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for part a) i get

$$ u = \partial_y \psi, \quad v = - \partial_x \psi$$

I need help with part d, if anyone can show me how to? thanks

Dmoreno
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italy
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  • whoever edited this, thats not the equations i get, its meant to be u = (partial psi) / (partial y) and v = - (partial psi)/(partial x) – italy Jun 10 '15 at 22:24
  • Hello Italy, this is what you typed: "u = partial psi \ partial y, v = - partial psi/partial x", which I believe matches what I wrote, doesn't it? Furthermore, it's correct. – Dmoreno Jun 10 '15 at 22:25
  • oh if its the same, than my bad! I have not seen it in this form before, but as long its the same – italy Jun 10 '15 at 22:26

1 Answers1

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Brute force tells us that if $\psi_y = u$, then:

$$ \psi(x,y) = \int u \, \mathrm{d}y + f(x) = \frac{x}{x^2+y^2} + f(x), $$ for some function $f$. Notice now that, on the other hand, if $\psi_x = -v$, then:

$$\psi(x,y) = - \int v \, \mathrm{d}x + g(y) = \frac{x}{x^2+y^2} + g(y), $$ for some function $g$. The computation of the integrals is left to you (check!).

What can we conclude about $f$ and $g$? Is there any obvious choice for them?

Cheers!

Dmoreno
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    f and g must be the same constant? but this doesn't really prove it does it? I was thinking to differentiate each of u and v, than equate the two ? – italy Jun 10 '15 at 22:31
  • Of course, $f = g = c$! Another question, does the value of $c$ affect to the velocity field? – Dmoreno Jun 10 '15 at 22:32
  • No I dot think it it does. – italy Jun 10 '15 at 22:40
  • Please, @italy, consider accepting the answer or let me know if you need any more help. I'll be happy to help. – Dmoreno Jun 11 '15 at 01:01