# Thread: Partial derivatives, equation help

1. ## Partial derivatives, equation help

The velocity components for a flow field are as follows;

$\displaystyle v_{x} = \frac{\partial \psi}{ \partial y} = a( x^{2} - y^{2})$

and

$\displaystyle v_{y} = -\frac{\partial \psi}{\partial x} = -2axy$

a) Prove that it statifies the continuity equation ; $\displaystyle \frac{\partial v_{x}}{\partial x} + \frac{\partial v_{y}}{\partial y} = 0$

b) Determine the stream function $\psi$

Not sure if my method for part 'a' is correct, I just started rearranging the equations

$\displaystyle v_{x} = \frac{\partial \psi}{ \partial y}$

$\displaystyle v_{x} \partial y = \partial \psi$

sub this into $\displaystyle v_{y} = -\frac{\partial \psi}{\partial x}$

$\displaystyle v_{y} = -\frac{v_{x} \partial y}{\partial x}$

$\displaystyle \frac{ v_{y}}{\partial y} = -\frac{v_{x}}{\partial x}$

$\displaystyle \frac{ v_{y}}{\partial y} + \frac{v_{x}}{\partial x} = 0$

It's not the exact desired equation but I am not sure how to get the $v_{x} and v_{y}$ as partial derivatives?

Also any help on part 'b', do I integrate?

Thank you.

2. you need to find $v_x$ and $v_y$ ....

$\displaystyle v_{x} = \frac{\partial \psi}{ \partial y} \Rightarrow v_x = \int a(x^2-y^2) dx$

$\displaystyle v_{y} = -\frac{\partial \psi}{\partial x} = -2axy \Rightarrow v_y = \int -2axy dy$

then you will do partial derivation on v_x and v_y to do this

$\displaystyle \frac{ v_{y}}{\partial y} + \frac{v_{x}}{\partial x} = 0$

3. I think you're making it more complicated than it needs to be. Just compute

$\dfrac{\partial v_{x}}{\partial x}=2ax.$

Do the same for $\dfrac{\partial v_{y}}{\partial y}.$ The result is immediate.

4. Originally Posted by yeKciM
you need to find $v_x$ and $v_y$ ....

$\displaystyle v_{x} = \frac{\partial \psi}{ \partial y} \Rightarrow v_x = \int a(x^2-y^2) dx$

$\displaystyle v_{y} = -\frac{\partial \psi}{\partial x} = -2axy \Rightarrow v_y = \int -2axy dy$

then you will do partial derivation on v_x and v_y to do this

$\displaystyle \frac{ v_{y}}{\partial y} + \frac{v_{x}}{\partial x} = 0$

Thank you, but I dont understand why you are integrating? How does that make v_{x} and v_{y} into partial derivatives?

5. For the integration, just follow the general idea for solving exact ODE's. That is, start with

$\displaystyle\frac{\partial\psi}{\partial y}=a(x^{2}-y^{2})\quad\rightarrow\quad \psi=ax^{2}y-a\frac{y^{3}}{3}+f(x)\quad\rightarrow\quad\frac{\p artial\psi}{\partial x}=2axy+f'(x).$

Compare this with the other expression you have, and go from there.

6. Originally Posted by Ackbeet

$\displaystyle\frac{\partial\psi}{\partial y}=a(x^{2}-y^{2})\quad\rightarrow\quad \psi=ax^{2}y-a\frac{y^{3}}{3}+f(x)\quad\rightarrow\quad\frac{\p artial\psi}{\partial x}=2axy+f'(x).$

Compare this with the other expression you have, and go from there.
after integrating I get

$v_{y} = axy^{2} + C$

$v_{x} = ax^{2}y - \frac{ay^{3}}{3} + C$

where dO I go from here?

Thanks

7. I think you're confusing yourself with the $v$'s versus the $\psi$'s. For part b, lose the $v$'s entirely and work just with the function $\psi.$ Does that clear away a few things?

8. Originally Posted by Ackbeet
I think you're confusing yourself with the $v$'s versus the $\psi$'s. For part b, lose the $v$'s entirely and work just with the function $\psi.$ Does that clear away a few things?

I am not on part 'b' yet, I still can't do part 'a'/

or is my orginal post correct for part a?

9. No, you can't throw around partial differentials like you can total differentials. Your entire approach is, I'm afraid, doomed to failure. Just follow my hint in post # 3.

10. ^^ oh right, now I get it. I really was over complicating it.

11. So for part 'b' to find the function $\psi$
I have to integrate $\frac{\partial \psi}{\partial y} = a(x^{2} - y^{2} )$
and the same for $- \frac{\partial \psi}{\partial x} = -2axy$