Question

Find the central second difference of u in y-direction using the Taylor series expansion.

(a) \(\frac{u_{i,j+1}+2u_{i,j}+u_{i,j-1}}{(\Delta y)^2}\)

(b) \(\frac{u_{i,j+1}-2u_{i,j}+u_{i,j-1}}{(\Delta y)^2}\)

(c) \(\frac{u_{i,j+1}-2u_{i,j}-u_{i,j-1}}{(\Delta y)^2}\)

(d) \(\frac{u_{i,j+1}+2u_{i,j}-u_{i,j-1}}{(\Delta y)^2}\)

The question was posed to me during an interview for a job.

Question is taken from Finite Difference Method topic in section Finite Difference Methods of Computational Fluid Dynamics

Answer 1

The correct answer is (b) \(\frac{u_{i,j+1}-2u_{i,j}+u_{i,j-1}}{(\Delta y)^2}\)

The explanation: To get the second difference,

\(u_{i,j+1}+u_{i,j-1}=2 u_{i,j}+(\frac{\partial^2 u}{\partial y^2})_{i,j}(\Delta y)^2+⋯\)

\((\frac{\partial^2 u}{\partial y^2})_{i,j}=\frac{u_{i,j+1}-2 u_{i,j}+u_{i,j-1}}{(\Delta y)^2} +⋯\)

After truncating,

\((\frac{\partial^2 u}{\partial y^2})_{i,j}=\frac{u_{i,j+1}-2 u_{i,j}+u_{i,j-1}}{(\Delta y)^2}\).