Question

The system in Problem 19, like the system in (2), can be solved with no advanced knowledge. Solve for \(x(t)\) and \(y(t)\) and compare their graphs with your sketches in Problem 19. Determine the limiting values of \(x(t)\) and \(y(t)\) as \(t \to \infty \). Explain why the answer to the last question makes intuitive sense.

Short answer

The solution is

\(\begin{aligned}{l}x(t) = {c_1}{e^{\sqrt {\frac{{{k^2}\left( {{V_A} - {V_B}} \right)}}{{{V^2}_A{V_B}}}} t}} + {c_2}{e^{ - \sqrt {\frac{{{k^2}\left( {{V_A} - {V_B}} \right)}}{{{V^2}_A{V_B}}}t} }}\\y(t) = {c_1}\sqrt {\frac{{{V_A}}}{{{V_B}}}} {e^{\sqrt {\frac{{{k^2}\left( {{V_A} - {V_B}} \right)}}{{{V^2}_A{V_B}}}} t}} - {c_2}\sqrt {\frac{{{V_A}}}{{{V_B}}}} {e^{ - \sqrt {\frac{{{k^2}\left( {{V_A} - {V_B}} \right)}}{{{V^2}_A{V_B}}}} t}}\end{aligned}\)v

Step-by-step solution

Given information

The limiting values of \(x(t)\) and \(y(t)\) as \(t \to \infty \).

Find second order differential equation

The two differential equations describe two compartments A and B with volumes \({V_A}\) and \({V_B}\) and nutrient concentrations \(x(t)\) and \(y(t)\).

\(\begin{aligned}{l}\frac{{dx}}{{dt}} = \frac{k}{{{V_A}}}\left( {y - x} \right)...\left( 1 \right)\\\frac{{dy}}{{dt}} = \frac{k}{{{V_B}}}\left( {x - y} \right)...\left( 2 \right)\end{aligned}\)

Which are

\(\begin{aligned}{l}\left( {\frac{k}{{{V_A}}} + \frac{d}{{dt}}} \right)x = \frac{k}{{{V_A}}}y...\left( 3 \right)\\\frac{k}{{{V_B}}}x = \left( {\frac{k}{{{V_A}}} + \frac{d}{{dt}}} \right)y...\left( 4 \right)\end{aligned}\)

Then we must use the following strategy to solve for x and y.

Multiply the first equation shown in (3) by \(\left( {\frac{k}{{{V_A}}} + \frac{d}{{dt}}} \right)\) and the second equation shown in (4) by \( - \frac{k}{{{V_A}}}\), then we have

\(\begin{aligned}{l}{\left( {\frac{k}{{{V_A}}} + \frac{d}{{dt}}} \right)^2}x = \frac{k}{{{V_A}}}\left( {\frac{k}{{{V_A}}} + \frac{d}{{dt}}} \right)...\left( 5 \right)\\ - \frac{{{k^2}}}{{{V_A}{V_B}}}x = - \frac{k}{{{V_A}}}\left( {\frac{k}{{{V_A}}} + \frac{d}{{dt}}} \right)...\left( 6 \right)\end{aligned}\)

Add equation (5) to equation (6)

\(\begin{aligned}{l}{\left( {\frac{k}{{{V_A}}} + \frac{d}{{dt}}} \right)^2}x - \frac{{{k^2}}}{{{V_A}{V_B}}}x = \frac{k}{{{V_A}}}\left( {\frac{k}{{{V_A}}} + \frac{d}{{dt}}} \right)y - \frac{k}{{{V_A}}}\left( {\frac{k}{{{V_A}}} + \frac{d}{{dt}}} \right)y\\{\left( {\frac{k}{{{V_A}}} + \frac{d}{{dt}}} \right)^2}x - \frac{{{k^2}}}{{{V_A}{V_B}}}x = 0\\\frac{{{d^2}x}}{{d{t^2}}} + \frac{{2k}}{{{V_A}}}\frac{d}{{dt}} + \frac{{{k^2}}}{{{V^2}_A}}x - \frac{{{k^2}}}{{{V_A}{V_B}}}x = 0\end{aligned}\)

\(\frac{{{d^2}x}}{{d{t^2}}} + \frac{{2k}}{{{V_A}}}\frac{{dx}}{{dt}} + \frac{{{k^2}\left( {{V_B} - {V_A}} \right)}}{{{V^2}_A{V_B}}}x = 0\)

Second order differential equation for the concentration \(x\).

Find general solution

To solve the differential equation in equation (5), we must first assume that \(x = {e^{mt}}\), and then differentiate with respect to \(t\) as follows:

\(\begin{aligned}{l}\frac{{dx}}{{dt}} = m{e^{mt}}....\left( a \right)\\\frac{{{d^2}x}}{{d{t^2}}} = {m^2}{e^{mt}}....\left( b \right)\end{aligned}\)

Substitute with equation (a) and (b) into equation (7)

\(\begin{aligned}{l}{m^2}{e^{mt}} + \frac{{2k}}{{{V_A}}}\left( {m{e^{mt}}} \right) + \frac{{{k^2}\left( {{V_B} - {V_A}} \right)}}{{{V^2}_A{V_B}}}{e^{mt}} = 0\\{m^2}{e^{mt}} + \frac{{2km}}{{{V_A}}}\left( {{e^{mt}}} \right) + \frac{{{k^2}\left( {{V_B} - {V_A}} \right)}}{{{V^2}_A{V_B}}}{e^{mt}} = 0\\\left( {{m^2} + \frac{{2km}}{{{V_A}}} + \frac{{{k^2}\left( {{V_B} - {V_A}} \right)}}{{{V^2}_A{V_B}}}} \right){e^{mt}} = 0\end{aligned}\)

Since \({e^{mt}}\) cannot equal to \(0\).

The auxiliary equation is

\({m^2} + \frac{{2km}}{{{V_A}}} + \frac{{{k^2}\left( {{V_B} - {V_A}} \right)}}{{{V^2}_A{V_B}}} = 0\)

Then we have

\(\begin{aligned}{l}{m_{1,2}} = \frac{{ - \frac{{2k}}{{{V_A}}} \pm \sqrt {\frac{{4{k^2}}}{{{V^2}_A}} - \frac{{4{k^2}\left( {{V_B} - {V_A}} \right)}}{{{V^2}_A{V_B}}}} }}{2}\\ = \frac{{ - \frac{{2k}}{{{V_A}}} \pm \frac{{2k}}{{{V_A}}}\sqrt {1 - \frac{{\left( {{V_B} - {V_A}} \right)}}{{{V_B}}}} }}{2}\\ = - \frac{k}{{{V_A}}} \pm \frac{k}{{{V_A}}}\sqrt {\frac{{{V_A}}}{{{V_B}}}} \end{aligned}\)

General solution for concentration \(x\) as

\(\begin{aligned}{l}x(t) = {c_1}{e^{{m_1}t}} + {c_2}{e^{{m_2}t}}\\ = {c_1}{e^{\frac{k}{{{V_A}}}\left( {\sqrt {\frac{{\left( {{V_A} - {V_B}} \right)}}{{{V_B}}}} } \right)t}} + {c_2}{e^{ - \frac{k}{{{V_A}}}\left( {\sqrt {\frac{{\left( {{V_A} - {V_B}} \right)}}{{{V_B}}}} } \right)t}}\\ = {c_1}{e^{\sqrt {\frac{{{K^2}\left( {{V_A} - {V_B}} \right)}}{{{V^2}_A{V_B}}}} t}} + {c_2}{e^{ - \sqrt {\frac{{{K^2}\left( {{V_A} - {V_B}} \right)}}{{{V^2}_A{V_B}}}} t}}\end{aligned}\)

Find solution for concentration

After that, we must insert the answer for \(x\) given in equation (8) into equation (1) to discover the solution for concentration \(y(t)\).

\(\frac{d}{{dt}}\left( {{c_1}{e^{\sqrt {\frac{{{k^2}\left( {{V_A}{V_B}} \right)}}{{V_A^2{V_B}}}} t}} + {c_2}{e^{ - \sqrt {\frac{{{k^2}\left( {{V_A}{V_B}} \right)}}{{V_A^2{V_B}}}} t}}} \right) = \frac{k}{{{V_A}}}y\)

\( - \frac{k}{{{V_A}}}\left( {{c_1}{e^{\sqrt {\frac{{{k^2}\left( {{V_A}{V_B}} \right)}}{{V_A^2{V_B}}}} t}} + {c_2}{e^{ - \sqrt {\frac{{{k^2}\left( {{V_A} + {V_B}} \right)}}{{V_A^2{V_B}}}} t}}} \right)\)

\({c_1}\sqrt {\frac{{{k^2}\left( {{V_A} - {V_B}} \right)}}{{V_A^2{V_B}}}} {e^{\sqrt {\frac{{{k^2}\left( {{V_A} - {V_B}} \right)}}{{V_A^2{V_B}}}} t}} - {c_2}\sqrt {\frac{{{k^2}\left( {{V_A} + {V_B}} \right)}}{{V_A^2{V_B}}}} {e^{ - \sqrt {\frac{{{k^2}\left( {{V_A} + {V_B}} \right)}}{{V_A^2{V_B}}}} t}} = \frac{k}{{{V_A}}}y\)

\( - \frac{k}{{{V_A}}}\left( {{c_1}{e^{\sqrt {\frac{{{k^2}\left( {{V_A} - {V_B}} \right)}}{{V_A^2{V_B}}}} t}} + {c_2}{e^{ - \sqrt {\frac{{{k^2}\left( {{V_A}{V_B}} \right)}}{{V_A^2{V_B}}}} t}}} \right)\)

\(\begin{aligned}{l}{c_1}\frac{k}{{{V_A}}}\sqrt {\frac{{{V_A} - {V_B}}}{{{V_B}}}} {e^{\sqrt {\frac{{{k^2}\left( {{V_A} - {V_B}} \right)}}{{V_A^2{V_B}}}} t}} + {c_1}\frac{k}{{{V_A}}}{e^{\sqrt {\frac{{{k^2}\left( {{V_A} - {V_B}} \right)}}{{V_A^2{V_B}}}} t}}\\ - {c_2}\frac{k}{{{V_A}}}\sqrt {\frac{{{V_A} + {V_B}}}{{{V_B}}}} {e^{ - \sqrt {\frac{{{k^2}\left( {{V_A} + {V_B}} \right)}}{{V_A^2{V_B}}}} t}} + {c_2}\frac{k}{{{V_A}}}{e^{ - \sqrt {\frac{{{k^2}\left( {{V_A} + {V_B}} \right)}}{{V_A^2{V_B}}}} t}} = \frac{k}{{{V_A}}}y\end{aligned}\)

\(\begin{aligned}{l}{c_1}\frac{k}{{{V_A}}}\sqrt {\frac{{{V_A} - {V_B}}}{{{V_B}}}} e\sqrt {\frac{{{k^2}\left( {{V_A} - {V_B}} \right)}}{{V_A^2{V_B}}}} t + {c_1}\frac{k}{{{V_A}}}e\sqrt {\frac{{{k^2}\left( {{V_A}{V_B}} \right)}}{{V_A^2{V_B}}}} t\\ - {c_2}\frac{k}{{{V_A}}}\sqrt {\frac{{{V_A} + {V_B}}}{{{V_B}}}} {e^{ - \sqrt {\frac{{{k^2}\left( {{V_A} + {V_B}} \right)}}{{V_A^2{V_B}}}} t}} + {c_2}\frac{k}{{{V_A}}}{e^{ - \sqrt {\frac{{{k^2}\left( {{V_A} + {V_B}} \right)}}{{V_A^2{V_B}}}} t}} = \frac{k}{{{V_A}}}y\end{aligned}\)’

\({c_1}\frac{k}{{{V_A}}}\sqrt {\frac{{{V_A}}}{{{V_B}}}} {e^{\sqrt {\frac{{{k^2}\left( {{V_A}{V_B}} \right)}}{{V_A^2{V_B}}}} t}} - {c_2}\frac{k}{{{V_A}}}\sqrt {\frac{{{V_A}}}{{{V_B}}}} {e^{ - \sqrt {\frac{{{k^2}\left( {{V_A} + {V_B}} \right)}}{{V_A^2{V_B}}}} t}} = \frac{k}{{{V_A}}}y\)

Then we have

\(y(t) = {c_1}\sqrt {\frac{{{V_A}}}{{{V_B}}}} {e^{\sqrt {\frac{{{k^2}\left( {{V_A} - {V_B}} \right)}}{{V_A^2{V_B}}}} t}} - {c_2}\sqrt {\frac{{{V_A}}}{{{V_B}}}} {e^{ - \sqrt {\frac{{{k^2}\left( {{V_A} + {V_B}} \right)}}{{V_A^2{V_B}}}} t}} - - - - (9)\)

is the solution for concentration \(y.\)

Now we must substitute \(t = \infty \) into equations (8) and (9) to find the limiting values of \(x(t)\) and \(y(t)\), respectively.

\(\begin{aligned}{l}\mathop {lim}\limits_{t \to \infty } x = {c_1}{e^\infty } + {c_1}{e^{ - \infty }}\\ = \infty - 0\\ = \infty \end{aligned}\)

\(\begin{aligned}{l}\mathop {lim}\limits_{t \to \infty } y = {c_1}\sqrt {\frac{{{V_A}}}{{{V_B}}}} {e^\infty } + {c_2}\sqrt {\frac{{{V_A}}}{{{V_B}}}} {e^{ - \infty }}\\ = \infty - 0\\ = \infty \end{aligned}\)

Conclusion

The solution is

\(\begin{aligned}{l}x(t) = {c_1}{e^{\sqrt {\frac{{{k^2}\left( {{V_A} - {V_B}} \right)}}{{{V^2}_A{V_B}}}} t}} + {c_2}{e^{ - \sqrt {\frac{{{k^2}\left( {{V_A} - {V_B}} \right)}}{{{V^2}_A{V_B}}}t} }}\\y(t) = {c_1}\sqrt {\frac{{{V_A}}}{{{V_B}}}} {e^{\sqrt {\frac{{{k^2}\left( {{V_A} - {V_B}} \right)}}{{{V^2}_A{V_B}}}} t}} - {c_2}\sqrt {\frac{{{V_A}}}{{{V_B}}}} {e^{ - \sqrt {\frac{{{k^2}\left( {{V_A} - {V_B}} \right)}}{{{V^2}_A{V_B}}}} t}}\end{aligned}\)