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.

  • $\begingroup$ 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? $\endgroup$ – Carl Christian Mar 6 '19 at 18:38
  • $\begingroup$ 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 $u_{j+1/2}^n=\xi^ne^{i\beta(j+1/2)\Delta x}$ $\endgroup$ – 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.

  • $\begingroup$ I have never seen this relation before. Would you be able to provide me with a derivation from the standard Taylor series? $\endgroup$ – MRT Mar 7 '19 at 9:44
  • $\begingroup$ 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. $\endgroup$ – Lutz Lehmann Mar 7 '19 at 11:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.