Area under decay curve exp(-0.6969t/h) where t is the time (with t=0 initially) and h is a constant "half life" is analytically integrable, but what if the half life is increasing with time? I. e. if h(t) = H + at. (Note exp(-0.6969) is not exactly 0.5 but close and easy to remember.) Can you integrate / find the area under decay curve exp[-0.6969t/h(t)] where t is the time in years and half life is given by: h(t) = 12.4(1+ 0.3t) with t = 0 on 1/1/2015. Probably only get it numerically? This problem arises when trying to evaluate the total blocking of a "puff" green house gas methane CH4 is released at t = 0. CH4 in the air is mainly destroyed by the OH- radical. OH- production rate is limited by the flux of harsh solar UV. For at least the 800,000 years this production rate was greater than the release of CH4, so CH4 concentration remained low, but that is no longer true. - - - - - - - Here is why I want to integrate with h(t) = 12.4(1+ 0.3t) with t in years and t = 0 on 1/1/2013: For the last few decades, the CH4 release rate has been faster than the OH- is formed. I.e. now the CH4 is destroying the OH- so the half life of CH4 is increasing. The CH4 half life was 9.6 years in 2003 and 12.4 in 2013, but some give it as 12.6 so I use 0.3t increase. As CO2 has a half life of about a 1000 years, other green house gas puffs are usually compare to it. I.e. How much more global warming a puff of CH4 will make compared to an equal mass of CO2 released at the same time, during the next X years. Here are some results: t = 1/365: 120 times more than the CO2 did the first day. t = 10: 104 times more in first 10 years, etc. t = 20: 83.8 times more t = 50: 48.4 times more t = 100: 28.5 times more t = 500: 8.1 times more Please also do numerically integral for h = 12.4 (that was assumed for compared to CO2 data above) for at least t = 20, 50 & 100. With that and the above data I can convert your results into a new "compared to CO2" table for the years you give both. I. e. if you numerical integration for 50 year integral with h = 12.4 gives X and with h(t) it gives Y, then the increasing half life increases the CH4 problem by factor of Y/X or compared to CO2, the damage done in for example 50 years becomes (Y/X)times 48.4 greater than CO2 does in 50 years. With three (or more if you do more) I can "curve fit" thru the three (and no difference for the first day point)
\(\int_0^x \left( \frac{1}{2} \right)^{\frac{t}{h}} dt = \frac{h-h 2^{-\frac{x}{h}}}{ \ln 2 }\) \(\begin{array}{c|cc} x & \frac{h-h 2^{-\frac{x}{h}}}{ \ln 2 } \\ \hline \\ h & \frac{h}{2 \ln 2 } \\ 2h & \frac{3 h}{4 \ln 2 } \\ 3h & \frac{7 h}{8 \ln 2 } \\ 4h & \frac{15 h}{16 \ln 2 } \\ 5h & \frac{31 h}{32 \ln 2 } \end{array}\) I don't think that you are integrating the correct formula. Just as \(f(t) = \left( \frac{1}{2} \right)^{\frac{100 t}{1240}}\) is a solution to \(f'(t) = - \frac{100 \ln 2}{1240} f(t), f(0) = 1\), I think you want to solve \(g'(t) = - \frac{100 \ln 2}{1240 + 372 t} g(t), g(0) = 1\), with solution \( g(t) = 2^{- \frac{25}{93} \ln \frac{10 + 3 t}{10} } \). The analytic integral of g(t) can be done for positive x, \(\int_0^x 2^{- \frac{25}{93} \ln \frac{10 + 3 t}{10} } dt = \frac{ (93 x + 310) 2^{- \frac{25}{93} \ln \frac{10 + 3 x}{10} } -310}{93 - 25 \ln 2} \) \(\begin{array}{c|cc} x & \int_0^x \left( \frac{1}{2} \right)^{\frac{100 t}{1240}} dt & \int_0^x \left( \frac{1}{2} \right)^{\frac{100 t}{1240 + 372 t}} dt & \int_0^x 2^{- \frac{25}{93} \ln \frac{10 + 3 t}{10} } dt \\ \hline \\ 10 & 7.6605 & 9.05212 & 8.5597 \\ 20 & 12.041 & 17.6458 & 15.859 \\ 30 & 14.545 & 26.1318 & 22.578 \\ 40 & 15.977 & 34.5681 & 28.926 \\ 50 & 16.796 & 42.9759 & 35.004 \\ 75 & 17.619 & 63.925 & 49.363 \\ 100 & 17.823 & 84.8183 & 62.877 \end{array}\)
Thanks. No, make that double thanks. Not only did you do what I asked, but you understood what I should have asked to be integrated and did that too! I. e. there is some curve that has value unity at t = 0 and steadily declines with time telling the number of decaying particle still existing at some later t. The damage the particles due is proportional to the area under this curve and not under a curve passing thru multiple half life points as I thought. I. e. the half life is just a point on the curve I want the area under with value 0.5 and likewise two half lives are a point on the curve I need the area of where that curve has value 0.25, etc. My math is quite rusty, but with about an hour of effort, I was able to follow and check you post down to the point where you gave the right side of g(t) = …. I. e. I could not differentiate that right side to show it was the g'(t) you gave (the curve I need integrated to get the cumulative damage done by CH4 as function of time with its half life increasing.), but would bet my last dollar you are correct. It is interesting to note that the middle column of the three you give at end of your post (area under my wrong curve) is greater than the area under the correct g(t) curve given in the right most column. I think I understand why. But to the main point, using your data for 20 year to illustrate how: Some months ago you did the harder problem of finding the “Warming Potential,” WP, (how much more damage in set of years CH4 does compared to CO2). Your result for CH4's WP20 was 83.8 and you new data show the increasing life time increases that by the factor 15.859 / 12.041 so a more realistic value for WP20 is: 83.8(15.859 / 12.041) = 110.37 or 31.7% more damage is to be expected than everybody doing analysis of the CH4 threat is suggesting, but in fact it is even worse because the rate of half life increase is not a constant (0.3 years per year) but an accelerating rate as the concentration of the destroying OH- falls. Doing the problem still more correctly would be very hard as OH- is not the only wayCH4 is removed from the air and how much CH4's release rate will increase in positive feed back with global (or just Arctic) temperature rise is not known. Also, in adition to some model of the growing CH4 release rate, with a small part subtracted off to represent the non-OH radical removal processes, one would need to model the falling OH- concentration, at least down to where the removal by OH- is less than these other removal processes. I. e. I ignore them now, and eventually one could ignore the OH- removal process, which is now dominate. I strongly recommend you publish your results as CH4 is more of a problem than currently thought to be. I don't need to be mentioned, even in a footnote in your paper. You have done all the work. All I did was to interest you (I hope) in the analysis of this important problem. At the very least you should make the scientists of the Arctic Methane Emergency Group aware of what you have done. If you don't want to bother with that, tell me and I will try to.
By the method illustrated in third paragraph from end of post 3, that found the GW20 = 110.37 or 31.7% greater than the constant 12.4 half live's value of 83.8, I now post the results for WP0, WP10, WP20, WP50 & WP100: WP0 = 120 & 0% increase by more realistic growing half life WP10 = 116.20 & increased by 11.4% WP20 = 110.37 & increased by 31.7% WP50 = 106.04 & increased by 219.09% WP100 = 100.54 & increased by 352.79% One could "curve fit" thru these five points to get the general (but a lower bound) functional form for how much more CH4's Warming potential is than it usually is considered to be. A simple linear fit accurate at 0 and 100 years is: WP(t) = 120 - 0.1946t and that is probably good enough (within the error limits) without any quadratic term. I. e. gives WP50 = 110.27 instead of 110.37 or "perfect" by my standards. As the rate of half life increase assumed was linear, but in fact is growing faster than linear the following is also even a more perfect estimate: WP(y) = 120 - 0.156y and y in years after "CH4 puff" release. Want to know why I used 0.156? Well that is because then: WP(w) = 120 - 0.003w where w is the number of weeks after puff release. Now all we need to learn how serious CH4 being released is a model for it R(w) where R the number of tons of CH4 released in some week., w, after a start week with known total tons of CH4 already in the air, Co. I think what we want to integrate, if we knew R(w) and Co is: Co(120-0.003w) + R(w)(120-0.003w) or {Co + R(w)} (120-0.003w) dw. That is the product of an increasing function and a decreasing one over the range 0 to N weeks. R(w) probably is "quasi-linear" (linear to best accuracy we could hope for) plus a relative small harmonic modulation with period of 52 weeks. If that is reasonable model of R(w) then the integration is simple. Hell even I might be able to do it with aid of table of {aw sin(bw + c)} etc. integration results.
No. Co(120-0.003w) + R(w)(120-0.003w) or {Co + R(w)} (120-0.003w) dw is not what needs to be integrated. Perhaps, it is [Co(120-0.003{w-Wo}) + R(w)(120-0.003{w-Wo})] dw, Or, [Co + R(w)](120-0.003{w-Wo})dw where Wo is the number of the week in which Co was known.
