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Heather's questions at Yahoo! Answers regarding the modeling of herbicide absorption with an IVP

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MarkFL

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Feb 24, 2012
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Here is the question:

Mathematical Model Question?

A simple model for the absorption of a herbicide placed on the surface of a leaf is that it will be absorbed at a rate proportional to the difference in concentration between that on the surface and that on the interior. Assume the rate constant is α.

(a) If a layer with constant concentration Ca is placed on the leaf, write an equation for the concentration of herbicide in the leaf, C(t) as time progresses.

(b) Solve the equation for concentration C(t) if there is no herbicide in the leaf to begin.

(c) Now suppose the plant circulatory system (xylem and phloem) disperses the herbicide throughout the plant with rate proportional to concentration in the leaf and rate constant β. Modify your equation accordingly and hence your solution.

(d) What is the difference in outcome between cases in (b) and (c)?
I have posted a link there to this topic so the OP can see my work.
 
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MarkFL

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Feb 24, 2012
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Hello Heather,

(a) The time rate of change of herbicide in the leaf is proportional to the difference between the constant concentration on the surface and the variable concentration in the leaf, so we may state:

\(\displaystyle \frac{dC}{dt}=\alpha\left(C_a-C \right)\) where \(\displaystyle 0<\alpha\)

(b) We are given the initial value \(\displaystyle C(0)=0\) and asked to solve the resulting initial value problem (IVP).

To solve the ordinary differential equation (ODE), we may either separate variables or write in standard linear form, and use in integration factor. I will demonstrate both methods.

i) Separate variables:

We may thus write the ODE as:

\(\displaystyle \frac{1}{C-C_a}\,dC=-\alpha\,dt\)

Integrate, switching dummy variables and using the boundaries as the limits of integration:

\(\displaystyle \int_0^{C}\frac{1}{u-C_a}\,du=-\alpha\int_0^t\,dv\)

Applying the fundamental theorem of calculus (FTOC), we obtain:

\(\displaystyle \left[\ln\left|u-C_a \right| \right]_0^C=-\alpha\left[v \right]_0^t\)

\(\displaystyle \ln\left|\frac{C-C_a}{-C_a} \right|=-\alpha t\)

Convert from logarithmic to exponential form:

\(\displaystyle \frac{C-C_a}{-C_a}=e^{-\alpha t}\)

Solving for $C(t)$, we obtain:

\(\displaystyle C(t)=C_a\left(1-e^{-\alpha t} \right)\)

ii) Express as linear ODE

\(\displaystyle \frac{dC}{dt}+\alpha C=\alpha C_a\)

Compute the integrating factor:

\(\displaystyle \mu(t)=e^{\alpha\int\,dt}=e^{\alpha t}\)

Multiply the ODE by this factor:

\(\displaystyle e^{\alpha t}\frac{dC}{dt}+\alpha e^{\alpha t}C=\alpha C_ae^{\alpha t}\)

Rewrite the left side as the differentiation of a product:

\(\displaystyle \frac{d}{dt}\left(e^{\alpha t}C \right)=\alpha C_ae^{\alpha t}\)

Integrate with respect to $t$

\(\displaystyle \int\,d\left(e^{\alpha t}C \right)=\alpha C_a\int e^{\alpha t}\,dt\)

\(\displaystyle e^{\alpha t}C=C_ae^{\alpha t}+c_1\)

Solve for $C(t)$:

\(\displaystyle C(t)=C_a+c_1e^{-\alpha t}\)

Use the initial values to determine the parameter $c_1$ (the constant of integration):

\(\displaystyle C(0)=C_a+c_1=0\,\therefore\,c_1=-C_a\)

And so we find the solution satisfying the given conditions is:

\(\displaystyle C(t)=C_a\left(1-e^{-\alpha t} \right)\)

(c) Here we are told herbicide is also leaving the leaf at a rate proportional to $C(t)$, and so the ODE becomes:

\(\displaystyle \frac{dC}{dt}=\alpha\left(C_a-C \right)-\beta C\) where \(\displaystyle C(0)=0\) and \(\displaystyle 0<\alpha,\beta\)

While we could still use both methods to solve the resulting IVP, let's just use the linear method. So, let's express the ODE in standard linear form:

\(\displaystyle \frac{dC}{dt}+(\alpha+\beta)C=\alpha C_a\)

Computing the integrating factor, we find:

\(\displaystyle \mu(t)=e^{(\alpha+\beta)\int\,dt}=e^{(\alpha+\beta)t}\)

And so the ODE becomes:

\(\displaystyle e^{(\alpha+\beta)t}\frac{dC}{dt}+(\alpha+\beta)e^{(\alpha+\beta)t}C=\alpha C_ae^{(\alpha+\beta)t}\)

Expressing the left side as the differentiation of a product, we obtain:

\(\displaystyle \frac{d}{dt}\left(e^{(\alpha+\beta)t}C \right)=\alpha C_ae^{(\alpha+\beta)t}\)

Integrate with respect to $t$:

\(\displaystyle \int\,d\left(e^{(\alpha+\beta)t}C \right)=\alpha C_a\int e^{(\alpha+\beta)t}\,dt\)

\(\displaystyle e^{(\alpha+\beta)t}C=\frac{\alpha C_a}{\alpha+\beta}e^{(\alpha+\beta)t}+c_1\)

Solving for $C(t)$, we get:

\(\displaystyle C(t)=\frac{\alpha C_a}{\alpha+\beta}+c_1e^{-(\alpha+\beta)t}\)

Use the initial values to determine the parameter $c_1$ (the constant of integration):

\(\displaystyle C(0)=\frac{\alpha C_a}{\alpha+\beta}+c_1=0\,\therefore\,c_1=-\frac{\alpha C_a}{\alpha+\beta}\)

And so we find the solution satisfying the given conditions is:

\(\displaystyle C(t)=\frac{\alpha C_a}{\alpha+\beta}\left(1-e^{-(\alpha+\beta)t} \right)\)

(d) The main difference I see is in the limiting concentration in the leaf, that is the concentration as $t\to\infty$:

For the model in part (b), we have:

\(\displaystyle \lim_{t\to\infty}C(t)=C_a\)

And for the model in part (c), we have:

\(\displaystyle \lim_{t\to\infty}C(t)=\frac{\alpha C_a}{\alpha+\beta}\)

Notice if \(\displaystyle \beta=0\), then the two models are equivalent, as we should expect.

Another difference is that the second model approaches its limiting value more rapidly, since the herbicide is being dispersed, or removed from the leaf, resulting in a slower rate of change. Here is a plot of two possible curves, letting $C_a=\alpha=\beta=1$:

heather.jpg