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Before I get to the heat equation I'd like to talk about the advection equation. Descritize that with FD in time and BD in space: \begin{equation} \dfrac{u^{n+1}_i - u^{n}_i}{\Delta t} + v \dfrac{u^{n}_{i} - u^{n}_{i-1}}{\Delta x}= 0 \end{equation} there are higher order terms which were cancelled out so let's bring those back in ie. trucnation errors: \begin{equation} \dfrac{\partial u}{\partial t} + \dfrac{\Delta t}{2}\dfrac{\partial^2 u}{\partial t^2} + v \dfrac{\partial u}{\partial x} -v\dfrac{\Delta x}{2}\dfrac{\partial^2 u}{\partial x^2} = 0 \end{equation} Move things around a lil: \begin{equation} \dfrac{\partial u}{\partial t} + v \dfrac{\partial u}{\partial x} = -v\dfrac{\Delta x}{2}\dfrac{\partial^2 u}{\partial x^2} + \dfrac{\Delta t}{2}\dfrac{\partial^2 u}{\partial t^2} \end{equation} with some vodo magic \begin{equation} \dfrac{\partial u}{\partial t} + v \dfrac{\partial u}{\partial x} = -v\dfrac{\Delta x}{2}\dfrac{\partial^2 u}{\partial x^2} + v^2\dfrac{\Delta t}{2}\dfrac{\partial^2 u}{\partial^2 x} \end{equation} and some compression \begin{equation} \dfrac{\partial u}{\partial t} + v \dfrac{\partial u}{\partial x} = \dfrac{v}{2} \left(- \Delta x + v \Delta t \right)\dfrac{\partial^2 u}{\partial^2 x} + \text{higher order terms} \end{equation} When the convection equation is discretized we don't solve the original equation but the equation written above. If the RHS coefficient $-\Delta x + v \Delta t$

  • Positive $-\Delta x + v \Delta t > 0 \implies \dfrac{v \Delta x}{\Delta t}<1$ we get the CFL condition and a positive diffusion coefficient which is why explicit schemes are explicit. A low CFL would imply more diffusion.
  • negative $-\Delta x + v \Delta t > 0$ implies a negative coefficient for second order derivative which is why the solution blows up.

Now I am trying to apply this rational to the heat equation. Descritize that with FD in time and CD in space: \begin{equation} \dfrac{u^{n+1}_i - u^{n}_i}{\Delta t} = D \dfrac{u^{n}_{i+1} -2 u^{n}_{i} + u^{n}_{i-1}}{\Delta x^2} \end{equation} the truncated equation: \begin{equation} \dfrac{\partial u}{\partial t} + \dfrac{\Delta t}{2}\dfrac{\partial^2 u}{\partial t^2} = D \dfrac{\partial^2 u}{\partial x^2} + D \dfrac{\Delta x^2}{12}\dfrac{\partial^4 u}{\partial x^2} \end{equation} and finally \begin{equation} \dfrac{\partial u}{\partial t} = D \dfrac{\partial^2 u}{\partial x^2} -D^2 \dfrac{\Delta t}{2}\dfrac{\partial^4 u}{\partial x^4} + D \dfrac{\Delta x^2}{12}\dfrac{\partial^4 u}{\partial x^4} \end{equation} compress a lil: \begin{equation} \dfrac{\partial u}{\partial t} = D \dfrac{\partial^2 u}{\partial x^2} +\left (-D^2 \dfrac{\Delta t}{2} + D \dfrac{\Delta x^2}{12} \right) \dfrac{\partial^4 u}{\partial x^4} \end{equation} I wouldn't want a negative coefficient next to the fourth order derivative because that would mean that I would be solving an equation that blows up (I guess) so I want to guarantee that:

  • $-D^2 \dfrac{\Delta t}{2} + D \dfrac{\Delta x^2}{12} > 0 \implies \dfrac{D \Delta t }{\Delta x^2}< \dfrac{1}{6} $
    which is not consistent with the Von Neumann stability condition for this scheme which is:
  • $\dfrac{D \Delta t }{\Delta x^2}< \dfrac{1}{2} $

any hints where I went wrong?

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    $\begingroup$ For a Fourier mode exp(ikx -iwt), the second derivative produces $(ik)^2=-k^2$, so for the mode growth rate to be negative we want $D (-k^2)$ to be negative, that's why we need positive diffusion coefficient D. For hyper-diffusion (the 4th derivative operator) it would be $D_4 k^4$, so the coefficient $D_4$ in front should be negative. But your equation has both 2nd and 4th derivative on RHS, so the proper analysis should be done by putting a Fourier mode there and combining all terms to make the mode growth rate negative, that should be easy to do. $\endgroup$ Aug 22 at 3:34
  • $\begingroup$ Looks like I missed the diffusion part. I still would like to do this without resorting to fourier analysis and stick to truncation error analysis $\endgroup$ Aug 22 at 21:54
  • $\begingroup$ But note that for the 4th derivative term the condition of not blowing up should have sign opposite to what you have there, based on those Fourier analysis arguments. $\endgroup$ Aug 23 at 2:06

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