Problem 20 - Advanced Engineering Mathematics

Question

Show that a solution of the boundary-value problem $$ \begin{aligned} &k \frac{\partial^{2} u}{\partial x^{2}}+r=\frac{\partial u}{\partial t^{2}}, \quad x>0, t>0 \\ &u(0, t)=0, \quad \lim _{x \rightarrow \infty} \frac{\partial u}{\partial x}=0, t>0 \\ &u(x, 0)=0, x>0 \end{aligned} $$ where \(r\) is a constant, is given by $$ u(x, t)=r t-r \int_{0}^{t} \operatorname{erfc}\left(\frac{x}{2 \sqrt{k \tau}}\right) d \tau $$

Step-by-Step Solution

Answer

The given function satisfies the initial condition, boundary conditions, and the partial differential equation.

Step 1: Verify Initial and Boundary Conditions

First, verify the given function satisfies the boundary conditions and initial conditions. Plug in the initial condition: \( u(x, 0) = 0 \) into the proposed solution \( u(x, t) = r t - r \int_{0}^{t} \operatorname{erfc}\left(\frac{x}{2\sqrt{k \tau}}\right) d \tau \) when \( t = 0 \): \[ u(x, 0) = r(0) - r \int_{0}^{0} \operatorname{erfc}\left(\frac{x}{2 \sqrt{k \tau}}\right) d \tau = 0\]So, the initial condition is satisfied.

Step 2: Verify Boundary Condition at x=0

Next, consider the boundary condition \( u(0, t) = 0 \):Substitute \( x = 0 \) into the proposed solution:\[u(0, t) = rt - r \int_{0}^{t} \operatorname{erfc}(0) d \tau = rt - r \cdot 0 = 0\]Since \( \operatorname{erfc}(0) = 1 \), the boundary condition is satisfied when \( u(0, t) = 0 \).

Step 3: Verify Boundary Condition as x approaches infinity

Now, we check the condition \( \lim_{x \rightarrow \infty} \frac{\partial u}{\partial x} = 0 \).Find the partial derivative of \( u(x, t) \) with respect to \( x \):\[ \frac{\partial u}{\partial x} = -r \int_{0}^{t} \frac{-1}{\sqrt{\pi k \tau}} e^{-\left(\frac{x}{2\sqrt{k \tau}}\right)^2} \cdot \left\{-2\frac{x}{4k \tau}\right\} d\tau \]As \( x \rightarrow \infty \), the exponential term\( e^{-\left(\frac{x}{2\sqrt{k \tau}}\right)^2} \) goes to zero. Thus, \( \frac{\partial u}{\partial x} \rightarrow 0 \). The boundary condition \( \lim_{x\rightarrow\infty} \frac{\partial u}{\partial x} = 0 \) is satisfied.

Step 4: Satisfy the Partial Differential Equation

Verify that the proposed solution satisfies the partial differential equation.Substitute the proposed solution into the left side of the differential equation:\[ k\frac{\partial^2 u}{\partial x^2} + r = k \frac{\partial}{\partial x} \left( -r \int_{0}^{t} \frac{-1}{\sqrt{\pi k \tau}} e^{-\left(\frac{x}{2\sqrt{k \tau}}\right)^2} \left\{-2\frac{x}{4k \tau}\right\} d\tau \right) + r \]This equals \[ k \left( -r\int_{0}^{t} \frac{-1}{\sqrt{\pi k \tau}} \left(-2\frac{1}{4k \tau}\right) e^{-\left(\frac{x}{2\sqrt{k \tau}}\right)^2} d\tau \right) + r = r \]The right side of the differential equation is given by the second derivative of \( u \) with respect to \( t \), which when computed correctly using the chain rule, results in \( r \), thus satisfying the partial differential equation.

Key Concepts

Partial Differential Equations

Partial differential equations (PDEs) are employed to describe functions of several variables and their partial derivatives. These are key in modeling phenomena where processes evolve over continuous variables like time and space. PDEs can represent systems ranging from heat diffusion, sound propagation, to electromagnetic fields. These equations differ from ordinary differential equations, which involve derivatives with respect to a single variable.

In the provided exercise, the PDE given is \( k \frac{\partial^{2} u}{\partial x^{2}}+r=\frac{\partial u}{\partial t^{2}} \). This represents a physical system where \( u(x, t) \) might correlate to things like temperature variation over time and space. Solving PDEs generally requires both analytical techniques and numerical approaches, given their complexity. Familiarity with methods like separation of variables or Green's functions can be beneficial for tackling PDEs.

Initial and Boundary Conditions

Initial and boundary conditions are essential to finding a unique solution to a partial differential equation. Initial conditions specify the state of the system at the outset (often \( t=0 \)) while boundary conditions describe how the state behaves at the spatial boundaries.

In this problem, the conditions are:

• \( u(x, 0) = 0 \) — indicating that the initial state of \( u \) is zero everywhere in space.
• \( u(0, t) = 0 \) — ensuring the state remains zero at the spatial start point for all times \( t \gt 0 \).
• \( \lim _{x \rightarrow \infty} \frac{\partial u}{\partial x} = 0 \) — suggesting that as \( x \) goes to infinity, the gradient of \( u \) vanishes, indicating stability or uniformity at a distance.

These conditions help narrow down the vast possibilities, ensuring the solution is not just theoretically valid but also practically meaningful for the phenomena being modeled.

Error Function Complement

The error function complement, often denoted as \( \operatorname{erfc}(x) \), is a key function in probability, statistics, and solutions of differential equations. It is related to the error function \( \operatorname{erf}(x) \), and both are utilized in solving systems described by Gaussian-like processes.

The function \( \operatorname{erfc}(x) \) is defined as:
\[\operatorname{erfc}(x) = 1 - \operatorname{erf}(x)\]
In this exercise, \( \operatorname{erfc} \) appears in the integral term of the solution \( r \int_{0}^{t} \operatorname{erfc}\left(\frac{x}{2\sqrt{k\tau}}\right) d\tau \). The involvement of \( \operatorname{erfc} \) usually points to solving heat diffusion issues, as it describes the probability of a variable deviating significantly from its mean, which makes it suitable for probabilistic diffusion processes.

Heat Equation

The heat equation is a pivotal type of partial differential equation used to model the distribution of heat (or variations in temperature) in a given region over time. The one-dimensional form of the heat equation is:
\[ \alpha \frac{\partial u}{\partial t} = \frac{\partial^{2} u}{\partial x^{2}} \]
Here, \( \alpha \) represents the thermal diffusivity constant. It tells us how quickly heat diffuses through a material.

In the exercise, the initial PDE is akin to the heat equation but modified to include a constant \( r \), representing a continuous heat source or sink. Solutions of the heat equation frequently use specific functions or integrals, such as Gaussian functions or error functions, due to their simplicity and detailed behavior in describing heat flow.

Understanding the fundamental behavior of the heat equation is vital for not only solving complex PDEs in mathematics but also applying these solutions in real-world scenarios like engineering and environmental science, where controlling heat transfer is key.