# About the the stability of using an explicit scheme on the heat equation

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: $$$$\dfrac{u^{n+1}_i - u^{n}_i}{\Delta t} + v \dfrac{u^{n}_{i} - u^{n}_{i-1}}{\Delta x}= 0$$$$ there are higher order terms which were cancelled out so let's bring those back in ie. trucnation errors: $$$$\dfrac{\partial u}{\partial t} + \dfrac{\Delta t}{2}\dfrac{\partial^2 u}{\partial t^2} + v \dfrac{\partial u}{\partial x} -v\dfrac{\Delta x}{2}\dfrac{\partial^2 u}{\partial x^2} = 0$$$$ Move things around a lil: $$$$\dfrac{\partial u}{\partial t} + v \dfrac{\partial u}{\partial x} = -v\dfrac{\Delta x}{2}\dfrac{\partial^2 u}{\partial x^2} + \dfrac{\Delta t}{2}\dfrac{\partial^2 u}{\partial t^2}$$$$ with some vodo magic $$$$\dfrac{\partial u}{\partial t} + v \dfrac{\partial u}{\partial x} = -v\dfrac{\Delta x}{2}\dfrac{\partial^2 u}{\partial x^2} + v^2\dfrac{\Delta t}{2}\dfrac{\partial^2 u}{\partial^2 x}$$$$ and some compression $$$$\dfrac{\partial u}{\partial t} + v \dfrac{\partial u}{\partial x} = \dfrac{v}{2} \left(- \Delta x + v \Delta t \right)\dfrac{\partial^2 u}{\partial^2 x} + \text{higher order terms}$$$$ When the convection equation is discretized we don't solve the original equation but the equation written above. If the RHS coefficient $$-\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: $$$$\dfrac{u^{n+1}_i - u^{n}_i}{\Delta t} = D \dfrac{u^{n}_{i+1} -2 u^{n}_{i} + u^{n}_{i-1}}{\Delta x^2}$$$$ the truncated equation: $$$$\dfrac{\partial u}{\partial t} + \dfrac{\Delta t}{2}\dfrac{\partial^2 u}{\partial t^2} = D \dfrac{\partial^2 u}{\partial x^2} + D \dfrac{\Delta x^2}{12}\dfrac{\partial^4 u}{\partial x^2}$$$$ and finally $$$$\dfrac{\partial u}{\partial t} = D \dfrac{\partial^2 u}{\partial x^2} -D^2 \dfrac{\Delta t}{2}\dfrac{\partial^4 u}{\partial x^4} + D \dfrac{\Delta x^2}{12}\dfrac{\partial^4 u}{\partial x^4}$$$$ compress a lil: $$$$\dfrac{\partial u}{\partial t} = D \dfrac{\partial^2 u}{\partial x^2} +\left (-D^2 \dfrac{\Delta t}{2} + D \dfrac{\Delta x^2}{12} \right) \dfrac{\partial^4 u}{\partial x^4}$$$$ 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?

• 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. Aug 22 at 3:34
• 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 Aug 22 at 21:54
• 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. Aug 23 at 2:06