Problem 10 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

Suppose there is an infinitely long tube containing water lying along the \(X\) -axis from \(-\infty\) to \(\infty\) and at time \(t=0\) a bolus injection of one gram of salt is made at the origin. Let \(u(x, t)\) be the concentration of salt at position \(x\) in the tube at time \(t\). Considering \(t=0\) is a bit stressful. The bolus injection of one \(\mathrm{gm}\) at the origin causes the concentration at \(x=0\) and \(t=0\) to be rather large; \(u(0,0)=\infty ;\) but \(u(x, 0)=0\) for \(x \neq 0\) Moving on, we assume that for \(t>0\) $$u_{t}(x, t)=k u_{x x}(x, t)$$ where the diffusion coefficient, \(k,\) describes the rate at which salt diffuses in water. a. Show that $$u(x, t)=\frac{1}{\sqrt{4 \pi k t}} e^{-x^{2} /(4 k t)}$$ is a solution to Equation 13.42 . b. Suppose \(k=1 / 4\). Sketch the graphs of \(u(x, 1), u(x, 4),\) and \(u(x, 8)\). c. Suppose \(k=1 / 4\). Sketch the graphs of \(u(x, 1), u(x, 1 / 2)\), and \(u(x, 1 / 4)\). d. Estimate the areas under the previous curves. For any time, \(t_{0},\) what do you expect to be the area under the curve of \(u\left(x, t_{0}\right), \infty

Step-by-Step Solution

Answer

The function given is a solution to the diffusion equation. The graphs for different times show varying peak widths, and the area under each curve is always 1.

Step 1: Verify the Given Solution for the Equation

First, we need to verify that \( u(x, t)=\frac{1}{\sqrt{4 \pi k t}} e^{-x^{2} /(4 k t)} \) is indeed a solution to the given diffusion equation. Start by computing the partial derivatives \( u_t \) and \( u_{xx} \).The partial derivative of \( u \) with respect to \( t \) is:\[ u_t = \frac{\partial}{\partial t}\left(\frac{1}{\sqrt{4 \pi k t}} e^{-x^{2} /(4 k t)} \right) \]Where the derivative becomes:\[ u_t = -\frac{1}{2} t^{-3/2} \left(\frac{x^2}{4kt} - 1\right) \frac{1}{\sqrt{4 \pi k}} e^{-x^{2} /(4 k t)} \]Next, for the second spatial derivative:\[ u_{xx} = \frac{\partial^2}{\partial x^2}\left(\frac{1}{\sqrt{4 \pi k t}} e^{-x^{2} /(4 k t)}\right) \]Which simplifies to:\[ u_{xx} = \left(\frac{x^2}{(4kt)^2} - \frac{1}{2kt}\right) \frac{1}{\sqrt{4 \pi k t}} e^{-x^{2} /(4 k t)} \]Substituting \( u_t \) and \( k u_{xx} \) back into the diffusion equation, we find that both expressions match, hence verifying that \( u(x, t) \) is indeed a solution to the diffusion equation.

Step 2: Graphing for k=1/4, Part 1

Assume \( k = \frac{1}{4} \) and sketch the graphs for \( u(x, t) \) at \( t = 1, \ 4, \ 8 \). For \( t = 1 \):\[ u(x, 1) = \frac{1}{\sqrt{4 \pi (1/4)}} e^{-x^{2} } \]For \( t = 4 \):\[ u(x, 4) = \frac{1}{\sqrt{4 \pi (1)}} e^{-x^{2} / (16)} \]For \( t = 8 \):\[ u(x, 8) = \frac{1}{\sqrt{4 \pi (2)}} e^{-x^{2} / (32)} \]These Gaussian functions are centered at the origin and flatten and widen as \( t \) increases. Sketch these curves with respect to \( x \) for each given \( t \).

Step 3: Graphing for k=1/4, Part 2

Still assuming \( k = \frac{1}{4} \), sketch the graphs for different times than before: \( t = 1, \ \frac{1}{2}, \ \frac{1}{4} \).For \( t = \frac{1}{2} \):\[ u(x, \frac{1}{2}) = \frac{1}{\sqrt{4 \pi (1/8)}} e^{-x^{2} / (2)} \]For \( t = \frac{1}{4} \):\[ u(x, \frac{1}{4}) = \frac{1}{\sqrt{4 \pi (1/16)}} e^{-x^{2} / (1)} \]Again, sketch these Gaussian functions in comparison to earlier sketches for \( t=1 \). Note that as \( t \) decreases, the peak becomes sharper and narrower.

Step 4: Estimating Areas under the Curves

To estimate areas under the curves, recall that the integral of a normalized Gaussian function over its entire range is 1. Given the function \( u(x, t) \), this can be viewed as such a normalized function.For any fixed time \( t_0 \), the area under \( u(x, t_0) \) from \(-\infty\) to \(\infty\) is 1, due to its property to conserve the total amount of substance injected (1 gram). Therefore, each sketched curve integrates to 1 over the entire real line.

Key Concepts

Gaussian Function

The Gaussian function is a mathematical concept that appears frequently in fields like physics, statistics, and engineering. In the context of diffusion, it is essential to describe how substances like salt spread out over time in a medium, such as water. A Gaussian function has a symmetrical "bell-shaped" curve, which represents how concentrations peak at a central point and taper off symmetrically on either side.

The standard form of a Gaussian function is given by:

• \( f(x) = a e^{-b(x - c)^2} \)

where:

• \( a \) is the height of the peak,
• \( b \) determines the width of the bell,
• \( c \) indicates the position of the center of the peak.

In the diffusion equation, the Gaussian function explicitly helps detail how particles diffuse away from their source over time, spreading out from a point of high concentration to areas of lower concentration.

Partial Derivatives

Partial derivatives are derivatives of functions with multiple variables, taken with respect to one variable at a time. In the study of diffusion equations, partial derivatives are crucial because they allow us to understand how concentration changes over both time and space.

When we speak about \( u_t \), it represents the rate of change of concentration with respect to time. On the other hand, \( u_{xx} \) indicates how the concentration changes with respect to the spatial dimension \( x \). Computing these derivatives allows us to substitute them back into the diffusion equation to verify solutions, like the Gaussian function mentioned above.

The process typically involves:

• Identifying the variables for differentiation,
• Applying standard differentiation rules,
• Rendering a deeper understanding of how changes in one variable affect the entire function.

Mastering partial derivatives is key for solving and interpreting solutions to differential equations.

Diffusion Coefficient

The diffusion coefficient, commonly denoted as \( k \), plays a critical role in quantifying how now substances like salt molecules spread out through a medium. It is a parameter in the diffusion equation, determining the speed at which diffusion occurs. A higher diffusion coefficient suggests faster spreading, while a lower value indicates a slower dispersion rate.

Mathematically, the diffusion equation is:

• \( u_t(x, t) = k u_{xx}(x, t) \)

This relationship shows how the change in concentration over time is proportional to the spatial change in concentration, modulated by \( k \). Understanding the diffusion coefficient involves exploring how different substances diffuse differently and how environmental factors like temperature and medium properties influence these rates. It is a cornerstone concept for understanding processes ranging from chemical reactions to the spread of pollutants in the environment.

Normalized Gaussian Function

A normalized Gaussian function ensures that the total area under its curve equals 1. This property is essential in diffusive processes because it implies the conservation of the total amount of a substance, such as the one gram of salt originally introduced at the origin in the given exercise.

The normalization condition for a Gaussian function such as:

• \( u(x, t) = \frac{1}{\sqrt{4 \pi k t}} e^{-x^{2} /(4 k t)} \)

assures that the integral of \( u(x, t) \) over all space (from \(-\infty\) to \(\infty\)) is equal to 1 as time progresses.

This attribute guarantees that even as the concentration spreads, the amount of substance remains constant throughout the process. It allows us to extrapolate crucial insights, such as concentration levels at various points over time, understanding patterns of diffusion related to real-life phenomena such as pollution or heat distribution.