We are interested in solving the advection equation u t = u x where 0 ≤ x &lt; 1, t ≥ 0with periodic boundary conditions and u ( x , 0 ) = f ( x ) , f ( x ) = f ( x + 1 ) In the grid x 0 , x 1 , … x N , t 0 , t 1 , … t N .using the forward euler method, or v ( x j , t n + 1 ) − v ( x j , t n ) Δ t = v ( x j + 1 , t n ) − v ( x j − 1 , t n ) 2 Δ x where v is the approximation of u at the grid points.What I don't understand:We are essentially using v ( x j − 1 , t n ) , v ( x j , t n ) , v ( x j + 1 , t n ) to calculate v ( x j , t n + 1 ). Initially this is fine because v ( x j , 0 ) is known for all j, but how would we calculate v ( x 0 , t 1 )? we can't, since that would require us to know v ( x − 1 , t 0 ) which doesn't exist. And this issue propagates, each new t n we can calculate one less point than t n − 1 .What am I missing?

Question

We are interested in solving the advection equation       u    t    =      u    x   where   0  ≤  x  &amp;lt;  1,   t  ≥  0with periodic boundary conditions and   u  (  x  ,  0  )  =  f  (  x  )  ,  f  (  x  )  =  f  (  x  +  1  ) In the grid       x    0    ,      x    1    ,  …      x    N  ,       t    0    ,      t    1    ,  …      t    N  .using the forward euler method, or             v      (              x        j            ,              t                  n          +          1                    )      −      v      (              x        j            ,              t        n            )              Δ      t        =            v      (              x                  j          +          1                    ,              t        n            )      −      v      (              x                  j          −          1                    ,              t        n            )              2      Δ      x       where   v is the approximation of   u at the grid points.What I don&#039;t understand:We are essentially using   v  (      x          j      −      1        ,      t    n    )  ,  v  (      x    j    ,      t    n    )  ,  v  (      x          j      +      1        ,      t    n    ) to calculate   v  (      x    j    ,      t          n      +      1        ). Initially this is fine because   v  (      x    j    ,  0  ) is known for all   j, but how would we calculate   v  (      x    0    ,      t    1    )? we can&#039;t, since that would require us to know   v  (      x          −      1        ,      t    0    ) which doesn&#039;t exist. And this issue propagates, each new       t    n   we can calculate one less point than       t          n      −      1      .What am I missing?

trivialaxxf · Accepted Answer

Your recurrence relation for the forward time, central space scheme is      u          j              n      +      1        =  −  γ      u          j      −      1              n        +      u          j              n        +  γ      u          j      +      1              n      where   γ  =  Δ  t      /    (  2  Δ  x  ). Writing the first few terms out (for   j  =  1  ,  .  .  ,  M  −  1) yields                              u                      1                                n            +            1                                              =        −        γ                  u                      0                                n                          +                  u                      1                                n                          +        γ                  u                      2                                n                                                        u                      2                                n            +            1                                              =        −        γ                  u                      1                                n                          +                  u                      2                                n                          +        γ                  u                      3                                n                                              ⋮                                      u                      M            −            1                                n            +            1                                              =        −        γ                  u                      M            −            2                                n                          +                  u                      M            −            1                                n                          +        γ                  u                      M                                n                              which we can write as a matrix                    (1)                              (                                                                      u                                      1                                                        n                    +                    1                                                                                                                        u                                      2                                                        n                    +                    1                                                                                                                        u                                      3                                                        n                    +                    1                                                                                                      ⋮                                                                                      u                                      M                    −                    1                                                        n                    +                    1                                                                                )                =                  (                                                    1                                            γ                                                                                                              −                γ                                            1                                            γ                                                                                                −                γ                                            1                                            γ                                                                                                                                                                  −                γ                                            1                                              )                          (                                                                      u                                      1                                                        n                                                                                                                        u                                      2                                                        n                                                                                                                        u                                      3                                                        n                                                                                                      ⋮                                                                                      u                                      M                    −                    1                                                        n                                                                                )                +                  (                                                    −                γ                                  u                                      0                                                        n                                                                                                      0                                                                    0                                                                    ⋮                                                                    γ                                  u                                      M                                                        n                                                                                )                    To answer your question &#039;how would we calculate   v  (      x          0        ,      t          1        )?&#039;, periodic boundary conditions imply       u          0              n        =      u          M      −      1              n       and       u          1              n        =      u          M              n       for all   n (time). Therefore, we can rewrite the matrix equation (1) as                    (2)                              (                                                                      u                                      1                                                        n                    +                    1                                                                                                                        u                                      2                                                        n                    +                    1                                                                                                                        u                                      3                                                        n                    +                    1                                                                                                      ⋮                                                                                      u                                      M                    −                    1                                                        n                    +                    1                                                                                )                =                  (                                                    1                                            γ                                                                        −                γ                                                                    −                γ                                            1                                            γ                                                                                                −                γ                                            1                                            γ                                                                                                                        γ                                                                        −                γ                                            1                                              )                          (                                                                      u                                      1                                                        n                                                                                                                        u                                      2                                                        n                                                                                                                        u                                      3                                                        n                                                                                                      ⋮                                                                                      u                                      M                    −                    1                                                        n                                                                                )                    We also know that the initial condition   u  (  x  ,  0  )  =  f  (  x  )    ⟹        u          j              0        =      f          j       where       f          j        =  f  (      x          j        ) is   f  (  x  ) evaluated at the spatial grid points       x          j      . So for the first iteration we need to solve      (                                        u                          1                                      1                                                                        u                          2                                      1                                                                        u                          3                                      1                                                            ⋮                                                  u                          M              −              1                                      1                                            )    =      (                            1                          γ                                          −          γ                                      −          γ                          1                          γ                                                      −          γ                          1                          γ                                                                  γ                                          −          γ                          1                      )        (                                        f                          1                                                                        f                          2                                                                        f                          3                                                            ⋮                                                  f                          M              −              1                                            )  and thereafter we solve (2).

Laila Murphy · Accepted Answer

When you rewrite your equation                    (1)                                          v            (                          x                              j                                      ,                          t                              n                +                1                                      )            −            v            (                          x                              j                                      ,                          t                              n                                      )                                Δ            t                          =                              v            (                          x                              j                +                1                                      ,                          t                              n                                      )            −            v            (                          x                              j                −                1                                      ,                          t                              n                                      )                                2            Δ            x                              you get this                    (2)                    v        (                  x                      j                          ,                  t                      n            +            1                          )        −        v        (                  x                      j                          ,                  t                      n                          )        =                              Δ            t            (            v            (                          x                              j                +                1                                      ,                          t                              n                                      )            −            v            (                          x                              j                −                1                                      ,                          t                              n                                      )            )                                2            Δ            x                                                  (3)                    v        (                  x                      j                          ,                  t                      n            +            1                          )        =        v        (                  x                      j                          ,                  t                      n                          )        +                              Δ            t            (            v            (                          x                              j                +                1                                      ,                          t                              n                                      )            −            v            (                          x                              j                −                1                                      ,                          t                              n                                      )            )                                2            Δ            x                              you said we have the initial condition where       t          n        =  0 . The grid is only between   0  ≤  x  ≤  1 then you discretize it so       x          j      −      1       is the first point in the array. Forward euler is the most basic runge kutte method.

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