Let f ( t ) = α e − β t , where α , β are constantsLet g ( t ) = y ( t )Then the resulting convolution f ∗ g is: f ∗ g = ∫ 0 t α e − β ( t − τ ) y ( τ ) d τDoes anyone know how one would take the derivative of this expression?In general, are there rules for taking derivative of f ∗ g, for some given f , g?

Name: Let f ( t ) = α e − β t , where α , β are constantsLet g ( t ) = y ( t )Then the resulting convolution f ∗ g is: f ∗ g = ∫ 0 t α e − β ( t − τ ) y ( τ ) d τDoes anyone know how one would take the derivative of this expression?In general, are there rules for taking derivative of f ∗ g, for some given f , g?
Uploaded: 2022-02-23T17:22:57+00:00
Description: Let f ( t ) = α e − β t , where α , β are constantsLet g ( t ) = y ( t )Then the resulting convolution f ∗ g is: f ∗ g = ∫ 0 t α e − β ( t − τ ) y ( τ ) d τDoes anyone know how one would take the derivative of this expression?In general, are there rules for taking derivative of f ∗ g, for some given f , g?

Question

Let   f  (  t  )  =  α      e          −      β      t      , where   α  ,  β are constantsLet   g  (  t  )  =  y  (  t  )Then the resulting convolution   f  ∗  g is:  f  ∗  g  =      ∫    0    t    α      e          −      β      (      t      −      τ      )        y  (  τ  )  d  τDoes anyone know how one would take the derivative of this expression?In general, are there rules for taking derivative of   f  ∗  g, for some given   f  ,  g?

cismadmec · Accepted Answer

It suffices that one of the two functions is in       L    1   and the other is in       L          ∞        ∩      C    1   and its derivative is in       L          ∞      . More generally:If   f  ∈      L    1    (            R        d    ) and   g  ∈      C    k    (            R        d    ), such that       D          α        g  ∈      L          ∞        (            R        d    ) for all       |    α      |    ≤  k, then   f  ⋆  g  ∈      C    k    (            R        d    ) and:       D          α        (  f  ⋆  g  )  =  f  ⋆      D          α        gWhere   α  =  (      α    1    ,  …  ,      α    d    )  ∈            N        d  ,       |    α      |    =      α    1    +  ⋯  +      α    d  , and:      D          α        g  :=                    ∂                              |                    α                      |                              g              ∂              x        1                              α            1                              ⋯      ∂              x        d                              α            d                              This can be proved by induction on k. To prove the case k=1, we differentiate under the integral sign.

Yaretzi Mcconnell · Accepted Answer

If f and g are             C        1   and       f    ′    ,      g    ′   are both integrable, then:  (  f  ∗  g      )    ′    =      f    ′    ∗  g  =  f  ∗      g    ′    .Basically, whenever       f    ′    ∗  g and   f  ∗      g    ′   are well-defined, one has the equality above.

f g=int_0^(t) alpha e^(-beta(t−tau)) y(tau)d tau. Does anyone know how one would take the derivative of this expression? In general, are there rules for taking derivative of f g, for some given f,g?

Answered question

Answer & Explanation

New Questions in Differential Equations