# A Differential Equation Governing Methane Concentration in the Atmosphere

Often one hears that methane (CH4) is many times worse than carbon dioxide (CO2) as a greenhouse gas. For example the Wikipedia article on atmospheric methane (https://en.wikipedia.org/wiki/Atmospheric_methane) states that over 20 year period, CH4 is 84 times more potent than CO2 as a greenhouse gas. However, over 100 years, this potency drops to 28 times stronger.

What is really going on here mathematically? That is the subject of this article.

There is a chemical reaction that converts CH4 to CO2 and H2O in the atmosphere at some rate k1. Let y be the concentration of methane in the atmosphere. Let k2 be the rate of emission.

The governing differential equation for this problem is:

dy/dt = -k1*y + k2

where t is time. That is, the rate of change of the concentration of methane in the atmoshere has two contributions:
1. The chemical reaction that converts CH4 to CO2, which is proportional to the concentration of methane
2. The emission rate of methane

The solution to this differential equation is:
y(t) = k2/k1 + (y0 - k2/k1)e^(-k1*t)

With differential equations, it is often hard to find the solution, but fairly easy to check them. To review, the derivative (rate of change) of a constant is zero. The derivative of e^x with respect to x is e^x! If you want to take the derivative of e^x(t) with respect to time, you need to use the chain rule. Thus:

d(e^x(t))/dt = e^x(t)*dx/dt

So the derivative of y(t), as given by the equation above, is

dy/dt = 0 +(y0 - k2/k1)e^(-k1*t)*(-k1)

dy/dt = -k1*(y0 - k2/k1)e^(-k1*t)

dy/dt = -k1*(y(t) - k2/k1)

or dy/dt = -k1*y +k2

which is the original differential equation. Hence the solution indeed satisfies the differential equation.

This solution says that the long term concentration of methane, under a stable emssions scenario, is k2/k1. Any deviation from this concentration will decay away.

Thus the math shows that the equilibrium concentration of methane is proportional to the emission rate k2, and inversely proportional to the decay rate k1. Any deviation from equilibrium follows an exponential decay law.

So what is the take away from the mathematics?

Let us assume that the decay rate k1 is roughly constant. In reality, k1 depends on things like temperature and the concentration of O2, CO2, H2O, and hydroxyl radicals (https://en.wikipedia.org/wiki/Hydroxyl_radical) in the atmosphere, but lets ignore these effects for simplicity and think of k1 as the average decay rate.

In any case, the way people can affect the amount of methane in the atmosphere is through the emission rate k2. Keep k2 the same, and the temperature of the planet due to CH4 will not increase, except for the fact that CH4 decays to less potent CO2. In contrast, if k2 increases for example due to human activities like methane leasks from fracking, pipelines and agriculture, then the Earth will come to a new warmer equilibrium. Conversely, if k2 decreases via plugging methane leaks, then the Earth will cool because the new equilibrium concentration of methane will be less!

The conclusion is that, in a business as usual scenario (constant emmisions), global warming due to methane is not an increasing problem, except for the fact that methane decays to CO2. But this does not mean we shouldn’t take action on methane, because any reduction in the methane emission rate will cool the planet.

Stay tuned for a future article that will estimate the values of k1 and k2, and investigate the application of this mathematical solution to estimating the potency of methane as a greenhouse gas over various time frames.

Just a former physicist turned applied mathemetician

## More from Matthew Fulkerson

Just a former physicist turned applied mathemetician