Problem 63 - Calculus for Biology and Medicine

Question

Show that $$ c(x, t)=\frac{1}{\sqrt{8 \pi t}} \exp \left[-\frac{x^{2}}{8 t}\right] $$ solves $$ \frac{\partial c(x, t)}{\partial t}=2 \frac{\partial^{2} c(x, t)}{\partial x^{2}} $$

Step-by-Step Solution

Answer

The function satisfies the diffusion equation as shown by matching derivatives.

Step 1: Identify Partial Derivatives

We begin by identifying the partial derivatives of the function \( c(x, t) = \frac{1}{\sqrt{8 \pi t}} \exp \left[-\frac{x^{2}}{8 t}\right] \). We need \( \frac{\partial c}{\partial t} \) and \( \frac{\partial^{2} c}{\partial x^{2}} \).

Step 2: Find the Partial Derivative with respect to Time

To compute \( \frac{\partial c}{\partial t} \), apply the product rule and the chain rule:\[ c(x, t) = \frac{1}{\sqrt{8 \pi t}} \cdot \exp \left(-\frac{x^2}{8t}\right) \]Calculate the partial derivative:\[ \frac{\partial c}{\partial t} = \left( -\frac{1}{2} \right) \left( \frac{1}{t^{3/2} \sqrt{8 \pi}} \right) \exp \left(-\frac{x^2}{8t}\right) + \frac{1}{\sqrt{8 \pi t}} \left( \frac{x^2}{8t^2} \right) \exp \left(-\frac{x^2}{8t}\right) \]Simplifying, we have:\[ \frac{\partial c}{\partial t} = c(x,t) \left( -\frac{1}{2t} + \frac{x^2}{8t^2} \right) \]

Step 3: Find the Second Partial Derivative with respect to Space

First, calculate \( \frac{\partial c}{\partial x} \):\[ \frac{\partial c}{\partial x} = \frac{1}{\sqrt{8 \pi t}} \cdot \left( -\frac{x}{4t} \right) \exp \left(-\frac{x^2}{8t}\right) = -\frac{x}{4t} c(x,t) \]Next, calculate \( \frac{\partial^{2} c}{\partial x^{2}} \):\[ \frac{\partial^{2} c}{\partial x^{2}} = \frac{\partial}{\partial x} \left( -\frac{x}{4t} c(x,t) \right) = -\frac{1}{4t} c(x,t) + \frac{x^2}{16t^2} c(x,t) \]Simplifying gives:\[ \frac{\partial^{2} c}{\partial x^{2}} = c(x,t) \left( -\frac{1}{4t} + \frac{x^2}{16t^2} \right) \]

Step 4: Apply the Diffusion Equation

Substitute the expressions for the derivatives into the given differential equation:\[ \frac{\partial c}{\partial t} = 2 \frac{\partial^{2} c}{\partial x^{2}} \]We derived:\[ \frac{\partial c}{\partial t} = c(x,t) \left( -\frac{1}{2t} + \frac{x^2}{8t^2} \right) \]\[ 2 \frac{\partial^2 c}{\partial x^2} = 2 \cdot c(x,t) \left( -\frac{1}{4t} + \frac{x^2}{16t^2} \right) = c(x,t) \left( -\frac{1}{2t} + \frac{x^2}{8t^2} \right) \]Both expressions match, verifying the solution.

Key Concepts

Diffusion Equation

The diffusion equation is fundamental in describing how substances spread over time. It's usually expressed as \( \frac{\partial c}{\partial t} = D \frac{\partial^2 c}{\partial x^2} \), where:

• \( c(x, t) \) is the concentration at a position \( x \) and time \( t \).
• \( D \) is the diffusion coefficient, a constant that characterizes the rate of diffusion.

In our exercise, the equation given is \( \frac{\partial c}{\partial t} = 2 \frac{\partial^2 c}{\partial x^2} \), showcasing a simple case where the diffusion coefficient \( D \) is 2. The aim here is to verify that the solution for \( c(x, t) \) indeed satisfies this equation. When solving such equations, they model phenomena like the spread of heat in a metal rod or the dispersal of a drop of ink in water. Understanding this process helps us explore a multitude of real-world applications related to diffusion.

Partial Derivatives

Partial derivatives are a type of derivative that shows how a function changes as just one of its variables is varied, holding the others constant. For multivariable functions, this is an important tool. In the problem, we are working with a function of two variables, \( c(x, t) \). Therefore, we consider derivatives with respect to time \( t \) and space \( x \).

• The partial derivative with respect to time, \( \frac{\partial c}{\partial t} \), measures how the concentration changes as time progresses.
• The second partial derivative with respect to space, \( \frac{\partial^2 c}{\partial x^2} \), highlights how the concentration changes with spatial position.

Determining these derivatives accurately is key to demonstrating that the function \( c(x, t) \) satisfies the diffusion equation.

Chain Rule

The chain rule is a fundamental principle in calculus used to differentiate compositions of functions. When working with partial derivatives, the chain rule allows us to connect changes in one variable with changes in another variable through an intermediary. In the given problem, the chain rule helps in computing \( \frac{\partial c}{\partial t} \), where the function involves the exponential of a ratio \( x^2 / 8t \). To differentiate this expression with respect to \( t \):

• Recognize that the expression inside the exponential, \( -\frac{x^2}{8t} \), is itself a function of \( t \) while also being multiplied by \( \frac{1}{\sqrt{8\pi t}} \).
• Apply the chain rule to appropriately manage the dependency of the exponential function on \( t \).

Effectively using the chain rule supports the process of finding the correct rate of change for more complex, layered functions.

Product Rule

The product rule is a powerful tool in calculus used when differentiating a product of two functions. It states that for two functions \( u \) and \( v \), the derivative of their product is given by \( (uv)' = u'v + uv' \). In our solution, the concentration function \( c(x, t) \) is expressed as a product of two separate components:

• \( u(t) = \frac{1}{\sqrt{8\pi t}} \)
• \( v(x, t) = \exp \left( -\frac{x^2}{8t} \right) \)

For \( \frac{\partial c}{\partial t} \), applying the product rule is crucial because it involves differentiating each part with respect to \( t \) and summing those derivatives as per the rule's requirement. This provides a methodical approach to unweaving complex expressions into solvable pieces. Understanding the product rule ensures that no part of a function is overlooked while taking derivatives, leading to the correct application of the diffusion equation.