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})

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}).