second order approximation of third derivative

I'm following some free online material from MIT about differential equations and I'm trying to solve problem 1 - question 2 of this assignment https://ocw.mit.edu/courses/aeronautics-and-astronautics/16-920j-numerical-methods-for-partial-differential-equations-sma-5212-spring-2003/assignments/ps2.pdf

It asks to develop a second-order accurate finite difference approximation of:

$$6 u u_{x} - u_{xxx}$$

And I'm a bit confused. I know how to do a second-order accurate finite difference of the second derivative with $\frac{u_{i-1} - 2 u_i + u_{i+1}}{h^2}$ but I'm not sure if that's even possible for a third order derivative $u_{xxx}$. How should I approach this problem?

Combine it with a central difference quotient, $$u_{xxx}(x_i)\approx\frac{\frac{u_{i+2}-2u_{i+1}+u_i}{h^2}-\frac{u_i-2u_{i-1}+u_{i-2}}{h^2}}{2h}$$ As this is symmetric, the error term has no first order component, is thus second order.
• No, they say that the linearization and thus its eigenvalues are dominated by the third derivative term. Kind of like $Δt/Δx$ is dominated by $Δt/Δx^3$. The eigenvectors of $u_t=-u_{xxx}$ are harmonic functions, there is a maximal frequency in $x$ direction with minimal wavelength $2Δx$ for the given discretization, that $\lambda=\frac\pi{Δx}$ you insert into $z=\lambdaΔt$. See for example chebfun.org/examples/ode-linear/Regions.html or the textbook of your choice. – LutzL Jun 16 '18 at 6:41
• A-stability relates to the question if the solver gives for $y_t=λy$ where $Re(λ)<0$ a numerical solution that converges to zero. Easy calculation with explicit Euler show that this fails dramatically if the step size is too large. The first step is to recognize that rescaling the time scale changes also $λ$, one could normalize the time so that $λ=1$ or $Δt=1$. But it is somewhat easier to keep the time scale and see that the convergence depends on a function in the time-scale invariant product $z=λΔt$, the condition has the form $|f(z)|<1$. For explicit RK methods $f$ is a polynomial. – LutzL Jun 16 '18 at 6:49