http://www.egr.msu.edu/~somerton/Nusselt/
here you can find some formulas for calculating Nusselt, Prandtl, Reynolds, Rayleigh and Grasshoff numbers. Those are important for evaluating conditions in different systems. Numbers will tell you which state of convection is around your geometry (natural, forced, laminar, turbulent, external, internal). For each case there exist some correlation.
More info about Nusselt number and others you can find eg. on Wikipaedia.
After calculating convenient numbers, you can calculate heat transfer coefficient zumbeispiel from Nusselt nuber.
Some notes to get you on your way:
Power in full sunlight at sea level $\approx 1kW/m^2$ - remember to adjust for relative angle of sun and solar panel.
Specific heat capacity of water: 4.186 J/g/K
Typical collector efficiency: somewhere in the range 50-90%.
Let's do a quick crude static calculation. It sounds like you're going to do a dynamic microsim - great stuff, go for it; if we can make a static estimate first, that'll give you an idea of how the numbers connect to each other, and give you a ballpark number to check the output of your simulation against.
Let's take the sun to be shining directly on the panel, along the normal to the plate, so the power hitting the plate is $1000 W/m^2$. Let's assume a 60% efficiency, so we're putting $600 W/m^2$ into the heat transfer fluid. Let's ignore any anti-freeze for the moment, and just assume our heat-transfer fluid is pure water, thus having a specific heat capacity of 4.186 J/g/K. Now, if you had 1 litre of water collecting heat per square metre, then you'd raise its temperature by
$$\frac{600 W}{1000 g \times 4.186 J /g/K} \approx 0.14 K/s$$
Now, here's a quick calculation for a whole day. Let's take a ball-park figure of 6 full sun-hours (that's 16 hours of summer daylight, derated by guesstimate to account for the variation of angle between panel and sun), and a 200-litre thermal storage tank; then our increase in temperature of the tank in a day, from a single square metre of collector would be, to the first order, be:
$$\Delta T_{tank} = \frac{0.14 K/s \times 3600 s/h \times 6 h}{200} \approx 15 K$$
So a $4m^2$ system would give you $\Delta T_{tank} \approx 60 K$ . Then that would get derated, based on system losses, including thermal losses from the collector plate and the thermal store.
Heat loss from the panel itself will (broadly) be proportional to the difference between the heat-transfer fluid, and the ambient air temperature - that gives the tapering you mentioned.
And you'll need a figure for the rate at which heat can be exchanged between the heat-transfer fluid that passes through the collector, and the thermal store.
(See also Appendix H of SAP 2009 which starts on pdf p73 - but bear in mind that that's a coarse static approximation of a solar thermal system in the UK, not a dynamic simulation - but it has some figures to get you started on collector efficiency and thermal losses)
Best Answer
The one-dimensional heat equation for a solid can be written as: $$ \rho C_p\frac{\partial T}{\partial t}= -\frac{\partial}{\partial x} \left( k\frac{\partial T}{\partial x} \right) +\sigma $$ where $\sigma$ is the source term and $ \dot q =-k\frac{\partial T}{\partial x}$ is the diffusive heat flux. At the boundary the temperature and the flux must be continuous (if we consider contact resistance negligible, otherwise a gap in temperature could be possible) that is: $$T_1=T_2$$ $$\dot q_1=\dot q_2 $$