Problem 4 - 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.

Step-by-Step Solution

Answer

(a) Maximum at \(x = 0\). (b) \(c(0, t)\) is decreasing. (c) Limit is \(\infty\) at \(x=0\), 0 otherwise. (d) Integral equals 1. (e) Total substance conserved, initially at origin.

Step 1: Determine Local Maximum

To find a local maximum of the function \( c(x, t) \), we first compute the first partial derivative with respect to \( x \):\[ \frac{\partial c(x, t)}{\partial x} = \frac{\partial}{\partial x} \left( \frac{1}{\sqrt{4 \pi D t}} \exp \left[-\frac{x^{2}}{4 D t}\right] \right). \]Applying the chain rule, we get\[ \frac{\partial c(x, t)}{\partial x} = \left( -\frac{x}{2Dt} \right) \frac{1}{\sqrt{4 \pi D t}} \exp \left[-\frac{x^{2}}{4 D t}\right]. \]Setting this derivative equal to zero to find critical points, we solve:\[ -\frac{x}{2Dt} = 0 \Rightarrow x = 0. \]The second derivative test with \( \frac{\partial^2 c(x, t)}{\partial x^2} \) will confirm \( x = 0 \) as a maximum since it yields a negative value due to the concavity when evaluated at \( x = 0 \).

Step 2: Decreasing Nature of c(0, t)

Evaluate \( c(0, t) \) to show it decreases with \( t \):\[ c(0, t) = \frac{1}{\sqrt{4 \pi D t}}. \]Take the derivative with respect to \( t \):\[ \frac{d}{dt} c(0, t) = -\frac{1}{2} (4 \pi D)^{-1/2} t^{-3/2}. \]Since this derivative is negative for all \( t > 0 \), \( c(0, t) \) is a decreasing function of \( t \).

Step 3: Limits as t Approaches Zero

Consider the limits as \( t \to 0^+ \):1. When \( x = 0 \): \[ \lim_{t \to 0^+} c(0, t) = \lim_{t \to 0^+} \frac{1}{\sqrt{4 \pi D t}} = \infty. \]2. When \( x eq 0 \): \[ \lim_{t \to 0^+} c(x, t) = \lim_{t \to 0^+} \frac{1}{\sqrt{4 \pi D t}} \exp \left[-\frac{x^{2}}{4 D t}\right] = 0, \] since the exponential term tends to zero very rapidly.

Step 4: Evaluation of the Gaussian Integral

The integral of \( c(x, t) \) over all space must equal 1:\[ \int_{-\infty}^{\infty} c(x, t) \ dx = \int_{-\infty}^{\infty} \frac{1}{\sqrt{4 \pi D t}} \exp \left[-\frac{x^{2}}{4 D t}\right] \ dx. \]Change variables by setting \( u = \frac{x}{\sqrt{4Dt}} \), \( du = \frac{1}{\sqrt{4Dt}} \, dx \). The limits are unaffected:\[ \int_{-\infty}^{\infty} \exp(-u^2) \, du = \sqrt{\pi} = \int_{-\infty}^{\infty} c(x, t) \, dx = 1, \]confirming the solution normalizes to 1.

Step 5: Interpretation and Initial Concentration

The integral \( \int_{-\infty}^{\infty} c(x, t) \, dx = 1 \) implies the total probability (or concentration) over space is conserved.Initially, at \( t = 0 \), the limit analysis shows that \( c(0, t) \) approaches infinity while \( c(x, t) \to 0 \) for \( x eq 0 \). This suggests all substance was initially concentrated at the origin (\( x = 0 \)), and diffusion spreads it over time according to a Gaussian distribution.

Key Concepts

Partial Differential Equations

Partial Differential Equations (PDEs) play a crucial role in mathematical modeling, particularly when describing phenomena like diffusion, heat flow, and wave propagation. The diffusion equation itself, \( \frac{\partial c(x, t)}{\partial t} = D \frac{\partial^2 c(x, t)}{\partial x^2} \), represents how substances like heat or particles spread over space and time.
Partial implies multiple variables are involved. Here, \( x \) for space and \( t \) for time. These equations describe how a function like concentration \( c(x, t) \) changes. PDEs can be quite complex, but they break down problems into manageable parts.
Key points about PDEs:

• Involves multiple variables and their partial derivatives.
• Describes changes over space and time.
• Used in modeling various physical phenomena.

Understanding PDEs is essential for grasping how natural processes and engineering problems are expressed mathematically, providing a basis for predicting behavior under different conditions.

Gaussian Functions

Gaussian Functions describe the probability distribution of variables over continuous ranges and are often visualized as bell-shaped curves. In the context of diffusion, the solution \( c(x, t) = \frac{1}{\sqrt{4 \pi D t}} \exp \left[-\frac{x^2}{4 D t}\right] \) is a Gaussian function, relating to how concentration spreads over time.
This function features a few important properties:

• Centered at zero, indicating the highest concentration at the origin.
• Symmetric about the mean, \( x = 0 \), reflecting equal probability of dispersion in all directions.
• Standard deviation grows over time, meaning dispersion increases.

In diffusion, the Gaussian function shows how an initially concentrated substance gradually spreads outward, decreasing in concentration at the center over time, while covering a larger area. This characteristic symmetry and smoothness makes Gaussian functions a perfect model for various natural distributions.

Initial Conditions in PDEs

Initial Conditions in Partial Differential Equations are key to accurately solving for a particular situation or scenario. They specify the state at the beginning of the process, such as concentration or temperature at time \( t = 0 \). In the given diffusion problem, initial conditions influence how the solution evolves over time.
For instance, the solution indicates that initially, all the substance is concentrated at the origin, as shown by the limit \( \lim_{t \to 0^+} c(x, t) \) yielding different results based on \( x \):

• As \( x = 0 \), \( c(x, t) \to \infty \), summarizing infinite concentration.
• When \( x eq 0 \), \( c(x, t) \to 0 \), showing no concentration elsewhere.

These conditions show where and how substances start in models, crucial for realistic simulations and predictions. Accurate initial conditions allow predictions on how a phenomenon like diffusion behaves immediately before transitions start spreading the effect through the modeled space.