# Stream Function

1. Mar 27, 2006

I'm having some trouble trying to decihper the notation used for the stream function in two dimensions.

Say we have a velocity field:
$$\vec V(x,y)$$

The fluid is incompressible, thus Laplaces equation must be satisfied.
$$\nabla^2 u = 0$$
Where: $$u = \Nabla \vec V$$
Thus: $$u_x = V_1$$
$$u_y = V_2$$

Where $u_x [/tex] is short hand for the partial derivative of [itex] u(x,y) [/tex] So now here comes the stream function. Is it a vector function? It has to be right? The definition I have is that the stream function satisfies: $$u = -\nabla \times s(x,y)\hat k$$ (1) Now the curl is supposed to return a vector right? So how is this satisfied with (1). I'm guessing that it must deal with the [itex] \hat k [/tex] But if someone could help me clear this up that would be cool. Also please, note that we are ONLY dealing with 2 dimensions for right now. Thanks Last edited by a moderator: Mar 28, 2006 2. Mar 28, 2006 ### StatusX I assume you mean for the fluid to also be irrotational, and for u to be the fluid potential. Is that right? Then you should have [itex]\vec V = \nabla u$, although from the next two lines it seems like you know this, and the above was a typo.

Technically the curl can only be taken of a vector field, and it is another vector field, and this operation is only defined in 3D. You can use the curl in 2D if you think of your 2D space as a plane in a larger 3D space. Then the curl of any vector field in the plane is perpendicular to the plane (ie, something times $\hat k$), and so may be treated as a scalar (this happens with the vorticity $\vec \omega$). Conversely, if you have a vector field that is everywhere normal to the plane, it may be treated as a scalar, and its curl is a vector field in the plane. This latter case is what happens with the stream function. Also, I think you want to set the velocity, not the potential, equal to the curl of the stream function.

Last edited: Mar 28, 2006