Problem 37 - Advanced Engineering Mathematics

Question

The molecular concentration \(C(x, t)\) of a liquid is given by \(C(x, t)=t^{-1 / 2} e^{-x^{2} / k t}\). Verify that this function satisfies the diffusion equation: $$ \frac{k}{4} \frac{\partial^{2} C}{\partial x^{2}}=\frac{\partial C}{\partial t} $$

Step-by-Step Solution

Answer

The function satisfies the diffusion equation.

Step 1: Differentiate with respect to x

First, we need to find the second partial derivative of \( C(x, t) \) with respect to \( x \). Start by differentiating \( C(x, t) \) once:\[ \frac{\partial C}{\partial x} = t^{-1/2} \cdot \frac{d}{dx}\left(e^{-x^2 / kt}\right) = t^{-1/2} \cdot \left(-\frac{2x}{kt}\right) e^{-x^2 / kt} = -\frac{2x}{kt^{3/2}} e^{-x^2 / kt}. \]

Step 2: Differentiate again with respect to x

Differentiate \( \frac{\partial C}{\partial x} \) with respect to \( x \) to find the second derivative:\[ \frac{\partial^2 C}{\partial x^2} = \frac{d}{dx} \left(-\frac{2x}{kt^{3/2}} e^{-x^2 / kt} \right). \]Using the product rule gives:\[ = \left( -\frac{2}{kt^{3/2}} e^{-x^2 / kt} \right) + \left( -\frac{2x}{kt^{3/2}} \right) \left( -\frac{2x}{kt} \right) e^{-x^2 / kt} = -\frac{2}{kt^{3/2}} e^{-x^2 / kt} + \frac{4x^2}{k^2t^{5/2}} e^{-x^2 / kt}. \]

Step 3: Differentiate with respect to t

Now, differentiate \( C(x, t) \) with respect to \( t \):\[ \frac{\partial C}{\partial t} = \frac{d}{dt} \left( t^{-1/2} e^{-x^2 / kt} \right). \]Using the product rule:\[ = \left( -\frac{1}{2} t^{-3/2} e^{-x^2 / kt} \right) + \left( t^{-1/2} \right) \left( \frac{x^2}{kt^2} \right) e^{-x^2 / kt} = -\frac{1}{2t^{3/2}} e^{-x^2 / kt} + \frac{x^2}{kt^{5/2}} e^{-x^2 / kt}. \]

Step 4: Check the diffusion equation

Substitute the derivatives into the diffusion equation to check equality:\[ \frac{k}{4} \cdot \left( -\frac{2}{kt^{3/2}} e^{-x^2 / kt} + \frac{4x^2}{k^2t^{5/2}} e^{-x^2 / kt} \right) = -\frac{1}{2t^{3/2}} e^{-x^2 / kt} + \frac{x^2}{kt^{5/2}} e^{-x^2 / kt}. \]Expanding the left side yields:\[ -\frac{1}{2t^{3/2}} e^{-x^2 / kt} + \frac{x^2}{kt^{5/2}} e^{-x^2 / kt}, \] which matches the right side.

Key Concepts

Partial Derivatives

Partial derivatives are a fundamental concept in calculus, especially when working with functions that have more than one variable. In our exercise, you might notice we have a function of two variables: position \( x \) and time \( t \). This function, \( C(x, t) \), is actually representing the molecular concentration of a liquid.
When we perform a partial derivative on \( C(x, t) \) with respect to \( x \), we treat \( t \) as a constant. This is because we're interested in understanding how changes in \( x \) affect \( C \), while keeping \( t \) fixed. Similarly, when differentiating with respect to \( t \), \( x \) is held constant.
To verify that the concentration function satisfies the diffusion equation, you must carefully compute these partial derivatives. Firstly, we calculated \( \frac{\partial C}{\partial x} \), which helps in finding \( \frac{\partial^2 C}{\partial x^2} \), the second partial derivative with respect to \( x \). Each step helps unravel the relationship between the variables and the concentration.
Understanding how partial derivatives apply is crucial, as they show us how variables interact within the function, enabling us to solve complex equations like the diffusion equation.

Molecular Concentration

Molecular concentration refers to the amount of a particular molecule present in a given volume of solution. This concept is key in the diffusion equation exercise, as we're analyzing how concentration changes over time and space.
The given function \( C(x, t) = t^{-1/2} e^{-x^2 / kt} \) describes how the concentration of molecules at a point \( x \) changes as time \( t \) evolves. Let's break this down:

• \( t^{-1/2} \): This term suggests that the concentration decreases over time. As \( t \) increases, \( t^{-1/2} \) decreases, indicating a dilution effect typically observed in diffusion processes.
• \( e^{-x^2 / kt} \): This exponential term represents how concentration varies spatially with \( x \). It predicts lower concentration away from the origin, reaffirming that diffusion causes the molecules to spread out over time.

Grasping how such functions represent molecular concentrations helps in understanding a range of physical and chemical processes, including heat conduction and the spreading of particles within a fluid.

Product Rule

The product rule is an essential tool in calculus used when differentiating products of two functions. In the context of the diffusion equation, it plays a crucial role in obtaining partial derivatives.
For a product of two functions \( u(x) \) and \( v(x) \), the product rule states that their derivative is \( \frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x) \).
When applied to our exercise, the product rule comes into play while differentiating terms involving both \( x \) and \( t \). For instance, when differentiating \( -\frac{2x}{kt^{3/2}} e^{-x^2 / kt} \) in step 2, the product rule helps distribute the differentiation across the product of a polynomial and an exponential term.

• The first part involves differentiating the coefficient and exponential separately, resulting in \( \left(-\frac{2}{kt^{3/2}} e^{-x^2 / kt}\right) \).
• Simultaneously, the rule allows us to differentiate the terms \( \left(-\frac{2x}{kt^{3/2}} \right) \) and \( \left(-\frac{2x}{kt}\right) \), as seen in the final result.

Knowing how to apply the product rule can significantly simplify the process of differentiating complex expressions involving multiple variables, thereby confirming the diffusion equation.