Local truncation error for the forward-difference method I need to show that if $\gamma=\frac{K\tau}{h^2}=\frac{1}{6}$, then in the explicit forward-difference method $\frac{w_{k,j+1}-w_{k,j}}{\tau}-K\frac{w_{k+1,j}-2w_{k,j}-w_{k-1,j}}{h^2}=0$ the local truncation error is $O(\tau^2)$ or similarly $O(h^4)$. 
My calculations (through substituting Taylor expansions) keep leading me to $\tau_{k,j}=\frac{\partial u}{\partial t}(x_k,t_j)-K\frac{\partial^2 u}{\partial^2 x}(x_k,t_j)+\frac{\tau}{2}\frac{\partial^2 u}{\partial^2 t}(x_k,\xi_1)-K\frac{h^2}{12}\frac{\partial^4 u}{\partial^4 x}(\xi_2,t_j)=O(\tau)+O(h^2)=O(\tau+h^2)$
Do I need to take the Taylor Series to higher powers or am I miscalculating?
Also, I have calculated this without using $\gamma$, so where would this come from?
 A: We have 
\begin{align}
w_{k,j+1} &= w_{k,j} +\tau\partial_t w_{k,j}+ \frac{\tau^2}{2} \partial_{tt} w_{k,j}+\partial_{ttt} \frac{\tau^3}{6} w_{k,j}+ O(\tau^4)
\\
w_{k+1,j}&=w_{k,j} +h\partial_x w_{k,j}+ \frac{h^2}{2} \partial_{xx} w_{k,j}+\partial_{xxx} \frac{h^3}{6} w_{k,j}+ \partial_{xxxx}\frac{h^4}{24} +\partial_{5x}\frac{h^5}{120} +O(h^6) \\
w_{k-1,j}&=w_{k,j} -h\partial_x w_{k,j}+ \frac{h^2}{2} \partial_{xx} w_{k,j}-\partial_{xxx} \frac{h^3}{6} w_{k,j}-\partial_{xxxx}\frac{h^4}{24}+\partial_{5x}\frac{h^5}{120} +O(h^6) \\
\end{align}
From that it follows,that 
\begin{align}
w_{k,j+1}-w_{k,j} = \tau\partial_t w_{k,j}+ \frac{\tau^2}{2} \partial_{tt} w_{k,j}+\partial_{ttt} \frac{\tau^3}{6} w_{k,j}+ O(\tau^4)
\end{align}
and
\begin{align}
w_{k+1,j}-2w_{k,j}+w_{k-1,j}= h^2\partial_{xx}w_{kj}+\partial_{xxxx}\frac{h^4}{12} +O(h^5)
\end{align}
And now, with $\tau=\frac{h^2}{6K}$ we have 
\begin{align}
\frac{w_{k,j+1}-w_{k,j}}{\tau} &= \partial_t w_{k,j}+ \frac{\tau}{2} \partial_{tt} w_{k,j}+\partial_{ttt} \frac{\tau^2}{6} w_{k,j}+ O(\tau^3)\\
&= \partial_t w_{k,j}+ \frac{h^2}{12K} \partial_{tt} w_{k,j}+\partial_{ttt} \frac{h^4}{216K^2} w_{k,j}+ O(h^6)
\end{align}
and together with
\begin{align}
K\frac{w_{k+1,j}-2w_{k,j}+w_{k-1,j}}{h^2}= K\partial_{xx}w_{kj}+K\frac{h^2}{12}\partial_{xxxx} w_{kj}+O(h^4)
\end{align}
We get
\begin{align}
&\frac{w_{k,j+1}-w_{k,j}}{\tau}+K\frac{w_{i+1,j}-2w_{i,j}+w_{i-1,j}}{h^2}\\&=
\partial_t w_{kj}+ \frac{h^2}{12K} \partial_{tt} w_{k,j}+\partial_{ttt} 
 \frac{h^4}{216K^2} w_{k,j}+ O(h^6) \\
&+ K\partial_{xx}w_{kj}+K\frac{h^2}{12}\partial_{xxxx}w_{kj}+O(h^4) \\
&= \partial_t w_{k,}+K\partial_{xx}w_{kj}+(\frac{h^2}{12K}\partial_{tt} w_{k,j}+K\frac{h^2}{12}\partial_{xxxx}w_{kj})+O(h^4) \\
&= \partial_t w_{k,}+K\partial_{xx}w_{kj} +\frac{h^2}{12K} (\partial_{tt} w_{k,j}+K^2\partial_{xxxx}w_{kj}) +O(h^4)
\end{align}
And from now on I am not 100% sure, this is correct. I would really appreciate, if someone would double check my logic. But as we want to solve $u_t +Ku_{xx} = 0$ we can differentiate this term with respect to  $t$ and get $u_{tt} + Ku_{xxt} = 0$. But we could also differentiate w.r.t to $x$ twice and would get $u_{txx}+Ku_{xxxx}=0$. Combining the two, we would get $u_{tt}= -K^2u_{xxxx}$.
Thus $\frac{h^2}{12K} (\partial_{tt} w_{k,j}+K^2\partial_{xxxx}w_{kj})=0$. And with this we get fourth order accuracy in $h$.
Edit: As pointed out correctly, I missread your question and used $u_t+Ku_{xx}$ instead of $u_t-Ku_{xx}$. I don't want to correct everything here, so I define $K =-\tilde K$. Where $K$ is "my" $K$ and $\tilde K$ is used in your question.
