Q8E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

If p(t) is a Malthusian population that diminishes according to (2), then\({\bf{p(}}{{\bf{t}}_{\bf{2}}}{\bf{) - p(}}{{\bf{t}}_{\bf{1}}}{\bf{)}}\)is the number of individuals in the population whose lifetime lies between\({{\bf{t}}_{\bf{1}}}\,\,{\bf{and}}\,\,{{\bf{t}}_{\bf{2}}}\). Argue that the average lifetime of the population is given by the formula\(\frac{{\int_{\bf{0}}^\infty  {{\bf{t}}\left| {\frac{{{\bf{dp(t)}}}}{{{\bf{dt}}}}} \right|{\bf{dt}}} }}{{{\bf{p(0)}}}}\)and show that this equals.

Step-by-Step Solution

Answer

For the required result follow the steps.

Step 1: Find the average lifetime result.

The average lifetime of the population p is given by\(\int_0^\infty  {{\bf{t}}{\bf{.w(t)dt}}} \).

 

Let’s, find the probability density for this problem. The result is 

 

\({\bf{W(}}{{\bf{t}}_{\bf{1}}}{\bf{,}}{{\bf{t}}_{\bf{2}}}{\bf{) = }}\frac{{{\bf{p(}}{{\bf{t}}_{\bf{2}}}{\bf{) - p(}}{{\bf{t}}_{\bf{1}}}{\bf{)}}}}{{{\bf{p(0)}}}}\)

 

Because the number of the alive individuals in the population decreases with the.

times, so that,

 

\({\bf{W(}}{{\bf{t}}_{\bf{1}}}{\bf{,}}{{\bf{t}}_{\bf{2}}}{\bf{) = }}\int_{{{\bf{t}}_{\bf{1}}}}^{{{\bf{t}}_{\bf{2}}}} {{\bf{w(t)}}} {\bf{dt}}\)

 

From Newton’s-Leibniz formula;

 

\(\begin{aligned}{c}{\bf{p(}}{{\bf{t}}_{\bf{2}}}{\bf{) - p(}}{{\bf{t}}_{\bf{1}}}{\bf{) = }}\int_{{{\bf{t}}_{\bf{1}}}}^{{{\bf{t}}_{\bf{2}}}} {\frac{{{\bf{dp(t)}}}}{{{\bf{dt}}}}} {\bf{dt}}\\{\bf{p(}}{{\bf{t}}_{\bf{2}}}{\bf{) - p(}}{{\bf{t}}_{\bf{1}}}{\bf{) =  - }}\int_{{{\bf{t}}_{\bf{1}}}}^{{{\bf{t}}_{\bf{2}}}} {\frac{{{\bf{dp(t)}}}}{{{\bf{dt}}}}} {\bf{dt}}\\{\bf{p(}}{{\bf{t}}_{\bf{2}}}{\bf{) - p(}}{{\bf{t}}_{\bf{1}}}{\bf{) = }}\int_{{{\bf{t}}_{\bf{1}}}}^{{{\bf{t}}_{\bf{2}}}} {\left| {\frac{{{\bf{dp(t)}}}}{{{\bf{dt}}}}} \right|} {\bf{dt}}\end{aligned}\)

 

Then, 

 

\(\frac{{\bf{1}}}{{{\bf{p(0)}}}}\int_{{{\bf{t}}_{\bf{1}}}}^{{{\bf{t}}_{\bf{2}}}} {\left| {\frac{{{\bf{dp(t)}}}}{{{\bf{dt}}}}} \right|} {\bf{dt = }}\int_{{{\bf{t}}_{\bf{1}}}}^{{{\bf{t}}_{\bf{2}}}} {{\bf{w(t)}}} {\bf{dt}}\)

 

The function under the integral is equal to\({\bf{w(t) = }}\frac{{\bf{1}}}{{{\bf{p(0)}}}}\left| {\frac{{{\bf{dp(t)}}}}{{{\bf{dt}}}}} \right|\).

 

Substitute this into the formula for the average lifetime of the population, so, 

 

\(\frac{1}{{p(0)}}\int_0^\infty  {{\bf{t}}\left| {\frac{{{\bf{dp(t)}}}}{{{\bf{dt}}}}} \right|} {\bf{dt}}\)

Step2: Find the value of p

Now,

 

\(\begin{aligned}{c}\frac{{{\bf{dp(t)}}}}{{{\bf{dt}}}}{\bf{ =  - kp}}\\{\bf{p(t) = p(0)}}{{\bf{e}}^{{\bf{ - kt}}}}\end{aligned}\)

 

Integrating by parts and substitution.

 

\(\begin{aligned}{c}\frac{1}{{{\bf{p}}(0)}}\int_0^\infty  {{\bf{t}}\left| {\frac{{{\bf{dp(t)}}}}{{{\bf{dt}}}}} \right|} {\bf{dt = }}\int_\infty ^0 {{\bf{t}}\frac{{{\bf{dp(t)}}}}{{{\bf{dt}}}}} {\bf{dt}}\\{\bf{ = }}\frac{1}{{{\bf{p}}(0)}}\int_0^\infty   {{\bf{p(t}})} {\bf{dt}}\\{\bf{ = }}\int_0^\infty  {p\left( 0 \right){{\bf{e}}^{{\bf{ - kt}}}}} {\bf{dt}}\\{\bf{ =  - }}\frac{{\bf{1}}}{{\bf{k}}}\end{aligned}\)

 

This is the required result.