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BluzMan
Russian Federation4235 Posts
This time I need to find the general solution for this pair of equations: ![[image loading]](http://upload.wikimedia.org/math/e/9/1/e9155dcf8f9e3e3a58d34f757b121766.png) ![[image loading]](http://upload.wikimedia.org/math/2/3/3/23322e14d5f47462fd56523d8a5824b0.png) It seems quite easy at first, but I'm having some trouble trying to find the exact form of the solution. Help please, I need to get it done tomorrow or I will have troubles graduating.  | ||
Chill
Calgary25990 Posts
Edit: Please start the problem, or indicate the areas you are having trouble with. "SOLVE THIS OMG OLOLOL" is not an acceptable OP. | ||
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MasterReY
Germany2708 Posts
EPIC SMILEY ! to topic: a agree with chill  | ||
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jgad
Canada899 Posts
∂V/∂t = v∇V - 1/ρ∇P + ƒ The second term is independent of V (so it seems), so you just have to solve the basic form ∂V/∂t = v∇V - C Which should be pretty straightforward. | ||
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Cambium
United States16368 Posts
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Maenander
Germany4926 Posts
On October 28 2008 22:25 jgad wrote: Strange to write the divergence as V•∇ - it's usually written with the gradient first, but dot product is commutative so it should be zero in any case. That makes the whole first term zero, so you essentially have to solve ∂V/∂t = v∇V - 1/ρ∇P + ƒ The second term is independent of V (so it seems), so you just have to solve the basic form ∂V/∂t = v∇V - C Which should be pretty straightforward. dot product is commutative for normal vectors, which contain (commutative) scalars and not derivatives, which are of course not commutative. I wish I could help you Bluzman, but I did this stuff long time ago and I am not going to delve into it again for this one problem, it takes too much time. Maybe you can simplify it by using the epsilon tensor (Levi-Civita symbol) and writing it out. | ||
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jgad
Canada899 Posts
dot product is commutative for normal vectors, which contain (commutative) scalars and not derivatives, which are of course not commutative. Yeah, I thought of this too, but it would seem that the first term would end up being -(Vx∂/∂x + Vy∂/∂x + Vz∂/∂z)V anyway, which would come out as : -V(∇•V) so I think -(V•∇)V = -V(∇•V) = 0 (given the second equation) I could be wrong. | ||
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Maenander
Germany4926 Posts
On October 28 2008 23:12 jgad wrote: Yeah, I thought of this too, but it would seem that the first term would end up being -(Vx∂/∂x + Vy∂/∂x + Vz∂/∂z)V anyway, which would come out as : -V(∇•V) so I think -(V•∇)V = -V(∇•V) = 0 (given the second equation) I could be wrong. You are wrong, it is -(Vx∂/∂x + Vy∂/∂x + Vz∂/∂z)Vx, -(Vx∂/∂x + Vy∂/∂x + Vz∂/∂z)Vy, -(Vx∂/∂x + Vy∂/∂x + Vz∂/∂z)Vz, which is a vector in itself and that vector does not equal (0,0,0), atleast not because ∇•V=0 (scalar!) | ||
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Chill
Calgary25990 Posts
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jgad
Canada899 Posts
On October 28 2008 23:18 Maenander wrote: You are wrong, it is -(Vx∂/∂x + Vy∂/∂x + Vz∂/∂z)Vx, -(Vx∂/∂x + Vy∂/∂x + Vz∂/∂z)Vy, -(Vx∂/∂x + Vy∂/∂x + Vz∂/∂z)Vz, which is a vector in itself and that vector does not equal (0,0,0), atleast not because ∇•V=0 (scalar!) Yeah, you're right. So the first coordinate in the first term reduces to -(Vx∂/∂x + Vy∂/∂y + Vz∂/∂z)Vx = -Vx∂/∂x(Vx) - Vy∂/∂y(Vx) - Vz∂/∂z(Vx) = -Vx(1) - Vy(0)- Vz(0) = -Vx and the same for the rest (y and z), so -(V•∇)V = -V, it would seem? Anyway, I just noticed that the second term in the OP is ∆V and not ∇V - this seems confusing as it's either a typo or we need to know what ∆V is a delta of (V(t2) - V(t1)) or (V(x2) - V(x1)) - it could be anything. | ||
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Maenander
Germany4926 Posts
And I don´t see why -Vx∂/∂x(Vx) - Vy∂/∂y(Vx) - Vz∂/∂z(Vx) = -Vx(1) - Vy(0)- Vz(0) should be correct. edit: And I agree with chill here, Bluzman should show us his work, so we can point out errors. Again I am too rusty and too lazy to solve this (I should do other things anyway ^^). | ||
micronesia
United States24747 Posts
On October 28 2008 23:24 Chill wrote: lol i got owned I don't see you getting pwned. How did you get pwned? | ||
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BuGzlToOnl
United States5918 Posts
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jgad
Canada899 Posts
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jgad
Canada899 Posts
On October 28 2008 23:50 Maenander wrote: ∆ is the Laplace-Operator, I guess in the US you would write ∇ ^2. Weird, and yeah, I've only ever seen it as ∇^2. Anyway, I agree that more information is needed about what these fields are supposed to be - if these are all general quantities which could be anything then I think the solution is going to be really messy. And I don´t see why -Vx∂/∂x(Vx) - Vy∂/∂y(Vx) - Vz∂/∂z(Vx) = -Vx(1) - Vy(0)- Vz(0) should be correct. Yeah, I'm trying to all of this too quickly. It occured to me just after I posted it, but yeah, there's no reason why V can't be Vx(x,y,z), Vy(x,y,z), Vz(x,y,z), in which case this all gets really horrible. | ||
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BluzMan
Russian Federation4235 Posts
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BuGzlToOnl
United States5918 Posts
On October 29 2008 00:10 jgad wrote: looks like ~third year electrodynamics. Incompressible vector fields (∇•V=0) define the magnetic field or the electric field in regions of no space charge, for example. Fun, fun. | ||
cgrinker
United States3824 Posts
On October 28 2008 23:24 Chill wrote: lol i got owned I thought the smiley was quite nice actually. Aren't you only supposed to police the strat forums though? | ||
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cgrinker
United States3824 Posts
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jgad
Canada899 Posts
Got $1m? | ||
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