The GATE Grind

GATE 2019 CE (CE1) – Question 37

Engineering Mathematics · PDE: Fourier series, separation of variables, diffusion, wave and Laplace equations · 2 marks · Multiple choice

A one-dimensional domain is discretized into N subdomains of width Δx, with nodes i = 0, 1, …, N. Time steps have width Δt. Which is the forward-time centred-space approximation at node i and time step n for $\partial v/\partial t=\beta\partial^2v/\partial x^2$?

  1. $\frac{v_i^{n+1}-v_i^n}{\Delta t}=\beta\frac{v_{i+1}^n-2v_i^n+v_{i-1}^n}{(\Delta x)^2}$
  2. $\frac{v_{i+1}^{n+1}-v_i^n}{\Delta t}=\beta\frac{v_{i+1}^n-2v_i^n+v_{i-1}^n}{2\Delta x}$
  3. $\frac{v_i^n-v_i^{n-1}}{\Delta t}=\beta\frac{v_{i+1}^n-2v_i^n+v_{i-1}^n}{(\Delta x)^2}$
  4. $\frac{v_i^n-v_i^{n-1}}{2\Delta t}=\beta\frac{v_{i+1}^n-2v_i^n+v_{i-1}^n}{2\Delta x}$

Practise this question in The GATE Grind →

Show answer and explanation

Correct answer: (A) $\frac{v_i^{n+1}-v_i^n}{\Delta t}=\beta\frac{v_{i+1}^n-2v_i^n+v_{i-1}^n}{(\Delta x)^2}$

Explanation

Use a forward time difference at time n and a centred second spatial difference: $$\frac{v_i^{n+1}-v_i^n}{\Delta t}=\beta\frac{v_{i+1}^n-2v_i^n+v_{i-1}^n}{(\Delta x)^2}.$$ Answer **A**. The second derivative needs squared spacing; option C uses a backward time difference.