(Single) Diagonally Implicit RK Method

Let us now focus attention on the s x s Butcher tableau for the Implicit method. Hairer & Wanner quote Roger Alexander’s paper, where Alexander suggests the use of a lower-triangularized form of the table. By using the lower triangular form, the need for iterations vanishes and it becomes a single-step process.

In doing so, Alexander suggests that the term ai,j becomes 0 for all i < j. He refers to such a process as a Diagonally Implicit RK method.

The Implicit RK formulation necessitates the solution of ki’s via Newton-type iterations at each time step. A reformulation of the ki’s can be written as follows (Hairer & Wanner):

Rearranging the terms to a form that has the independent variable ki on the left hand side, the common term that arises in the iteration is:

The partial derivative is equivalent to the Jacobian.

Alexander proceeds further to state that if all the ai,i terms are equal, (i.e., the outer diagonal terms of the lower triangular matrix), then the above matrix with the Jacobian needs to be evaluated only once at each time step. Such a formulation is referred to as the Single Diagonally Implicit RK method, or the SDIRK.

The Butcher tableau for the SDIRK method thus takes the form:

Implicit RK Methods:

My previous post touched on the differences between Implicit & Explicit methods. Implicit methods (such as the backward Euler’s method) require an iterative procedure at each step to arrive at a solution before proceeding to the next time step. The advantage, however, is that implicit methods are truly stable methods. Irrespective of the time step, convergence with Implicit methods are much better compared to Explicit ones.

A higher order Implicit-RK (IRK) method can be represented in the following form (Atkinson et al):

The result of the summation of all cj terms for each ki is a square-matrix of ALL terms (as opposed to just the lower triangular matrix with the case of an Explicit RK method). The Butcher tableau for the Implicit method thus takes the form:

 

For example, the ki terms for a 2-stage IRK method would be:

 

A Newton-type iterative solver would have to be employed to solve the above set of equations to obtain k1 & k2. In general, an s-order system will require the solution of ns coupled equations (Hairer & Wanner). On equations of several orders of magnitude, this can significantly slow down the process. Unless the set of equations to be solved are considerably stiff, where it becomes a choice between stability vs execution time.

 

Solving the Burgers’ Equation with a “double-step” initial condition.

As the last illustration for the Burgers’ equation, case study-1 presented in Wouwer et al has been considered. The initial & boundary values are defined by the equation (which is also an exact solution of the Burgers’ equation):

 where

The initial values at time t=0, and the boundary conditions for t>0 can be calculated from the above equations. The Method of Lines can then be used to solve the discretized ODE form of the Burgers’ equations. A 120-element model was used to solve the “double-step” input. I have referred to it as the double-step, as it will be evident from the solution. At t=0, the drop from boundary condition of 1 (at x=0) to 0.10 (at x=1) comprises of two step changes (as opposed to the single step change in the previous post).

Numerical Solution of the Burgers’ Equation for a Sine Input Function

The numerical integration of the Burgers’ equation with a half-sine wave at time t=0 is shown below. 120 elements were used to integrate the solution, with Re=1000. The MOL approach was used to convert the PDE’s to ODE’s. The ODE’s were integrated using the Rosenbrock technique with the VELDS coefficients.