Given the relations: F 1 ( x , y , u , v ) = u e y + ( y + 1 ) e v x + x y + 1 = 0 F 2 ( x , y , u , v ) = x + y + x u + e y + v − 1 = 0 find the linear approximation of ( u , v ) in terms of ( x , y ) near ( x , y , u , v ) = ( 0 , 0 , − 2 , 0 ).

Name: Given the relations: F 1 ( x , y , u , v ) = u e y + ( y + 1 ) e v x + x y + 1 = 0 F 2 ( x , y , u , v ) = x + y + x u + e y + v − 1 = 0 find the linear approximation of ( u , v ) in terms of ( x , y ) near ( x , y , u , v ) = ( 0 , 0 , − 2 , 0 ).
Uploaded: 2022-05-06T15:56:33+00:00
Description: Given the relations: F 1 ( x , y , u , v ) = u e y + ( y + 1 ) e v x + x y + 1 = 0 F 2 ( x , y , u , v ) = x + y + x u + e y + v − 1 = 0 find the linear approximation of ( u , v ) in terms of ( x , y ) near ( x , y , u , v ) = ( 0 , 0 , − 2 , 0 ).

Question

Given the relations:                                    F          1                (        x        ,        y        ,        u        ,        v        )        =        u                  e          y                +        (        y        +        1        )                  e                      v            x                          +        x        y        +        1        =        0                                            F          2                (        x        ,        y        ,        u        ,        v        )        =        x        +        y        +        x        u        +                  e                      y            +            v                          −        1        =        0            find the linear approximation of   (  u  ,  v  ) in terms of   (  x  ,  y  ) near   (  x  ,  y  ,  u  ,  v  )  =  (  0  ,  0  ,  −  2  ,  0  ).

arbotsck8sg · Accepted Answer

First of all you have to adapt your notation to the multivariable case. Since your (implicit) function   (  u  ,  v  )  =  f  (  x  ,  y  ) is   f  :            R        2    →            R        2   the linear approximation is  f  (  x  ,  y  )  ≈  f  (  a  ,  b  )  +      J    f    (  a  ,  b  )  ⋅  (  x  −  a  ,  y  −  b      )    ⊤  where       J    f   is the jacobian matrix, i.e. the matrix of partial derivatives:      J    f    =      (                                                                      ∂                u                                            ∂                x                                                                                                        ∂                u                                            ∂                y                                                                                                                    ∂                v                                            ∂                x                                                                                                        ∂                v                                            ∂                y                                                          )  Now, the point is that partial derivatives can be obtained from the implicit function theorem without explicitly writing   (  u  ,  v  ) as a function of   (  x  ,  y  ). So, differentiating both equations with respect to   x gives:                                          ∂                          F              1                                            ∂            x                                              =                  e          y                                                            ∂              u                                      ∂              x                                      +        v                (        y        +        1        )                          e                      v            x                          +        x                (        y        +        1        )                          e                      v            x                                                                      ∂              v                                      ∂              x                                      +        y        =        0                                                  ∂                          F              2                                            ∂            x                                              =        1        +        u        +        x                                                    ∂              u                                      ∂              x                                      +                  e                      y            +            v                                                                      ∂              v                                      ∂              x                                      =        0            and you can obtain both partial derivatives (observe that now the system is linear). Repeating the process with   y you obtain the other two. Finally, substituting for the point with the given values of   (  u  ,  v  ,  x  ,  y  ) gives the answer.

Given the relations: <mtable columnalign="right left right left right left right left right lef

Answered question

Answer & Explanation

New Questions in Differential Equations