Problem 14 - Advanced Engineering Mathematics

Question

In Problems 13 and 14, use the substitution \(u(r, t)=v(r, t)+\psi(r)\) to solve the given boundary-value problem.$$ \begin{aligned} &\frac{\partial^{2} u}{\partial r^{2}}+\frac{1}{r} \frac{\partial u}{\partial r}+\beta=\frac{\partial u}{\partial t}, \quad 00, \quad \beta \text { a constant } \\ &u(1, t)=0, \quad t>0 \\ &u(r, 0)=0, \quad 0

Step-by-Step Solution

Answer

Use substitution to break down the PDE, solve \( \psi(r) \) for terms to cancel, then solve for \( v(r,t) \) and apply boundary conditions.

Step 1: Apply Substitution

Start by applying the given substitution \( u(r, t) = v(r, t) + \psi(r) \). Substitute into the original boundary-value problem:

Step 2: Differentiate Using the Substitution

Substitute \( u(r, t) = v(r, t) + \psi(r) \) into the PDE:\[\frac{\partial^2 u}{\partial r^2} = \frac{\partial^2 v}{\partial r^2} + \psi''(r), \quad \frac{\partial u}{\partial r} = \frac{\partial v}{\partial r} + \psi'(r) \]Substitute these into the equation.

Step 3: Simplify the PDE

The PDE becomes:\[\frac{\partial^2 v}{\partial r^2} + \psi''(r) + \frac{1}{r}(\frac{\partial v}{\partial r} + \psi'(r)) + \beta = \frac{\partial v}{\partial t} + \psi'(r)\]Rearrange to group the terms involving \( v \) on one side and those involving \( \psi \) on the other, to get the PDE in terms of \( v(r,t) \).

Step 4: Choose a Suitable \( \psi(r) \)

Choose \( \psi(r) \) such that \( \psi''(r) + \frac{1}{r}\psi'(r) + \beta = 0 \). This ensures the \( \psi(r) \) terms cancel out from one side of our equation. Solve this ordinary differential equation to find \( \psi(r) \).

Step 5: Solve the Simplified PDE for \( v(r, t) \)

After ensuring \( \psi \) terms cancel, the resulting PDE simplifies to:\[\frac{\partial^2 v}{\partial r^2} + \frac{1}{r}\frac{\partial v}{ partial r} = \frac{\partial v}{\partial t}\]Now solve this PDE subject to the boundary conditions.

Step 6: Apply Boundary Conditions

Use the given boundary conditions:- \( v(1, t) = -\psi(1) \)- \( v(r, 0) = -\psi(r) \)Apply these conditions to solve for \( v(r, t) \).

Step 7: Construct the Solution for \( u(r, t) \)

Combine \( v(r, t) \) and \( \psi(r) \) to construct the final solution for \( u(r, t) \), using \( u(r, t) = v(r, t) + \psi(r) \).

Key Concepts

Partial differential equations

Partial differential equations, or PDEs, are a type of mathematical equation used to describe various phenomena such as heat, sound, and fluid dynamics in fields like physics and engineering. Unlike ordinary differential equations, which involve derivatives of a function with just one variable, PDEs involve partial derivatives of functions with multiple variables. These equations can be quite complex as they describe the rate of change concerning multiple variables simultaneously.
PDEs are crucial for modeling systems where changes happen in more than one dimension, for example:

• Temperature changes over time and across an area.
• Wave propagation across the surface of water.

In the given boundary-value problem, the PDE involves the function u(r, t), which changes with respect to both 'r' and 't'. Here, 'r' could represent a spatial variable while 't' could represent time.

Substitution method

The substitution method is a powerful technique used to simplify complex differential equations, making them easier to solve. By substituting one equation into another, you often convert a partial differential equation into simpler forms, sometimes reducing complexity significantly.
In this problem, we use the substitution \( u(r, t) = v(r, t) + \psi(r) \) to transform the original PDE. This substitution was chosen because it separates the function into two parts:

• \( v(r, t) \), which typically simplifies the partial differential aspect.
• \( \psi(r) \), which handles conditions best solved as an ordinary differential equation.

This separation allows us to focus on solving simpler equations rather than tackling the entire complex PDE at once. Substitution is effective because it redefines the problem in more manageable terms.

Boundary conditions

Boundary conditions are essential in solving PDEs as they define the values that the solution must satisfy on the boundaries of the domain. These conditions ensure that the solution is unique and physically meaningful.
In this exercise, two boundary conditions are provided:

• \( u(1, t) = 0 \) specifies the solution's behavior at the boundary \( r = 1 \).
• \( u(r, 0) = 0 \) defines the solution's starting condition at \( t = 0 \).

Boundary conditions are applied after substituting the initial function to ensure the modified equations respect these constraints. By determining how \( \psi(r) \) and \( v(r, t) \) satisfy these boundaries, we can construct a complete solution for \( u(r, t) \). For example, knowing \( v(1, t) = -\psi(1) \) and \( v(r, 0) = -\psi(r) \) helps validate the solution over the specified domain.

Ordinary differential equation

An ordinary differential equation (ODE) involves functions with derivatives of only one independent variable. ODEs are usually simpler and are used to describe one-dimensional dynamic systems or processes.
In this boundary-value problem, after substituting \( u(r, t) = v(r, t) + \psi(r) \), solving for \( \psi(r) \) becomes an ODE:

• \( \psi''(r) + \frac{1}{r}\psi'(r) + \beta = 0 \)

This ODE captures the behavior of \( \psi(r) \), independent of \( t \), and must be solved separately. Solving the ODE for \( \psi(r) \) allows the elimination of certain terms in the PDE when substituted back, simplifying the problem to find \( v(r, t) \). This systematic breakdown into smaller, more manageable parts underscores the elegance and power of mathematical methods in solving complex problems.