Given u m ( x ) = 8 R 3 sin ⁡ π m 1 x 1 R sin ⁡ π m 2 x 2 R sin ⁡ π m 3 x 3 R with the boundary δ Ω = { x ∣ x 1 = R , 0 &lt; x 2 &lt; R , 0 &lt; x 3 &lt; R } and m is a vector with m = ( m 1 , m 2 , m 3 ) , m i ∈ N Then calculates the normale derivative ∂ ∂ n u m ( x ) on the boundary δ Ω and derives (1) ∂ ∂ n u m ( x ) = 8 π m 1 R 4 sin ⁡ π m 2 x 2 R sin ⁡ π m 3 x 3 R I'm very new to the topic of normal derivatives and I've never seen it calculated on "strange" boundaries like δ Ω. Could someone briefly describe how we can derive (1)?

Question

Given      u    m    (  x  )  =      8          R      3          sin  ⁡            π              m        1                    x        1              R      sin  ⁡            π              m        2                    x        2              R      sin  ⁡            π              m        3                    x        3              R  with the boundary   δ  Ω  =  {  x  ∣      x    1    =  R  ,  0  &amp;lt;      x    2    &amp;lt;  R  ,  0  &amp;lt;      x    3    &amp;lt;  R  } and m is a vector with   m  =  (      m    1    ,      m    2    ,      m    3    )  ,      m    i    ∈      N  Then calculates the normale derivative       ∂          ∂      n            u    m    (  x  ) on the boundary   δ  Ω and derives                    (1)                              ∂                      ∂            n                                    u          m                (        x        )        =                              8            π                          m              1                                            R            4                                  sin        ⁡                              π                          m              2                                      x              2                                R                        sin        ⁡                              π                          m              3                                      x              3                                R                    I&#039;m very new to the topic of normal derivatives and I&#039;ve never seen it calculated on &quot;strange&quot; boundaries like   δ  Ω. Could someone briefly describe how we can derive (1)?

Brogan Navarro · Accepted Answer

u      m        (    x    )    =          8              R        3                  sin    ⁡                  π                  m          1                          x          1                    R            sin    ⁡                  π                  m          2                          x          2                    R            sin    ⁡                  π                  m          3                          x          3                    R            ∇          u      m        (    x    )    =                  (                            8        π                  m          1                            R        4                  cos    ⁡                  π                  m          1                          x          1                    R            sin    ⁡                  π                  m          2                          x          2                    R            sin    ⁡                  π                  m          3                          x          3                    R        ,                  ∂                  u          m                (        x        )                    ∂                  x          2                      ,                  ∂                  u          m                (        x        )                    ∂                  x          3                                    )            The given boundary is   δ  Ω  =  {  x  ∣      x    1    =  R  ,  0  &amp;lt;      x    2    &amp;lt;  R  ,  0  &amp;lt;      x    3    &amp;lt;  R  }As you can see the boundary is a square in plane       x    1    =  R and the unit normal vector to this plane is either (1,0,0) or (−1,0,0).Please note that,                    ∂                  u          m                (        x        )                    ∂        n              =                  n        ^              ⋅    ∇          u      m        (    x    )    =    ±                  8        π                  m          1                            R        4                  cos    ⁡    π          m      1            sin    ⁡                  π                  m          2                          x          2                    R            sin    ⁡                  π                  m          3                          x          3                    R             (as       x    1    =  R)Given       m    1    ∈      N    ,  cos  ⁡  π      m    1    =  ±  1

Given u_m(x)=(8)/(R^3)sin (pi m_1 x_1)(R)sin(pi m_2 x_2)/(R) sin (pi m_3 x_3)(R) with the boundary delta Omega={x|x_1=R,0<x_2<R,0<x_3<R} and m is a vector with m=(m_1,m_2,m_3),m_i in NN

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