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.
Given that $u$ is sufficiently regular / smooth / continuously often differentiable, you can interchange differentiations and thus write \begin{align}u_{tt} &= \partial_t u_t = \partial_t \Bigg[ \frac{\partial^2}{\partial x^2} \bigg(u^3 - u - \frac{\partial^2 u}{\partial x^2} \bigg) + S \Bigg] \\ &= \frac{\partial^2}{\partial x^2} \Bigg( \partial_t u^3 - \partial_t u - \partial_{xx} \partial_t u \Bigg) + 0 \\ &= \frac{\partial^2}{\partial x^2} \Bigg( 3u^2 \partial_t u - \partial_t u - \partial_{xx} \partial_t u \Bigg) \\ &= \frac{\partial^2}{\partial x^2} \Big[\big( 3u^2 - 1 - \partial_{xx} \big) \partial_t u \Big]\\ &= \frac{\partial^2}{\partial x^2} \Bigg[ \big( 3u^2 - 1 - \partial_{xx} \big) \Bigg( \frac{\partial^2}{\partial x^2} \bigg(u^3 - u - \frac{\partial^2 u}{\partial x^2} \bigg) + S \Bigg) \Bigg] \end{align} which is a nasty expression.