Question

(a) Suppose \({\bf{a = b = 1}}\) in the Gompertz differential equation (7). Since the DE is autonomous, use the phase portrait concept of Section \({\bf{2}}{\bf{.1}}\) to sketch representative solution curves corresponding to the cases \({{\bf{P}}_{\bf{0}}}{\bf{ > e}}\) and \({\bf{0 < }}{{\bf{P}}_{\bf{0}}}{\bf{ < e}}\).

(b) Suppose \({\bf{a = 1,b = - 1}}\) in (7). Use a new phase portrait to sketch representative solution curves corresponding to the cases \({{\bf{P}}_{\bf{0}}}{\bf{ > }}{{\bf{e}}^{{\bf{ - 1}}}}\) and \({\bf{0 < }}{{\bf{P}}_{\bf{0}}}{\bf{ < }}{{\bf{e}}^{{\bf{ - 1}}}}\).

(c) Find an explicit solution of (7) subject to \({\bf{P(0) = }}{{\bf{P}}_{\bf{0}}}\).

Short answer

(a) So, the graph is shown as below:

(b) So, the sketch is shown as below:

(c) So, the required solution is \(P = {e^a}{e^{\left( {{\bf{n}}{P_0} - \frac{a}{b}} \right){e^{ - bc}}}}\).

Step-by-step solution

Gompertz differential equation

For the three parts of this problem we will deal with the Gompertz differential equation

\(\frac{{dP}}{{dt}} = P(a - b\ln P)\)

(a)Step 2: Evaluation

Let \(a = b = 1\)

Gompertz differential equation becomes

\(\frac{{dP}}{{dt}} = P(1 - \ln P)\)

We create the phase portrait for differential equation (1) by first solving \(\frac{{dP}}{{dt}} = 0\).

\(0 = P(1 - \ln P)\)

Recall \(\ln e = 1\)

Therefore, we have \(P = 0\) and \(P = e\).

This creates two intervals, \(0 < P < e\)and \(e < P\).

Plot the graph

Now we take values of \(P\) that are in each of the intervals. Substitute \(P = 1\) into the differential equation (1),

\(\frac{{dP}}{{dt}} = 1(1 - \ln 1)\)

\( = 1(1 - 0)\)

\( = 1\)

In the first step we use the fact that \(\ln 1 = 0\). Now we substitute \(P = 3\) into the differential equation (1),

\(\frac{{dP}}{{dt}} = 3(1 - \ln 3)\)

\( = 3(1 - 1.099)\)

\( = - 0.296\)

The resulting phase diagram and representative solution curves are shown below:

(b)Step 4: Evaluation

Let \(a = 1\) and \(b = - 1\)

Gompertz differential equation becomes

\(\frac{{dP}}{{dt}} = P(1 + \ln P)\)

We create the phase portrait for differential equation (2) by first solving \(dP/dt = 0\).

\(0 = P(1 + \ln P)\)

Recall \(\ln e = 1\). Using properties of logarithms we have \(\ln (1/e) = - 1\).

Therefore, we have \(P = 0\) and \(P = 1/e\).

This creates two intervals, \(0 < P < 1/e\) and\(1/e < P\).

Use properties of logarithm

Now we take values of \(P\) that are in each of the intervals. Substitute \(P = 1/{e^2}\) into the differential equation (2),

\(dP = \frac{1}{{{e^2}}}\left( {1 + \ln \frac{1}{{{e^2}}}} \right)\)

\( = \frac{1}{{{e^2}}}(1 + ( - 2))\)

\( = - \frac{1}{{{e^2}}}\)

We use properties of logarithms to evaluate the differential equation above and below. Now we substitute \(P = e\) into the differential equation (1),

\(\frac{{dP}}{{dt}} = e(1 + \ln e)\)

\( = e(1 + 1)\)

\( = 2e\)

Plot the graph

The resulting phase diagram and representative solution curves are shown below:

Evaluation

We can solve Gompertz differential equation,

\(\frac{{dP}}{{dt}} = P(a - b\ln P)\)

by separation of variables. Separating the expression yields,

\(\frac{{dP}}{{P(a - b\ln P)}} = dt{\rm{ }}\)

\(\int {\frac{{dP}}{{P(a - b\ln P)}}} = \int d t\)

In order to integrate the left hand side we use \(u\)-substitution with \(u = a - b\ln P\).

The result is \( - \frac{1}{b}\ln \mid a - b\ln P = t + c\).

Solve for constant

We use properties of logarithms and exponentials to solve for \(P\).

\(a - b\ln P = c{e^{ - bt}}\)

\(\ln P = \frac{a}{b} - {c_2}{e^{ - bt}}\)

\(P = {e^b}{e^{ - c{e^{ - b}}}}\)

Now we use the initial condition \(P(0) = {P_0}\) to solve for the constant \({c_2}\).

\({P_0} = {e^a}{e^{ - {c_2}{e^{ - (0)}}}}\)

\({P_0} = {e^{{b^{ - {c_2}}}}}\)

\(\ln {P_0} = \frac{a}{b} - {c_2}\)

\({c_2} = \frac{a}{b} - \ln {P_0}\)

We substitute the constant \({c_2}\) in order to get the final solution to the differential equation.

\(P = {e^a}{e^{\left( {{\bf{n}}{P_0} - \frac{a}{b}} \right){e^{ - bc}}}}\)