# Finite difference stability problem

My apologies for the title I'm not quite sure how to title a problem like this.

I need to show the following result:

$$u_j^{n+1} = e^{\Delta t\partial/\partial t}u_j^n$$

Where $$u_j^n$$ is the numerical solution to the PDE: $$u_t = u_{xx}$$ for point $$(x_j,t^n)$$.

I was given the advice to try a Taylor series expansion about the point $$u_j^{n+1}$$ which is as follows: $$u_j^{n+1} = u_j^n + \Delta tu_t + (\Delta t)^2\frac{u_{tt}}{2} + (\Delta t)^3\frac{u_{ttt}}{6} + \cdots$$

From the course we have the following relation when performing stability analysis on the numerical method which is: $$u_j^n = \xi^ne^{i\beta j\Delta x}$$

Note that the $$n$$ is the $$n$$-th time step and $$j$$ is the $$j$$-th step in space. $$\xi$$ is sometimes referred to as the amplitude factor.

If I then set these two things equal to each other I have: $$u_j^{n+1} = \xi^{n+1}e^{i\beta j\Delta x} = u_j^n + \Delta tu_t + (\Delta t)^2\frac{u_{tt}}{2} + (\Delta t)^3\frac{u_{ttt}}{6} + \cdots$$

And I suppose you could do another substitution to get: \begin{align} u_j^{n+1} &= \xi^{n+1}e^{i\beta j\Delta x} = \xi^ne^{i\beta j\Delta x} + \Delta tu_t + (\Delta t)^2\frac{u_{tt}}{2} + (\Delta t)^3\frac{u_{ttt}}{6} + \cdots\\ &= \xi e^{i\beta j\Delta x} = \Delta tu_t + (\Delta t)^2\frac{u_{tt}}{2} + (\Delta t)^3\frac{u_{ttt}}{6} + \cdots \end{align}

We also know that $$u_t = u_{xx}$$ and $$u_{tt} = u_{xxxx}$$ so: $$u_j^{n+1} = \xi e^{i\beta j\Delta x} = \Delta tu_{xx} + (\Delta t)^2\frac{u_{4x}}{2} + (\Delta t)^3\frac{u_{6x}}{6} + \cdots$$

After trying all of this I don't know what more to do so that I have the desired result. I also have the hint that $$u_j^{n+1} = e^{\Delta t\partial/\partial t}u_j^n = e^{\Delta t\partial^2/\partial x^2}u_j^n = \cdots$$

and that $$\frac{\partial}{\partial x} = \frac{1}{\Delta x}\left(\delta_x - \frac{1}{24}\delta_x^3 + \frac{3}{640}\delta_x^5 + \cdots\right) = \frac{2}{\Delta x}\sinh^{-1}\left[\frac{1}{2}\delta_x\right],$$ $$\delta_xu_j^n = u_{j+\frac{1}{2}}^n - u_{j-\frac{1}{2}}^n$$

if that helps at all. Hopefully that is someone out there who is familiar with what I speak.

• There are many schemes for solving the heat equation. Which one are you studying? How do you define the exponential function applied to an operator? – Carl Christian Mar 6 '19 at 18:38
• I have included how the $\delta_x$ operator is defined at the end of my post. We have been looking at the Crank-Nicolson Method recently. I am unsure how to define the exponential but the $u_{j+1/2}^n=\xi^ne^{i\beta(j+1/2)\Delta x}$ – MRT Mar 6 '19 at 19:57

You can in general formally abbreviate the Taylor series as $$f(t+Δt)=[e^{Δt∂/∂t}f](t),$$ this is valid whenever the function is analytical.
From this one also infers that in your text the $$u^n_j$$ are the samples $$u(x_j,t_n)$$ of the exact solution.
• There is no derivation necessary, just formally expand the exponential series $f(t)+\sum_{k>0}\frac{∂^kf}{∂t^k}(t)\frac{Δt^k}{k!}$ and you should recognize the Taylor expansion. – Lutz Lehmann Mar 7 '19 at 11:11