# Finite approximation of PDEs

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## Homework Statement

Given u_tt = F(x,t,u,u_x, u_xx), give the finite difference approximation of the pde (ie using u_x = (u(x + dx; t) - u(x - dx; t))/(2dx) etc.)

## Homework Equations

Well, clearly, u_x = (u(x + dx; t) - u(x - dx; t))/(2dx)

## The Attempt at a Solution

I really have no idea how that formula applies, but I do know that u_tt = u(x,t+dt) - 2u(x,t) + u(x,t-dt) / 2(dt)^2. How the non-homogeneous term F applies, I have no clue.

I'd be eternally grateful for any help anyone has to provide...considering I was the only person to reply to my last thread (on PDEs.)

The finite difference approximation of u_tt = F(x,t,u,u_x, u_xx) is given by:u_tt = (F(x, t + dt, u(x, t + dt), (u(x + dx; t + dt) - u(x - dx; t + dt))/(2dx), (u(x + 2dx; t + dt) - 2u(x; t + dt) + u(x - 2dx; t + dt))/(4dx^2)) - 2F(x, t, u(x, t), (u(x + dx; t) - u(x - dx; t))/(2dx), (u(x + 2dx; t) - 2u(x; t) + u(x - 2dx; t))/(4dx^2)) + F(x, t - dt, u(x, t - dt), (u(x + dx; t - dt) - u(x - dx; t - dt))/(2dx), (u(x + 2dx; t - dt) - 2u(x; t - dt) + u(x - 2dx; t - dt))/(4dx^2))---------------------------------------------------------------------------------------------------------------------------/ 2(dt)^2