if it satisfy my initial condition, it is the solution

ichamail · November 4, 2023, 15:43

Quote:

Originally Posted by FMDenaro

The key is that you want to start assuming z(t)=0. Actually, we have only the equation and the initial condition at t0, no more conditions are assumed.

I never assumed tha z(t) = 0.
I said z(t)=0 is a solution, because it satisfies the equation.
If z(t)=0 satisfies the initial condition too, it is the solution

FMDenaro · November 4, 2023, 15:44

Quote:

Originally Posted by ichamail

I never said that.

I know saying $\vec{\zeta}(t) = \vec{0}$ is a trivial solution. But if it satisfy my initial condition, it is the solution

No, the mathematical problem is: PDE+intial conditions+boundary conditions. You cannot get a solution only if you satisfy the intial conditions.

The vorticity equation for ideal fluid is hyperbolic and requires BCs on an open space-time domain. Your trivial solution must satisfy also the BC on the frontiers.

ichamail · November 4, 2023, 15:48

Quote:

Originally Posted by FMDenaro

No, the mathematical problem is: PDE+intial conditions+boundary conditions. You cannot get a solution only if you satisfy the intial conditions.

The vorticity equation for ideal fluid is hyperbolic and requires BCs on an open space-time domain. Your trivial solution must satisfy also the BC on the frontiers.

Ok, I see...what type of bc do we need. We know that vorticity of undisturbed flow at boundaries is zero for all time. Can we use that as BC?

FMDenaro · November 4, 2023, 16:14

Of course, if you set a zero initial condition and zero BC on the frontier.

ichamail · November 4, 2023, 16:23

Quote:

Originally Posted by FMDenaro

Of course, if you set a zero initial condition and zero BC on the frontier.

I had in my mind flows around bodies, so such a BC would always be true I guess.

Thinking of a body starting to move at t=0, then z is everywhere zero at t=0 and after that is zero far away from the body at all directions for t>0.

FMDenaro · November 4, 2023, 16:30

Quote:

Originally Posted by ichamail

I had in my mind flows around bodies, so such a BC would always be true I guess.

Thinking of a body starting to move at t=0, then z is everywhere zero at t=0 and after that is zero far away from the body at all directions for t>0.

You are in the field of the classic aerodynamics, now. Since you have slip condition on a body, the vorticity on a wall is not regular.

You have to think to inviscid flow problems that are solved also with vorticity distribution.

Lift is generated this way.

LuckyTran · November 4, 2023, 17:41

Quote:

Originally Posted by ichamail

I never assumed tha z(t) = 0.
I said z(t)=0 is a solution, because it satisfies the equation.
If z(t)=0 satisfies the initial condition too, it is the solution

The thing is...this is not a proof. You are assuming the conclusion as a premise which is circular reasoning. It is a common mistake in inductive arguments so don't feel too bad about it.

Btw the proof is not actually complicated nor profound. Filippo has basically demonstrated it for you, it falls right out after you apply the fundamental theorem of calculus. The crux is that you need to prove that the lagrangian derivative is zero for all time.

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November 3, 2023, 16:33		#21
LuckyTran Senior Member Lucky Join Date: Apr 2011 Location: Orlando, FL USA Posts: 5,761 Rep Power: 66	So now you have the initial value problem. If the initial value is 0. The change when the initial value is 0 is 0. Hence, when it is 0, it stays 0. There are numerous analogues. The simple harmonic oscillator for example is y''=y. If you put y(t=0)=0, then y''=y'=y=0. A simple harmonic oscillator doesn't oscillate if there is no initial oscillation even though the system is capable of oscillating.

November 3, 2023, 18:36		#24
LuckyTran Senior Member Lucky Join Date: Apr 2011 Location: Orlando, FL USA Posts: 5,761 Rep Power: 66	So now it is suddenly obvious how y(t=0)=0 does not necessarily imply y'(t=0)=0? Interesting..... So do you get it now?

November 4, 2023, 14:11		#26
LuckyTran Senior Member Lucky Join Date: Apr 2011 Location: Orlando, FL USA Posts: 5,761 Rep Power: 66	That point is that you don't know y'(t=0) is anything until you apply it to the PDE regardless of what y(t=0). You are saying it... I really don't get why you don't understand your own words and why you're confusing yourself. Note that these two statements are not equivalent: $\vec{\zeta}(t) = \vec{0} \implies \frac{d \vec{\zeta} }{dt}(t) = 0$ is trivial because as you write $\vec{\zeta}(t) = \vec{0} \implies \frac{d \vec{0} }{dt}(t) = 0$ . It's trivial because this is true , everywhere, all the time and has nothing to do with PDE's. The derivative of a constant function is 0. What you want to write is: $\vec{\zeta}(t=t_0) = \vec{0} \implies \frac{d \vec{\zeta} }{dt}(t) = 0$ for any/all real t. This statement is only true due to the vorticity transport equation for this particular case of inviscid irrotational flow and under the condition that the vorticity was zero at least once. I know I didn't fully write out in detailed explicit notation what was meant by my words, but you cannot fancy it up and improperly insert notation because you can end up writing something inconsistent.

November 4, 2023, 15:25		#32
FMDenaro Senior Member Filippo Maria Denaro Join Date: Jul 2010 Posts: 6,896 Rep Power: 73	The key is that you want to start assuming z(t)=0. Actually, we have only the equation and the initial condition at t0, no more conditions are assumed.

November 4, 2023, 16:14		#37
FMDenaro Senior Member Filippo Maria Denaro Join Date: Jul 2010 Posts: 6,896 Rep Power: 73	Of course, if you set a zero initial condition and zero BC on the frontier.