# Stream function derivation and derivative problem???

I am having a problem with understanding the stream function and its derivation. I understand the following:

Continuity equation has to hold for a stream line, thus: $$u \,dy - v \,dx = 0$$ Now I assume that u is a function of x, and v is a function of y, both describing velocity.

Now if we integrate this expression, we can arrive at a constant, which as to my understanding is the stream function: $$\int u \,dy - v \,dx = \psi$$

Now my confusion starts: I see the following expression around: $$u = \frac{\delta \psi}{\delta \,y}$$ how can the derivation of a constant not lead to zero??? And next, how is that expression found anyways? Here my attempt: $$\int u(x) \, dy = \psi + \int v(y) \, dx$$ taking the derivative with respect to y: $$u(x) = \frac{\delta \psi}{\delta \,y} + \frac{\delta }{\delta y} v(y) x$$ So I am left still with an extra term... What is going on? Thanks!

## 1 Answer

What is going on here is simple and somewhat obfuscated by the notation. The equation

$$u(x,y) dx + v(x,y) dy = 0$$

can be understood through the corresponding planar autonomous system

$$u(x,y) \frac{dx}{dt} + v(x,y) \frac{dy}{dt} = 0.$$

This equation says that the trajectories of the autonomous system are perpendicular to the vector field $\begin{bmatrix} u(x,y) \\ v(x,y) \end{bmatrix}$. If $\frac{\partial u}{\partial y} = \frac{\partial v}{\partial x}$, then it turns out that $\begin{bmatrix} u(x,y) \\ v(x,y) \end{bmatrix}$ is the gradient of a function. This function is called the stream function; let's call it $\psi$. Because the trajectories of the autonomous system are perpendicular to $\nabla \psi$, $\psi$ is constant along the trajectories of the autonomous system. That is, $\psi(x(t),y(t))=\psi(x(0),y(0))$ for all $t$.

The statement "$\begin{bmatrix} u(x,y) \\ v(x,y) \end{bmatrix}$ is the gradient of $\psi$" amounts to saying that $\frac{\partial \psi}{\partial x} = u$ and $\frac{\partial \psi}{\partial y} = v$. There is no contradiction in these being nonzero, because a small perturbation only in the $x$ or $y$ direction in general will not keep us on the trajectory of the system (i.e. on the level set of $\psi$). Only $\frac{d \psi}{dt}$ should be expected to be zero.

• Aaaaahhh! Thanks a lot! Written out it now all makes so much more sense. – user3604362 Nov 25 '15 at 8:38
• Actually, one question... Where do you get the expression $\frac{du}{dx}$ ? I would only arrive to an expression as $\frac{u}{dx}$ – user3604362 Nov 25 '15 at 9:30
• @user3604362 If you stop thinking in terms of differentials, everything makes a lot more sense. The point is that you have a smooth vector field $(u,v)$, and the differential equation tells you that your trajectories are perpendicular to this vector field. You can differentiate this field and its components however you like. – Ian Nov 25 '15 at 12:24
• @user3604362 We wanted to check $\frac{\partial u}{\partial y}=\frac{\partial v}{\partial x}$ because if $(u,v)=\nabla \psi$ then this is the same as $\frac{\partial^2 \psi}{\partial y \partial x} = \frac{\partial^2 \psi}{\partial x \partial y}$, which we should have if we anticipate the existence of a stream function. – Ian Nov 25 '15 at 12:24
• That I understand. But it seems to me that smth is wrong in the reasoning. I was told that due to the nature of a stream line, $V \cross ds = 0$... And thus we arrive to the expression $u dy - vdx = 0$... But from this equation one never arrives to the expression you are giving me, which is $\frac{du}{dx} + \frac{dv}{dy} = 0$. This expression can only come from the comtinuity equation... But then, what happens to the stream line definition? Smth is wrong here... – user3604362 Nov 26 '15 at 16:04