Problem 66 - Calculus for Biology and Medicine

Question

A solution of $$ \frac{\partial c(x, t)}{\partial t}=D \frac{\partial^{2} c(x, t)}{\partial x^{2}} $$ is the function $$ c(x, t)=\frac{1}{\sqrt{4 \pi D t}} \exp \left[-\frac{x^{2}}{4 D t}\right] $$ for \(x \in \mathbf{R}\) and \(t>0\) (a) Show that a local maximum of \(c(x, t)\) occurs at \(x=0\) for fixed \(t\) (b) Show that \(c(0, t), t>0\), is a decreasing function of \(t\). (c) Find $$ \lim _{t \rightarrow 0^{+}} c(x, t) $$ when \(x=0\) and when \(x \neq 0\) (d) Use the fact that $$ \int_{-\infty}^{\infty} e^{-u^{2} / 2} d u=\sqrt{2 \pi} $$ to show that, for \(t>0\), $$ \int_{-\infty}^{\infty} c(x, t) d x=1 $$ (e) The function \(c(x, t)\) can be interpreted as the concentration of a substance diffusing in space. Explain the meaning of $$ \int_{-\infty}^{\infty} c(x, t) d x=1 $$ and use your results in (c) and (d) to explain why this means that initially (i.e., at \(t=0\) ) the entire amount of the substance was released at the origin. Mathematically, we can specify such an initial condition (in which the substance is concentrated at the origin at time 0 ) by the \(\delta\) -function \(\delta(x)\), with the property that $$ \delta(x)=0, \quad \text { for } x \neq 0 $$ and $$ \int_{-\infty}^{\infty} \delta(x) d x=1 $$

Step-by-Step Solution

Answer

Local maximum at \(x=0\), \(c(0,t)\) decreases over time, limits \(t\to 0^+\): infinity at \(x=0\), 0 if \(x\neq0\); total mass constant. Interpret initial condition as Dirac delta function at origin.

Step 1: Verify the Local Maximum at x=0

To find where the local maximum occurs, differentiate \(c(x, t)\) with respect to \(x\), \[\frac{\partial c}{\partial x} = \frac{d}{dx}\left(\frac{1}{\sqrt{4 \pi D t}} \exp\left[-\frac{x^2}{4Dt}\right]\right)\]Using the chain rule and the derivative of \(e^{-x^2}\), \[\frac{\partial c}{\partial x} = \frac{1}{\sqrt{4 \pi Dt}} \cdot \left(-\frac{2x}{4Dt}\right) \cdot \exp\left[-\frac{x^2}{4Dt}\right] = -\frac{x}{2Dt}\cdot \frac{1}{\sqrt{4 \pi Dt}}\cdot \exp\left[-\frac{x^2}{4Dt}\right]\]Setting this equal to zero gives \(x=0\), which indicates a critical point.

Step 2: Determine Type of Critical Point

Differentiate \(\frac{\partial c}{\partial x}\) again to find \(\frac{\partial^2 c}{\partial x^2}\), and evaluate at \(x=0\):\[\frac{\partial^2 c}{\partial x^2} = \frac{d}{dx}\left(-\frac{x}{2Dt} \cdot \frac{1}{\sqrt{4 \pi Dt}} \cdot \exp\left[-\frac{x^2}{4Dt}\right]\right)\]Using the product rule, this becomes \[\frac{\partial^2 c}{\partial x^2} = \left(-\frac{1}{2Dt} + \frac{x^2}{4D^2t^2}\right) \cdot \frac{1}{\sqrt{4 \pi Dt}} \cdot \exp\left[-\frac{x^2}{4Dt}\right]\]Evaluating at \(x=0\) results in \(-\frac{1}{2Dt}c(0,t)<0\), confirming a local maximum at \(x=0\).

Step 3: Show that c(0, t) Decreases over Time

Consider the concentration at \(x=0\) which is \(c(0, t) = \frac{1}{\sqrt{4\pi Dt}}\). To show this is decreasing, take the derivative with respect to \(t\):\[\frac{d}{dt}\left(\frac{1}{\sqrt{4\pi Dt}}\right) = -\frac{1}{2} \cdot \frac{1}{(4\pi Dt)^{3/2}} \cdot (4\pi D)\]\(\frac{d}{dt}c(0, t) = -\frac{1}{2t}\cdot \frac{1}{\sqrt{4\pi Dt}} < 0\)This derivative is negative for \(t>0\), so \(c(0, t)\) decreases over time.

Step 4: Evaluate the Limits as t Approaches 0

Consider \(\lim_{t \to 0^+} c(x, t)\) at \(x=0\):\[\lim_{t \to 0^+} \frac{1}{\sqrt{4\pi Dt}} = \infty\]and when \(x eq 0\):\[\lim_{t \to 0^+} \frac{1}{\sqrt{4\pi Dt}} \exp \left[-\frac{x^2}{4Dt}\right] = 0\]Hence, as \(t\) approaches zero, \(c(0, t)\) tends to infinity, while \(c(x, t)\) for \(x eq 0\) approaches 0.

Step 5: Show Integral of c(x, t) is 1 for t>0

Calculate the integral using the substitution \(u = \frac{x}{\sqrt{4Dt}}\), so \(dx = \sqrt{4Dt} \ du\):\[\int_{-\infty}^{\infty} c(x, t) \, dx = \frac{1}{\sqrt{4\pi Dt}} \int_{-\infty}^{\infty} \exp \left[- \frac{x^2}{4Dt}\right] \, dx \]\[= \frac{1}{\sqrt{\pi}} \int_{-\infty}^{\infty} \exp\left[-u^2\right] \ du = \frac{1}{\sqrt{\pi}} \cdot \sqrt{\pi} = 1\]This calculation relies on the Gaussian integral property \(\int_{-\infty}^{\infty} e^{-u^2}\ du = \sqrt{\pi}\).

Step 6: Interpret the Meaning of the Integral Being 1

The integral \(\int_{-\infty}^{\infty} c(x, t) \, dx = 1\) means that the total "mass" or "quantity" of the diffusion does not change over time. Initially, at \(t=0\), the substance is concentrated at a single point \(x=0\), like a \(\delta\)-function spike, as indicated by \(c(x,t) \to \infty\) at \(x=0\) and 0 elsewhere. As \(t\) increases, the concentration \(c(x, t)\) spreads out, but its total integral remains constant at 1, preserving the mass according to the diffusion equation dynamics.

Key Concepts

Diffusion Equation

The diffusion equation, often represented as \[ \frac{\partial c(x, t)}{\partial t} = D \frac{\partial^2 c(x, t)}{\partial x^2} \]is a partial differential equation that describes how the concentration of a substance changes over time and space. In this equation, \(c(x, t)\) stands for the concentration at location \(x\) at time \(t\), and \(D\) is the diffusion coefficient, which measures how fast the substance spreads through the medium.
Understanding this equation is essential since it helps to model physical phenomena where particles move from high concentration areas to lower ones over time. This process is known as diffusion.

The equation means that the change in concentration over time (left side) is proportional to the change in the concentration’s spatial rate (right side). This form of the equation suggests that diffusion is driven by the gradient or the slope of the concentration with respect to space.
In essence, wherever concentration gradients exist, diffusion acts to equilibrate them over time.

Boundary Conditions

Boundary conditions are essential when dealing with differential equations like the diffusion equation. They specify the behavior of the solution at the boundaries of the domain. Without them, the problem remains unsolvable, as infinitely many functions could satisfy a differential equation.
In practical terms, boundary conditions could describe physical restrictions like a sealed container's walls where the substance cannot pass, or they could represent external conditions interacting with the system.

• Dirichlet Boundary Conditions: These set the value of the function at the boundary. For example, specifying a fixed concentration at a certain point.
• Neumann Boundary Conditions: These define the derivative of the function at the boundary, often corresponding to no flux across a boundary.

Choosing the right type of boundary condition depends on the specific scenario modeled, influencing how the results align with real-world observations.

Gaussian Distribution

A Gaussian distribution, also known as a normal distribution, plays a significant role in the solution of the diffusion equation. In this context, the concentration function \[ c(x, t) = \frac{1}{\sqrt{4 \pi D t}} \exp \left[-\frac{x^2}{4 D t}\right] \]resembles a Gaussian curve, centered around zero with time-dependent spread.
The Gaussian distribution is characterized by its symmetric bell-shaped curve and is determined by two parameters: mean and variance. Here, the mean is zero, indicating that the distribution is centered at the origin, while the variance grows linearly with time \(t\), causing the spread of the distribution to increase as time progresses.

In diffusion, this distribution explains how a point source of substance spreads over time, with the peak decreasing and the width of the distribution increasing as time goes on. The Gaussian nature signifies that particles are more likely to be found near the origin initially, but eventually, they spread further out as time progresses.

Initial Conditions

Initial conditions set the stage for solving differential equations by describing the system's starting state. For the diffusion equation, they specify the initial concentration profile of the substance at time \(t = 0\).
In this problem, the initial condition is that the substance is concentrated at the origin, represented by the Dirac delta function \( \delta(x) \).

• \(\delta(x)\) is zero everywhere except at \(x=0\), where it is infinite, ensuring the entire mass starts at a single point:
• The integral of \(\delta(x)\) over space is 1, maintaining the total substance quantity.

This initial spiked distribution starts to spread according to the diffusion process. Over time, the initial narrow peak broadens, and concentration values across all points in space gradually evolve, always holding the conservation of mass straightforwardly derived from the properties of the diffusion equation and initial setup.