User:Dushyantkpathak/sandbox

201701062 with the aid of existing and innovative science of calculus, this is an attempt to formulate a fatal human problem of epidemic into a calculus real analysis and with the aid of differential equations,try an analysis that helps in the eradication of the same. %This is a LaTeX template for homework assignments \documentclass{article} \usepackage[utf8]{inputenc} \usepackage{amsmath}

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%\section*{Assignment Title} %Name: \line(1,0){120} %you can change the length of the lines by changing %the number in the curly brackets %\\Date: \line(1,0){120}--- {\LARGE{\textbf{Project on Epidemic Spread}}}\\ Aim: \\ Considering a small group of people suffering from an infectious disease,inserted into a large population which is capable of catching the disease,we have to formulate a differential equation model and analyze it.\\ Formulation:\\

\begin{enumerate} \item Identify the variables taken into consideration. \item State basic assumptions including the law describing the spread of the epidemic. \item Define the differential equation model to formulate the aforementioned problem. \end{enumerate} Step 1: We divide the entire number into four disjoint sets:\\ A)The exposed(E):Those persons who are not as of the time that                 has been considered been affected by the                      epidemic,but are capable of catching the                      disease.\\          B)The Dormant(Y):Those who are currently infected but not yet                  capable of either detection,or transmission                   of the disease to the population.\\ C)The Ill(I): Persons who are ill and are capable of transmitting it to the others.\\        D)The Immune(M): Those who have had the disease and are either dead,or have acquired permanent immunity from the disease,or are quarantined(isolated from the community).\\ Step 2:the assumptions involved are:\\ A)the variables are treated as continuous values,which is acceptable as these are integers and the total population N is in lakhs or crores.\\        B)the total population, other than the deaths due to the   disease as included in M,remains constant,that is the number of births equals the number of deaths,which is again an acceptable assumption as the time frame is limited and a little    difference  can be neglected.\\ C)All the variables show change as functions of time,ignoring other factors.\\    Step 3:Analysis:\\         A)EY model(exposed-dormant) \begin{itemize} \item Declarations:\\ 1)E(t)+Y(t)+I(t)+L(t)=N……………1\\                 Total population is defined as the sum of all the above categories of people.\\               2)dE/dt=B I(t) E(t)………………..2\\ This declaration takes into consideration the fact that change in exposed(E) is only when people from the category ill(I) move into it,and that people from no other category can move into E.\\ And from Eqn.1 as N is conserved, increase in E is compensated by a decrease in I.\\ 3)For the particular case we take\\                    Y(t)=0 and L(t)=0\\                Hence from equation 1\\                    we get\\                      E(t)+I(t)=0 for t>0\\       \item Formation:\\                       Let I(0)=I0,E(0)=E0=N-I0\\ Now E+I=N\\ Differentiating both sides with respect to time,\\ dE/dt+dI/dt=dN/dt=0 (as population N is assumed constant.)\\ dI/dt refers to the change in ill people which can only be increased due to intake from E to I,considering Y and L to be 0.Now ill people can only be changed when exposed move to ill,hence\\ dI/dt is proportional to E and I\\ hence dI/dt=B I (N-I)\\ dS/dt=B S (N-S)\\ \item solution:\\ I(N-I)=c eNBt\\ \end{enumerate} Initial conditions application gives value of the arbitrary constant c.\\ Finally we get\\ (Ie-BNt/(N-I))=1/(N/I0-1)\\ Hence I(t)= [N/{1+(N/I0-1)e-BNt}]\\ E(t)=N-I(t)=[N(N/I0-1)/eBNt+(N/I0-1)}]\\ Equation A and its derivative equations are thus found out,and the interpretations made are as follows:\\ 1)I(t) is an increasing function of time with I0>>0 and **lim [I(t)]=N\\ 2)a. I’(t)>0 implies N-2I>0 implies IN/2\\ c.I’t=0 implies N=2I implies I=N/2\\ hence the graph is concave upwards for IN/2 and has a point of inflexion at I=N/2.\\ Similarly we can find out the graph of S(t) vs t which is like a mirror image of I(t).\\ Graph of I’(t)vs.t(the epidemic curve)\\ It is so called because the rate of change of I(t),measures the spread of the epidemic.\\ From equations **\\ Which imply that I’(t) is maximum when I=N/2\\ Then from equations **\\ t=1/BN[log(N/I0-1)]=tmax\\ [I’(t)]max=BN2/4\\ Then\\ I’(tmax+k)=N2Be**\\ It follows that the epidemic curve is symmetrical about the ordinate\\ T=tmax=1/BN [log(N/I0-1)]\\ The highest point on the curve is given by (tmax,I’max)\\ B)EYE model\\ in this modified version we assume that the ill individual can recover and become exposed againat at a rate which dependsonly on I,i.e. M.I where M is a positive constant.\\ now I can be changed by either entry from E or by going back to E.\\ hence dI/dt=I(N-I)-MI\\ dS/dt=-E(N-E) + MI\\ dI/dt= I[(N-M/B)-I]\\ Solution:\\  I/N-I=Ce{Bt}\\   application of initial conditions gives\\   C=[1/a/I{0}-1]\\   thus [I e{-Bat}/(a-I)]=[1/a/I{0}-1]\\   I(t)={a/[1+(a/I{0}-1)e^{-Bat}]}\\ S(t)=N-I(t)=[(N-a)+N(a/(I0-1))e-Bat]/[1+(a/I0-1)e-Bat]\\ Interpretation\\ a=N-(M/B)\\ Case 1)let a>0 i.e. B>M/N\\ Lim I(t)=a/1=N-M/B\\ Case 2)let a<0 i.e. BM/N\\        = N       B<M/N \\ C)EIL model\\ We assume that the individuals are removed from the ill class at a rate proportional to I. \\ dL/dt=G I \\ G is a positive constant called the removal rate. \\ Removal may be due to death,permanent immunity or quarantine. \\ Hence\\ E(t)+I(t)+L(t)=N\\ Diff. w.r.t. t on b.s.\\ dI/dt=-dE/dt-dL/dt\\ =BEI-GI\\ Differential equation model in this case is given by\\ dE/dt=-BEI\\ dI/dt=BEI-GI\\ dL/dt=GI\\ we consider ,in this model that the group of ill increases by introduction from the E and decreases by removal to the L group.\\ Solution\\ dE/dL=(dE/dt)/(dL/dt)=-BE/G=-E/P\\ where P=G/B\\ Also\\ dI/dE=-1+(P/E)\\ solving,we get\\ S(t)=S0e—R/P\\ I(t)=(N-S)+P log(S/S0)\\ Interpretation\\ I)Ultimate behavior of I(t),E(t) and L(t):\\ The qualitative behaviours of E,L,and I are as follows\\ i)	**limL(t)=Lin<=N\\ ii)	E(t) =E>=0\\ iii)	I(t) will increase only if\\ iv)	dI/dt>0 E>P P=G/B\\ v)	P is called the relative removal rate.\\ vi)	Iin=(N-Lin-Ein)\\ II) threshold phenomena\\ If the initial population of the exposed is below P then that means that dI/dt is negative,that is ,the number of ill is falling down which implies the epidemic is reducing.thus there is a critical value of P which the initial E population must exceed in order for the existence of the epidemic.\\ This is called the threshold phenomena.\\ III)Ultimately some persons will always escape epidemic ,i.e. Ein>0\\ Ein(t)=E0 e-R(t)/P\\ Since 0E0e-N/P\\ Since e-N/P is strictly positive,\\ Ein(t)>0\\ Which implies,that there will always be some people in the community who escape the disease.\\ IV)Relation between I(t) and E(t) LimE->E0I(t)=N-E0=I0 LimE->0+ I(t)=-inf** dI/ds=0 =>S=P d2I/dS2=-P/S2<0 thus we can see that the graph of I vs S starts with the point (S0,I0) and is traced from the right to left with concavity downwards and the highest point given by S=P,it crosses the axis I=0 at a point (Sin,0). In view of the threshold phenomena ,we have the following conclusion :if the initial point P0(S0,I0) lies to the left of the ordinate P then there will be no epidemic;if P0 lies to the right of the ordinate S=P,the curve will start at (S0,I0) trace itself from right to left,initially rising upto the point H and then fall till it meets the axis I=0 at F(Sin,0).
 * dI/[I(N-I)]=B**dt +constant\\
 * 1/N[dI/I+dI/N-I]=Bt+k\\
 * (aI)/I(a-I)=B**dt +constant\\
 * lim e-Bat =0 therefore\\

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