Original Article
On the effectiveness of the search and find method to suppress spread of SARS-CoV-2
2021 年 97 巻 1 号 p. 22-49
詳細
2021 年 97 巻 1 号 p. 22-49
Search and find methods*) such as cluster tracing1)–6) or large-scale PCR testing**) of those who exhibit no symptoms or only mild symptoms of COVID-19 is shown by data analysis to be a powerful means to suppress the spread of COVID-19 instead of, or in addition to, lockdown of the entire population. Here we investigate this issue by analyzing the data from some cities and countries and we establish that search and find method is as powerful as lockdown of a city or a country. Moreover, in contrast to lockdown, it neither causes inconvenience to citizens nor does it disrupt the economy. Generally speaking, it is advisable that both social distancing and increased test numbers be employed to suppress spread of the virus. The product of the total test number with the rate of positive cases is the crucial index.
We consider first the case of Tokyo to illustrate our method of analysis. In this introductory section, we focus on the number of those who are infected, but the origin of whose infections cannot be traced to a “cluster”: The Japanese government has been initially concentrating on “cluster tracing”, in which they assumed that the major part of the infections originate from a “cluster”. But the subsequent rapid increase in infections without clear origin indicates a weakness of concentrating only on clusters. In later sections we do not distinguish between infections with or without the knowledge of their sources. Both cases are taken into account by a single parameter σ to be defined later. “Cluster tracing” increases σ by increasing the positivity rate of tests, and large-scale PCR testing increases σ by increasing the number of tests.
In Tokyo, the daily number of those infected by the virus whose origin of infection cannot be traced, between March 25 and April 4 (https://stopcovid19.metro.tokyo.lg.jp/en/), is;
| \begin{equation} \{13,22,20,22,24,7,50,40,38,59,80\}. \end{equation} | [1] |
How can we achieve it? Obviously, fitting by a linear polynomial is not sufficient.
Number of those who are infected per day in Tokyo between March 25 and April 4.
We develop a model on the following naive considerations and justify it later in Appendix 1 based on the standard model of epidemiology, i.e., the SIR model.8),9)
We use the following notation and derive a simple formula taking the Tokyo case as an example:
1. NT = the number of infections in Tokyo on the date T;
2. nT = those who were officially recognized among NT and therefore isolated;
3. UT = NT − nT: those who stay in town unrecognized as infected;
4. κUT = total increase of number of infected per day (negative κ means the decrease). Note that those who are officially recognized as infected do not contribute to this number.
5. Suppose we put T = March 25, then the number of those infected after the m-th day becomes,
| \begin{align} \prod_{n = 1}^{m}(1 + \kappa (n))U_{25} &= \exp\left[\sum_{n = 1}^{m}\log(1 + \kappa(n))\right]U_{25} \\ &\simeq \exp \left[\sum_{n = 1}^{m}\kappa (n) \right]U_{25}. \end{align} | [2] |
6. This can be written in the continuous limit as,
| \begin{equation} \exp\left[\int_{0}^{x}\kappa (t)dt \right]U_{25}. \end{equation} | [3] |
| \begin{equation} dU/dx = \kappa (x)U. \end{equation} | [4] |
7. The above data shown in Eq. [1] and in Fig. 1 are for dn/dx and not for U. How are they related?
Before answering this question, we point out that κ is composed of three components, as derived in Appendix 1:
| \begin{equation} \kappa = K - \rho - \sigma, \end{equation} | [5] |
The parameters K, ρ and σ in Eq. [5] are explained as follows:
K: the infection rate, which can be written as K = λf2 with λ the infection rate at normal “individual contact rate”. If we reduce the contact rate by a factor f $ \leqslant $ 1 the rate becomes K = λf2. The contact rate of general public can be reduced only by lockdown of the entire city or by an official order to citizens to stay home and/or wear masks and maintain social distancing. We may call f2 the “herd contact rate”.
Another way to explain this is the following:
The “herd contact rate” is proportional to the square of the entire population N when N is large (contact number is proportional to N(N − 1)/2 ∝ N2 if N is large). We regard f ∝ N − Njoin, where, when the lockdown is started, Njoin denotes the number of individuals locked down. In our model, the entire population is not locked down right away.
In later analysis we use the following fact: the joining population Njoin is a function of time x and analytically it can be written as,
| \begin{equation*} N_{\text{join}} = ax + bx^{2} + cx^{3} + \cdots. \end{equation*} |
ρ: Natural recovery and/or cure rate of COVID-19. It is not known accurately so far but we can estimate it as will be shown later.
σ: Rate of finding the infected among the symptomatic individuals. The infected found must be separated from the public and included in nT.
σ can be defined and estimated as follows:
| \begin{equation} \sigma = \frac{\text{number of tested individuals per day} \times \text{positivity rate of the test}}{\text{number of infections}}. \end{equation} | [6] |
“Tested individuals” in Eq. [6] refers in practice to individuals tested by PCR, for which the efficiency is discussed in Appendix 6.
To increase the value of σ there are at least two distinct approaches, since it is proportional to number of tests and to the positivity rate of the test. For example, China and South Korea adopted the policy of increasing the number of tests while Japan instead opted to increase the positivity rate by tracing the “clusters”. Either way, they were relatively successful in suppressing the initial outbreak of the COVID-19 compared to other countries.
We must aim at
| \begin{align} \kappa \leqslant 0 &\to \lambda f^{2} \leqslant \rho + \sigma \\ &\to \text{reproduction number $R_{t}$} = \frac{\lambda f^{2}}{\rho + \sigma} \leqslant 1. \end{align} | [7] |
Incidentally, although the parameter Rt is the most commonly used and convenient parameter to discuss the spread of virus, other parameters have been proposed in the literature as described in Appendix 4.
8. Rough estimation of ρ: Here we use the data from the cruise ship Diamond Princess (https://www.statista.com/statistics/1099517/japan) that harbored in Yokohama with infected passengers. The total number of those infected, and who remained in Japan and disembarked early March, is 672. Among them, 638 recovered by March 19. This proportion exceeds the conventional estimate of about 80%, but we adopt the value above. Then we obtain 638/20 ≈ 32 as the number of those cured per day. Therefore, we get,
| \begin{equation} \rho = 32/672 = 0.05. \end{equation} | [8] |
| \begin{equation} \kappa = \lambda f^{2} - \rho - \sigma. \end{equation} | [5] |
| \begin{equation} \lambda = \kappa + \rho = 0.2. \end{equation} | [9] |
| \begin{equation} \kappa = 0.2f^{2} - 0.05 = 0, \end{equation} | [10] |
| \begin{equation} f = \sqrt{\frac{0.05}{0.2}} = 0.5. \end{equation} | [11] |
9. From the above definition of σ we can answer the question “how are U and dn/dx related?”, namely,
| \begin{equation} \frac{dn}{dx} = \sigma U = \sigma U_{0}\exp\left[\int_{0}^{x}\kappa (t)dt\right]. \end{equation} | [12] |
In passing we note that the analyses presented in the following sections do not distinguish between symptomatic and asymptomatic infections, which may not always be appropriate. In Appendix 5, we show an analysis that accounts for this distinction.
We are now ready to analyze the situation of COVID-19 in several geographical locations based on our model described in this section. We first analyze three large cities and then several countries.
From March 23 to April 23 the numbers of new infections per day (https://stopcovid19.metro.tokyo.lg.jp/en/) are:
| \begin{align} & \{16,17,41,47,40,63,68,13,78,66,97,89,116,143,\\ &\quad 83,79,144,178,187,197,166,91,161,126,148,\\ &\quad 201,181,107,102,123,132,134\}\ (\text{Fig. 3}). \end{align} | [13] |
Infection per day of Tokyo from March 23 to April 23.
We analyze all new infections per day without distinguishing among infections with or without identified sources. The same will be true for other data we analyze (Los Angeles and New York) later in this paper. The unique Japanese effort to trace the source of infection as a cluster is captured as an increase in the efficiency of the test in Eqs. [5] and [6], thus increasing the value of σ.
First, we point out that σ is proportional to the number of tests per day as its defining formula Eq. [6] shows. For Tokyo, the number of tested individuals per day at the time of this initial analysis is approximately 300, independent of time. We see later that Los Angeles has an increasing number of tested individuals per day and New York has a constant value of 8900 per day. The value of σ is quite interesting when compared to New York data, which we analyze later in this note.
We fit this Tokyo time series using the Mathematica least squares method, assuming the form of an exponential of a third order polynomial:
| \begin{align} &\sigma U_{0}\exp\left[\int_{0}^{x}\kappa (t)dt \right] \\ &\quad = \exp[2.75961 + 0.196892x + 0.0044258x^{2} \\ &\qquad + 6.67625 \times 10^{-6}x^{3}], \end{align} | [14] |
| \begin{equation} \kappa = 0.196892 - 0.00885x + 2.003 \times 10^{-5}x^{2}. \end{equation} | [15] |
| \begin{equation} \kappa = \lambda (1 - ax)^{2} - 0.05 - \sigma. \end{equation} | [16] |
| \begin{equation} a = 0.0045, \end{equation} | [17a] |
| \begin{equation} \lambda = 0.98, \end{equation} | [17b] |
| \begin{equation} \sigma = 0.73. \end{equation} | [17c] |
If we incorporate the time delay between infection and presentation of symptoms, which is estimated to be 3 to 5 days with the highest activity 8 days after the infection, we must consider the time shift between dn/dx and U. Elaborate discussion of the time gap between infection and appearance of symptoms is given in Appendix 2. We ignore this effect for the time being but it must be included when we discuss more accurate timing.
The flattening of U occurs somewhat after we reach κ = 0. This corresponds to the maximum of the daily infection number.
| \begin{equation} \kappa = 0.196892 - 0.00885x + 2.003 \times 10^{-5}x^{2} = 0. \end{equation} | [18] |
The number of infections in Tokyo flattened around April 17. At that time,
| \begin{equation} f^{2} = 0.8, \end{equation} | [19] |
We plot here our fit to the number of daily infections in Tokyo in Fig. 4.
We plot in Fig. 5 the flattening of Tokyo infection number n, which may occur somewhat later according to the above model. Flattening does not require strengthening of social distancing.
Tokyo total infection flattening in 50 days.
Incidentally, for κ = 0:
| \begin{equation} \lambda (1 - ax)^{2} - 0.05 - \sigma = 0. \end{equation} | [20] |
At this point we have
| \begin{equation} R_{t} = \frac{\lambda f^{2}}{\gamma + \sigma} = \frac{\lambda (1 - ax)^{2}}{\gamma + \sigma} = 1. \end{equation} | [21] |
Have we reached the government goal of f2 = 0.2 by May 2? The answer is the following:
Assume that f had reached the value of 0.7 already on March 23, a 30% decrease in the contact rate f. Then f2 was 0.49 at the date.
| \begin{equation} (1 - ax)^{2} = (1 - 0.0045 \times 40)^{2} = (0.82)^{2} \approx 0.67. \end{equation} | [22] |
| \begin{align} &\text{$f^{2}$ on May 2 is $0.49 \times 0.67$}\\ &\quad = 0.33 - \text{not too far from 0.2}, \end{align} | [23] |
| \begin{equation} R_{t} = \frac{\lambda f^{2}}{\gamma + \sigma} = \frac{0.98 \times 0.33}{0.05 + 0.73} = 0.41. \end{equation} | [24] |
| \begin{equation} R_{t} = \frac{\lambda f^{2}}{\gamma + \sigma} = \frac{0.98 \times 0.67}{0.05 + 0.73} = 0.84. \end{equation} | [25] |
We now plot the following function in Fig. 6,
| \begin{align} dn/dx& = \exp[2.75961 + 0.196892x - 0.0044258x^{2}\\ &\quad + 6.67625 \times 10^{-6}x^{3}], \end{align} | [26] |
dn/dx from March 24 to May 10.
x = 0 corresponds to March 24 and x = 46 to May 10. The value of May 10 announced by the local Tokyo Government is a bit above 20, which agrees remarkably well with this diagram and confirms the applicability of our parametrization.
We solve
| \begin{align} \log[dn/dt] &= 2.75961 + 0.196892x - 0.0044258x^{2} \\ &\quad + 6.67625 \times 10^{-6}x^{3} = 0. \end{align} | [27] |
| \begin{equation} x = 60.3 \to \text{24 of May}. \end{equation} | [28] |
In this subsection we analyze the Tokyo data from May 24 to June 26 with special attention to the relaxing of social distancing on May 25.
The daily new infection number from May 24 to June 26 is (https://stopcovid19.metro.tokyo.lg.jp/en/):
| \begin{align} & \{14,8,10,11,15,21,14,5,13,34,12,34,12,\\ &\quad 28,20,26,14,13,12,18,22,25,47,48,27,\\ &\quad 16,41,35,39,35,29,31,55,48,54\}. \end{align} | [29] |
Tokyo daily new infections from May 24 to June 26.
There seems to be no way to fit this data unless we take into account the effect of asymptomatic infections, because if we only consider infections with symptoms it would appear that new daily infections have died away towards the end of May.
The value of λ on May 25 is calculated to be,
| \begin{equation} \lambda_{25} = 0.18(1 - 61 \times 0.0045)^{2} = 0.095. \end{equation} | [30] |
Therefore, we fit the data with,
| \begin{equation} \kappa = \lambda_{25}(1 + a_{25}x)^{2} - 0.05. \end{equation} | [31] |
This gives,
| \begin{equation} dn/dx = 6\exp[0.045x + 0.023ax^{2} + 0.015a^{2}x^{3}]. \end{equation} | [32] |
dn/dx curve in Eq. [32].
The analysis above indicates that there exists a realistic danger of breakout of COVID-19 spread in Tokyo. The only way to suppress the breakout seems to be to apply the search and find method for the asymptomatic infections, if one wants to avoid another lockdown.
The pooling method for PCR swabs may be suitable if the number of tests becomes large: combine swabs from ten individuals, for example, and test them as one swab. Only when the test is positive individual swabs should be put under PCR test.
Next, we analyze the data from Los Angeles (https://corona-virus.la/data).
The data shows that the daily new infection numbers from March 27 to April 13 are the following:
| \begin{align} &\{257,344,332,342,548,513,534,521,711,663,\\ &\quad 420,550,620,429,475,456,323,239\}. \end{align} | [33] |
| \begin{align} p(x) &= 5.41365 + 0.17285x - 0.00541924x^{2} \\ &\quad - 0.000205763x^{3}. \end{align} | [34] |
| \begin{equation} \kappa = K - \rho - \sigma, \end{equation} | [5] |
| \begin{equation} K = \lambda (1 - ax)^{2},\quad \rho = 0.05. \end{equation} | [35] |
Number of infection from March 27 to April 13 in LA.
Daily infection number fitted by exp[p(x)] (top part is partly cut off. See Fig. 10).
The data (https://corona-virus.la/data) shows that the accumulated test numbers during this period are given as:
| \begin{align} \text{test number} &=\{6741,10028,14033,16646,21372,\\ &\qquad 21746,24685,28931,33082,\\ &\qquad 37430,41189,42215\}. \end{align} | [36] |
| \begin{equation} 3958.73 + 3131.95x - 3.30969x^{2} + 1.38928x^{3}. \end{equation} | [37] |
| \begin{equation} Y = 3131.95 - 6.62x + 4.2x^{2}. \end{equation} | [38] |
| \begin{align} \sigma& = (\text{number of tests}) \times (\text{positivity rate})\\ &\quad /(\text{those who have symptoms}). \end{align} | [39] |
| \begin{align} \sigma &= \frac{\alpha}{3132}(3131.95 - 6.62x + 4.2x^{2}) \\ &= \alpha (1 - 0.0021135\,x + 0.00134\,x^{2}), \end{align} | [40] |
This gives,
| \begin{align} \kappa &= \lambda (1 - ax)^{2} - 0.05 \\ &\quad - \alpha (1 - 0.0021135\,x + 0.00134\,x^{2}). \end{align} | [41] |
| \begin{align} &A\exp\bigg[\int_{0}^{x}dx\{\lambda (1 - ax)^{2} - 0.05 \\ &\quad - \alpha (1 - 0.00211\,x + 0.00134\,x^{2})\}\bigg]. \end{align} | [42] |
We then have,
| \begin{align} \kappa & = dp/dx = 0.17285 - 0.0108x - 0.000618x^{2}\\ & = \lambda (1 - ax)^{2} - 0.05 \\ &\quad - \alpha (1 - 0.0021135x + 0.00134x^{2}). \end{align} | [43] |
| \begin{equation} \alpha = 0.483, \end{equation} | [44a] |
| \begin{equation} \lambda = 0.70428, \end{equation} | [44b] |
| \begin{equation} a = 0.0084. \end{equation} | [44c] |
This result reveals various important facts. We point out, as the most important fact, that the policy of so-called social distancing by lockdown of a city does not seem to be efficient enough without increasing the number of tests in the above LA data analysis.
4.A. Analysis of Los Angeles data from March 27 to May 12.The data is given (https://corona-virus.la/data) as:
| \begin{align} dn/dx& = \{257,344,332,342,548,513,534,521,711,\\ &\qquad 663,420,550,620,429,475,456, 323,\\ &\qquad 239,670,472,399,567,642,334,1491,\\ &\qquad 1400,1318,1081,1035,607,440,900,597,\\ &\qquad 1541,733,1065,691,781,568,1638,\\ &\qquad 851,815,883,1011,454,591,961\}. \end{align} | [45] |
dn/dx of LA from March 27 to May 12.
We can fit this by the following function,
| \begin{align} dn/dx &= \exp[5.94616 + 0.002852x \\ &\quad + 0.001579x^{2} - 0.0000294x^{3}]. \end{align} | [46] |
dn/dx fit for LA data March 27 to May 12.
We solve,
| \begin{align} &\lambda (1 - ax)^{2} - 0.05 - \alpha (1 - 0.0021135x + 0.00134x^{2}) \\ &\quad= 0.00285 + 0.00316x - 0.0000883x^{2}, \end{align} | [47] |
| \begin{equation} \lambda = 0.131,\quad a = - 0.0114,\quad \alpha = 0.07862. \end{equation} | [48] |
Next, we analyze the data from New York State.
From March 23 to April 15 the daily number of the infection (https://www1.nyc.gov/site/doh/covid/covid-19-data.page) is:
| \begin{align} & \{3477,4365,4705,4872,4953,3350,3424,5973,\\ &\quad 5136,4883,5533,5396,3666,3565,6118,\\ &\quad 5706,5229,4647,3979,3113\}. \end{align} | [49] |
We show it in Fig. 15, which is rather similar to the Los Angeles data except for the scale.
Daily infection number of NY from March 23 to April 15.
Another set of data is available on the accumulated number of test frequency from March 23 to April 21:
| \begin{align} & \{51406,63752,70857,78936,83000,87414,\\ &\quad 97409,104096,110606,127788,135861,\\ &\quad 148519,162464,187166,178917,184122,\\ &\quad 191868,197273,210610,219737,229040,\\ &\quad 237474,243074,251840\}. \end{align} | [50] |
We plot this in Fig. 16. This looks linear and we fit it as follows:
| \begin{equation} \text{accumulated test number} = 40411.3 + 8944.52x. \end{equation} | [51] |
Accumulated number of tests in NY from March 23 to April 15.
This means that the daily test number is 8944.52.
We fit the daily infection number depicted in Fig. 15. The contribution of the test to κ is just a constant times 8944.52 which is also a constant. We designate it as σ. The infection rate λf2 is put to be λ(1 − ax)2 as in the case of LA but we assume (ax)2 ≪ ax which will be justified later.
Then κ = λ(1 − 2ax) − 0.005 − σ and the daily number is $\sigma \exp \int_{0}^{t}\kappa dx $.
Fitting was done using the Mathematica least squares method and we get,
| \begin{equation} \kappa = 0.0611162 - 0.00566x. \end{equation} | [52] |
| \begin{align} \kappa &= \lambda (1 - 2ax) - 0.005 - \sigma \\ &= 0.0611162 - 0.00566x. \end{align} | [53] |
| \begin{equation} \lambda = 0.704. \end{equation} | [54] |
| \begin{equation} \sigma = 0.594,\quad a = 0.0040. \end{equation} | [55] |
The reason we ignored the (ax)2 is that fitting the Fig. 15 data by a third order polynomial yields a negative coefficient for the third order term x3, which is impossible unless we have a x2 term in γ. The latter can be achieved by including the increasing medical effort and/or increasing death rate. Both of these contribute to reducing the daily detected infection number dn/dx because the coefficient of x2 in λ(1 − ax)2 is positive thus contributing a positive number to x3 term to the daily infection number. We simplified the process by ignoring the x3 term in the fitting. The consequence is that we have one free parameter in our fit. We fixed it by adopting the Los Angeles value for λ, which could be more or less common within a country.
5.A. Analysis of New York data from March 23 to May 11.The new infections per day (https://www1.nyc.gov/site/doh/covid/covid-19-data.page) is,
| \begin{align} & \{3477,4365,4705,4872,4953,3350,3424,5973,\\ &\quad 5136,4883,5533,5396,3666,3565,6118,5706,\\ &\quad 5229,4647,3979,3113,3526,3555,2117,2351,\\ &\quad 3768,3059,3457,2832,2465,1589,1010,2304,\\ &\quad 2712,2334,2003,1863,1047,776,1532,1503,\\ &\quad 1382,927,1040,612,401,586\}. \end{align} | [56] |
NY data of daily infection from March 23 to May 11.
We fit this by our original function which was used to analyze the data from March 23 to April 17:
| \begin{equation} dn/dx = \exp[8.1789 + 0.0611162x - 0.00283x^{2}]. \end{equation} | [57] |
The fit is qualitatively good but it seems it is substantially lower than the actual value after mid-April. Therefore, we try to fit the whole data set by a new function of exponential of a certain third order polynomial:
The result is,
| \begin{align} dn/dx &= \exp[8.2887 + 0.0347x - 0.0019x^{2} \\ &\quad + (5.8924)^{-6}x^{3}]. \end{align} | [58] |
Fitting function to daily infection: Eq. [58].
Solving the equation,
| \begin{align} \kappa& = \lambda (1 - ax)^{2} - 0.05 - \sigma \\ &= 0.0347 - 0.00386x + 0.0000177x^{2}. \end{align} | [59] |
| \begin{equation} a = 0.00917, \end{equation} | [60a] |
| \begin{equation} \lambda = 0.210, \end{equation} | [60b] |
| \begin{equation} \sigma = 0.125. \end{equation} | [60c] |
We compare the situation of three cities in Tables 1.a, 1.b and 1.c. The effect of Δt is remarkable. The value of λ (infection rate) is the smallest in Tokyo and larger in LA and in NY. The value of σ is comparable between Tokyo and New York but Los Angeles has a much larger value when taking into account its time dependence.
| Tokyo | Los Angeles | New York | |
|---|---|---|---|
| infection | 3500 | 18500 | 147000 |
| λ | 0.98 | 0.704 | LA value |
| a | 0.0045 | 0.0084 | 0.0040 |
| σ(α) | 0.73 | 0.483 | 0.594 |
| Tokyo | Los Angeles | New York | |
|---|---|---|---|
| infection | 4997 | 34428 | 185206 |
| λ | 0.98 | 0.131 | 0.219 |
| a | 0.0045 | −0.0114 | 0.00917 |
| σ(α) | 0.73 | 0.0786 | 0.12 |
| Tokyo | Los Angeles | New York | |
|---|---|---|---|
| infection | 4997 | 34428 | 185206 |
| λΔt | 0.18 | 0.726 | 0.54 |
| aΔt | 0.0045 | −0.0114 | 0.00917 |
| σΔt (αΔt) | 0.127 | 0.053** | 0.071 |
**This is the value of αΔt. σΔt = 0.053 (1 − 0.0021135x + 0.00134x2) and increases rapidly. For example, on May 12 the value of σ is 0.269 much larger than the value of Tokyo or New York indicating the effort of county of Los Angeles to increase the PCR test.
The most significant difference between Tokyo and the above two U.S. cities is the scale of infection. In particular, NY has 42 times more infections than Tokyo. To understand it we note that,
The data is given (https://www.worldometers.info/coronavirus/country/japan/) as follows:
| \begin{align} & \{96,112,194,173,87,225,206,233,303,351,383,\\ &\quad 252,351,511,579,658,743,507,390,455,482,\\ &\quad 585,628,566,390,368,377,423,469,441,\\ &\quad 353,206,191,276,236,193,263\}. \end{align} | [61] |
Japan data of daily infection from March 26 to May 1.
Fitting is done with Mathematica least squares method. The result is:
| \begin{align} dn/dx &= \exp[4.29867 + 0.20722x \\ &\quad - 0.00616699x^{2} + 0.00003387x^{3}]. \end{align} | [62] |
Combining this result with the histogram in Fig. 23, we have Fig. 25.
Fitting the Japan number of daily infection (top part is partially cut off. See Fig. 23).
We derive,
| \begin{align*} \kappa& = 0.20722 - 0.0123x + 0.00010x^{2} \\ &= \lambda (1 - 2ax + a^{2}x^{2}) - 0.05 - \sigma. \end{align*} |
| \begin{equation} a = 0.016, \end{equation} | [63a] |
| \begin{equation} \lambda = 0.384, \end{equation} | [63b] |
| \begin{equation} \sigma = 0.127. \end{equation} | [63c] |
(1) Date of κ = 0.
| \begin{equation} 0.207 - 0.012x + 0.00010x^{2} = 0, \end{equation} | [64] |
| \begin{equation} x = 20 \to \text{April 15}. \end{equation} | [65] |
(2) Behavior of Rt.
| \begin{align} R_{t} &= \frac{\lambda f^{2}}{\gamma + \sigma} = \frac{0.384(1 - 0.016x)^{2}}{0.05 + 0.127} \\ &= 2.16(1 - 0.016x)^{2}. \end{align} | [66] |
Behavior of Rt during the period of March 26 to May 1.
(3) The current (May 1) value of f2.
| \begin{equation} f^{2}(\text{May 1}) = (1 - 0.016 \times 36)^{2} = 0.180. \end{equation} | [67] |
Total Japan infection from March 26 till 50 days later (May 15).
For this purpose, we consider two cases:
(1) f2 reaches 0.4 and stays there afterwards. We first calculate on which day this was realized in our model:
| \begin{equation} (1 - 0.016x)^{2} = 0.4. \end{equation} | [68] |
| \begin{align} \kappa& = \lambda f^{2} - \gamma - \sigma \\ &= 0.384 \times 0.40 - 0.05 - 0.127 = - 0.023. \end{align} | [69] |
| \begin{align} 501& = \exp[4.39 + 0.207 \times 23 - 0.0062 \\ &\quad \times 23^{2} + 0.000034 \times 23^{3}]. \end{align} | [70] |
Daily Japanese infection after April 18 when we reached f2 = 0.4.
When can we reach dn/dx < 1 meaning no new infection?
| \begin{equation} 501\exp[- 0.023x] = 1. \end{equation} | [71] |
(2) f2 reaches the value of 0.2 and stays there afterwards. This happens on the day given by,
| \begin{equation} (1 - 0.016x)^{2} = 0.2. \end{equation} | [72] |
| \begin{align} \kappa& = \lambda f^{2} - \gamma - \sigma \\ &= 0.384 \times 0.2 - 0.05 - 0.127 = - 0.10. \end{align} | [73] |
| \begin{align} dn/dx &= \exp[4.29 + 0.2072 \times 35 - 0.00616 \\ &\quad \times 35^{2} + 0.0000338 \times 35^{3}] = 233. \end{align} | [74] |
We plot the function 233 exp[−0.1x] in Fig. 29.
Daily new infection of Japan after April 30.
We find the day when dn/dx goes below 1,
| \begin{equation} 233\exp[- 0.1x] = 1, \end{equation} | [75] |
| \begin{equation} x = 54.5. \end{equation} | [76] |
We now double the value of σ from 0.127 to 0.254 and see how it changes the situation.
Case (1) f2 = 0.4.
| \begin{equation} \kappa = \lambda f^{2} - \gamma - \sigma = - 0.023 - 0.127 = - 0.150. \end{equation} | [77] |
f2 is 0.4 but with σ = 0.254 double the value of current σ.
We get the date of dn/dx = 1,
| \begin{equation} 501\exp[- 0.150\,x] = 1. \end{equation} | [78] |
Case (2) f2 = 0.2.
| \begin{equation} \kappa = \lambda f^{2} - \gamma - \sigma = - 0.150 - 0.127 = - 0.277. \end{equation} | [79] |
f2 = 0.2 together with double σ.
The date of dn/dx = 1 is given by,
| \begin{equation} 233\exp[- 0.277x] = 1. \end{equation} | [80] |
We have dn/dx = 1 on 20 days after April 30, that is, May 20, showing that it is possible to terminate the social distancing policy before the end of May. Maybe we should stress that this is the only case among the four cases considered above in which it is safely possible to terminate the Japanese governmental emergency status at the end of May. Collaboration of social distancing and the increase of PCR test numbers is essential. A premature relaxation of the social distancing will lead to further spreading of the virus. The actual date depends on how large the time gap between the infection and the appearance of symptoms as will be discussed in Appendix 2 and mentioned in section 4.
The following two diagrams of Fig. 32 and Fig. 33 illustrate the effect of doubling of the value of σ to the daily new infections of Tokyo. To be more precise about the dates we must subtract the incubation period of average 7 days in each case.
Case of f2 = 0.4 with (sharp drop) or without (slow drop) doubling of σ.
Case of f2 = 0.2 with (sharp drop) or without (slow drop) doubling of σ.
We summarize the above result with this subtraction included in the Table 2. Table 2 shows that safely lifting of emergency status at the end of May is not possible irrespective of whether f2 has reached the value of 0.4 or 0.2 although, if we adopt more moderate criterion than dn/dt = 1, we may lift the emergency at the end of May if f2 = 0.2. Whichever value of f2, we can escape from the emergency situation towards the end of May or early June in case of f2 = 0.4 provided that we double the value of σ. In fact, in case of f2 = 0.2 with doubled σ, we did not have to extend the emergency status for the month of May, if we had doubled the value of σ before this time.
| σ | 0.127 | 0.127 | 0.254 | 0.254 |
| f2 | 0.4 | 0.2 | 0.4 | 0.2 |
| date of no new infection | December 15 | June 16 | June 5 | May 13 |
The reader is reminded that to achieve the doubling of σ requires doubling of:
| \begin{align*} \sigma& \propto \text{number of test cases times positivity}\\ &\quad \text{rate of the test}. \end{align*} |
We extend our polynomial fit to early May. x = 60 corresponds to May 5 in Fig. 34. Then we solve,
| \begin{equation} 4.298 + 0.207x - 0.00616x^{2} + 0.000033x^{3} = 0, \end{equation} | [81] |
Extended dn/dt polynomial fit to May 5 (x = 60).
We find,
| \begin{equation*} x = 71.29 \to \text{5 of June}. \end{equation*} |
We assume we can increase the value of σ on the date x → x + Δt as follows:
We take into account the time lag between the infection and the appearance of the symptom in this analysis. We have from Appendix 2,
| \begin{equation} \frac{\sigma_{\varDelta t}}{1 - \sigma_{\varDelta t}\,\varDelta t} = \sigma\quad \text{or}\quad \sigma_{\varDelta t} = \frac{\sigma}{1 + \varDelta t\,\sigma}. \end{equation} | [82] |
| \begin{equation} \sigma_{\varDelta t} = 0.046. \end{equation} | [83] |
| \begin{equation} \sigma (\text{after date $x$}) = \frac{1.2\sigma_{\varDelta t}}{1 - 1.2\sigma_{\varDelta t}\varDelta t} = 0.243. \end{equation} | [84] |
After the date t we must add −(0.243–0.127)x to the polynomial,
| \begin{align} P &= 4.298 + 0.207x - 0.00616698x^{2} \\ &\quad + 0.00003387x^{3}. \end{align} | [85] |
| \begin{align} P_{T} &= 4.2987 + 0.091x - 0.006167x^{2} \\ &\quad + 0.00003387x^{3}. \end{align} | [86] |
| \begin{align} dn/dx &= 32.3/0.097 \times \exp[4.2987 + 0.091x \\ &\quad - 0.006167x^{2} + 0.00003387x^{3}]. \end{align} | [87] |
Modified curve with 20% increase of σΔt on May 15.
We obtain the date when dn/dx < 1 by solving:
| \begin{align} 1 &= (32.3/0.097) \times \exp[4.2987 + 0.091x \\ &\quad - 0.006167x^{2} + 0.0000339x^{3}]. \end{align} | [88] |
| \begin{equation} x = 62.41 \to \text{27 of May}. \end{equation} | [89] |
We now proceed to analyze the U.S. data.
The daily new infection of U.S. is given (https:// www.cdc.gov/coronavirus/2019-ncov/cases-updates/cases-in-us.html) as:
| \begin{align} & \{2797,3419,4777,3528,5836,8821,10934,10115,\\ &\quad 13487,16916,17965,19332,18251,22635,22562,\\ &\quad 27043,26315,18819,9338,63455,43438,34347,\\ &\quad 31534,31705,33251,33288,29145,24157,26385,\\ &\quad 27259,29164,29002,29916,26008,29468,26527,\\ &\quad 25868,37144,30226,33119,29355,23459,\\ &\quad 23901,26512,30787,30326,29671\}. \end{align} | [90] |
U.S. daily new infections from March 17 to May 2.
It is clear that we can divide the U.S. data into two periods: March 17 to April 4 and April 5 to May 2. Something happened on April 5 which should be investigated.
(1) Analysis of the first period. The initial period (March 17 to April 4) is shown in Fig. 37. This can be fitted by,
| \begin{align} &\exp[7.87537 + 0.0891674x + 0.0198095x^{2} \\ &\quad - 0.00106288x^{3}], \end{align} | [91] |
| \begin{align} \kappa &= \lambda (1 - ax)^{2} - 0.05 - \sigma (x) \\ &= 0.0891674 + 0.04x - 0.0032x^{2}. \end{align} | [92] |
New infections per day in the first period.
We put,
| \begin{equation} \sigma (x) = s + ux + vx^{2}. \end{equation} | [93] |
Then we get,
| \begin{equation} \lambda - 0.05 - s = 0.089, \end{equation} | [94a] |
| \begin{equation} - 2a\lambda - u = 0.04, \end{equation} | [94b] |
| \begin{equation} \lambda a^{2} - v = - 0.0032. \end{equation} | [94c] |
In this case, we may try to fit with the exponential of a second order polynomial, with the result,
| \begin{equation} dn/dx = \exp[7.38432 + 0.350848x - 0.0120767x^{2}], \end{equation} | [95] |
| \begin{equation} \kappa = \lambda (1 - 2ax) - 0.05 - \sigma = 0.351 - 0.036x, \end{equation} | [96] |
| \begin{equation} 2a\lambda = 0.036, \end{equation} | [97] |
| \begin{equation} \lambda - \sigma = 0.401. \end{equation} | [98] |
To estimate roughly these parameters, we use the Los Angeles value for λ assuming this parameter should be rather universal:
| \begin{equation} \lambda = 0.704. \end{equation} | [99] |
| \begin{equation} a = 0.026, \end{equation} | [100] |
| \begin{equation} \sigma = 0.303. \end{equation} | [101] |
(2) Analysis of the second period. The daily infection number is shown in Fig. 38. If we ignore bins of the first two days, April 5 and April 6 when something must have happened (actually several days before taking into account the time delay) the bins are more or less flat with the value around 30000 infections per day.
New infections per day in the second period.
This implies
| \begin{equation} d^{2}n/dx^{2} = \sigma dU/dx = \kappa \sigma U = 0, \end{equation} | [102] |
| \begin{equation} \kappa = \lambda f^{2} - \gamma - \sigma = 0, \end{equation} | [103] |
| \begin{equation} R_{t} = \frac{\lambda f^{2}}{\gamma + \sigma} = 1. \end{equation} | [104] |
This indicates the following:
(1) U.S. data shows that the f2 reached the non-zero minimum value and is staying with that value in the whole second period.
(2) Since there is no way to decrease f2 any further we can think of two ways to decrease Rt:
In order to roughly estimate the value of f2 we use the values obtained in analyzing the first period:
| \begin{equation} \lambda = 0.704,\quad \gamma = 0.05,\quad \sigma = 0.303. \end{equation} | [105] |
| \begin{equation} \kappa = \lambda f^{2} - \gamma - \sigma = 0.704f^{2} - 0.05 - 0.304 = 0. \end{equation} | [106] |
Towards the end of May, U.S. data shows slight decline in the number of daily new infections. We analyze this situation:
The data is given (https://www.cdc.gov/coronavirus/2019-ncov/cases-updates/cases-in-us.html) as:
| \begin{align} & \{2797,3419,4777,3528,5836,8821,10934,10115,\\ &\quad 13487,16916,17965,19332,18251,22635,22562,\\ &\quad 27043,26315,18819,9338,63455,43438,34347,\\ &\quad 31534,31705,33251,33288,29145,24157,26385,\\ &\quad 27259,29164,29002,29916,26008,29468,26527,\\ &\quad 25868,37144,30226,33119,29355,23459,23901,\\ &\quad 26512,30787,30326,29671,29763,19138,22303,\\ &\quad 23366,30861,25996,26660,23792,18106,21467,\\ &\quad 20869,27191,22977,31967,13284,24481,23405,\\ &\quad 22860,20522,24266,26229,15342,24958,16429,\\ &\quad 19680,21304\}. \end{align} | [107] |
Later period of U.S. data of daily new infection.
We fit this by an exponential of the third order polynomial,
| \begin{equation} p = 7.984 + 0.1876x - 0.004495x^{2} + 0.0000318x^{3}, \end{equation} | [108] |
Fitting of the U.S. data of new infection by Eq. [108].
From this we obtain,
| \begin{align} \kappa &= \lambda (1 - ax)^{2} - 0.05 - \sigma \\ & = 0.188 - 0.009x + 0.000096\,x^{2}. \end{align} | [109] |
| \begin{equation} a = 0.021, \end{equation} | [110a] |
| \begin{equation} \lambda = 0.214, \end{equation} | [110b] |
| \begin{equation} \sigma = - 0.024, \end{equation} | [110c] |
| \begin{align} R_{t} &= 1 + \frac{\kappa}{\sigma + \rho} \\ &= 1 + (0.188 - 0.009x + 0.000096\,x^{2}) \times 38.5. \end{align} | [111] |
Among the major European countries Italy, France and Germany are moving towards the recovery. But, as far as the data shows, U.K. and Sweden are still struggling with the constant or even increase of the new infections. U.K. is the worst and it is very similar to the situation in U.S. Sweden started with controlled initial stage and the total number did not increase as much as other countries but the current situation does not seem to be going well.
Daily new infection number from April 1 to April 30 is given (https://www.worldometers.info/coronavirus//country/sweden/) by,
| \begin{align} & \{512,519,612,365,387,376,487,726,722,544,466,\\ &\quad 332,465,497,482,613,676,606,563,392,545,\\ &\quad 682,751,812,610,463,286,695,681,790\}. \end{align} | [112] |
Swedish daily new infection from April 1 to April 30.
We fit this by a function exp[linear function of the date]. The result is,
| \begin{equation} dn/dx = \exp[6.13547 + 0.0096879x]. \end{equation} | [113] |
| \begin{equation} \kappa = \lambda - 0.05 - \sigma = 0.0097. \end{equation} | [114] |
Fit to the Swedish daily new infections (top part is cut off. See Fig. 41).
We understand that Sweden adopted the policy, at least partially, taking into account what is called “herd immunity model” which we discuss in Appendix 3. The above analysis implies that its success is not confirmed. Rather, its difficulty is implied in some serology test (antibody test) performed in many parts of the world.
Data is available in the form of histogram from WHO (https://covid19.who.int). The rough numbers are taken from the histogram every 2 days apart starting from March 31 to April 28:
| \begin{align} & \{\text{March}\ 31, \text{April}\ 2,4,6,8,10,12,14,16,18,20,22,\\ &\quad 24,26,28\} \to\{300,330,400,400,450,570,350,\\ &\quad 600,650,760,800,1000,1000,800,780\}. \end{align} | [115] |
Infection in African continent from March 31 to April 28.
This is fitted by the following exponential of a polynomial with the help of Mathematica:
| \begin{equation} dn/dx = \exp(5.542201 + 0.1318x - 0.003069x^{2}). \end{equation} | [116] |
| \begin{align} \kappa &= \lambda (1 - 2ax)^{2} - 0.05 - \sigma \\ &= 0.132 - 0.00614x, \end{align} | [117] |
| \begin{equation} 2a\lambda = 0.006, \end{equation} | [118] |
| \begin{equation} \lambda - 0.05 - \sigma = 0.132. \end{equation} | [119] |
| \begin{equation} a = 0.004,\quad \sigma = 0.522. \end{equation} | [120] |
Fitting curve and the histogram of daily African infection.
Among the Latin American countries Brazil has the largest case of infections followed by Peru and Mexico. Brazil is still in the state of sharp increase of new infections per day as the following data (https://www.worldometers.info/coronavirus/country/brazil/) shows:
| \begin{align} & \{\text{March}\ 27, 30, \text{April}\ 2,5,8,11,14,17,20,23,26,29,\\ &\quad \text{May}\ 2,5\} \to\{502,323,1076,852,133,1089,\\ &\quad 1832,217,1927,3735,3379,6276,4970,9493\}. \end{align} | [121] |
New infection number of Brazil.
This can be fitted by an exponential of a polynomial:
| \begin{equation} dn/dx = \exp[6.24 - 0.0775x + 0.0213x^{2}]. \end{equation} | [122] |
If we adopt the Los Angeles value of λ = 0.704, we get,
| \begin{equation} a = - 0.03, \end{equation} | [123a] |
| \begin{equation} \sigma = 0.576. \end{equation} | [123b] |
In the Eurasian continent Russia and India are as dangerous as Brazil in the data of new infections. China seems to have controlled the situation which is unique and we analyze the Chinese situation here. The daily new infections are given from January 30 to March 1 (https://www.worldometers.info/coronavirus/country/china/) as:
| \begin{align} & \{1500,1772,2066,2275,2622,2901,3067,3250,\\ &\quad 3260,3281,3171,2904,2015,14108,5090,\\ &\quad 2641,2008,2048,1888,1749,391,889,823,\\ &\quad 943,746,554,560,480,423,413,411,356\}. \end{align} | [124] |
Daily new infection in China from January 30 to March 1.
We fit this with exp[polynomial] and the result is,
| \begin{align} dn/dx &= \exp[6.7945 + 0.3398x \\ &\quad - 0.0230x^{2} + 0.000359x^{3}]. \end{align} | [125] |
Fit to the dn/dx shown in Fig. 47.
We can now calculate the parameters of our model:
| \begin{align} \kappa &= \lambda (1 - ax)^{2} - 0.05 - \sigma \\ &= 0.3398 - 0.046x + 0.00108x^{2}. \end{align} | [126] |
| \begin{equation} a = 0.047, \end{equation} | [127a] |
| \begin{equation} \lambda = 0.49, \end{equation} | [127b] |
| \begin{equation} \sigma = 0.101. \end{equation} | [127c] |
We now discuss some policy issues involved.
(1) Why do some countries have smaller numbers of infections? The number of infections increases exponentially, and human policy/culture can affect only the argument of this exponential function. This indicates that a small difference in the policy/culture can make a huge difference in the consequences. To explain the situation, we take a simple case of linear exponent approximation:
Assume that total infection N ∼ eax and a ∼ 0.8/day in one place A and 0.5/day in another place B at the initial stage. After 20 days we have,
| \begin{equation} N_{A}/N_{B} \sim e^{(0.8 - 0.5) \times 20} = e^{6} \sim 403. \end{equation} | [128] |
(2) What happens following the reopening after lockdown? We take Tokyo case as an example, where we have,
| \begin{equation} \lambda = 0.98, \end{equation} | [129a] |
| \begin{equation} a = 0.004, \end{equation} | [129b] |
| \begin{equation} \sigma = 0.73. \end{equation} | [129c] |
| \begin{equation} e^{(\lambda - \rho - \sigma)x} = e^{0.2x}\quad \text{and}\quad e^{(\lambda_{\Delta t} - \rho - \sigma_{\Delta t})x} = e^{0.003x}. \end{equation} | [130] |
Above is the case of total opening but we may have a partial opening with finite f in K = λf2. In the above case, to make
| \begin{equation} \kappa = 0.98f^{2} - 0.05 - 0.73 = 0, \end{equation} | [131] |
| \begin{equation} f = 0.9. \end{equation} | [132] |
In case of NY we have,
| \begin{equation} \kappa = 0.219f^{2} - 0.05 - 0.125 = 0, \end{equation} | [133] |
| \begin{equation} f = 0.89. \end{equation} | [134] |
10% reduction of social (or physical) contact from the normal time may or may not be possible. In case of whole Japan, we have,
| \begin{equation} \lambda = 0.384,\quad \sigma = 0.127, \end{equation} | [135] |
| \begin{equation} \kappa = 0.384f^{2} - 0.05 - 0.127 = 0. \end{equation} | [136] |
| \begin{equation} f = 0.68. \end{equation} | [137] |
This is especially true when non-symptomatic infection is dominating the spread as is shown in case of Tokyo in section 3.A.
It is efficient to strengthen the test opportunities without decreasing its positivity rate and isolate those who were tested positive for SARS-CoV-2 rather than isolate everyone by adopting the policy of social distancing, although social distancing is very helpful.
This is especially true because the number of asymptomatic infections is possibly much larger than that with symptom as demonstrated in New York or Los Angeles data and discussed in Appendix 5 of this paper. Unless we perform relatively larger scale test (first antibody test and then antigen or PCR test), we will be forced to lock down a city or a country again and again until efficient vaccine is invented.
As shown in Eq. [7] in section 2, the effect of testing and of recovery rate appears in the denominator of Rt implying the importance of these two effects. This is probably the most important aspect of our model that is distinct from the conventional interpretation of SIR model.
The analysis shows that U.S. is in trouble unless it can increase the test numbers substantially or some effective medicine is invented. U.K. also needs substantial increase of testing although we do not show the analysis here and Sweden needs it to a lesser degree. The Swedish policy of herd immunity is discussed in Appendix 3. The African continent seems to be responding reasonably well so far but ambiguity remains. Brazil seems to be heading for a disaster unless some decisive measure is taken. In Eurasian continent, Russia and India are still far from controlling the situation although we do not show the analysis in this paper.
I appreciate greatly a friend of mine Jiro Arafune (former director general of Institute for Cosmic Ray Research of the University of Tokyo) for discussions and suggestions. I also thank Professor Toshimichi Ikemura who was one of my colleagues at The Graduate University for Advanced Studies and an expert of virus research. I would also like to express my deep appreciation to Nobuhiro Go and Koichi Yazaki both of whom are highly recognized scientists and classmates of mine from the Physics Department of the University of Tokyo. They gave me excellent advice and constructive guidance. Other classmates, Yasushi Arai, Reiji Sano and Yuko Tokieda, encouraged me strongly for which I express my sincere gratitude. Finally, I would like to thank Jonathan Miller of OIST for his various useful comments and his suggestion to publish the draft paper.
Edited by Toshimitsu YAMAZAKI, M.J.A.
Correspondence should be addressed: H. Sugawara, KEK, 1-1 Oho, Tsukuba, Ibaraki 305-0801, Japan (e-mail: sugawara@post.kek.jp).
*) The Japanese government adopted this policy at the onset of COVID-19 infection in Japan (https://www.mhlw.go.jp/content/10906000/000599837.pdf (in Japanese)). This policy is particularly effective when there are “superspreaders”, for which there is some evidence as observed in Refs. 1–6.
**) See Appendix 6, and the references cited there, on the sensitivity of the test. The importance of PCR testing for suppression of the spread of SARS-CoV-2 is stressed, for example, in a message from OECD (https://www.oecd.org/coronavirus/policy-responses/testing-for-covid-19-a-way-to-lift-confinement-restrictions-89756248/). Some scientific analyses include the research by Takashi Odagaki (see Ref. 7).
Here I explain how we can reach the above model starting from the “standard model” in the field of mathematical epidemiology. We use standard notations rather than the our notations used in the above main sections: ρ → γ, x → t, U → It and n → Ih.
SIR model is defined to be a set of differential equations for S: susceptible, I: infected and R: recovered:
| \begin{equation} dS/dt = - \beta\,IS/N, \end{equation} | [1.1] |
| \begin{equation} dI/dt = \beta\,IS/N - \gamma I, \end{equation} | [1.2] |
| \begin{equation} dR/dt = \gamma I. \end{equation} | [1.3] |
Here β IS/N is the rate of decrease of the susceptible S and γ is the recovery rate and S + I + R = N. We assume N is a constant (total population of a body we want to discuss). We modify the model in the following way:
(1) Approximation. At least in case of COVID-19 we can safely assume that S ≫ I or R. Therefore, we can safely put,
| \begin{equation} S/N \simeq 1. \end{equation} | [1.4] |
| \begin{equation} dS/dt = - \beta\,I, \end{equation} | [1.5] |
| \begin{equation} dI/dt = \beta\,I - \gamma I, \end{equation} | [1.6] |
| \begin{equation} dR/dt = \gamma I. \end{equation} | [1.7] |
(2) Modification.
2.1) In SIR model it is usually assumed that β and γ are independent of time. But β depends on the social distancing, for example, and therefore depends on time. γ can be time dependent through increasing medical effort although natural recovery power is independent of time.
We can think of β as follows: it can be written as λf2 where λ is the infection rate at certain time and f is the contact rate of all the members of the population we consider. Both parameters can depend on time. The other contribution comes from the test.
2.2) More important modification is the following: We divide I into two parts: one is the infected but not recognized so and is at large. We write this as Iu. The other one is the infected and segregated.
| \begin{equation} I = I_{u} + I_{h}. \end{equation} | [1.8] |
| \begin{equation} dS/dt = - \beta\,I_{u}, \end{equation} | [1.9] |
| \begin{equation} dI/dt = \beta\,I_{u} - \gamma I, \end{equation} | [1.10] |
| \begin{equation} dR/dt = \gamma I. \end{equation} | [1.11] |
Ih comes out of I by the PCR test, and assuming test capacity can cover the entire symptomatic population, we have
| \begin{equation} dI_{h}/dt = \sigma I_{u}. \end{equation} | [1.12] |
Equation dI/dt = β Iu − γI becomes,
| \begin{align} &dI_{u}/dt + dI_{h}/dt = dI_{u}/dt + \sigma I_{u} \\ &\quad = \beta\,I_{u} - \gamma (I_{u} + I_{h}) = (\beta - \gamma)I_{u} - \gamma I_{h}. \end{align} | [1.13] |
| \begin{equation} dI_{u}/dt = (\beta - \sigma - \gamma)I_{u} - \gamma \int_{0}^{t}\sigma I_{u}\,dt. \end{equation} | [1.14] |
When we can ignore the last term, we reach the equation:
| \begin{equation} dI_{u}/dt = (\beta - \sigma - \gamma)I_{u}. \end{equation} | [1.15] |
One can include the effect of herd immunity (±something else that prevents infection) by assuming S ≠ N as follows:
| \begin{equation} S/N = 1 - m_{i}/N \equiv 1 - \phi, \end{equation} | [1.16] |
Then the equation for I becomes,
| \begin{equation} dI/dt = \beta\,I(1 - \phi) - \gamma I. \end{equation} | [1.17] |
| \begin{equation} \beta' = \beta (1 - \phi). \end{equation} | [1.18] |
| \begin{equation} (\beta - \sigma - \gamma) \to (\beta' - \sigma - \gamma) = (\beta (1 - \phi) - \sigma - \gamma). \end{equation} | [1.19] |
The time dependence of the herd immunity is given by the first equation of the SIR model:
| \begin{equation} dS/dt = - \beta\,I_{u}. \end{equation} | [1.5] |
| \begin{equation} \frac{Nd\phi}{dt} = \beta\,I_{u} = \beta\,I_{u0}\exp\left[\int_{0}^{x}\kappa (t)dt\right]. \end{equation} | [1.20] |
| \begin{equation} I_{u} = I_{u0}\exp\left[\int_{0}^{x}\kappa (t)dt\right]. \end{equation} | [4] |
Here we discuss how to incorporate the interval of time subsequent to infection until onset of symptoms. We use t as the time variable instead of x.
We start with,
| \begin{equation} dn/dt = \sigma U(t - \varDelta t). \end{equation} | [2.1] |
Then the equation,
| \begin{equation} dU(t)/dt = \kappa (t)\,U(t) = (K - \sigma - \gamma)U(t), \end{equation} | [2.2] |
| \begin{align} dU(t)/dt& = \kappa (t)\,U(t) = (K - \gamma)U(t) - \sigma U(t - \varDelta t) \\ &= \kappa U(t) - \sigma (U(t - \varDelta t) - U(t)). \end{align} | [2.3] |
| \begin{equation} U(t) = C(t)\exp\left[\int_{0}^{t}\kappa (t)dt\right]. \end{equation} | [2.4] |
| \begin{align} &(dC/dt)\exp\left[\int_{0}^{t}\kappa (t)dt\right] = \sigma (U(t) - U(t - \varDelta t))\\ & = \sigma U_{0}\left(\exp\left[\int_{0}^{t}\kappa (t)dt\right] - \exp\left[\int_{0}^{t - \Delta t}\kappa (t)dt\right]\right). \end{align} | [2.5] |
| \begin{equation} C = 1 + \sigma U_{0}\int_{0}^{t}dt\exp\left(1 - \exp \left[- \int_{t - \Delta t}^{t}\kappa (t)dt\right]\right), \end{equation} | [2.6] |
| \begin{align} dn/dt &= \sigma U(t - \varDelta t)\\ & = \sigma \exp\left[\int_{0}^{t - \Delta t}\kappa (t)dt \right]\\ & \quad \times \bigg\{1 + \sigma U_{0}\int_{0}^{t - \Delta t}dt\\ & \quad \times\exp\left(1 - \exp \left[- \int_{t - \Delta t}^{t}\kappa (t)dt\right]\right)\bigg\}. \end{align} | [2.7] |
| \begin{align} &dn/dt = \sigma U(t - \varDelta t) \\ &= \sigma \exp\left[\int_{0}^{t - \Delta t}\kappa (t)dt\right]\\ & \quad \times \left\{1 + \sigma U_{0}\varDelta t\int_{0}^{t - \Delta t}\exp \kappa (t)\,dt\right\}. \end{align} | [2.8] |
| \begin{equation} \kappa (t) = \lambda (1 - at)^{2} - \sigma - \gamma. \end{equation} | [2.9] |
Following analysis was performed by Koichi Yazaki. We start from Eq. [2.3]:
| \begin{equation} dU(t)/dt = \kappa U(t) - \sigma (U(t - \varDelta t) - U(t)). \end{equation} | [2.10] |
| \begin{equation} (1 - \sigma \varDelta t)dU(t)/dt = \kappa U(t). \end{equation} | [2.11] |
| \begin{equation} dU(t)/dt = \kappa' U(t), \end{equation} | [2.12] |
| \begin{equation} \kappa' \equiv \kappa/(1 - \sigma \varDelta t), \end{equation} | [2.13] |
| \begin{equation} U(t) = \exp\left[\int_{0}^{t}\kappa' dt\right]. \end{equation} | [2.14] |
On the other hand, from Eq. [2.1], we have,
| \begin{align} dn/dt &= \sigma U(t - \varDelta t) = \sigma \exp\left[\int_{0}^{t - \Delta t}\kappa' dt\right] \\ &\equiv \langle \sigma \rangle \exp[F(t - \varDelta t)]. \end{align} | [2.15] |
| \begin{equation} \kappa'(t - \varDelta t) = dF(t - \varDelta t)/dt \equiv dP(t)/dt. \end{equation} | [2.16] |
| \begin{align} dP/dt & = \kappa'(t - \varDelta t) = \kappa (t - \varDelta t)/(1 - \sigma \varDelta t)\\ & = \lambda (1 - at)^{2} - \gamma - \sigma = \lambda_{\varDelta t}(1 - a_{\varDelta t}(t - \varDelta t))^{2} \\ &\quad - \gamma_{\varDelta t} - \sigma_{\varDelta t}/(1 - \sigma_{\varDelta t}\varDelta t)\\ & = \frac{\lambda_{\varDelta t}(1 + a_{\varDelta t}\varDelta t)^{2}}{1 - \sigma_{\varDelta t}\varDelta t}\left(1 - \frac{a_{\varDelta t}t}{1 + a_{\varDelta t}\varDelta t}\right)^{2} \\ &\quad - \frac{\gamma_{\varDelta t} + \sigma_{\varDelta t}}{1 - \sigma_{\varDelta t}\varDelta t}, \end{align} | [2.17] |
| \begin{equation} \lambda_{\varDelta t}(1 + a_{\varDelta t}\varDelta t)^{2}/(1 - \sigma_{\varDelta t}\varDelta t) = \lambda, \end{equation} | [2.18a] |
| \begin{equation} a_{\varDelta t}/(1 + a_{\varDelta t}\varDelta t) = a, \end{equation} | [2.18b] |
| \begin{equation} (\gamma_{\varDelta t} + \sigma_{\varDelta t})/(1 - \sigma_{\varDelta t}\varDelta t) = \gamma + \sigma. \end{equation} | [2.18c] |
| \begin{equation} a_{\varDelta t} = a/(1 - \Delta t\,a), \end{equation} | [2.19a] |
| \begin{equation} \sigma_{\varDelta t} = \sigma/(1 + \varDelta t\,\sigma), \end{equation} | [2.19b] |
| \begin{equation} \gamma_{\varDelta t} = \gamma/(1 + \varDelta t\,\sigma), \end{equation} | [2.19c] |
| \begin{equation} \lambda_{\varDelta t} = \frac{(1 - \varDelta t\,a)^{2}}{1 + \varDelta t\,\sigma}\lambda. \end{equation} | [2.19d] |
This correction was pointed out in section 5. We can also continue to use λ, a, γ, σ rather than λΔt, aΔt, γΔt, σΔt in our formula.
We list the values of these parameters in the form of table created by Koichi Yazaki (Tables 3.a and 3.b).
| Δt | λ | a | σ |
|---|---|---|---|
| 0 | 0.3743 | 0.01648 | 0.1171 |
| 2 | 0.2886 | 0.01704 | 0.08775 |
| 4 | 0.2331 | 0.01764 | 0.06824 |
| 6 | 0.1914 | 0.01828 | 0.05519 |
| 8 | 0.1614 | 0.01898 | 0.04560 |
| 10 | 0.1380 | 0.01973 | 0.03828 |
| 12 | 0.1190 | 0.02054 | 0.03253 |
| 14 | 0.1933 | 0.02142 | 0.02792 |
| Δt | λ | a | σ |
|---|---|---|---|
| 0 | 0.9781 | 0.004523 | 0.7312 |
| 2 | 0.4123 | 0.004566 | 0.2854 |
| 4 | 0.2665 | 0.004608 | 0.1708 |
| 6 | 0.1996 | 0.004651 | 0.1185 |
| 8 | 0.1612 | 0.004695 | 0.08863 |
| 10 | 0.1362 | 0.004740 | 0.06939 |
| 12 | 0.1187 | 0.004785 | 0.05601 |
| 14 | 0.1057 | 0.004831 | 0.04621 |
Here we discuss a model which is in sharp contrast to our model described above:
Herd immunity model.This model asserts that acquiring herd immune of the entire population is the best way to stop the spreading of the virus. The policy was loosely taken in Sweden based on this model and, as was shown above, it is not a big success at least up to now. If Sweden will continue this policy remains to be seen. Here we describe this model and its hidden or explicit assumptions.
First, we point out that, when we justified our model starting from standard SIR model, we assumed that the susceptible population S is large compared to I (infected) or R (recovered). The herd immunity model asserts that S can be reduced substantially by the herd immunity in sharp contradiction with our model assumption.
We explain the model based on a recent paper by a Kyoto group10):
The points they are making are summarized as follows:
Here is the summary of assumptions made in the above analysis.
As for the item 6, we already have some data from New York and Santa Clara county in California: New York gives 14.9% (https://www.6sqft.com/new-york-covid-antibody-test-preliminary-results/) and Santa Clara (https://www.sccgov.org/sites/covid19/Pages/covid19) gives 2.8 to 4.2%, far below the value 55% adopted in the above paper. I interpret the importance of these numbers not by excluding the entire herd immune population from those susceptible but rather by including a part of them in the unidentified infection (U in my notation). This part still keeps the virus inside together with the antibody. The ratio of this part to the entire herd immunity is not known and the combined test of antibody and PCR is a way to find out the ratio.
In conclusion, both New York data and Santa Clara data justify our assumption of assuming the susceptible ≫ the infected or recovered up to 10% level.
The effect of reduction of susceptible parameter due to herd immunity is included in our scheme as discussed at the end of Appendix 1.
Appendix 4. Some parameters other than Rt11)Some authors11) use different indices to characterize the dynamics of COVID-19. For example, some researchers suggest as a possible index (time variable is t in the following):
| \begin{align} K_{d} &= \frac{N(d) - N(d - 7)}{N(d)} \\ &= \cfrac{\displaystyle\int_{0}^{d}\cfrac{dN}{dt}dt - \int_{0}^{d - 7}\cfrac{dN}{dt}dt}{\displaystyle\int_{0}^{d}\cfrac{dN}{dt}dt} = \cfrac{\displaystyle\int_{d - 7}^{d}\cfrac{dN}{dt}dt}{\displaystyle\int_{0}^{d}\cfrac{dN}{dt}dt}. \end{align} | [4.1] |
We have positive dN/dt (daily new infection). It is obvious that this number is between 0 and 1. This non-local (in time variable) index may have its own meaning. Here we compare this index with commonly used indices as follows:
We observe that the dangerous character of COVID-19 is that dN/dt seems to behave like exp[polynomial] and the polynomial is higher than the first order. It increases rapidly and decreases rapidly, too.
The indicator must clearly show how this polynomial is changing as a function of time. The above expression integrates over this exp[polynomial] and obscures the contribution of the exponentiated term, although it may reflect some other feature of COVID-19.
Suppose dN/dt = exp P and, instead of the above expression, we use,
| \begin{equation} \cfrac{\displaystyle\int_{d - 1}^{d}\cfrac{d^{2}N}{dt^{2}}dt}{\displaystyle\int_{0}^{d}\cfrac{d^{2}N}{dt^{2}}dt}. \end{equation} | [4.2] |
| \begin{equation} \kappa = \frac{d^{2}N/dt^{2}}{dN/dt} = dP/dt. \end{equation} | [4.3] |
| \begin{equation} dN/dt = \sigma U = \sigma \exp\left[\int_{0}^{t}\frac{dP}{dt}dt\right] = \sigma \exp\left[\int_{0}^{t}\kappa\,dt\right]. \end{equation} | [4.4] |
| \begin{equation} \kappa = \lambda f^{2} - \gamma - \sigma. \end{equation} | [4.5] |
| \begin{equation} R_{t} = \frac{\lambda f^{2}}{\gamma + \sigma}, \end{equation} | [4.6] |
Japanese government decided that the measure of exit from the emergency is,
| \begin{equation} \int_{d - 7}^{d}\frac{dn}{dt}dt \leq 0.5 \times 10^{-5}\,N. \end{equation} | [4.7] |
| \begin{equation} \int_{d - 7}^{d}\frac{dn}{dt}dt = \int_{d - 7}^{d}dt\,\sigma \exp\left[\int_{0}^{t}\kappa\,dt\right] \leq 0.5 \times 10^{-5}\,N, \end{equation} | [4.8] |
This can be written using the Kd discussed above:
| \begin{equation} K_{d} = \cfrac{\displaystyle\int_{d - 7}^{d}\cfrac{dn}{dt}dt}{\displaystyle\int_{0}^{d}\cfrac{dn}{dt}dt} \leq \cfrac{0.5 \times 10^{-5}\,N}{\displaystyle\int_{0}^{d}\cfrac{dn}{dt}dt} = \frac{0.5 \times 10^{-5}\,N}{n(d)/N}. \end{equation} | [4.9] |
Here n(d)/N is the ratio of total visible infection on d-th day divided by the population. Clearly Kd is between 0 and 1 but n(d)/N is not guaranteed to be larger than 0.5 × 10−5.
Anyway, the point here is that, in terms of Kd the criterion is dependent on the target area we consider. The question is whether to take Kd as a universal index or a target dependent parameter.
Appendix 5. Asymptomatic infections (This appendix is by Jiro Arafune with some comments of h.s.)We have unidentified infections (U in our notation; see for example https://www.fukuishimbun.co.jp/articles/-/1071784, http://www.kansensho.or.jp/uploads/files/topics/2019ncov/covid19_casereport_200409_5.pdf (both in Japanese)). There are two kinds among them: those officially with symptoms and those officially without. The latter can be further divided into two groups. One kind is those who have infection with symptoms too mild to satisfy the official criterion for testing. It depends on the strictness of the criteria for testing. For example, Japan had a very strict criterion such as 4 consecutive days of body temperature higher than 37.5 °C. Under such a strict criterion we have large number of individuals belonging to U who don’t contribute to n(t), the group of officially identified infections. Another kind is the infection of those who have developed the immunoglobulin against the SARS-CoV-2 but harbor the virus transiently before completely eliminating it or maintaining it permanently in the body. According to the data from New York and a part of California there are at least several percent of the population who belong to this category. It is possible that they transmit the virus to large numbers of people before they are no longer infectious.
We take two examples to estimate the number of unidentified symptomatic infections that is designated as U in our model: Japan and Tokyo. We use t as the time variable.
(1) Japan case. We calculate σΔt = σ/(1 + Δtσ) with σ = 0.127 (section 6) and Δt = 10.
We find σΔt = 0.056 and this gives,
| \begin{align} U &= \frac{1}{\sigma_{\varDelta t}}\frac{dn}{dt} = \frac{1}{0.056}\frac{dn}{dt} \\ &= \frac{1}{0.056}\exp[4.29867 + 0.20722t \\ &\quad - 0.00616699t^{2} + 0.00003387t^{3}], \end{align} | [5.1] |
Infection numbers U in Japan.
(2) Tokyo case. σΔt = 0.127 (section 5.A.) gives,
| \begin{align} U &= \frac{1}{\sigma_{\varDelta t}}\frac{dn}{dt} = \frac{1}{0.127}\frac{dn}{dt} \\ &= \frac{1}{0.127}\exp[2.75961 + 0.196892t\\ &\quad - 0.0044258t^{2} + 6.67625 \times 10^{-6}\,t^{3}]. \end{align} | [5.2] |
Infection number U in Tokyo.
This plot shows that there were about 100 unidentified symptomatic infections on May 1 but it will disappear on May 24 when dn/dt becomes less than 1. In fact, taking into account the time lag, unidentified infection with symptoms may have disappeared already by May 14.
But these numbers do not seem to include the asymptomatic infections as we will show later in this appendix. There may be more non-symptomatic cases.
It is important to realize that the remaining unidentified infections without symptoms can cause a lot of damage without further intervention. Therefore, we treat this issue here. We can divide U(t) into two groups; those with symptoms, and those without symptoms (due to two reasons as given above) even after infected.
We denote the number of the former and that of the latter, as U1 and U2, respectively (we neglect the effect of the finiteness of the time gap between the infection and the test).
We modify the original differential equations dU/dt = κU in the following:
| \begin{equation} dU_{1}/dt = (K_{11} - \rho_{1} - \sigma_{1})U_{1} + K_{12}\,U_{2}, \end{equation} | [5.3a] |
| \begin{equation} dU_{2}/dt = (K_{22} - \rho_{2} - \sigma_{2})U_{2} + K_{21}\,U_{1}. \end{equation} | [5.3b] |
| \begin{equation} dn/dt = \sigma_{1}\,U_{1} + \sigma_{2}\,U_{2}. \end{equation} | [5.4] |
| \begin{equation} K_{11} = K_{22} = K_{12} = K_{21} \equiv K. \end{equation} | [5.5] |
| \begin{equation} \sigma_{2} = 0. \end{equation} | [5.6] |
| \begin{equation} \frac{dU_{1}}{dt} = (K - \rho_{1} - \sigma_{1})U_{1} + KU_{2} \equiv \kappa_{1}U_{1} + KU_{2}, \end{equation} | [5.7a] |
| \begin{equation} \frac{dU_{2}}{dt} = (K - \rho_{2} - \sigma_{2})U_{2} + KU_{1} \equiv \kappa_{2}U_{2} + KU_{1}. \end{equation} | [5.7b] |
| \begin{align} &d^{2}U_{1,2}/dt^{2} - (\kappa_{1} + \kappa_{2})dU_{1,2}/dt \\ &\quad - (K^{2} - \kappa_{1}\kappa_{2})U_{1,2} = 0. \end{align} | [5.8] |
| \begin{equation} U_{1,2} = \xi_{1,2}e^{\lambda_{+}t} + \zeta_{1,2}e^{\lambda_{-}t}. \end{equation} | [5.9] |
| \begin{align} \lambda_{\pm}& = \frac{1}{2}\Big(2K - \rho_{1} - \sigma_{1} - \rho_{2} \\ &\quad \pm \sqrt{(\rho_{1} + \sigma_{1} - \rho_{2})^{2} + 4K^{2}}\Big). \end{align} | [5.10] |
| \begin{equation} \xi_{1,2} = \frac{K}{\lambda_{+} - \kappa_{1,2}}\xi_{2,1}, \end{equation} | [5.11a] |
| \begin{equation} \zeta_{1,2} = \frac{K}{\lambda_{-} - \kappa_{1,2}}\zeta_{2,1}. \end{equation} | [5.11b] |
(1) Initially, when infections are increasing, λ+ is positive; λ− can be positive or negative.
(2) Later when the U1 infection is decreasing,
| \begin{equation} \lambda_{+} \leq 0\quad \text{and}\quad \lambda_{-} \leq 0. \end{equation} | [5.12] |
λ− is more negative than λ+ so that later behavior is dominated by λ+.
Therefore, we get,
| \begin{align} \frac{U_{2}}{U_{1}} &\sim \frac{\xi_{2}e^{\lambda_{+}t}}{\xi_{1}e^{\lambda_{+}t}} \\ &= \frac{\xi_{2}}{\xi_{1}} = \frac{\lambda_{+} - \kappa_{1}}{K} \\ &= \cfrac{\cfrac{1}{2}\Big({- \kappa_{1}} + \kappa_{2} + \sqrt{(\kappa_{1} - \kappa_{2})^{2} + 4K^{2}}\Big)}{K}, \end{align} | [5.13] |
| \begin{align} - \kappa_{1} + \kappa_{2} &= - (K - \rho_{1} - \sigma_{1}) + (K - \rho_{2}) \\ &= \rho_{1} + \sigma_{1} - \rho_{2}. \end{align} | [5.14] |
| \begin{equation} \frac{U_{2}}{U_{1}} = \frac{\rho_{1} + \sigma_{1} - \rho_{2} + \sqrt{(\rho_{1} + \sigma_{1} - \rho_{2})^{2} + 4K^{2}}}{2K}. \end{equation} | [5.15] |
We consider two cases:
(1) ρ2 >= ρ1 + σ1. This case corresponds to the assumption that those without symptoms have much stronger resistance to the infection due to stronger personal immunity which may also result in faster recovery.
In this case we have,
| \begin{equation} \frac{U_{2}}{U_{1}} \sim \frac{- \rho_{2} + \sqrt{\rho_{2}{}^{2} + 4K^{2}}}{2K}. \end{equation} | [5.16] |
| \begin{equation} \frac{U_{2}}{U_{1}} \sim \frac{K}{\rho_{2}}. \end{equation} | [5.17] |
| \begin{equation} \frac{U_{2}}{U_{1}} \sim 1. \end{equation} | [5.18] |
| \begin{equation} \lambda_{+} = K\frac{U_{2}}{U_{1}} + \kappa_{1} = K\frac{U_{2}}{U_{1}} + K - \rho_{1} - \sigma_{1}, \end{equation} | [5.19] |
| \begin{equation} \approx K - \rho_{1} - \sigma_{1}{:}\ \text{during lockdown}, \end{equation} | [5.20] |
| \begin{equation} \approx 2K - \rho_{1} - \sigma_{1}{:}\ \text{after opening lockdown}. \end{equation} | [5.21] |
(2) If natural recovery does not change very much between U1 and U2, as was shown in at least some results of the study (https://www.fukuishimbun.co.jp/articles/-/1071784, http://www.kansensho.or.jp/uploads/files/topics/2019ncov/covid19_casereport_200409_5.pdf (both in Japanese)), we may put,
| \begin{equation} \rho_{1} = \rho_{2}. \end{equation} | [5.22] |
| \begin{equation} \frac{U_{2}}{U_{1}} = \frac{\sigma_{1} + \sqrt{\sigma_{1}{}^{2} + 4K^{2}}}{2K}. \end{equation} | [5.23] |
| \begin{equation} \frac{U_{2}}{U_{1}} \sim \frac{\sigma_{1}}{K}. \end{equation} | [5.24] |
| \begin{equation} \lambda_{+} = K\frac{U_{2}}{U_{1}} + \kappa_{1} = K\frac{U_{2}}{U_{1}} + K - \rho_{1} - \sigma_{1} = K - \rho_{1}, \end{equation} | [5.25] |
Also, the transition time of reopening after lockdown and the time lag between infection and expression of symptoms must be taken into account. Here we considered extreme cases to demonstrate the importance of asymptomatic infections.
This result means two things:
There exist large number of studies on this subject. For example, there is a report from FDA (Food and Drug Administration) entitled, “Accelerated Emergency Use Authorization (EUA) Summary Covid-19 RT-PCR Test (Laboratory Corporation of America)” (https://www.fda.gov/media/136151/download).
On the subject we refer to a few which serve for our purpose.
(1) Research in vitro. The main purpose of this research is to determine the minimum number of SARS-CoV-2 “molecules” per µL that can be detected by the Polymerase Chain Reaction (PCR) method. It depends on which part of the RNA sequence of the virus is used and how much of the complementary primer is used. It also depends on the duration of the polymerase chain reaction that multiplies copies of the sequence. Here we quote one study from the abovementined FDA report, which claims that 6.5 copies of RNA/µL can be detected.
(2) Statistical analysis. Sensitivity in the field is significantly different from the above studies in vitro. It depends on where the swab is taken: nasal, throat or sputum. It also depends crucially on the time when the test is performed: how many days after onset of symptoms. Statistical research by the Oxford group12) based on 298 tests across 30 patients (150 nasal and 148 throat swabs) shows the following:
