Problem 64 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

The function \( c(x, t) \) satisfies the given partial differential equation.

Step 1: Differentiate c(x, t) with respect to t

First, we need to calculate \( \frac{\partial c(x, t)}{\partial t} \). Start by differentiating the expression \( c(x, t)=\frac{1}{\sqrt{2 \pi t}} \exp \left[-\frac{x^{2}}{2t}\right] \) with respect to \( t \).Use the product and chain rules. The first part, \( \frac{1}{\sqrt{2 \pi t}} \), differentiates to:\[-\frac{1}{2} \frac{1}{t^{3/2} \sqrt{2 \pi}} \exp \left[-\frac{x^{2}}{2t}\right]\]The second part, \( \exp \left[-\frac{x^{2}}{2t}\right] \), differentiates to:\[ \frac{x^{2}}{2t^2 \sqrt{2\pi t}} \exp \left[-\frac{x^{2}}{2t}\right] \]Combining these, we get: \[ \frac{\partial c(x, t)}{\partial t} = \left( -\frac{1}{2t} + \frac{x^{2}}{2t^{2}} \right) \frac{1}{\sqrt{2\pi t}} \exp \left[-\frac{x^{2}}{2t}\right] \].

Step 2: Differentiate c(x, t) twice with respect to x

Next, compute \( \frac{\partial^{2} c(x, t)}{\partial x^{2}} \). Start by finding the first derivative \( \frac{\partial c(x, t)}{\partial x} \), and then differentiate it again.First derivative of \( c(x, t) \) with respect to \( x \):\[ \frac{\partial}{\partial x} \left( \frac{1}{\sqrt{2 \pi t}} \exp \left[-\frac{x^{2}}{2t} \right] \right) = -\frac{x}{t \sqrt{2\pi t}} \exp \left[-\frac{x^{2}}{2t} \right] \]Now, differentiate again with respect to \( x \) for the second derivative:\[ \frac{\partial^2 c(x, t)}{\partial x^2} = \left(\frac{x^{2}}{t^{2}} - \frac{1}{t} \right) \frac{1}{\sqrt{2\pi t}} \exp \left[-\frac{x^{2}}{2t}\right] \].

Step 3: Verify the equality of derivatives

Substitute the expressions from Step 1 and Step 2 into the given partial differential equation (PDE):\[ \frac{\partial c(x, t)}{\partial t} = \frac{1}{2} \frac{\partial^2 c(x, t)}{\partial x^2} \]From Step 1:\[ \frac{\partial c(x, t)}{\partial t} = \left( -\frac{1}{2t} + \frac{x^{2}}{2t^{2}} \right) \frac{1}{\sqrt{2\pi t}} \exp \left[-\frac{x^{2}}{2t}\right] \]From Step 2:\[ \frac{1}{2} \frac{\partial^2 c(x, t)}{\partial x^2} = \frac{1}{2} \left( \frac{x^{2}}{t^{2}} - \frac{1}{t} \right) \frac{1}{\sqrt{2\pi t}} \exp \left[-\frac{x^{2}}{2t} \right]\]Both expressions are equal, thus confirming that \( c(x, t) \) solves the PDE.

Key Concepts

Heat Equation

The heat equation is a type of partial differential equation (PDE) that models how heat diffuses through a medium over time. In mathematical terms, it is often written as:

• \( \frac{\partial u}{\partial t} = \alpha \frac{\partial^{2} u}{\partial x^{2}} \)

where \( u \) is a function of space \( x \) and time \( t \), and \( \alpha \) represents the diffusivity constant. The equation describes how the temperature distribution \( u \, (x, t) \) changes in space and time.

In the exercise, we've seen a special solution to a version of the heat equation where \( \alpha = \frac{1}{2} \). This specific form, when solved correctly, helps predict how heat or temperature spreads through a substance over time, perfect for physics or engineering applications.

Differentiation

Differentiation is a fundamental concept in calculus that involves finding the rate at which a function changes at any point. It essentially answers the question of how something changes instantaneously. In our context, we are dealing with partial differentiation, where we differentiate a function with respect to one variable while keeping others constant.

In the exercise, this technique was used extensively:

• First, differentiating the function \( c(x, t) \) with respect to \( t \), which gives us the rate of change of the function concerning time.
• Then, differentiating twice with respect to \( x \) to find how the function changes as the position \( x \) varies.

By practicing these steps, you better understand how to apply differentiation to solve partial differential equations like the heat equation.

Chain Rule

The chain rule is a differentiation rule used when dealing with composite functions. Specifically, it helps find the derivative of a function that is composed of other functions. This rule is particularly powerful in solving problems involving partial differentiation.

In our exercise, the chain rule was used to differentiate exponential expressions with respect to \( t \) and \( x \). For example:

• When differentiating \( \exp \left[-\frac{x^{2}}{2t}\right] \), the chain rule allows us to efficiently compute its partial derivatives by addressing the inner function, \(-\frac{x^{2}}{2t}\).

This technique shows the beauty of calculus in simplifying complex derivatives by breaking them down into more manageable parts.

Exponential Function

The exponential function, \( e^{x} \), is a crucial mathematical function characterized by its constant growth rate and arises naturally in various scientific contexts. It is known for its unique property where the derivative and the function itself are essentially the same, making it particularly interesting in calculus and differential equations.

For our given problem, the function:

• \( c(x, t) = \frac{1}{\sqrt{2 \pi t}} \exp \left[-\frac{x^{2}}{2t}\right] \)

incorporates this exponential form. When differentiated, the exponential component shows how functions can maintain consistency even under differentiation due to the special nature of the \( e \)-based formula.

Understanding this allows deeper insights into solving equations like the heat equation, where solutions often involve exponential functions due to their naturally decaying nature over time.