Problem 16 - 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. $$ \left.\frac{\partial u}{\partial x}\right|_{x=0}=-f(t), \quad \lim _{x \rightarrow \infty} u(x, t)=0, \quad u(x, 0)=0 $$

Step-by-Step Solution

Answer

Apply the Laplace transform, solve the transformed equation using boundary conditions, then find the inverse Laplace for the solution.

Step 1: Apply the Laplace Transform

We apply the Laplace transform with respect to time (t) to the heat equation \(u_{xx} = u_t\). Let \(U(x,s)\) be the Laplace transform of \(u(x,t)\), i.e., \(U(x, s) = \mathcal{L}\{u(x, t)\} = \int_{0}^{\infty} u(x, t) e^{-st} dt\). The transformed equation becomes \(u_{xx} - sU(x, s) = -u(x, 0)\). Given \(u(x, 0) = 0\), it simplifies to \(u_{xx} - sU(x, s) = 0\).

Step 2: Solve the Transformed ODE

The equation \(U_{xx} - sU = 0\) is a second-order ordinary differential equation in \(x\). The general solution for such an equation is \(U(x, s) = A(s) e^{\sqrt{s} x} + B(s) e^{-\sqrt{s} x}\), where \(A(s)\) and \(B(s)\) are functions determined by boundary conditions.

Step 3: Apply Boundary Conditions

Use the boundary conditions to find \(A(s)\) and \(B(s)\). The condition \(\lim_{x \to \infty} u(x, t) = 0\) implies that \(B(s) = 0\) because \(e^{\sqrt{s}x}\) would diverge otherwise. Thus, \(U(x, s) = A(s)e^{-\sqrt{s} x}\).

Step 4: Apply Neumann Boundary Condition

The Neumann boundary condition \(\left.\frac{\partial u}{\partial x}\right|_{x=0} = -f(t)\) when transformed becomes \(\left.\frac{\partial U}{\partial x}\right|_{x=0} = -\mathcal{L}\{f(t)\}\). Therefore, \(-\sqrt{s}A(s) = -F(s)\), so \(A(s) = \frac{F(s)}{\sqrt{s}}\), where \(F(s) = \mathcal{L}\{f(t)\}\).

Step 5: Express Complete Solution in Laplace Domain

Substitute \(A(s)\) into the general solution to get \(U(x, s) = \frac{F(s)}{\sqrt{s}} e^{-\sqrt{s} x}\). This expression describes the solution in the Laplace domain.

Step 6: Use Inverse Laplace Transform to Find Solution

To find \(u(x, t)\), apply the inverse Laplace transform to obtain \(u(x, t) = \mathcal{L}^{-1}\left\{\frac{F(s)}{\sqrt{s}} e^{-\sqrt{s} x}\right\}\). This step requires knowledge of specific inverse transforms or convolution theorems, depending on \(f(t)\).

Key Concepts

Heat Equation

The heat equation is a fundamental partial differential equation that describes how heat diffuses through a given region over time. It is often represented as \(u_t = u_{xx}\). In this notation, \(u(x, t)\) refers to the temperature at a

• specific location \(x\) in a medium
• and at a specific time \(t\).

The subscript \(t\) indicates differentiation with respect to time, and \(x\) denotes differentiation with respect to space.Solving the heat equation involves determining finding a function \(u(x, t)\) that satisfies the equation under specified conditions. These conditions can relate to the boundaries of the region (boundary conditions) or initial conditions at the start of observation (initial conditions).
Since heat transfer is ubiquitous in nature and engineering, understanding this equation is essential for a range of applications like predicting temperature changes in materials or optimizing heating processes. It also illustrates fundamental diffusion concepts applicable to various fields.

Boundary Conditions

Boundary conditions are crucial to solving differential equations like the heat equation, as they provide additional information needed to find a unique solution. In our specific problem, the boundary conditions are:

• \(\left.\frac{\partial u}{\partial x}\right|_{x=0}=-f(t)\) – a Neumann boundary condition specifying the rate of change of temperature at the boundary \(x=0\).
• \(\lim _{x \rightarrow \infty} u(x, t)=0\) – indicating that as \(x\) approaches infinity, the temperature tends to zero.
• \(u(x, 0) = 0\) – an initial condition stating at time zero, the temperature throughout the region is zero.

Each boundary condition plays a vital role in determining the functions \(A(s)\) and \(B(s)\) while solving the transformed ordinary differential equation. Applying these conditions involves translating the physical limitations and behaviors of the system into mathematical expressions. Understanding boundary conditions is vital for accurately modeling physical phenomena as they reflect how a system interacts with its environment.

Ordinary Differential Equation

An ordinary differential equation (ODE) involves functions of a single variable and their derivatives. In the process above, the heat equation was simplified through the Laplace transform, resulting in an ODE: \(u_{xx} - sU(x, s) = 0\). Solving this ODE is a crucial step in finding the solution to the initial partial differential equation.Ordinary differential equations are simpler than partial differential equations (PDEs) because they involve only derivatives with respect to one variable. This simplification allows mathematicians and engineers to use various established methods and techniques to find solutions, such as characteristic equations or separation of variables. The solution to an ODE often involves arbitrary constants or functions, which are then determined using boundary and initial conditions as we have seen with \(A(s)\) and \(B(s)\) in our examplified solution.

Inverse Laplace Transform

The inverse Laplace transform is a mathematical process used to revert a function from its Laplace-transformed form back to the time domain. After solving an equation in the Laplace domain, we often need to find the original function in terms of time to interpret it in a real-world context. In our example, we use the inverse Laplace transform \[\mathcal{L}^{-1}\left\{\frac{F(s)}{\sqrt{s}} e^{-\sqrt{s} x}\right\}\]to find \(u(x, t)\), which represents the temperature distribution as a function of space and time.This process often involves looking up transforms in tables or using mathematical software. Sometimes, it requires more advanced techniques such as convolution or complex integration. Understanding the inverse transform is crucial for completing the solution to differential equations and allowing hypotheses derived from mathematics to be applied back into their practical settings. The inverse Laplace transform ensures that the final solution is expressed in a form suitable for interpreting and predicting physical phenomena.