# LTE for the Cahn Hilliard Equation

I'm trying to Calculate the LTE of a first order finite difference (central difference) scheme for the following 1D equation:

$u_t = \frac{\partial^2}{\partial x^2}(u^3 - u - \frac{\partial^2 u}{\partial x^2}) + S$ where S is a predefined source term which independent of time.

Using the following definition for LTE:

$LTE = (\frac{1}{h})(u_{i}^{j+1}(x,t) - u_{i}^{j}(x,t+h))$

where $u_i^j$ represents the solution at $x$ increment $i$ and $t$ increment $j$.

$= (\frac{1}{h})(u_{i}^{j+1}(x,t) - \Sigma_{n=0}^{\infty} \frac{u^{(n)}}{n!} h^n)$

$\simeq(\frac{1}{h})\Big(6u\Big(\frac{u(x+k) - u(x-k)}{2k}\Big)^2 + (3u^2-1)\Big(\frac{u(x-k) - 2u(x) + u(x+k)}{k^2}\Big ) - \Big(\frac{u(x+2k) - 4u(x+k) + 6u - 4u(x-k) + u(x-2k)}{k^4}\Big) + \Big(\frac{u(t+h) - u(t)}{h}\Big) + S \Big)$

Taylor expansion of all terms in the above line to fourth order gives:

$LTE \simeq k^2(2uu_xu_{xxx} + (3u^2-1)u_{xxxx}) - h(\frac{1}{2}u_{tt}) + O(k^4, h^2)$

I'm unsure if my methods up to this point are justified, but assuming they are, I am now having trouble expressing $u_{tt}$ in terms of $x$ derivatives.

Another method to approximate LTE or help expressing $u_{tt}$ would be greatly appreciated.