How to solve these two second-order coupled PDE? 0 = A f ( x , y ) + B ∂ 2 f ( x , y ) ∂ x 2 + C ∂ 2 f ( x , y ) ∂ y 2 + D ∂ 2 g ( x , y ) ∂ x ∂ y 0 = A g ( x , y ) + B ∂ 2 g ( x , y ) ∂ y 2 + C ∂ 2 g ( x , y ) ∂ x 2 + D ∂ 2 f ( x , y ) ∂ x ∂ y

Question

How to solve these two second-order coupled PDE?  0  =  A  f  (  x  ,  y  )  +  B                    ∂        2            f      (      x      ,      y      )              ∂              x        2              +  C                    ∂        2            f      (      x      ,      y      )              ∂              y        2              +  D                    ∂        2            g      (      x      ,      y      )              ∂      x      ∂      y          0  =  A  g  (  x  ,  y  )  +  B                    ∂        2            g      (      x      ,      y      )              ∂              y        2              +  C                    ∂        2            g      (      x      ,      y      )              ∂              x        2              +  D                    ∂        2            f      (      x      ,      y      )              ∂      x      ∂      y

mar1nerne · Accepted Answer

(1)                    0        =        A        f        (        x        ,        y        )        +        B                                                            ∂                2                            f              (              x              ,              y              )                                      ∂                              x                2                                                    +        C                                                            ∂                2                            f              (              x              ,              y              )                                      ∂                              y                2                                                    +        D                                                            ∂                2                            g              (              x              ,              y              )                                      ∂              x              ∂              y                                      ,                                (2)                    0        =        A        g        (        x        ,        y        )        +        B                                                            ∂                2                            g              (              x              ,              y              )                                      ∂                              x                2                                                    +        C                                                            ∂                2                            g              (              x              ,              y              )                                      ∂                              y                2                                                    +        D                                                            ∂                2                            f              (              x              ,              y              )                                      ∂              x              ∂              y                                      ;            Having said the above, (1) and (2) can be uncoupled as follows set:                    (3)                    U        (        x        ,        y        )        =        f        (        x        ,        y        )        +        g        (        x        ,        y        )        ,                                (4)                    V        (        x        ,        y        )        =        f        (        x        ,        y        )        −        g        (        x        ,        y        )        ;            if we now add (1) and (2) we obtain  0  =  A  (  f  (  x  ,  y  )  +  g  (  x  ,  y  )  )  +  B                              ∂          2                (        f        (        x        ,        y        )        +        g        (        x        ,        y        )        )                    ∂                  x          2                      +  C                              ∂          2                (        f        (        x        ,        y        )        +        g        (        x        ,        y        )        )                    ∂                  y          2                      +  D                              ∂          2                (        f        (        x        ,        y        )        +        g        (        x        ,        y        )        )                    ∂        x        ∂        y                                (5)                    =        A        U        (        x        ,        y        )        +        B                                                            ∂                2                            U              (              x              ,              y              )                                      ∂                              x                2                                                    +        C                                                            ∂                2                            U              (              x              ,              y              )                                      ∂                              y                2                                                    +        D                                                            ∂                2                            U              (              x              ,              y              )                                      ∂              x              ∂              y                                      ,            that is,                    (6)                    A        U        (        x        ,        y        )        +        B                                                            ∂                2                            U              (              x              ,              y              )                                      ∂                              x                2                                                    +        C                                                            ∂                2                            U              (              x              ,              y              )                                      ∂                              y                2                                                    +        D                                                            ∂                2                            U              (              x              ,              y              )                                      ∂              x              ∂              y                                      =        0        ,            and similarly, subtracting yields                    (7)                    A        V        (        x        ,        y        )        +        B                                                            ∂                2                            V              (              x              ,              y              )                                      ∂                              x                2                                                    +        C                                                            ∂                2                            V              (              x              ,              y              )                                      ∂                              y                2                                                    −        D                                                            ∂                2                            V              (              x              ,              y              )                                      ∂              x              ∂              y                                      =        0        ,            where the sign of   D is negative to accomodate the fact that subtraction is &quot;asymmetric&quot;:   g  −  f  =  −  (  f  −  g  ). (6) and (7) are decoupled, and may be solved seperately, and then   f  (  x  ,  y  ) and   g  (  x  ,  y  ) may be recovered from                    (8)                    f        (        x        ,        y        )        =                              1            2                          (        U        (        x        ,        y        )        +        V        (        x        ,        y        )        )        ,                                (9)                    g        (        x        ,        y        )        =                              1            2                          (        U        (        x        ,        y        )        −        V        (        x        ,        y        )        )        .            Of course the above does not address the issues of boundary conditions, well-posedness, etc., but I think the boundary conditions for   U  (  x  ,  y  ),   V  (  x  ,  y  ) may follow a pattern similar to (3), (4). Worth looking at, though.

How to solve these two second-order coupled PDE? 0 = A f ( x , y ) +

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