Question

Chemical Reactions When certain kinds of chemicals are combined, the rate at which the new compound is formed is modeled by the autonomous differential equation $\frac{\mathbf{d}\mathbf{X}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathit{k}\mathbf{(}\mathit{\alpha }\mathbf{-}\mathit{X}\mathbf{)}\mathbf{(}\mathit{\beta }\mathbf{-}\mathit{X}\mathbf{)}$where k > 0 is a constant of proportionality and $\mathit{\beta }\mathbf{>}\mathit{\alpha }\mathbf{>}\mathbf{0}$. Here X(t) denotes the number of grams of the new compound formed in time t.

(a) Use a phase portrait of the differential equation to predict the behavior of X(t) as$\mathit{t}\mathbf{\to }\mathbf{\infty }$.

(b) Consider the case when$\mathit{\alpha }\mathbf{=}\mathit{\beta }$. Use a phase portrait of the differential equation to predict the behavior of X(t) as$\mathit{t}\mathbf{\to }\mathbf{\infty }$when$\mathit{X}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{<}\mathit{\alpha }$, When$\mathit{X}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{>}\mathit{\alpha }$.

(c) Verify that an explicit solution of the DE in the case when k = 1 and$\mathit{\alpha }\mathbf{=}\mathit{\beta }$is$\mathit{X}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\mathit{\alpha }\mathbf{-}\mathbf{1}\mathbf{/}\mathbf{(}\mathit{t}\mathbf{+}\mathit{c}\mathbf{)}$Find a solution that satisfies$\mathit{X}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathit{\alpha }\mathbf{/}\mathbf{2}$. Then find a solution that satisfies$\mathit{X}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathit{\alpha }$. Graph these two solutions. Does the behavior of the solutions as$\mathit{t}\mathbf{\to }\mathbf{\infty }$agree with your answers to part (b)?

Short answer

a)If the initial value$X_0<\beta$then,$\underset{x\to \infty }{lim}X\left(t\right)=\alpha$

If the initial value$X_0<\beta$then,localid="1667892600536" $\underset{x\to \infty }{lim}X\left(t\right)=\infty$

b)If the initial value$X_0<\alpha$then,$\underset{x\to \infty }{lim}X\left(t\right)=\alpha$

If the initial valuelocalid="1667892722631" $X_0>\alpha$then,$\underset{x\to \infty }{lim}X\left(t\right)=\infty$

c)If $X\left(0\right)=\frac{\alpha }{2}$, then$X\left(t\right)=\alpha -\frac{1}{t+\frac{2}{\alpha }}$

If $X\left(0\right)=2\alpha$, then$X\left(t\right)=\alpha -\frac{1}{t-\frac{1}{\alpha }}$

Step-by-step solution

Step 1(a): Predict the behavior of X(t) as

Let $\frac{dX}{dt}=0$, then

$k(\alpha -X)(\beta -X)=0$

It is given that $k>0$, then must be

$$\begin{gathered} \alpha -X=0,\beta -X=0 \\ X=\alpha ,X=\beta \end{gathered}$$

Let $X=X_0<\alpha$, then$\alpha -X>0$and $\beta -X>0$. So $\frac{dX}{dt}>0$.

Let $X=X_0,\beta >X_0>\alpha$, then role="math" localid="1667893385554" $\alpha -X<0$and $\beta -X>0$. So$\frac{dX}{dt}<0$

Let $X=X_0>\beta$, then$\alpha -X<0$and $\beta -X<0$. So $\frac{dX}{dt}>0$.

The phase portrait for part (a)

The phase portrait line is shown below,

From the phase portrait line, we see that

if the initial value ${\mathit{X}}_{\mathbf{0}}\mathbf{<}\mathit{\beta }$then,

$\underset{\mathbf{x}\mathbf{\to }\mathbf{\infty }}{\mathbf{l}\mathbf{i}\mathbf{m}}\mathbf{}\mathit{X}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathit{\alpha }$

If the initial value${\mathit{X}}_{\mathbf{0}}\mathbf{>}\mathit{\beta }$then,

$\underset{\mathbf{x}\mathbf{\to }\mathbf{\infty }}{\mathbf{lim}}\mathbf{}\mathit{X}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{\infty }$

Step 3(b): Predict the behavior of X(t) as t→∞ when α=β

Let $\frac{dX}{dt}=0$, then

role="math" localid="1667894283579" $$\begin{gathered} k(\alpha -X)(\beta -X)=0 \\ \alpha =\beta ,So \\ k(\alpha -X)(\alpha -X)=0 \\ k(\alpha -X{)}^2=0 \end{gathered}$$

It is given that $k>0$, then must be

$$\begin{gathered} \alpha -X=0 \\ X=\alpha \end{gathered}$$

The phase portrait for part (b)

Since X represents the number of grams of a new compound formed the range for X is $[0,\infty )$

The phase portrait line is shown below,

If , then localid="1668425149824" $(\alpha -X{)}^2>0$. So localid="1668425155856" $\frac{dX}{dt}>0$. The resulting phase portrait is shown below:

From the phase portrait line, we see that

If the initial valuelocalid="1668425162830" ${\mathit{X}}_{\mathbf{0}}\mathbf{<}\mathit{\alpha }$then,

localid="1668425168855" $\underset{\mathbf{x}\mathbf{\to }\mathbf{\infty }}{\mathbf{l}\mathbf{i}\mathbf{m}}\mathbf{}\mathit{X}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathit{\alpha }$

If the initial value${\mathit{X}}_{\mathbf{0}}\mathbf{>}\mathit{\alpha }$then,

$\underset{\mathbf{x}\mathbf{\to }\mathbf{\infty }}{\mathbf{lim}}\mathbf{}\mathit{X}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{\infty }$

Step 5(c): Verify that an explicit solution of the DE in the case when k = 1 and α=β

Substitute$k=1$and$\alpha =\beta$in $\frac{dX}{dt}=k(\alpha -X)(\beta -X)$, then

$$\begin{gathered} \frac{dX}{dt}=(\alpha -X)(\alpha -X) \\ \frac{dX}{dt}=(\alpha -X{)}^2 \end{gathered}$$

Separate the variables,

$\frac{dX}{{(\alpha -X)}^2}=dt$

Integrating both sides,

$$\begin{gathered} \int \frac{dX}{{(\alpha -X)}^2}=\int dt \\ \frac{1}{\alpha -X}=t+c \end{gathered}$$

Simplify,

$X=\alpha -\frac{1}{t+c}·······\left(1\right)$

Find the value of c when X(0)=α2

Substitute $X\left(0\right)=\frac{\alpha }{2}$in (1), to get c,

$$\begin{gathered} X\left(0\right)=\alpha -\frac{1}{0+c} \\ \frac{\alpha }{2}=\alpha -\frac{1}{c} \end{gathered}$$

Simplify,

$c=\frac{2}{\alpha }$

Now substitute the value of localid="1668425223446" role="math" $c=\frac{2}{\alpha }$in (1)

localid="1668425245296" $X=\alpha -\frac{1}{t+\frac{2}{\alpha }}$

The interval of definition for this solution must include localid="1668425296641" $t=0$and localid="1668425299385" $X\left(t\right)$is undefined for localid="1668425302672" $t=-2/\alpha$. So the interval of definition localid="1668425307031" $[0,\infty )$. As localid="1668425310585" $t\to \infty$the fractional component approaches zero, so localid="1668425314296" $X\left(t\right)\to \alpha$. The resulting graph is shown below:

Find the value of c when

Substitute $X\left(0\right)=2\alpha$in (1), to get c,

$$\begin{gathered} X\left(0\right)=\alpha -\frac{1}{0+c} \\ 2\alpha =\alpha -\frac{1}{c} \end{gathered}$$

Simplify,

$c=-\frac{1}{\alpha }$

Now substitute the value of $c=-\frac{1}{\alpha }$in (1)

$X=\alpha -\frac{1}{t-\frac{1}{\alpha }}$

The interval of definition for this solution must include $t=0$and X(t) is undefined for $t=\frac{1}{\alpha }$. So the interval of definition $[0,\frac{1}{\alpha })$. As $t\to {\left(\frac{1}{\alpha }\right)}^{-1}$the fractional component approaches $-\infty$, so $X\left(t\right)\to \infty$. The resulting graph is shown below:

Therefore, If $\mathit{X}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\frac{\mathbf{\alpha }}{\mathbf{2}}$, then$\mathit{X}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\mathit{\alpha }\mathbf{-}\frac{\mathbf{1}}{\mathbf{t}\mathbf{+}\frac{\mathbf{2}}{\mathbf{\alpha }}}$

If $\mathit{X}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathit{\alpha }$, then$\mathit{X}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\mathit{\alpha }\mathbf{-}\frac{\mathbf{1}}{\mathbf{t}\mathbf{-}\frac{\mathbf{1}}{\mathbf{\alpha }}}$