/sci/ - Science & Math

[Anonymous No.3988039]

Okay I give up

I'm trying to solve this problem. I'm given the field v(r) (picture top) and I'm supposed to find the stream lines (handily plotted in the pic too)

to do so i find the stream function (pic bottom) and set it to be constant

problem is, when I do this i find a terrible y(x). I repeated this a couple times and still get the same result so I don't know where could I be wrong.

Help?

[Anonymous No.3988048]

this is the analytic function i'm getting for the stream lines

to get the quadratic equation i used the trigonometric identity

tg (a-b)=(tg a - tg b)/(1+(tg a)*tg(b))

[Anonymous No.3988071]

>>3988048
whoops

"fi" would be the constant value for Phi

[Anonymous No.3988179]

aw man

just make a quick check guys!

[Anonymous No.3988330]

I'm not sure what you mean by stream function. The fact that you're suggesting a term u*x in \Phi makes me thing you mean a scalar function such that v=grad(\Phi). Is that so? and how did you get the function? Integrate the components respectively via mathematica? In any case, notice that your \Phi only contains Q together with the first term in your first pic. In the second pic you corrected this, it seems. Also, notice that in Mathematica, you can write \Pi and \Phi via "esc p esc" and "esc Ph esc" and so one. you should use this. Also Mathematica knows Grad, Div, Curl etc. Search for it in the help and I'd suggest you copy

DeclarePackage["VectorAnalysis`", {"start","VectorAnalysis","Div", "Grad", "Curl"}]

in your init.m file on your computer.

My naive integrations of the vector fields (if you're really interested in grad and not some curl variant, as wikipedias "stream function" article suggests) gives another term that your arctan, but no too different. different powers of y though. show me your complete (Mathematica) derivation.

[Josef crossed the Atlantic !!nUf2NflSAyw No.3988362]

Stream lines sounds like an integral curve to me, in your case that would be solving <span class="math">\partial_t\vec x(t)=v(t)[/spoiler] for some given initial condition <span class="math">\vec v(t_0)[/spoiler].

[Josef crossed the Atlantic !!nUf2NflSAyw No.3988370]

Stream lines sounds like an integral curve to me, in your case that would be solving <span class="math">\partial_t\vec x(t)=v(t)[/spoiler] for some given initial condition <span class="math">\vec x(t_0)[/spoiler]. That'll result in the stream line going through that point.

[Anonymous No.3988472]

>>3988330
its not quite a gradient? the stream function is such that ∂φ/∂x=v_y and ∂φ/∂y=-v_x

also yes I missed that Q/2π there, just a typo.

WHOOPS

i've suddenly stopped being dumb and it should be U*y instead of U*x

this changes EVERYTHING

possibly making it harder

[Anonymous No.3988484]

>>3988370
go to your campus right now

[Josef crossed the Atlantic !!nUf2NflSAyw No.3988488]

>>3988484
I am on my campus. Also, I've got an email address if you want to talk to me in person. (Yes, it's working.)

[GL !Bgpe2HycPk No.3988519]

>>3988488
He's not gonna send the pie via email.

[Anonymous No.3988672]

I played around a bit but I prerry much got what you got. And you can't easily solve for y. This doesn't really come as a surprise since it's fairly complicated at points as one sees in th plot..