Question

(a) Suppose a = b = 1in the Gompertz differential equation (7). Since the DE is autonomous, use the phase portrait concept of Section 2.1to sketch representative solution curves corresponding to the cases ${\mathit{P}}_{\mathbf{0}\mathbf{}}\mathbf{>}\mathit{e}$and $\mathbf{0}\mathbf{}\mathbf{<}{\mathit{P}}_{\mathbf{0}}\mathbf{}\mathbf{<}\mathit{e}$.

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

(c) Find an explicit solution of (7) subject to$\mathit{P}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}{\mathit{P}}_{\mathbf{0}}$.

Short answer

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

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

(c) So, the required solution is $\mathit{P}\mathbf{=}{\mathit{e}}^{\mathbf{a}}\mathbf{}{\mathit{e}}^{{\left[\begin{array}{cc}n{P}_0-& \frac{a}{b}\end{array}\right]}^{{\mathbf{e}}^{\mathbf{-}\mathbf{b}\mathbf{c}}}}$.

Step-by-step solution

Gompertz differential equation

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

$\frac{\mathbf{dP}}{\mathbf{dt}}\mathbf{=}\mathit{P}\mathbf{(}\mathit{a}\mathbf{-}\mathbf{}\mathit{b}\mathit{l}\mathit{n}\mathbf{\left(}\mathbf{P}\mathbf{\right)}\mathbf{)}$

(a)Evaluation

Let a = b =1

Gompertz differential equation becomes

$\frac{\mathbf{d}\mathbf{P}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathit{P}\mathbf{(}\mathbf{1}\mathbf{-}\mathbf{}\mathit{l}\mathit{n}\left(P\right)\mathbf{)}$

We create the phase portrait for differential equation (1) by first solving $\frac{\mathbf{d}\mathbf{P}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{0}$.

$\mathbf{0}\mathbf{=}\mathit{P}\mathbf{(}\mathbf{1}\mathbf{-}\mathit{l}\mathit{n}\left(P\right)\mathbf{)}$

Recall $\mathit{l}\mathit{n}\left(e\right)\mathbf{=}\mathbf{1}$

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),

$\frac{\mathbf{d}\mathbf{P}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{}\mathbf{1}\mathbf{(}\mathbf{1}\mathbf{-}\mathit{l}\mathit{n}\left(1\right)\mathbf{)}$

= 1( 1 - 0 )

= 1

In the first step we use the fact that $\mathit{l}\mathit{n}\left(1\right)\mathbf{=}\mathbf{0}$. Now we substitute P = 3 into the differential equation (1),

$\frac{\mathbf{d}\mathbf{P}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{3}\mathbf{(}\mathbf{1}\mathbf{-}\mathbf{}\mathit{l}\mathit{n}\left(3\right)\mathbf{)}$

= 3 ( 1 - 1.099 )

= - 0.296

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

(b) Evaluation

Let a = 1 and b = -1

Gompertz differential equation becomes

$\frac{\mathbf{d}\mathbf{P}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathit{P}\mathbf{(}\mathbf{1}\mathbf{+}\mathit{l}\mathit{n}\left(P\right)\mathbf{)}$

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

$\mathbf{0}\mathbf{=}\mathit{P}\mathbf{(}\mathbf{1}\mathbf{+}\mathit{l}\mathit{n}\left(P\right)\mathbf{)}$

Recall $\mathit{l}\mathit{n}\left(e\right)\mathbf{=}\mathbf{1}$. Using properties of logarithms we have $\mathit{l}\mathit{n}\left(\frac{1}{e}\right)\mathbf{=}\mathbf{-}\mathbf{1}$.

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 $\mathit{P}\mathbf{=}\mathbf{1}\mathbf{/}{\mathit{e}}^{\mathbf{2}}$into the differential equation (2),

$\mathit{d}\mathit{P}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{e}}^{\mathbf{2}}}\left(1+ln\left(\frac{1}{e^2}\right)\right)$

$\mathbf{=}\frac{\mathbf{1}}{{\mathbf{e}}^{\mathbf{2}}}\left(1+(-2\right)\mathbf{)}$

$\mathbf{=}\mathbf{-}\frac{\mathbf{1}}{{\mathbf{e}}^{\mathbf{2}}}$

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

$\frac{\mathbf{d}\mathbf{P}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathit{e}\mathbf{(}\mathbf{1}\mathbf{+}\mathit{l}\mathit{n}\left(e\right)\mathbf{)}$

= e ( 1 + 1 )

= 2 e

Plot the graph

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

(C) Evaluation

$\mathit{u}\mathbf{=}\mathit{a}\mathbf{-}\mathit{b}\mathit{l}\mathit{n}\left(P\right)$We can solve Gompertz differential equation,

$\frac{\mathbf{d}\mathbf{P}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathit{P}\mathbf{(}\mathit{a}\mathbf{-}\mathit{b}\mathit{l}\mathit{n}\left(P\right)\mathbf{)}$

by separation of variables. Separating the expression yields,

$\frac{\mathbf{d}\mathbf{P}}{\mathbf{P}\mathbf{(}\mathbf{a}\mathbf{-}\mathbf{b}\mathbf{l}\mathbf{n}\left(P\right)}\mathbf{=}\mathit{d}\mathit{t}$

$\mathbf{\int }\frac{\mathbf{d}\mathbf{P}}{\mathbf{P}\mathbf{(}\mathbf{a}\mathbf{-}\mathbf{b}\mathbf{l}\mathbf{n}\left(P\right)}\mathbf{=}\mathbf{\int }\mathit{d}\mathit{t}$

In order to integrate the left hand side we use -substitution with .

The result is $\mathbf{-}\frac{\mathbf{1}}{\mathbf{b}}\mathit{l}\mathit{n}\left(a-bln\left(P\right)\right)\mathbf{=}\mathit{t}\mathbf{+}\mathbf{}\mathit{c}$.

Solve for constant

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

$\mathit{a}\mathbf{-}\mathit{b}\mathit{l}\mathit{n}\left(P\right)\mathbf{=}\mathit{c}{\mathit{e}}^{\mathbf{-}\mathbf{b}\mathbf{t}}$

$\mathit{l}\mathit{n}\left(P\right)\mathbf{=}\frac{\mathbf{a}}{\mathbf{b}}\mathbf{-}{\mathit{c}}_{\mathbf{2}}{\mathit{e}}^{\mathbf{-}\mathbf{b}\mathbf{t}}$

$\mathit{P}\mathbf{=}{\mathit{e}}^{\mathbf{b}}{\mathit{e}}^{\mathbf{-}\mathbf{e}{\mathbf{c}}^{\mathbf{-}\mathbf{b}}}$

Now we use the initial condition $\mathit{P}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}{\mathit{P}}_{\mathbf{0}}$to solve for the constant ${\mathit{c}}_{\mathbf{2}}$.

${\mathit{P}}_{\mathbf{0}}\mathbf{=}{\mathit{e}}^{\mathbf{2}}{\mathit{e}}^{\mathbf{-}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{-}\mathbf{\left(}\mathbf{0}\mathbf{\right)}}}$

${\mathit{P}}_{\mathbf{0}}\mathbf{=}{\mathit{e}}^{{\mathbf{b}}^{\mathbf{-}{\mathbf{c}}_{\mathbf{2}}}}$

$\mathit{l}\mathit{n}\left(P_0\right)\mathbf{=}\frac{\mathbf{a}}{\mathbf{b}}\mathbf{-}{\mathit{c}}_{\mathbf{2}}$

${\mathit{c}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{a}}{\mathbf{b}}\mathbf{-}\mathit{l}\mathit{n}\left(P_0\right)$

We substitute the constant in order to get the final solution to the differential equation.

$\mathit{P}\mathbf{=}{\mathit{e}}^{\mathbf{a}}{\mathit{e}}^{\left(n{P}_0-\frac{a}{b}\right){\mathbf{e}}^{\mathbf{-}\mathbf{b}\mathbf{c}}}$