Perfect Gas Hodograph Equation

To evaluate M, we assume a perfect gas. We define a parameter $\tau$ as:

$$\tau\equiv\frac{\gamma-1}2\frac{V^2}{a_0}$$ (15)

Using the perfect gas relationship

$$a^2=a_0^2-\frac{\gamma-1}2V^2$$ (16)

we get $\tau$ in terms of M:

$$\tau=\frac{(\gamma-1)M^2/2}{1+(\gamma-1)M^2/2}$$ (17)

The parameter $\tau$ is somewhat like the characteristic Mach number; it varies from zero at M=0 to one at $M=\infty$. At Mach one, $\tau=(\gamma-1)/(\gamma-1)$. This value is written henceforth as $\tau^*$.

Solving Equation 17 for M² gives

$$M^2=\frac2{\gamma-1}\frac\tau{1-\tau}$$ (18)

And from Equation 15 we get

$$\frac{\partial}{\partial V} =\frac{2\tau}V\frac{\partial}{\partial\tau}$$ (19)

$$\frac{\partial^2}{\partial V^2} =\frac{\gamma-1}{a_0^2}\left(... ...}{\partial\tau} +2\tau\frac{\partial^2}{\partial\tau^2}\right)$$ (20)

Using Equations 18-20, Equation 14 transforms to:

$$\tau^2(1-\tau)\psi_{\tau\tau} +\tau\left(1+\frac{2-\gamma}{\g... ...+\frac14\left(1-\frac\tau{\tau^*}\right)\psi_{\theta\theta} =0$$ (21)