derivative of convolution integral

I'm confused for a derivation related to the derivative of convolution. Given that $$C_{im}(x,t)=\omega e^{-\omega t}*C_m(x,t)+C_{im}(x,0)e^{-\omega t}$$ By taking derivative of the above equation with respect to $t$, the result shown in paper is $$\frac{\partial C_{im}(x,t)}{\partial t}=\omega e^{-\omega t}*\frac{\partial C_m(x,t)}{\partial t}+\omega e^{-\omega t}C_m (x,0)-\omega e^{-\omega t}C_{im}(x,0)$$

where $C_m$ and $C_{im}$ are function of $x$ and $t$, and $C_m(x,0)$ and $C_{im}(x,0)$ are initial conditions. $*$ means convolution. I know the first and third term on the RHS of second equation are the derivatives of the two terms on the RHS of first equation, respectively. What I'm confused about is the term $\omega e^{-\omega t}C_m (x,0)$. Hope someone can help me figure it out. Thanks in advance.

• what is Cim(x,t) in case it matters? – Chill2Macht Jun 8 '16 at 0:25
• $C_{im}(x,t)$ is the concentration in immobile porous media. I have figured it out. Integration by parts for convolution is very tricky. I made a mistake when integrating for $\tau$ – kchen Jun 8 '16 at 3:06
• Would you mind posting your solution as an answer by the way? Otherwise I will vote to close the question – Chill2Macht Jun 8 '16 at 6:44
• This is the formula for derivative of convolution $$\frac{\partial c}{\partial t}*g=\int_0^t \frac{\partial c(t-\tau)}{\partial \tau}g(\tau) d\tau=c*\frac{\partial g}{\partial t}+cg_0-c_0g$$, where $c_0$ and $g_0$ are initial conditions. – kchen Jun 9 '16 at 20:13