How come ∇ 2 f = 1 r 2 ∂ ∂ r ( r 2 ∂ f ∂ r ) yet f = ∫ ∫ f ″ d r d r?In spherical coordinate, suppose f dependents only on r. Then ∇ 2 f = 1 r 2 ∂ ∂ r ( r 2 ∂ f ∂ r ). However, to integral f′′ back the f, ∫ ∫ f ″ d r d rHow could that be?

Question

How come       ∇    2    f  =      1          r      2            ∂          ∂      r        (      r    2              ∂      f              ∂      r        ) yet   f  =  ∫  ∫      f    ″    d  r  d  r?In spherical coordinate, suppose f dependents only on r. Then       ∇    2    f  =      1          r      2            ∂          ∂      r        (      r    2              ∂      f              ∂      r        ). However, to integral f′′ back the f,   ∫  ∫      f    ″    d  r  d  rHow could that be?

Tristian Clayton · Accepted Answer

It&#039;s not true that       ∇    2    f  (  r  )  =      f    ″    (  r  ) in 3D. In 1D we do have       ∇    2    f  (  x  )  =                    ∂        2            f      (      x      )              ∂              x        2              =      f    ″    (  x  ) and even if spherical symmetry is effectively 1D it&#039;s not as simple as to assume the 1D Laplacian with r instead of x (something changing radially is fundamentally different than something changing along one of the axes). The definition of the Laplacian as (here in 3D)      ∇    2    f  ≡                    ∂        2            f              ∂              x        2              +                    ∂        2            f              ∂              y        2              +                    ∂        2            f              ∂              z        2            is for Cartesian coordinates   (  x  ,  y  ,  z  ). If you start from this and change to spherical coordinates   (  r  ,  θ  ,  ϕ  ) and assume f only depends on r then you will get the expression you have in your question      ∇    2    f  (  r  )  =      f    ″    (  r  )  +      2    r        f    ′    (  r  )Try to derive this using the chain-rule             ∂      f      (      r      )              ∂      x        =      x    r              ∂      f      (      r      )              ∂      r       and so on. If you manage this then you can try to generalize it by doing it in n dimensions (for which the result is the same as above with the 2 above becoming n−1).Secondly if you have any (two times continuously differentiable) function of a single variable then we always have &quot;  f  (  z  )  =  ∬      f    ″    (  z  )      d    z      d    z&quot;, or more precisely  f  (  z  )  =  f  (  0  )  +      f    ′    (  0  )  z  +      ∫    0    z        (          ∫      0                        z          ′                            f      ″        (    w    )          d        w    )        d        z    ′  This expression doesn&#039;t care about which kind of coordinates it is, it&#039;s simply a consequence of the fundamental theorem of calculus.

How come grad^2 f=(1)/(r^2)(del)/(del r)(r^2 (del f)/(del r)) yet f=int int f"drdr?

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