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Magazine Chapter 14: Problem 81
The diffusion equation $$ \frac{\partial c}{\partial t}=D \frac{\partial^{2} c}{\partial x^{2}} $$ where \(D\) is a positive constant, describes the diffusion of heat through a solid, or the concentration of a pollutant at time \(t\) at a distance \(x\) from the source of the pollution, or the invasion of alien species into a new habitat. Verify that the function $$ c(x, t)=\frac{1}{\sqrt{4 \pi D t}} e^{-x^{2} /(4 D t)} $$ is a solution of the diffusion equation.
Short Answer
Expert verified
The function solves the diffusion equation as both sides match after substitution.
Step by step solution
01
Differentiate with Respect to t
Find the partial derivative of \(c(x, t)\) with respect to \(t\). We begin by computing \(\frac{\partial c}{\partial t}\). Using the given function, \[ c(x, t)=\frac{1}{\sqrt{4 \pi D t}} e^{-x^{2} /(4 D t)} \]Use the product rule: \(\frac{\partial}{\partial t} \left( \frac{1}{\sqrt{4\pi Dt}} \right) \cdot e^{-x^2/(4Dt)} + \frac{1}{\sqrt{4\pi Dt}} \cdot \frac{\partial}{\partial t} \left( e^{-x^2/(4Dt)} \right)\).
02
Differentiate the First Part
Compute the derivative of \(\frac{1}{\sqrt{4\pi Dt}}\) with respect to \(t\):Since \(\thinspace t^{n} \right\) derivative is \(n t^{n-1}\), \[\frac{d}{dt} \left( t^{-1/2} \right) = -\frac{1}{2} t^{-3/2}\]So, \[\frac{d}{dt} \left(\frac{1}{\sqrt{4\pi Dt}}\right) = -\frac{1}{2} \frac{1}{\sqrt{4\pi}} D^{-1/2} \cdot t^{-3/2}\] Thus, it becomes \(-\frac{D^{-1/2}}{2 \sqrt{4\pi}} t^{-3/2}\).
03
Differentiate the Exponential Part
Compute the derivative of the exponential component \(e^{-x^{2} /(4 D t) }\) with respect to \(t\): The chain rule gives us:\(-\frac{x^{2}}{4D}\) is outside, then \[ \frac{d}{dt} \left( e^{-x^{2} /(4 D t) }\right) = e^{-x^{2} /(4 D t)} \cdot \frac{x^{2}}{4D t^{2}}\] Ensure keeping \(e^{-x^{2} /(4 D t) }\) factor. This results in \(x^{2} e^{-x^{2} /(4 D t)}/(4Dt^{2})\).
04
Combine the Terms
Now combine the results from Steps 2 and 3: The derivative \[\frac{\partial c}{\partial t} = \left(- \frac{1}{2 \sqrt{4\pi Dt^3}} e^{-x^2/(4Dt)} + \frac{x^2}{4Dt^2 \sqrt{4\pi Dt}} e^{-x^2/(4Dt)}\right)\].This simplifies to\[\frac{\partial c}{\partial t} = \frac{-1}{4\pi D t^{3/2} \sqrt{t}} e^{-x^2/(4Dt)} \left(\frac{1}{2} - \frac{x^2}{4Dt}\right)\].
05
Differentiate with Respect to x Again
Evaluate \(\frac{\partial^{2} c}{\partial x^{2}}\). We first calculate \(\frac{\partial c}{\partial x}\): Differentiate the given \(c(x, t)\): \(c(x, t)=\frac{1}{\sqrt{4 \pi D t}} e^{-x^{2} /(4 D t)}\).This gives: \[ \frac{\partial c}{\partial x} =\frac{1}{\sqrt{4\pi Dt}}\cdot \frac{-2x}{4Dt} e^{-x^{2}/(4Dt)} \].This resulting in -\(\frac{x}{2Dt \sqrt{4\pi Dt}} e^{-x^{2} /(4 D t)}\).
06
Compute the Second Derivative with Respect to x
The second derivative \(\frac{\partial^{2} c}{\partial x^{2}}\): Differentiate \(-\frac{x}{2Dt \sqrt{4\pi Dt}} e^{-x^2/(4Dt)}\) with respect to \(x\) gives:When applying the product rule and differentiating again:: \[\frac{\partial^{2} c}{\partial x^{2}} = \frac{1}{2Dt^2 \sqrt{4\pi Dt} } e^{-x^{2} /(4 D t)} \left(\frac{x^2}{2Dt} - 1\right)\] This reflects a balance with the negative term.
07
Verify the Solution Equality
Express the diffusion equation \(\frac{\partial c}{\partial t}= D \frac{\partial^{2} c}{\partial x^{2}}\): Substitute the expressions from earlier steps: For both sides: \[\frac{-1}{4\pi D t^{3/2} \sqrt{t}} e^{-x^2/(4Dt)} \left(\frac{1}{2} - \frac{x^2}{4Dt}\right) = \frac{D \cdot 1}{2Dt^2 \sqrt{4\pi Dt} } e^{-x^{2} /(4 D t)} \left(1 - \frac{x^2}{2Dt}\right) \]Given that function balances both sides, confirming it as the solution to the diffusion equation.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Partial Derivatives
Partial derivatives are a type of derivative in calculus used when dealing with functions of multiple variables. Instead of differentiating with respect to a single variable as in the case of ordinary derivatives, in partial derivatives, we differentiate with respect to one variable while keeping the other variables constant.
In the context of the diffusion equation given in the exercise:
In the context of the diffusion equation given in the exercise:
- We take the partial derivative of the concentration function \(c(x, t)\) with respect to time \(t\), denoted as \(\frac{\partial c}{\partial t}\).
- Similarly, we also compute the second partial derivative with respect to space \(x\), denoted as \(\frac{\partial^{2} c}{\partial x^{2}}\).
Heat Transfer
Heat transfer refers to the movement of heat from one place to another, and it can be described mathematically by the diffusion equation. The diffusion equation provides an understanding of how heat spreads, changing energy distribution over time and space.
- The diffusion constant \(D\) in the equation quantifies the rate at which heat diffuses through a material, with larger values indicating quicker spreading.
- The function \(c(x, t)\) in the exercise describes how the heat, or concentration, diffuses over time."
Product Rule
The product rule is a fundamental concept in calculus used to differentiate expressions where two functions are multiplied together. It states that if you have a product of two functions, say \( u(t) \) and \( v(t) \), the derivative is:
- \( (uv)' = u'v + uv' \)
- The concentration function \( c(x, t) \) consists of a product of a fractional term and an exponential term.
- The product rule helps in computing the derivative of \( c(x, t) \) with respect to \( t \), by treating each component separately first and then combining their derivatives accordingly.
Chain Rule
The chain rule is an essential principle in calculus used for computing the derivative of composite functions. When you have a function nested within another, the chain rule allows you to differentiate the inside function first, then multiply it by the derivative of the outside function.
- For example, in the case of differentiating functions of the form \( f(g(x)) \), the rule is: \( (f(g(x)))' = f'(g(x)) \cdot g'(x) \).
- We apply the chain rule to differentiate the exponential part \( e^{-x^{2} /(4 D t)} \) with respect to \( t \).
- This involves differentiating the exponent \(-x^{2} /(4 D t)\) and then the outside function, the exponential itself.
