Partial Derivative of Convolution

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
2 replies · 3K views
jstop
Messages
2
Reaction score
1
TL;DR
Calculating the partial derivative of a convolution
Hello, I am trying to calculate the partial derivative of a convolution. This is the expression:
##\frac{\partial}{\partial r}(x(t) * y(t, r))##​

Only y in the convolution depends on r. I know this identity below for taking the derivative of a convolution with both of the functions only depending on t:
##\frac{d}{dt}(x(t)*y(t) = (\frac{dx(t)}{dt}*y(t)) = (x(t)*\frac{dy(t)}{dt})##​

I'm not sure how this changes when taking a partial derivative with only one of the functions depending on the variable that the partial derivative is being taken with respect to. Any help/resources would be appreciated! (This is my first post here, please let me know if I need to improve my post formatting/structure).
 
on Phys.org
Your convolution is [itex](x * y)(t,r)=\int_{-\infty}^{\infty} x(\tau)y(t - \tau, r)\,d\tau[/itex], is it not? What happens if you differentiate that with respect to [itex]r[/itex]?
 
Reply
  • Like
Likes   Reactions: Twigg
Ah thanks, I think I see what you mean:
##\frac{\partial}{\partial r} (x * y)(t,r) = \int_{-\infty}^{\infty} x(\tau)\frac{\partial}{\partial r}y(t - \tau, r)d\tau = (x*\frac{\partial y}{\partial r})(t,r)##​
 
Reply
  • Like
Likes   Reactions: Twigg