Problem 7 - Calculus for Biology and Medicine

Question

The two-dimensional diffusion equation $$ \frac{\partial n(\mathbf{r}, t)}{\partial t}=D\left(\frac{\partial^{2} n(\mathbf{r}, t)}{\partial x^{2}}+\frac{\partial^{2} n(\mathbf{r}, t)}{\partial y^{2}}\right) $$ where \(n(\mathbf{r}, t), \mathbf{r}=(x, y)\), denotes the population density at the point \(\mathbf{r}=(x, y)\) in the plane at time \(t\), can be used to describe the spread of organisms. Assume that a large number of insects are released at time 0 at the point \((0,0)\). Furthermore, assume that, at later times, the density of these insects can be described by the diffusion equation (10.50). Show that $$ n(x, y, t)=\frac{n_{0}}{4 \pi D t} \exp \left[-\frac{x^{2}+y^{2}}{4 D t}\right] $$ satisfies \((10.50)\)

Step-by-Step Solution

Answer

The given function satisfies the diffusion equation and matches the initial condition.

Step 1: Verify the Initial Condition

The solution must satisfy the initial condition: a large number of insects released at point \((0,0)\) at time \(t=0\). For this, we consider \(n(x, y, 0)\) must approach a delta function \(n_0\delta(x)\delta(y)\). Initiate by verifying that \[ \lim_{t \to 0} n(x, y, t) = n_0 \delta(x) \delta(y) \].

Step 2: Verify the Given Function is Solution to Diffusion Equation

We need to substitute \(n(x, y, t) = \frac{n_0}{4 \pi D t} \exp \left[-\frac{x^2 + y^2}{4Dt}\right]\) into the given diffusion equation. First, calculate the partial derivatives with respect to time, \(t\), and spatial coordinates, \(x\) and \(y\).

Step 3: Calculate the Temporal Derivative

Compute \(\frac{\partial n}{\partial t}\) of the given function. Use the product rule and chain rule: \[\frac{\partial n}{\partial t} = -\frac{n_0}{4\pi D t^2}\exp \left[-\frac{x^2 + y^2}{4Dt}\right] + \frac{n_0}{4\pi D t} \left(\frac{x^2 + y^2}{4D t^2}\right) \exp \left[-\frac{x^2 + y^2}{4Dt}\right]\].

Step 4: Calculate the Spatial Derivatives

Now compute the second spatial partial derivatives \(\frac{\partial^2 n}{\partial x^2}\) and \(\frac{\partial^2 n}{\partial y^2}\). Consider: \[\frac{\partial n}{\partial x} = -\frac{n_0}{4\pi D t}\cdot\frac{2x}{4Dt} \exp\left[-\frac{x^2 + y^2}{4Dt}\right]\], \[\frac{\partial^2 n}{\partial x^2} = \frac{n_0}{8\pi D^2 t^2} \left( 1 - \frac{x^2}{2Dt} \right) \exp\left[-\frac{x^2 + y^2}{4Dt}\right]\]. Similar computations give \(\frac{\partial^2 n}{\partial y^2}\).

Step 5: Sum of Second Spatial Derivatives

Add the second spatial derivatives: \[ \frac{\partial^2 n}{\partial x^2} + \frac{\partial^2 n}{\partial y^2} = \frac{n_0}{4\pi D^2 t^2} \left( 1 - \frac{x^2 + y^2}{4Dt} \right) \exp\left[-\frac{x^2 + y^2}{4Dt}\right]\].

Step 6: Equate Time and Space Derivatives in Diffusion Equation

Substitute these derivatives into the diffusion equation: \[D\left(\frac{\partial^2 n}{\partial x^2} + \frac{\partial^2 n}{\partial y^2}\right) = \frac{\partial n}{\partial t}.\] Confirm both sides match, satisfying the equation: \[ -\frac{n_0}{4\pi D t^2}\exp \left[-\frac{x^2 + y^2}{4Dt}\right] + \frac{n_0}{4\pi D t} \left(\frac{x^2 + y^2}{4D t^2}\right) \exp \left[-\frac{x^2 + y^2}{4Dt}\right] \].

Step 7: Conclusion

Since the temporal and spatial derivatives satisfy the given diffusion equation, \(n(x, y, t) = \frac{n_0}{4 \pi D t} \exp \left[-\frac{x^2 + y^2}{4 D t}\right]\) is a solution. This solution also satisfies the initial condition described by a delta function, representing a source at \((0,0)\) at time \(t=0\).

Key Concepts

Partial Differential Equations

Partial differential equations (PDEs) are a type of mathematical equation used to describe how a quantity changes over space and time. In the case of the diffusion equation, we're dealing with how the density of a population, like insects in the exercise, spreads out in a two-dimensional space over time. The general form of the diffusion equation is \( \frac{\partial n(\mathbf{r}, t)}{\partial t} = D \left(\frac{\partial^{2} n(\mathbf{r}, t)}{\partial x^{2}} + \frac{\partial^{2} n(\mathbf{r}, t)}{\partial y^{2}}\right) \). Here, the term \( \frac{\partial n}{\partial t} \) represents how the population density \( n \) changes with time.
This equation models how substances or populations diffuse, or spread out, through a medium.

• Temporal Derivative \( \frac{\partial n}{\partial t} \): Describes the rate of change of density over time.
• Spatial Derivatives \( \frac{\partial^{2} n}{\partial x^{2}} \) and \( \frac{\partial^{2} n}{\partial y^{2}} \): Indicate how the density changes with respect to space, often capturing the phenomenon of spreading.
• Diffusion Coefficient \( D \): A constant that reflects how fast the diffusion process occurs.

By understanding these components, students can grasp how PDEs like the diffusion equation model complex real-world behaviors.

Population Density

Population density, denoted by \( n(x, y, t) \) in the diffusion equation, is a measure of how many individuals per unit area are present at a particular point and time. In ecological modeling, it provides insight into how organisms are distributed across space.

In this exercise, we see how the density of a population, initially concentrated at a single location, spreads out over time. The formula \( n(x, y, t) = \frac{n_{0}}{4 \pi D t} \exp \left[-\frac{x^{2}+y^{2}}{4 D t}\right] \) reflects this dispersion:

• The exponential term \( \exp \left[-\frac{x^{2}+y^{2}}{4 D t}\right] \) indicates that the density decreases as you move away from the initial release point \( (0,0) \).
• The distribution becomes wider over time \( t \), showing how the initial crowd of insects spreads into the surrounding area.
• As \( t \to 0 \), the formula approaches a delta function, a spike, at the release point, signifying that initially all organisms are concentrated at a point.

Understanding population density is crucial for visualizing how organisms expand their habitat or how substances like heat diffuse through a medium.

Initial Conditions

Initial conditions are fundamental in solving partial differential equations as they specify the state of the system at the start of the observation, which is at time \( t = 0 \) in many cases. For the diffusion process of our problem, the initial condition is critical because it represents the starting point of the population density's evolution.

In this exercise, the initial condition states that a large number of insects were released at the origin \( (0,0) \) at \( t=0 \). Mathematically, this situation is expressed using a delta function \( n_0 \delta(x) \delta(y) \), a mathematical concept used to represent an idealized concentration of mass or probability at a point.
Consider that:

• Delta Function: Serves as a spike that indicates an infinite density at one point but zero elsewhere initially.
• The function \( n(x, y, t) \) must respect this origination point as \( t \to 0 \), aligning with the physical scenario.

The initial conditions determine how the solution to the PDE unfolds over time, starting from this defined state. Without precisely defined initial conditions, it becomes challenging to accurately predict the future behavior of the system.