Problem 1 - 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 \( c(x, t) \) satisfies the PDE.

Step 1: Differentiate with respect to time

We begin by finding the partial derivative of \( c(x, t) \) with respect to \( t \). The given function is \[ c(x, t) = \frac{1}{\sqrt{8 \pi t}} \exp \left[-\frac{x^2}{8t}\right]. \]To differentiate, use the product rule where the first function is \( \frac{1}{\sqrt{8 \pi t}} \) and the second is the exponential term:\[\frac{\partial c(x, t)}{\partial t} = \frac{d}{dt}\left(\frac{1}{\sqrt{8 \pi t}}\right) \exp\left[-\frac{x^2}{8t}\right] + \frac{1}{\sqrt{8 \pi t}} \frac{d}{dt}\left(\exp\left[-\frac{x^2}{8t}\right]\right). \]The first part simplifies using the chain rule to \[ -\frac{1}{2} \frac{1}{t\sqrt{8 \pi t}} \exp\left[-\frac{x^2}{8t}\right]. \]For the second term, use:\[ \frac{d}{dt}\left(\exp\left[-\frac{x^2}{8t}\right]\right) = \exp\left[-\frac{x^2}{8t}\right]\left(\frac{x^2}{8t^2}\right), \] leading to:\[ \frac{x^2}{8t^2\sqrt{8 \pi t}} \exp\left[-\frac{x^2}{8t}\right].\] So the combined expression is:\[\frac{\partial c(x, t)}{\partial t} = \left(-\frac{1}{2t} + \frac{x^2}{8t^2}\right) \frac{1}{\sqrt{8 \pi t}} \exp \left[-\frac{x^2}{8t}\right].\]

Step 2: Differentiate with respect to space twice

Now, differentiate \( c(x, t) \) twice with respect to \( x \). The expression is \[ \frac{\partial c(x, t)}{\partial x} = \frac{1}{\sqrt{8 \pi t}} \frac{d}{dx}\left(\exp\left[-\frac{x^2}{8t}\right]\right). \]The derivative of the exponential function with respect to \( x \) is \[ -\frac{x}{4t} \exp \left[-\frac{x^2}{8t}\right]. \]Thus, \[ \frac{\partial c(x, t)}{\partial x} = -\frac{x}{4t\sqrt{8\pi t}} \exp \left[-\frac{x^2}{8t}\right].\]Next, differentiate again with respect to \( x \): \[\frac{\partial^2 c(x, t)}{\partial x^2} = \frac{d}{dx}\left(-\frac{x}{4t\sqrt{8 \pi t}} \exp \left[-\frac{x^2}{8t}\right]\right). \]Applying the product rule, the two components are:\[-\frac{1}{4t\sqrt{8\pi t}}\ \exp \left[-\frac{x^2}{8t}\right] + \frac{x^2}{16t^2}\frac{1}{\sqrt{8 \pi t}}\ \exp\left[-\frac{x^2}{8t}\right]. \]Thus, \[\frac{\partial^2 c(x, t)}{\partial x^2} = \left(-\frac{1}{4t} + \frac{x^2}{16t^2}\right) \frac{1}{\sqrt{8 \pi t}} \exp \left[-\frac{x^2}{8t}\right]. \]

Step 3: Verify the equality

Substitute the second derivative back into the original partial differential equation to check if \[ \frac{\partial c(x, t)}{\partial t} = 2 \frac{\partial^2 c(x, t)}{\partial x^2}\]holds true. From the expressions obtained, substitute:\[\frac{\partial c(x, t)}{\partial t} = \left( -\frac{1}{2t} + \frac{x^2}{8t^2} \right) \frac{1}{\sqrt{8\pi t}} \exp \left[-\frac{x^2}{8t}\right], \]and \[ 2 \frac{\partial^2 c(x, t)}{\partial x^2} = 2 \left( -\frac{1}{4t} + \frac{x^2}{16t^2} \right) \frac{1}{\sqrt{8\pi t}} \exp \left[-\frac{x^2}{8t}\right]\]\[ = \left( -\frac{1}{2t} + \frac{x^2}{8t^2} \right) \frac{1}{\sqrt{8\pi t}} \exp \left[-\frac{x^2}{8t}\right]. \]Since both expressions are equal, the function \( c(x, t) \) satisfies the PDE.

Key Concepts

Heat Equation

The Heat Equation is a classic example of a partial differential equation (PDE) used to describe the distribution of heat in a given region over time. This equation models how heat diffuses through a medium. Mathematically, it is expressed as \( \frac{\partial u}{\partial t} = \alpha abla^2 u \), where \( abla^2 \) is the Laplacian operator and \( \alpha \) is the thermal diffusivity constant.

In the given problem, we are interested in a specific form of the heat equation: \( \frac{\partial c(x, t)}{\partial t}=2 \frac{\partial^{2} c(x, t)}{\partial x^{2}} \). This equation indicates that the change in temperature over time \( \frac{\partial c}{\partial t} \) is proportional to the second spatial derivative \( \frac{\partial^2 c}{\partial x^2} \) with a factor of 2. This specific PDE suggests that the medium is homogeneous and isotropic.

When solving the heat equation, we look for functions that satisfy this relationship throughout the domain of interest. In this case, the solution \( c(x, t) \) is shown to sufficiently satisfy the equation by substituting into the PDE and confirming both sides are equal.

Exponential Function

The Exponential Function is a widely used mathematical function noted for its 'always positive' characteristic and smooth rate of change, which makes it valuable in modeling continuous growth or decay processes.

Our function \( c(x, t) = \frac{1}{\sqrt{8 \pi t}} \exp \left[-\frac{x^{2}}{8 t}\right] \) features an exponential term that describes how the function decays quickly with distance \( x \), influenced by time \( t \).

An exponential function \( \, a^x \) has the natural property that both its value and derivative are proportional to it. This property makes exponential functions suitable for describing natural processes like heat distribution, among others. In this context, the exponential function embodies the solution’s rapidly decreasing nature, as heat diffuses and distributes more uniformly across a medium.

• The term \( \exp \left[-\frac{x^{2}}{8t}\right] \) indicates a Gaussian profile, which helps achieve solutions for diffusion processes like heat transfers.

Understanding how the exponential function behaves and integrates within the heat equation is crucial to anticipating the behavior of the entire system being modeled.

Partial Derivatives

Partial Derivatives are at the core of understanding functions with several variables. They allow us to see how a function changes when altering one variable, keeping others constant. This is key in physics problems involving variables changing over time and space.

For our specific problem, you find partial derivatives using rules such as the chain rule and product rule, focusing on the time (\( t \)) and space (\( x \)) variables.

First, differentiating with respect to \( t \) (time) provides insight about how the function values evolve as time progresses. Using these tools, we calculate \( \frac{\partial c(x, t)}{\partial t} \) by applying the product and chain rule.

Secondly, taking partial derivatives with respect to \( x \) (space) twice reveals how the function values change as we move positionally within the medium. The second order derivative \( \frac{\partial^2 c(x, t)}{\partial x^2} \) gives us a measure of concavity related to space, enabling the analysis of diffusion rates.

• Applying these derivatives helps verify the solution against the heat equation, confirming its validity by demonstrating equivalency in expressions.
• Understanding partial derivatives in this context allows you to explore diverse phenomena involving rate changes in complex systems.

By mastering the technique of finding and interpreting partial derivatives, we can unlock deeper insights into the dynamics described by partial differential equations.