I need to demonstrate finding the LTE for the Explicit Euler scheme when i) $\mu=\frac{1}{6}$ and ii) $\mu\neq\frac{1}{6}$. I have been looking for a reference text/video and I looked at the lecturers notes but I really was struggling to follow how to even begin. I know I will have to carry out a Taylor expansion but not much beyond that.
The scheme: $\mathrm{U}_{i}^{n+1}=\mathrm{U}_{i}^{n}+\mu(\mathrm{U}_{i+1}^{n}-2\mathrm{U}_{i}^{n}+\mathrm{U}_{i-1}^{n})$
where: $1\le u\le M-1, n\ge 0$ $\mu=\frac{\bigtriangleup t}{(\bigtriangleup x)^2}$, $\mathrm{U}_{i}^{n}: 0\lt i \lt M $, $\mathrm{U}_{0}^{n}: n\ge 0$, $\mathrm{U}_{M}^{n}: n\ge 0$
We apply this to the heat equation using standard difference operators for the second derivative in space.
You get two error sources that might partially compensate, $$ u(x,t+Δt)-u(x,t)=Δt\,u_t+\frac12Δt^2\,u_{tt}+... $$ and $$ u(x+Δx,t)-2u(x,t)+u(x-Δx,t)=Δx^2\,u_{xx}+\frac1{12}Δx^4\,u_{xxxx}+... $$ Now use that also $u_{tt}=u_{xxxx}$