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Which of these equations defines the characteristic curve in an x – y plane?(a) \(\frac {dy}{dx_{char}} = \frac {- \frac {uv}{a^{2}} ± (\frac {u^2 + v^{2}}{a^{2}}) – 1}{1 – \frac {u^{2}}{a^{2}}}\)(b) \(\frac {dy}{dx_{char}} = \frac {- \frac {uv}{a^{2}} ± \sqrt {(\frac {u^{2} + v^{2}}{a^{2}})}}{\frac {u^{2}}{a^{2}}}\)(c) \(\frac {dy}{dx_{char}} = – \frac {uv}{a^{2}} ± \sqrt {(\frac {u^2 + v^{2}}{a^{2}})-1}\)(d) \(\frac {dy}{dx_{char}} = \frac {- \frac {uv}{a^{2}} ± \sqrt {(\frac {u^2 + v^{2}}{a^{2}})-1}}{1 – \frac {u^{2}}{a^{2}}}\)The question was posed to me at a job interview.My query is from Two Dimensional Irrotational Flow topic in division Numerical Techniques for Steady Supersonic Flow of Aerodynamics

Answer»

Correct answer is (d) \(\frac {DY}{dx_{char}} = \frac {- \frac {UV}{a^{2}} ± \sqrt {(\frac {u^2 + V^{2}}{a^{2}})-1}}{1 – \frac {u^{2}}{a^{2}}}\)

The explanation is: The approach of method of characteristics entails determining special curves, referred to as characteristics curves, along which the PDE transforms into a family of ordinary differential equations (ODE). If the ODEs have been identified, they can be solved along the CHARACTERISTIC curves to obtain ODE solutions, which can then be compared to the original PDE solution. The characteristic curve is given by:

\(\frac {dy}{dx_{char}} = \frac {- \frac {uv}{a^{2}} ± \sqrt {(\frac {u^2 + v^{2}}{a^{2}})-1}}{1 – \frac {u^{2}}{a^{2}}}\)


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