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The numerical problem

Of course, the 2-D Burgers equation cannot be solved analytically. We employ numerical methods based on finite-difference schemes.

We note i the space index in x direction, j the space index in y direction and n the time index. $N_x$ and $N_y$ are the numbers of point in each of these directions.

For that, one uses ADI Method (for Alternating Direction Implicit). Generally, this method is used to solve the compressible Navier-Stokes equations. When applied to the 2-D Burgers equation, this scheme becomes : 


 
\begin{displaymath}\begin{array}{ccc}\left[1+ \frac{\Delta t}{2} \left( u_{i,j......{ \delta}_x^2 \right) \right] u_{i,j}^{\ast} \\ \\\end{array}\end{displaymath} (3.1)
 here, $\bar{\delta}_x$ and $\bar{\delta}_y$ represent the first order central difference operator defined by :

\begin{displaymath}\bar{\delta}_x u_{i,j}^n = u_{i+1,j}^n - u_{i-1,j}^n \end{displaymath}
 

\begin{displaymath}\bar{\delta}_y u_{i,j}^n = u_{i,j+1}^n - u_{i,j-1}^n\end{displaymath}

and $\hat{ \delta}_y^2$ and $\hat{ \delta}_x^2$ represent the second order central difference operator defined by :

\begin{displaymath}\hat{\delta}_x^2 u_{i,j}^{n} = \frac{u^{n}_{i+1,j}-2u^{n}_{i,j}+u^{n}_{i-1,j}}{\Delta x^2} \end{displaymath}
 

\begin{displaymath}\hat{\delta}_y^2 u_{i,j}^{n} = \frac{u^{n}_{i,j+1}-2u^{n}_{i,j}+u^{n}_{i,j-1}}{\Delta y^2}\end{displaymath}

This method is first-order accurate with a truncature error of $O[\Delta t, (\Delta x)^2, (\Delta y)^2]$ and is unconditionally stable for the linear case. Obviously, a tridiagonal system of algebraic equations must be solved during each step. For that is used the gaussian elimination algoithm, sometimes refered to as the Thomas Algorithm (see Appendix A for more informations).

Principle of the ADI Method :
The algorithm initially consists in calculating on all the field the intermediate value $u_{i,j}^{\ast}$ starting from value $u_{i,j}^n$ calculated with the previous time step. In our case (arbitrary choice), we reverse the tridiagonal system with y fixed, and this for all including between 0 and 1 ($j=1,...,N_y-1$).

Then, once traversed all the grid (all x and all y), one calculates$u_{i, j}^{n+1}$ (solution sought with the time step n+1) from $u_{i,j}^{\ast}$ calculated with the fictitious time step $n+ \frac{1}{2}$.

The method thus consists in systematically making a double sweeping with each time step. The algorithm of calculation is given in a diagrammatic way in Appendix B.


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Next:ResultsUp:The 2-D Burgers Equation Previous:Resolution with a non   Contents
 
Alban Depoutre
2000-11-21