Problem 37 - Advanced Engineering Mathematics

Question

Verify that the given function satisfies the wave equation: $$a^{2} \frac{\partial^{2} u}{\partial x^{2}}=\frac{\partial^{2} u}{\partial t^{2}}$$ 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

Yes, the function satisfies the diffusion equation after verification.

Step 1: Calculate the First Partial Derivative with Respect to Time

Given the function \( C(x, t) = t^{-1/2} e^{-x^2 / k t} \), we first calculate the first partial derivative with respect to \( t \), denoted as \( \frac{\partial C}{\partial t} \).Using the chain rule and product rule:\[ \frac{\partial C}{\partial t} = \frac{\partial}{\partial t} \left( t^{-1/2} \right) e^{-x^2 / k t} + t^{-1/2} \frac{\partial}{\partial t} \left( e^{-x^2 / k t} \right) \]1. Differentiate \( t^{-1/2} \): \[ \frac{\partial}{\partial t} \left( t^{-1/2} \right) = -\frac{1}{2} t^{-3/2} \]2. Differentiate \( e^{-x^2 / k t} \): \[ \frac{\partial}{\partial t} \left( e^{-x^2 / k t} \right) = e^{-x^2 / k t} \frac{x^2}{k t^2} \] (using the chain rule)Combine these results:\[ \frac{\partial C}{\partial t} = -\frac{1}{2} t^{-3/2} e^{-x^2 / k t} + t^{-1/2} e^{-x^2 / k t} \frac{x^2}{k t^2} \]

Step 2: Calculate the First and Second Partial Derivatives with Respect to Space

Next, we find \( \frac{\partial^2 C}{\partial x^2} \). We first calculate the first partial derivative \( \frac{\partial C}{\partial x} \).Using the chain rule:\[ \frac{\partial C}{\partial x} = t^{-1/2} \frac{\partial}{\partial x} \left( e^{-x^2 / k t} \right) \]Differentiate \( e^{-x^2 / k t} \):\[ \frac{\partial}{\partial x} \left( e^{-x^2 / k t} \right) = -\frac{2x}{k t} e^{-x^2 / k t} \]Hence,\[ \frac{\partial C}{\partial x} = -\frac{2x}{k t^{3/2}} e^{-x^2 / k t} \]Now differentiate again to find \( \frac{\partial^2 C}{\partial x^2} \):Applying the product rule:\[ \frac{\partial^2 C}{\partial x^2} = t^{-1/2} \frac{\partial}{\partial x} \left( -\frac{2x}{k t} e^{-x^2 / k t} \right) \]Differentiate \( -\frac{2x}{k t} e^{-x^2 / k t} \):\[ \frac{\partial}{\partial x} \left( -\frac{2x}{k t} e^{-x^2 / k t} \right) = -\frac{2}{k t} e^{-x^2 / k t} + \frac{4x^2}{k^2 t^2} e^{-x^2 / k t} \]Thus,\[ \frac{\partial^2 C}{\partial x^2} = t^{-1/2} \left( -\frac{2}{k t} e^{-x^2 / k t} + \frac{4x^2}{k^2 t^2} e^{-x^2 / k t} \right) \]

Step 3: Substitute into Diffusion Equation and Simplify

Now substitute \( \frac{\partial^2 C}{\partial x^2} \) and \( \frac{\partial C}{\partial t} \) into the diffusion equation \( \frac{k}{4} \frac{\partial^2 C}{\partial x^2} = \frac{\partial C}{\partial t} \) to verify.The left side:\[ \frac{k}{4} \frac{\partial^2 C}{\partial x^2} = \frac{k}{4} t^{-1/2} \left( -\frac{2}{k t} + \frac{4x^2}{k^2 t^2} \right) e^{-x^2 / k t} \]Simplifying the expression:\[ \frac{k}{4} \times \left( -\frac{2}{k t} + \frac{4x^2}{k^2 t^2} \right) = -\frac{1}{2t} + \frac{x^2}{k t^2} \]The right side:\[ \frac{\partial C}{\partial t} = -\frac{1}{2} t^{-3/2} e^{-x^2 / k t} + t^{-1/2} \frac{x^2}{k t^2} e^{-x^2 / k t} \]Both sides are equivalent, confirming that \( C(x, t) \) satisfies the diffusion equation.

Key Concepts

Wave Equation

The wave equation is a fundamental concept in physics and engineering, used to describe wave motion of all kinds, from sound and light to water waves. It is a second-order linear partial differential equation. In its simplest form, it can be written as: \[a^2 u = a^2 \frac{\partial^2 u}{\partial t^2}\] Here, \( a^2 u \) is the Laplacian of \( u \), a measure of how \( u \) spreads out in space, and \( \frac{\partial^2 u}{\partial t^2} \) describes how \( u \) changes over time. The constant \( a^2 \) represents the square of the wave speed.

• Waves described by this equation include sound waves, where \( u \) might represent air pressure displacement, and electromagnetic waves, such as light.
• Solving the wave equation often requires a combination of techniques, including separation of variables, Fourier transform methods, or numerical approaches.

The wave equation is important in verifying if a given function truly represents a wave. We apply it to functions to verify consistency in how wave-like behaviors propagate over time and space.

Diffusion Equation

The diffusion equation describes processes where particles, energy, or other physical quantities spread out over space and time. It's frequently used in contexts like heat distribution (the heat equation) or concentration of substances in fluids. The form of the diffusion equation is: \[ \frac{k}{4} \frac{\partial^2 C}{\partial x^2} = \frac{\partial C}{\partial t} \] Here, \( \frac{\partial^2 C}{\partial x^2} \) reflects the change in curvature of the concentration profile \( C \) over space, indicating diffusion. \( \frac{\partial C}{\partial t} \) represents the rate of change of \( C \) over time. The constant \( k \) often relates to the diffusion coefficient of the substance.

• In practice, this equation tells us how quickly a substance will disperse in a medium.
• The challenge in using the diffusion equation lies often in solving it for complex initial or boundary conditions.

Verifying that a function satisfies the diffusion equation is akin to proving that it represents diffusion correctly, often requiring derivatives and simplifications for confirmation.

Chain Rule

The chain rule is a fundamental rule in calculus used for differentiating composite functions. When you have a function nested within another, the chain rule helps find the derivative efficiently. For functions \( y = f(g(x)) \), the chain rule is expressed as: \[ \frac{dy}{dx} = \frac{df}{dg} \cdot \frac{dg}{dx} \] Think of it as a way to "chain" together derivatives of the outer function and the inner function.
When it comes to partial differential equations, the chain rule is just as vital. Consider functions of multiple variables, where it is often necessary to keep track of how each part of a composite function changes.

• Involved in stepwise differentiation processes, such as calculating \( \frac{\partial}{\partial t} \) or \( \frac{\partial}{\partial x} \) for both wave and diffusion problems.
• Crucial for analyzing how changes in one variable can affect the entire system in dynamic scenarios.

Understanding and applying the chain rule is essential for tackling complex derivative expressions in partial differential equations.

Product Rule

The product rule provides a method for differentiating products of two functions, which is critical when working with complex expressions involving multiplication. When you have two functions, \( u(x) \) and \( v(x) \), their derivative according to the product rule is: \[\frac{d}{dx}[u(x) v(x)] = u(x) \frac{dv}{dx} + v(x) \frac{du}{dx} \] This ensures that both functions are accounted for in their rates of change. In the realm of partial differential equations, the product rule is frequently used when differentiating terms that involve a product of functions dependent on different variables.
For examples with diffusion or wave equations, you might see the product rule applied as part of:

• Deriving the second derivative in terms of space, where \( C(x, t) \) can be expanded, requiring each part to be differentiated individually.
• Making it possible to solve or verify equations where both time and space variables interact multiplicatively.

Mastering the product rule is key for efficient differentiation of multi-variable, multi-functional expressions in physics and engineering.