How to solve coupled linear 1st order PDEIt is fairly straight forward to solve linear 1st order PDEs by the method of characteristics. For example, if ∂ t f + a ∂ x f = b fwe have that d f d t = b f on the characteristic curve of d x d t = a. From this we deduce that f ( t , x ) = g ( C ) e b t where x = a t + C.Now, how does this work when f is multidimensional. Can I solve equations on the following form by characteristics, or by any other means? ∂ t f i ( t , x ) + ∑ j A i j ∂ x f j ( t , x ) = ∑ j B i j f j ( t , x )where the components of A and B might be dependent on x and t.In particular, I am trying to solve the following, { ∂ t f + c t ∂ x g = − ( a + 1 t ) f ∂ t g + c t ∂ x f = − ( b + 1 t ) g where f and g are functions of x and t , where t &gt; t 0 &gt; 0, c ≠ 0. Any help is highly appreciated.

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How to solve coupled linear 1st order PDEIt is fairly straight forward to solve linear 1st order PDEs by the method of characteristics. For example, if      ∂    t    f  +  a      ∂    x    f  =  b  fwe have that                     d        f                    d        t              =  b  f on the characteristic curve of                     d        x                    d        t              =  a. From this we deduce that   f  (  t  ,  x  )  =  g  (  C  )      e          b      t       where   x  =  a  t  +  C.Now, how does this work when   f is multidimensional. Can I solve equations on the following form by characteristics, or by any other means?      ∂    t        f    i    (  t  ,  x  )  +      ∑    j        A          i      j            ∂    x        f    j    (  t  ,  x  )  =      ∑    j        B          i      j            f    j    (  t  ,  x  )where the components of   A and   B might be dependent on   x and   t.In particular, I am trying to solve the following,      {                                        ∂            t                    f          +                                    c              t                                            ∂            x                    g          =          −                      (            a            +                                          1                t                                      )                    f                                                  ∂            t                    g          +                                    c              t                                            ∂            x                    f          =          −                      (            b            +                                          1                t                                      )                    g                        where   f and   g are functions of   x and   t , where   t  &amp;gt;      t    0    &amp;gt;  0,   c  ≠  0. Any help is highly appreciated.

Haleigh Vega · Accepted Answer

{                                        ∂            t                    f          +                                    c              t                                            ∂            x                    g          =          −                      (            a            +                                          1                t                                      )                    f                     .          .          .          .          .          .          (          1          )                                                  ∂            t                    g          +                                    c              t                                            ∂            x                    f          =          −                      (            b            +                                          1                t                                      )                    g                     .          .          .          .          .          .          (          2          )                        From   (  1  ),      ∂    t    f  +            c      t            ∂    x    g  =  −      (    a    +                  1        t              )    f            c      t            ∂    x    g  =  −      ∂    t    f  −      (    a    +                  1        t              )    f      ∂    x    g  =  −            t      c            ∂    t    f  −      (                            a          t                c              +                  1        c              )    f     .  .  .  .  .  .  (  3  )      ∂          x      t        g  =  −            t      c            ∂          t      t        f  −            1      c            ∂    t    f  −      (                            a          t                c              +                  1        c              )        ∂    t    f  −            a      c        f      ∂          x      t        g  =  −            t      c            ∂          t      t        f  −      (                            a          t                c              +                  2        c              )        ∂    t    f  −            a      c        f     .  .  .  .  .  .  (  4  )From   (  2  ),      ∂    t    g  +            c      t            ∂    x    f  =  −      (    b    +                  1        t              )    g      ∂          x      t        g  +            c      t            ∂          x      x        f  =  −      (    b    +                  1        t              )        ∂    x    g     .  .  .  .  .  .  (  5  )Put   (  3  ) and   (  4  ) into   (  5  ),  (  5  )  −            t      c            ∂          t      t        f  −      (                            a          t                c              +                  2        c              )        ∂    t    f  −            a      c        f  +            c      t            ∂          x      x        f  =  −      (    b    +                  1        t              )        (    −                  t        c                    ∂      t        f    −          (                                    a            t                    c                    +                        1          c                    )        f    )    −            t      c            ∂          t      t        f  −      (                            a          t                c              +                  2        c              )        ∂    t    f  −            a      c        f  +            c      t            ∂          x      x        f  =      (                            b          t                c              +                  1        c              )        ∂    t    f  +      (                            a          b          t                c              +                            a          +          b                c              +                  1                  c          t                      )    f            c      t            ∂          x      x        f  =            t      c            ∂          t      t        f  +        (                      (        a        +        b        )        t            c        +            3      c              )        ∂    t    f  +      (                            a          b          t                c              +                            2          a          +          b                c              +                  1                  c          t                      )    f      ∂          x      x        f  =                    t        2                    c        2                  ∂          t      t        f  +                    (        a        +        b        )                  t          2                +        3        t                    c        2                  ∂    t    f  +                    a        b                  t          2                +        (        2        a        +        b        )        t        +        1                    c        2              f

How to solve coupled linear 1st order PDE It is fairly straight forward to solve linear 1st order P

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