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#21 |
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Lucky
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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. |
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#22 |
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Filippo Maria Denaro
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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. |
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#23 | |
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But for the case of vorticity that you need only one initial vvalue I agree |
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#24 |
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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? |
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#25 | |
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![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() The point is that now I understand what you say, about knowing the initial value ![]() ![]() ![]() I just think, the way I would explain it now is this: we have this pde ![]() ![]() ![]() THOUGH, It doesn't satisfies EVERY initial condition. In case that the initial condition is ![]() So we can say ![]() |
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#26 |
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Lucky
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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: ![]() ![]() What you want to write is: ![]() 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. |
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#27 |
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Filippo Maria Denaro
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Why? The derivative is still a function of time, you cannot assess in general this conclusion. |
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#29 |
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Filippo Maria Denaro
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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. |
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#30 | |
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In the red highlighted part l didn't mean tha l came to this conclusion without the part bellow. |
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#31 |
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Filippo Maria Denaro
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You wrote:
Quote:
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. |
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#32 |
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Filippo Maria Denaro
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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.
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#33 |
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I know saying ![]() |
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#34 | |
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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 |
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#35 | |
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Filippo Maria Denaro
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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. |
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#36 | |
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#37 |
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Of course, if you set a zero initial condition and zero BC on the frontier.
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#38 | |
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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. |
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#39 | |
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Filippo Maria Denaro
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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. |
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#40 | |
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Lucky
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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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