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For the ADI in numerical method

$$\frac{U^{n+1/3}-U^n}{k/3} = \Delta^2_x U^{n+1/3} + \Delta^2_y U^n + \Delta^2_z U^{n+2/3}$$ $$....$$ $$....$$

don't like $U^{n+1/2} = \dfrac 1 2 (U^{n+1}+U^n),$ I can't find the definition of $U^{n+1/3},U^{n+2/3},$ can some one tell me he definition or it is just a intermedia variables?

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  • $\begingroup$ Could it be $U^{n+1/3}=\frac{2}{3}U^{n+1}+\frac{1}{3}U^{n}$ ? $\endgroup$ – Alex C Jan 24 '17 at 2:54
  • $\begingroup$ In general in FD literature, superscripts (here $n$) are time indices while subscripts (here $x$,$y$,$z$) reflect space variables. In your case, the method hence involves 2 intermediate time steps from $$n \to n+1/3 \to n+2/3 \to n+1$$ At each intermediate time step you will be "alternating directions": instead of directly differentiating $U$ with respect to $x$, $y$ and $z$, you'll first differentiate wrt to $x$ keeping other variables constant ($n \to n+1/3$), then $y$, then $z$ ($n+2/3 \to n+1$)). See the wikipage for an example in 2 dimensions, with half time step $n+1/2$. $\endgroup$ – Quantuple Jan 24 '17 at 10:45
  • $\begingroup$ @ Quantuple so, here $U^{n+1/3}$ is an unknown intermedia variable which should be solved from equation system with additional dimensions(can't directly obtained from $U^{n+1}$ and $U^n$) $\endgroup$ – A.Oreo Jan 24 '17 at 12:41
  • $\begingroup$ @A.Oreo Yes you first solve the implicit representation of the discretised PDE for $U^{n+1/3}$ knowing $U^n$. Then, knowing $U^{n+1/3}$ you look for $U^{n+2/3}$ again solving a linear system. Finally, you'll be able to solve find $U^{n+1}$ using $U^{n+2/3}$ (which will be a non-linear function of $U^n, U^{n+1/2}$) $\endgroup$ – Quantuple Jan 24 '17 at 15:17

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