Breaking up a Convolution Integral For a diffusion problem in a semi-infinite domain with a transient boundary condition, the temperature profile can be obtained from Duhamel's Principle as
$$
\theta(r,t)= \int_0^t \frac{\partial \phi}{\partial t}(r,t-\beta) f(\beta){\text d}\beta\tag{EQ. 1}
$$
in which $\phi(r,t)$ is the solution of the auxiliary problem (cf. Özışık, Chapter 5) with a time-invariant boundary condition, and is given by the inverse Laplace Transform
$$
\phi(r,t)=L^{-1}[\Phi(r,s)]
$$ 
where 
$$
\Phi(r,s)=\frac{1}{s}\frac{K_0(\sqrt{s}\hspace 0.1cm r)}{K_0(\sqrt{s})}
$$
and $K_0(s)$ is the Modified Bessel Function of the Second Kind, of Order $0$.
Now, if $f(t)$ which represents the transient variation of the boundary temperature is such that
$$
f(t) = f_{\text A} (t),  \hspace 1cm 0 \le t \le t_0\\
\hspace 0.2cm = f_{\text B} (t),  \hspace 1cm  t > t_0
$$
which satisfies $ f_{\text A} (t_0)= f_{\text B} (t_0)$, then the Duhamel Integral can be broken up as
$$
\theta(r,t)\!=\! 
\int_0^{t_0}\! \frac{\partial \phi}{\partial t}(r,t\!-\!\beta) f_{\text A}(\beta){\text d}\beta
\!+\!
\int_0^{t\!-\!t_0}\! \frac{\partial \phi}{\partial t}(r,t\!-\!t_0\!-\!\beta) f_{\text B}(\beta){\text d}\beta\tag{EQ. 2}
$$
where a variable shift has been used to cast the 2nd term on the RHS in a form amenable to the use of the Convolution Theorem upon Laplace transformation.
The problem is what to do with the first term on the RHS, for which the Convolution Theorem cannot be used.  The convolution kernel does not appear to lend itself to a form separable in $t$ and $t_0$ such that this term could be expressed as 
$$
\int_0^{t_0} \frac{\partial \phi}{\partial t}(r,t-\beta) f(\beta){\text d}\beta
=P(t)\int_0^{t_0} \frac{\partial \phi}{\partial t}(r,t_0-\beta)f(\beta){\text d}\beta
$$
which would solve the problem at once. Since this ins't the case however, I'm at an impasse. I do not see how I could get a tractable form of the First Term with Laplace Transformation, since the use of the Convolution Theorem is ruled out, and I'm not aware of any clever tricks to get around this.  
Any suggestions on a way forward would be deeply appreciated
A perfectly reasonable question to ask is why I don't just evaluate the Laplace Transform of the piecewise continuous function $f(t)$ and use the Convolution Theorem without breaking up the Duhamel Integral yielding
$$
\Theta(r,s)=L[\theta(r,t)]=\Phi(s)F(s)
$$
and invert the resulting transform. The reasons are


*

*This would only work for relatively trivial functions for which the Laplace Transform can be broken up and evaluated in closed form.

*(and more importantly), I don't know $f(t)$ a-priori. It evolves with the temperature profile as a consequence of a coupled system that in fact provides the boundary condition.
I hope this clarification helps to the problem in some more detail. If not, please let me know in the comments. 
 A: While I'm a bit disappointed that no one chimed in even with a comment, I would appreciate it if someone could at least comment on the answer that I've (hopefully correctly!) worked out
The term 
$$
\int_0^{t_0} \frac{\partial \phi}{\partial t}(r,t-\beta) f_{\text A}(\beta){\text d}\beta
$$
can be written as
$$
\int_0^{t_0} \frac{\partial \phi}{\partial t}(r,t-\beta) f_{\text A}(\beta){\text d}\beta
=
\int_0^{t} \frac{\partial \phi}{\partial t}(r,t-\beta) f_{\text A}(\beta){\text d}\beta
-
\int_{t_0}^{t} \frac{\partial \phi}{\partial t}(r,t-\beta) {f_{\text A}(\beta)}{\text d}\beta
$$
The second term on the RHS is zero since $f_A(t)$ is zero for all $t>t_0$. Therefore, substitution into EQ. 2 in the original post above gives the the temperature distribution as 
$$
\theta(r,t)= 
\int_0^{t} \frac{\partial \phi}{\partial t}(r,t-\beta) f_{\text A}(\beta){\text d}\beta
-
\int_{0}^{t-t_0} \frac{\partial \phi}{\partial t}(r,t-t_0-\beta) {f_{\text B}(\beta)}{\text d}\beta
$$
Now that the Convolution Theorem can be applied to both terms, Laplace transformation results in
$$
\Theta(r,s)= 
\Bigl[s\Phi(r,s)-\phi(r,0)\Bigr]F_A(s)
+
\Bigl[s\Phi(r,s)-\phi(r,t_0)\Bigr]F_B(s)
$$
UPDATE:Note, that I'm now unsure if the above is actually correct, since the integrals involved in the Laplace Transformation would use $e^{-st}$ for the first term and $e^{-s(t-t_0)}$ for the second term. I believe though, that this can be addressed by using $e^{-st}$ for both terms, so that
$$
\Theta(r,s)= 
\Bigl[s\Phi(r,s)-\phi(r,0)\Bigr]F_A(s)
+
e^{-st_0}\Bigl[s\Phi(r,s)-\phi(r,t_0)\Bigr]F_B(s)
$$
The solution of the auxiliary problem is based on a zero initial condition, therefore $\phi(r,0)=0$ and
$$
\Theta(r,s)= 
\Phi(r,s)f_A(t_0)+e^{-st_0}\Bigl[s\Phi(r,s)-\phi(r,t_0)\Bigr]F_B(s)
$$
which can be inverted to yield the temperature profile in the time domain.
In the interval $\{t_0,t\}$, the term $F_A(s)$ is zero, so that the Laplace Transform of $F_A(s)$ has to be evaluated not on the interval $(0,\infty)$, but according to
$$
\int_0^{t_0} e^{-st} f_{\text A}(t){\text d}t
$$
Please challenge each and every step above for correctness. Thanks!
