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

Find the central second difference of u in y-direction using the Taylo

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

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