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Derivative of a Convolution

by Shaybay92
Tags: convolution, derivative
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Shaybay92
#1
Sep2-12, 01:17 AM
P: 122
Hi,

I want to verify that the form of a particular solution satisfies the following ODE:

v' + (b/m)v = u/m

with

vpart= ∫e-(b/m)(t-r) (u(r)/m) dr

where the limits are from 0 to t

So I tried to differentiate v with respect to t, in order to substitute it back into the equation. But, how do you do that when the integral is with respect to r? Is there a need to change variables? How can you do this?

Cheers
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haruspex
#2
Sep2-12, 07:16 PM
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Quote Quote by Shaybay92 View Post
Hi,

I want to verify that the form of a particular solution satisfies the following ODE:

v' + (b/m)v = u/m

with

vpart= ∫e-(b/m)(t-r) (u(r)/m) dr

where the limits are from 0 to t. So I tried to differentiate v with respect to t,... How can you do this?
v(t) = ∫tf(t, r).dr
v(t+δt)= ∫t+δtf(t+δt, r).dr
= ∫tf(t+δt, r).dr + ∫tt+δtf(t+δt, r).dr
So v' = ∫t(d/dt)f(t, r).dr + f(t, t)
Shaybay92
#3
Nov8-12, 04:43 PM
P: 122
Sorry I'm not familiar with your method. I don't understand why you substitute "t+δt" for t. What approach are you using here? Could you elaborate or direct me to some further reading?

Cheers :)

haruspex
#4
Nov8-12, 06:50 PM
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Derivative of a Convolution

Quote Quote by Shaybay92 View Post
Sorry I'm not familiar with your method. I don't understand why you substitute "t+δt" for t. What approach are you using here? Could you elaborate or direct me to some further reading?

Cheers :)
The equation you posted for v(t) is generic - i.e. it's true for all t. So it's true both for a given t and for a later time t+δt. So you can write a second equation substituting t+δt for t consistently. Taking the difference, diving by δt, then letting δt tend to zero gives you v'. That is the standard process of differentiation.
Shaybay92
#5
Nov8-12, 06:54 PM
P: 122
Oh I see what you mean. Thanks for the clarification. I'm just not use to this notation :)


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