General Hodograph Equation

We begin by assuming inviscid, irrotational, isentropic, steady, two-dimensional flow. The flow is not necessarily incompressible. Because of the irrotationality condition, there exists a velocity potential $\phi$; and the steam function $\psi$, is everywhere perpendicular. That is, $\phi_x=\psi_y$ and $\phi_y=-\psi_x$.

In this derivation, the velocity vector ($\vec V$) is represented by its magnitude (V) and direction ($\theta$). This is simpler, in this case, than representing velocity by its components.

The stream function and velocity potential are functions of x and y, therefore the total differentials are:

  

$$d\phi \phi_x\,dx+\phi_y\,dy=V(\cos\theta\,dx+\sin\theta\,dy)$$ = (1)

$$d\psi \psi_x\,dx+\psi_y\,dy=\rho V(-\sin\theta\,dx+\cos\theta\,dy)$$ = (2)

The stream function and velocity potential can also be considered functions of V and $\theta$:

  

$$d\phi \phi_V\,dV+\phi_\theta\,d\theta$$ = (3)

$$d\psi \psi_V\,dV+\psi_\theta\,d\theta$$ = (4)

By eliminating $d\phi$ and $d\psi$ from Equations 1-4, we can get the total differentials dx and dy:

  

$$x_V\,dV + x_\theta\,d\theta$$ dx = (5)

$$y_V\,dV + y_\theta\,d\theta$$ dy = (6)

where

    

$$\frac1V\left(\phi_V\cos\theta- \frac1\rho\psi_V\sin\theta\right)$$ x_V = (7)

$$x_\theta \frac1V\left(\phi_\theta\cos\theta- \frac1\rho\psi_\theta\sin\theta\right)$$ = (8)

$$\frac1V\left(\phi_V\sin\theta- \frac1\rho\psi_V\cos\theta\right)$$ y_V = (9)

$$y_\theta \frac1V\left(\phi_\theta\sin\theta- \frac1\rho\psi_\theta\cos\theta\right)$$ = (10)

By setting $x_{V\theta}=x_{\theta V}$ and $y_{V\theta}=y_{\theta V}$, we can eliminate x and y from Equations 7-10. This results in the following equations:

  

$$\phi_V V\frac d{dV}\left(\frac1{\rho V}\right)\,\psi_\theta$$ = (11)

$$\phi_\theta \frac V\rho\,\psi_V$$ = (12)

We now eliminate $\phi$ by setting $\phi_{V\theta}=\phi_{\theta_V}$. The resulting equation is:

$$V\frac d{dV}\left(\frac1{\rho V}\right)\psi_{\theta\theta}= \frac d{dV}\left(\frac V\rho\right)\psi_V+ \frac V\rho\psi_{VV}$$ (13)

The equation can be simplified more by using Mach number to eliminate $\rho$ from the equations. Using the identity $M^2=\rho V\frac{d(1/rho)}{dV}$ (which is true for an isentropic flow) the hodograph equation becomes:

$$(M^2-1)\psi_{\theta\theta}=(M^2+1)V\psi_V+V^2\psi_{VV}$$ (14)

Equation 14 is the general equation for the stream function in the hodograph plane. The Mach number is not a function of $\psi$, therefore Equation 14 is linear with non-constant coefficients.