I know that we can find the first weak derivative for f ( x ) = | x |, which is w ( x ) = { − 1 x &lt; 0 0 x = 0 1 x &gt; 0 .But why is there no second weak derivative? If I try to find the first weak derivative of w ( x ), I get that ∫ R v ( x ) ψ ( x ) d x = − ∫ R w ( x ) ψ ′ ( x ) d x = 2 ψ ( 0 ) .What does this mean? Why is there no v such that D w = v? I feel like I'm missing something and I'm not sure how to interpret this result.

Question

I know that we can find the first weak derivative for   f  (  x  )  =  |  x  |, which is  w  (  x  )  =      {                            −          1                          x          &amp;lt;          0                                      0                          x          =          0                                      1                          x          &amp;gt;          0                          .But why is there no second weak derivative? If I try to find the first weak derivative of   w  (  x  ), I get that      ∫                  R              v  (  x  )  ψ  (  x  )    d  x  =  −      ∫                  R              w  (  x  )      ψ    ′    (  x  )    d  x  =  2  ψ  (  0  )  .What does this mean? Why is there no v such that   D  w  =  v? I feel like I&#039;m missing something and I&#039;m not sure how to interpret this result.

Lisbonaid · Accepted Answer

The notion of weak derivative is a subset of the notion of distribution derivative. There is indeed a distribution second derivative to this function and congratulations, you found it! It is 2      δ    0   where       δ    0   is the Dirac mass which verifies      ⟨          δ      0        ,    ψ    ⟩    =  ψ  (  0  )  .You should have a look on some reference on distribution theory, which contains your weak derivative theory.

I know that we can find the first weak derivative for f ( x ) = <mo fence="false"

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