I need to solve for $f(\tau)$ given by

$$ f(\tau) = A(\tau)+\gamma\phi(\tau;f(\tau)) $$

in each of the time intervals in the timeline below, where in the $k$th interval, $\tau = t-\bar t_{k-1}$ Note that the functional form of $A(\tau)$ is different in each of these intervals.


Closure is provided by the model for $\phi(\tau)$ derived from the solution of the transient diffusion equation in a semi-infinite medium which following Duhamel's Principle is

$$ \phi(t) = \int_{\beta=0}^{t} \Phi(t-\beta)f(\beta){\text d}\beta $$

*Note that the initial condition for the diffusion problem is at $t=0$ and not $\tau=0$ *. I don't have a closed form expression for the kernel $\Phi(t)$, but its Laplace Transform is

$$ \mathcal{L}\{\Phi(t)\}=\bar\Phi(s)=\sqrt s \frac{K_1(\sqrt s )}{K_0(\sqrt s)} $$

where $K_0()$ and $K_1()$ are the Modified Bessel Functions of the Second Kind of orders 0 and 1 respectively.

In order to couple the two equations, I propose breaking up the Duhamel Convolution Integral as

$$ \phi(\tau) \equiv \phi(t)=\int_{\beta=0}^{t} \Phi(t-\beta)f(\beta){\text d}\beta =\int_{\beta=0}^{\bar t_{k-1}} \Phi(t-\beta)f(\beta){\text d}\beta +\int_{\beta=0}^{\tau} \Phi(\tau-\beta)f(\beta){\text d}\beta $$

The specific problem with which I'm stuck is that I cannot apply the Convolution Theorem to the first term on the RHS of the equation above. Any ideas as to a possible workaround would be deeply appreciated.

I'm aware that if I solve the semi-infinite diffusion problem in each of the intervals, this issue disappears altogether and I'm working on that approach, but it is unfortunately problematic for a variety of reasons that really belong in a separate thread.

Thanks in advance.



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