Solve the system of partial differential equations: $$(1)\space\space \frac{\partial g}{\partial t} + v\frac{\partial g}{\partial x} = -k_1\left(g-h\right)$$
$$(2)\space\space \frac{\partial h}{\partial t} = k_2\left(g-h\right)$$
with initial conditions $g\left(x,0\right) = h\left(x,0\right) = 0$ and boundary condition $g\left(0, t\right) = f(t)$. $v$, $k_1$, and $k_2$ are positive constants.
This problem arises as a gas with temperature $g(x,t)$ passing through a fixed bed of solid material with temperature $h(x,t)$, although it can be transformed to be applied to other situations (adsorption of a contaminant in flowing fluid by a solid bed, for instance).
I'll denote the Laplace transforms of variables by the corresponding capital letter: for instance, the Laplace transform of $g\left(x,t\right)$ is $G\left(x,s\right)$.
The Laplace transform of the system of PDEs is: $$(1')\space\space sG+v\frac{\partial G}{\partial x}=-k_1\left(G-H\right)$$ $$(2')\space\space sH = k_2\left(G-H\right)$$
Solve (2') for $H$: $$(3)\space\space H = \frac{k_2G}{s+k_2}$$
Plug (3) into (1') and group like terms:
$$sG+v\frac{\partial G}{\partial x}=-k_1\left(G-\frac{k_2G}{s+k_2}\right)$$ $$(4)\space\space \frac{\partial G}{\partial x}+G\frac{1}{v}\left(s+k_1-\frac{k_1k_2}{s+k_2}\right)$$
The ODE (4) has solution $$(5)\space\space G = C\exp\left(-\tau\left(s+k_1-\frac{k_1k_2}{s+k_2}\right)\right)$$ where $\tau = x/v$ is the travel time for gas at the origin to reach point $x$ and $C$ is a function of $s$. At $x = 0$ (i.e. $\tau = 0$), $G\left(0,s\right) = C\left(s\right)$. But recall the boundary condition $g\left(0,t\right)=f\left(t\right)$, and therefore $G\left(0,s\right)=F\left(s\right)$. Therefore $C\left(s\right) = F\left(s\right)$, and this system can be stated in the form of a transfer function $W$: $$W\left(x,s\right) = \exp\left(-\tau\left(s+k_1-\frac{k_1k_2}{s+k_2}\right)\right)$$ The inverse Laplace transform of $W$ has an explicit solution. Factor the exponential as: $$ \exp\left(-\tau s\right)\exp\left(-\tau k_1\right)\exp\left(\frac{\tau k_1k_2}{s+k_2}\right)$$ The first factor represents a time shift: define $t' = t - \tau$ for later use. The second factor is constant in $s$. To handle the third factor, define $u = s+k_2$, giving $\exp\left(\frac{\tau k_1k_2}{u}\right)$. Using the fact that the Laplace transform of $I_0\left(2\sqrt t\right)$ is $\exp\left(1/s\right)/s$ and various properties of the Laplace transform, such as differentiation, frequency shifting, and time scaling gives us: $$w(x,t) = \exp(-k_2t'-k_1\tau)\left(\sqrt{\frac{\tau k_1 k_2}{t'}}I_1\left(2\sqrt{\tau k_1 k_2 t'}\right)+\delta(t')\right)$$ where $\delta(t)$ is the Dirac delta and $I_n$ is the modified Bessel function of order $n$.
Some initial conditions lead to useful solutions: