Given that u = arctan ⁡ ( x 3 + y 3 x − y ) , prove the following: x 2 ∂ 2 u ∂ x 2 + 2 x y ∂ 2 u ∂ x ∂ y + y 2 ∂ 2 u ∂ y 2 = ( 1 − 4 sin 2 ⁡ u ) sin ⁡ ( 2 u )Attempted incomplete solution: tan ⁡ ( u ) = x 3 + y 3 x − y = f (say) We note that f is a homogeneous function in x , y of degree 2 and hence, by a general result of Euler's Theorem, we have, x 2 ∂ 2 f ∂ x 2 + 2 x y ∂ 2 f ∂ x ∂ y + y 2 ∂ 2 f ∂ y 2 = 2 ( 2 − 1 ) f = 2 tan ⁡ ( u )

Question

Given that   u  =  arctan  ⁡      (                            x          3                +                  y          3                            x        −        y              )  , prove the following:      x    2                      ∂        2            u              ∂              x        2              +  2  x  y                    ∂        2            u              ∂      x      ∂      y        +      y    2                      ∂        2            u              ∂              y        2              =  (  1  −  4      sin    2    ⁡  u  )  sin  ⁡  (  2  u  )Attempted incomplete solution:  tan  ⁡  (  u  )  =                    x        3            +              y        3                    x      −      y        =  f         (say)  We note that   f is a homogeneous function in   x  ,  y of degree   2 and hence, by a general result of Euler&#039;s Theorem, we have,      x    2                      ∂        2            f              ∂              x        2              +  2  x  y                    ∂        2            f              ∂      x      ∂      y        +      y    2                      ∂        2            f              ∂              y        2              =  2  (  2  −  1  )  f  =  2  tan  ⁡  (  u  )

Kimberly Evans · Accepted Answer

Here is an outline of one way forward.Designate the argument of the arctangent by the new variable   t so that  t  (  x  ,  y  )  =                    x        3            +              y        3                    x      −      y      Therefore, we can write   u  (  t  (  x  ,  y  )  )  =  arctan  ⁡  (  t  (  x  ,  y  )  ). Then, we have                    (1)                    sin        ⁡        u                            =                  t                      1            +                          t              2                                                                        (2)                    sin        ⁡        (        2        u        )                            =                              2            t                                1            +                          t              2                                                                        (3)                              u          ′                (        t        )                            =                  1                      1            +                          t              2                                                                        (4)                              u          ″                (        t        )                            =                              −            2            t                                (            1            +                          t              2                                      )              2                                          And from the Chain Rule, we have                                          ∂            u                                ∂            x                                              =                  u          ′                (        t        )                              ∂            t                                ∂            x                                                                                  ∂            u                                ∂            y                                              =                  u          ′                (        t        )                              ∂            t                                ∂            y                                                            (5)                                                        ∂              2                        u                                ∂                          x              2                                                          =                  u          ″                (        t        )                              (                                          ∂                t                                            ∂                x                                      )                    2                +                  u          ′                (        t        )                                            ∂              2                        t                                ∂                          x              2                                                                        (6)                                                        ∂              2                        u                                ∂                          y              2                                                          =                  u          ″                (        t        )                              (                                          ∂                t                                            ∂                y                                      )                    2                +                  u          ′                (        t        )                                            ∂              2                        t                                ∂                          y              2                                                                        (7)                                                        ∂              2                        u                                ∂            x            ∂            y                                              =                  u          ″                (        t        )                  (                                    ∂              t                                      ∂              x                                                          ∂              t                                      ∂              y                                )                +                  u          ′                (        t        )                                            ∂              2                        t                                ∂            x            ∂            y                                            Using   (  5  )  −  (  7  ) and the general result of Euler&#039;s Theorem, we have      u    ″    (  t  )            (      x                        ∂          t                          ∂          x                    +      y                        ∂          t                          ∂          y                    )        2    +  2  t      u    ′    (  t  )  =  (  1  −  4      sin    2    ⁡  (  u  )  )  sin  ⁡  (  2  u  )Now, finish by calculating the partial derivatives of   t with respect to   x and   y and using   (  1  )  −  (  4  ).

varice2r · Accepted Answer

Specifically, your statement   tan  ⁡  u  =  f is clearer when written  tan  ⁡  u  (  x  ,  y  )  =  f  (  x  ,  y  )  .Now you can take the derivative of both sides with respect to   x.      sec    2    ⁡  (  u  (  x  ,  y  )  )            ∂      u              ∂      x        (  x  ,  y  )  =            ∂      f              ∂      x        (  x  ,  y  )  .Differentiating again gives  2      sec    2    ⁡  (  u  (  x  ,  y  )  )  tan  ⁡  (  u  (  x  ,  y  )  )            (                        ∂          u                          ∂          x                    (      x      ,      y      )      )        2    +      sec    2    ⁡  (  u  (  x  ,  y  )  )                    ∂        2            u              ∂              x        2              (  x  ,  y  )  =                    ∂        2            f              ∂              x        2              (  x  ,  y  )  .It gets a little messy with all the dependencies, especially on the derivatives, so you can drop them when you feel comfortable doing so.

Given that u=arctan((x^3+y^3)/(x−y)), prove the following: x^2 d^2u/dx^2+2xy d^2u/dxdy+y^2 d^2u/dy^2=(1-4sin^2usin(2u)

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