Problem 14 - Advanced Engineering Mathematics

Question

Use the Laplace transform to solve the heat equation \(u_{x x}=u_{t}, x>0, t>0\) subject to the given conditions. $$ \left.\frac{\partial u}{\partial x}\right|_{x=0}=u(0, t)-50, \quad \lim _{x \rightarrow \infty} u(x, t)=0, \quad u(x, 0)=0 $$

Step-by-Step Solution

Answer

The solution is \( u(x, t) = 50 e^{-(x^2/(4t))} \text{erfc} \left( \frac{x}{2\sqrt{t}} \right) \).

Step 1: Identify the Equation and Conditions

The heat equation is given by the partial differential equation (PDE) \( u_{xx} = u_t \). We have boundary conditions: \( \frac{\partial u}{\partial x}\big|_{x=0} = u(0, t) - 50 \), \( \lim_{x \rightarrow \infty} u(x, t) = 0 \), and the initial condition \( u(x, 0) = 0 \).

Step 2: Apply Laplace Transform in Time Domain

Take the Laplace transform of the heat equation with respect to time \( t \), which transforms \( u(x, t) \) into \( U(x, s) \). This gives us \( \mathcal{L}\{u_t\} = sU(x, s) - u(x, 0) = sU(x, s) \) because \( u(x, 0) = 0 \). So, the equation becomes \( u_{xx} = sU(x, s) \).

Step 3: Solve the Transformed Equation

The transformed equation \( u_{xx} = sU(x, s) \) is a second-order ordinary differential equation (ODE). The general solution for this ODE is \( U(x, s) = A e^{-\sqrt{s}x} + B e^{\sqrt{s}x} \).

Step 4: Apply Boundary Conditions at Infinity

Apply the boundary condition \( \lim_{x \rightarrow \infty} u(x, t) = 0 \). Since \( e^{\sqrt{s}x} \) grows without bound as \( x \to \infty \), we must have \( B = 0 \) to satisfy this condition. Thus \( U(x, s) = A e^{-\sqrt{s}x} \).

Step 5: Apply Boundary Condition at x=0

Use the boundary condition \( \frac{\partial u}{\partial x}\big|_{x=0} = u(0,t) - 50 \). In the Laplace domain, \( U_x(x, s) = -A\sqrt{s}e^{-\sqrt{s}x} \). At \( x = 0 \), \( U_x(0, s) = -A\sqrt{s} = U(0, s) - \frac{50}{s} \). Since \( U(0, s) = A \), this gives: \(-A\sqrt{s} = A - \frac{50}{s} \).

Step 6: Solve for Coefficient A

Solving the equation \(-A\sqrt{s} = A - \frac{50}{s} \) for \( A \), we have:\[ A + A\sqrt{s} = \frac{50}{s} \]\[ A(1 + \sqrt{s}) = \frac{50}{s} \]\[ A = \frac{50}{s(1 + \sqrt{s})} \].

Step 7: Invert the Laplace Transform

Substitute \( A = \frac{50}{s(1 + \sqrt{s})} \) back into \( U(x, s) = A e^{-\sqrt{s}x} \) to get:\[ U(x, s) = \frac{50}{s(1 + \sqrt{s})} e^{-\sqrt{s}x} \]Now invert the Laplace transform to find \( u(x, t) \). Recognize \( \frac{50}{s(1 + \sqrt{s})} e^{-\sqrt{s}x} \) as a convolution which can be evaluated using standard Laplace inverse techniques and tables.

Step 8: Compute the Inverse Transform

Recognizing \( \frac{1}{s(1 + \sqrt{s})} \) relates to a convolution involving exponential and complementary error function terms, compute \( u(x, t) \) using standard results. The inversion yields:\[ u(x, t) = 50 e^{-(x^2/(4t))} \text{erfc} \left( \frac{x}{2\sqrt{t}} \right) \], where \( \text{erfc} \) is the complementary error function.

Key Concepts

Heat Equation

The heat equation is a classic example of a partial differential equation (PDE) used to describe how heat diffuses through a given region over time. In one dimension, it is expressed as \( u_{xx} = u_t \). This equation signifies that the second spatial derivative equals the first time derivative, reflecting how temperature changes with respect to space and time.

Heat equations are crucial in physics and engineering, modeling heat distribution, and even financial mathematics in certain contexts.
By solving the heat equation, we can predict temperature variations and understand thermal behavior.

• The primary purpose: to model the conduction of heat.
• This PDE is used in various fields: physics, geometry, and beyond.

Solving it under given conditions helps find how initial temperature distributions evolve.

Partial Differential Equation

Partial Differential Equations (PDEs) are equations that include partial derivatives and involve functions of several variables. The heat equation, \( u_{xx} = u_t \), is a simple yet powerful PDE example.

PDEs serve as mathematical descriptions of various physical systems and phenomena. Understanding PDEs allows us to model complex behavior like temperature distribution (heat) or wave propagation (acoustics).

• Consist of partial derivatives with respect to multiple variables.
• Applications span physics, engineering, and beyond.
• Examples include heat equation, wave equation, and Laplace's equation.

Skills in solving PDEs are vital for scientists and engineers to predict system behavior under diverse conditions.

Boundary Conditions

Boundary conditions are essential in solving PDEs as they define the limits within which a solution exists. For the heat equation, we need to know what's happening at the boundaries to find a solution.

There are several types of boundary conditions:

• Dirichlet Conditions: Specify the function's value at the boundary, like \( u(x, 0) = 0 \).
• Neumann Conditions: Involve the derivative's value at the boundary, such as \( \frac{\partial u}{\partial x}\big|_{x=0} = u(0, t) - 50 \).

For our problem, the boundary conditions are:

- \( \lim_{x \rightarrow \infty} u(x, t) = 0 \) means the temperature approaches zero at infinity.
- Neumann boundary at \( x=0 \) relates to how heat flux is linked to temperature initially.
By using these conditions, solutions for the PDE can be narrowed down and determined accurately.

Complementary Error Function

The complementary error function, denoted as \( \text{erfc}(x) \), is a special function related to the Gaussian error function \( \text{erf}(x) \). It's useful in probability, statistics, and partial differential equations, particularly solutions involving diffusion processes.

The complementary error function is defined as:
\[ \text{erfc}(x) = 1 - \text{erf}(x) = \frac{2}{\sqrt{\pi}} \int_{x}^{\infty} e^{-t^2} \, dt \]
In the context of the heat equation, \( \text{erfc}(x) \) often appears when solving diffusion-related PDEs using methods like the Laplace or Fourier transforms.

• It represents the tail probability of the Gaussian distribution.
• Commonly appears in solutions to the heat equation.

The complementary error function helps express solutions neatly, as seen in the final result: \( u(x, t) = 50 e^{-(x^2/(4t))} \text{erfc} \left( \frac{x}{2\sqrt{t}} \right) \). It elegantly summarizes the temperature behavior with respect to space and time.