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Deriving of Duhamel Theorem

Deriving of Duhamel-theorem will be executed in the followings.

Let’s start from the weak derivative of convolution

$                      y(t) = \frac{\delta }{\delta t} \left \[ f(t)\ast h(t) \right \]

Apply the following denoting:
$                      f(t)=1(t)f(t) ;\quad h(t)=1(t)h(t)  ;\quad  f(t-\tau)=1(t-\tau)f(t-\tau) ;\quad  h(t-\tau)=1(t-\tau)h(t-\tau)

Let’s describe the convolution formula in the argument of weak derivative:

$                      y(t) = \frac{\delta }{\delta t}\int_0^tf(\tau)h(t-\tau)1(t-\tau)d\tau=\int_0^tf(\tau) \frac{\delta }{\delta t}\left [ h(t-\tau)1(t-\tau) \right ] d\tau

$                      y(t)= \int_0^tf(\tau) \left [ h(t-\tau) \frac{\delta 1(t-\tau)}{\delta t}+ 1(t-\tau) \frac{dh(t-\tau)}{dt} \right ] d\tau
$                      y(t)= \int_0^tf(\tau) \left [ \delta(t-\tau)h(t-\tau) +1(t-\tau) \frac{dh(t-\tau)}{dt} \right ] d\tau

$                      y(t)= \int_0^tf(\tau) \delta(t-\tau)h(t-\tau) d\tau + \int_0^tf(\tau) 1(t-\tau) \frac{dh(t-\tau)}{dt} d\tau
If $                      t>0 $                 \int_0^tf(\tau) \delta(t-\tau)h(t-\tau) d\tau=f(t)h(0) , and
$                   \int_0^tf(\tau) 1(t-\tau) \frac{dh(t-\tau)}{dt} d\tau=\int_0^tf(\tau) \frac{dh(t-\tau)}{dt} d\tau , where $                      0<\tau<t
Now, it is possible to get the final expression of Duhamel-theorem
$                      y(t)= f(t)h(0)   +  \int_0^tf(\tau) \frac{dh(t-\tau)}{dt} d\tau = f(t)h(0)   +  \int_0^tf(\tau) \dot{h}(t-\tau) d\tau , where $                   \frac{d}{dt} means the conventional derivative, which is denoted by “$               \dot{ }” as usual.


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