What are the implications of applying a direction-splitting within each stage of an SSP-RK scheme?

For instance, given a standard advective transport type equation: $$ \partial_{t}Q + \operatorname{div}(\mathbf{u}Q) = 0 $$ it's popular to make use of an explicit SSP-RK style time-stepping scheme, such as the well-known SSP-RK2 method: $$ Q^{(1)} = Q^{(n)} - \Delta t \operatorname{div}(\mathbf{u}Q^{(n)}) \\[1ex] Q^{(2)} = \tfrac{1}{2}Q^{(n)} + \tfrac{1}{2}Q^{(1)} - \tfrac{1}{2}\Delta t \operatorname{div}(\mathbf{u}Q^{(1)}) \\[1ex] Q^{n+1} = Q^{(2)} $$ But if the flux terms can only be evaluated via a direction-splitting: $$ \operatorname{div}(\mathbf{u}Q) = \partial_{x}(uQ) + \partial_{y}(vQ) $$ what are the implications of advancing each RK stage according to a simple direction split: $$ Q^{(1a)} = Q^{n} \\[1ex] Q^{(1b)} = Q^{(1a)} - \Delta t \, \partial_{x}(uQ^{(1a)}) \\[1ex] Q^{(1c)} = Q^{(1b)} - \Delta t \, \partial_{y}(vQ^{(1b)}) \\[2ex] Q^{(2a)} = \tfrac{1}{2}Q^{n} + \tfrac{1}{2}Q^{(1c)} \\[1ex] Q^{(2b)} = Q^{(2a)} - \tfrac{1}{2}\Delta t \, \partial_{y}(vQ^{(2a)}) \\[1ex] Q^{(2c)} = Q^{(2b)} - \tfrac{1}{2}\Delta t \, \partial_{x}(uQ^{(2b)}) \\[2ex] Q^{n+1} = Q^{(2c)} $$ Is the order of the method reduced? Does it loose its SSP property? Is it possible to combine such an approach with SSP-RK3 to get a third-order accurate method?

  • $\begingroup$ What is your reason for splitting? Why not just compute each of the directional derivative approximations and then apply them simultaneously to update the solution? $\endgroup$ – David Ketcheson Aug 12 '16 at 9:04

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