Correct answer is (b) Velocity potential equation
Best explanation: The velocity potential equation is given by:
(1 – \(\frac {Φ_x^2}{a^2}\)) Φxx + (1 – \(\frac {Φ_y^2}{a^2}\)) Φyy + (1 – \(\frac {Φ_z^2}{a^2}\)) Φzz – (\(\frac {2Φ_x Φ_y}{a^2}\)) Φxy – (\(\frac {2Φ_x Φ_z}{a^2}\)) Φxz – (\(\frac {2Φ_y Φ_z}{a^2}\)) Φyz
The total velocity potential is related to the perturbation velocity potential by:
Φx = V∞ + Φx, Φy = ϕy, Φz = ϕz
And its double derivative is given by
Φxx = ϕxx, Φyy = ϕyy, Φzz = ϕzz
Substituting these values in the velocity potential equation and multiplying it with a^2 we get
(a^2 – (V∞ + ϕx)^2)ϕxx + (a^2 – ϕy^2) ϕyy + (a^2 – ϕz^2) ϕzz – (2(V∞ + ϕx)ϕy)ϕxy – (2(V∞ + ϕx)ϕz)ϕxz – (2ϕyϕz) ϕyz
The above equation is known as the perturbation velocity potential equation which is a non linear equation.