Problem 28 - Advanced Engineering Mathematics

Question

In Problems 27 and 28 , show that the given partial differential equation possesses the indicated product solution. $$ \begin{aligned} &\frac{\partial^{2} u}{\partial r^{2}}+\frac{1}{r} \frac{\partial u}{\partial r}+\frac{1}{r^{2}} \frac{\partial^{2} u}{\partial \theta^{2}}=0 \\ &u=\left(c_{1} \cos \alpha \theta+c_{2} \sin \alpha \theta\right)\left(c_{3} r^{\alpha}+c_{4} r^{-\alpha}\right) \end{aligned} $$

Step-by-Step Solution

Answer

By substitution and simplification, the given solution satisfies the PDE as structured statements match correctly for all terms.

Step 1: Identify the form of the solution

Inside the partial differential equation (PDE) given, we are told that the function \( u \) is in the form of a product solution: \( u = (c_1 \cos \alpha \theta + c_2 \sin \alpha \theta)(c_3 r^\alpha + c_4 r^{-\alpha}) \). This is a clue that this problem can be solved using separation of variables.

Step 2: Express \( u \) as the product of functions of \( r \) and \( \theta \)

Let \( u(r, \theta) = R(r) \Theta(\theta) \), where \( R(r) = c_3 r^\alpha + c_4 r^{-\alpha} \) and \( \Theta(\theta) = c_1 \cos \alpha \theta + c_2 \sin \alpha \theta \). Now, we can separately work with functions of \( r \) and \( \theta \).

Step 3: Substitute \( u = R(r) \Theta(\theta) \) into the PDE

Substitute the product solution into the provided PDE: \( \frac{\partial^2 u}{\partial r^2} + \frac{1}{r} \frac{\partial u}{\partial r} + \frac{1}{r^2} \frac{\partial^2 u}{\partial \theta^2} = 0 \).

Step 4: Compute \( \frac{\partial^2 u}{\partial r^2} \) and \( \frac{1}{r} \frac{\partial u}{\partial r} \)

Firstly, find \( \frac{\partial u}{\partial r} = \Theta(\theta) (c_3 \alpha r^{\alpha-1} - c_4 \alpha r^{-\alpha-1}) \). Then, find \( \frac{\partial^2 u}{\partial r^2} = \Theta(\theta)(c_3 \alpha(\alpha-1) r^{\alpha-2} + c_4 \alpha(\alpha+1) r^{-\alpha-2}) \). Compute \( \frac{1}{r} \frac{\partial u}{\partial r} = \Theta(\theta) (c_3 \alpha r^{\alpha-2} - c_4 \alpha r^{-\alpha-2}) \).

Step 5: Evaluate \( \frac{1}{r^2} \frac{\partial^2 u}{\partial \theta^2} \)

Find \( \frac{\partial^2 u}{\partial \theta^2} = R(r) (-c_1 \alpha^2 \cos \alpha \theta - c_2 \alpha^2 \sin \alpha \theta) \). Therefore, \( \frac{1}{r^2} \frac{\partial^2 u}{\partial \theta^2} = R(r) \alpha^2 \Theta(\theta) \frac{1}{r^2} \).

Step 6: Substitute all derivatives back into the PDE

Replace each part into the original PDE: \[ \Theta\frac{1}{r^2} (c_3 \alpha^2 r^\alpha + c_4 \alpha^2 r^{-\alpha}) = -\Theta (c_3 \alpha(\alpha-1) r^{\alpha-2} + c_4 \alpha(\alpha+1) r^{-\alpha-2} + c_3 \alpha r^{\alpha-2} - c_4 \alpha r^{-\alpha-2}) \].

Step 7: Simplify and verify the equations

Simplify the equation from Step 6. You must show that both sides of this equation can be matched, satisfying all terms for every \( r \). As both sides can indeed be simplified into zero due to the structure of the components of coefficients of \( r^{\alpha-2} \) or \( r^{-\alpha-2} \), the original equation is satisfied, confirming the solution.

Key Concepts

Separation of Variables

Separation of variables is a powerful method used to solve partial differential equations (PDEs), especially when the equation can be expressed as a product of single-variable functions. It involves breaking down a complex PDE into simpler ordinary differential equations (ODEs) by assuming that the solution can be written as a product of functions, each depending only on a single variable.

In this problem, the equation is given in terms of polar coordinates, featuring variables \( r \) and \( \theta \). By setting the solution \( u(r, \theta) = R(r) \Theta(\theta) \), we express \( u \) as a product of functions, one depending solely on \( r \) and the other solely on \( \theta \).

• This method effectively reduces a PDE into separate ODEs, enabling easier manipulation and solution.
• The process relies on finding a common separation constant that enables the complete separation of variables within the PDE.

This approach greatly simplifies solving problems with specific boundary or initial conditions, often encountered in physics and engineering applications.

Product Solution

A product solution is an expression for the solution of a PDE that is achieved through the separation of variables. In this context, it refers to the representation of the solution as a product of two or more functions, each concerning a different variable of the PDE.

For the given problem, the product solution for \( u \) is: \[ u = (c_1 \cos \alpha \theta + c_2 \sin \alpha \theta)(c_3 r^\alpha + c_4 r^{-\alpha})\]The product solution method harnesses the advantage of handling different aspects of the solution separately. Each component of the product addresses a different part of the equation.

• The first part of the product \( (c_1 \cos \alpha \theta + c_2 \sin \alpha \theta) \) covers the \( \theta \)-dependence.
• The second part \( (c_3 r^\alpha + c_4 r^{-\alpha}) \) covers the \( r \)-dependence.

By focusing on parts separately, complex interactions among variables in PDEs become manageable through simpler functions responsible for distinct parts.

Laplace's Equation

Laplace's equation is a second-order PDE that is widely used in potential theory, fluid dynamics, and electrodynamics. The form of Laplace's equation is:\[ abla^2 u = 0\]in Cartesian coordinates, or \[ \frac{\partial^2 u}{\partial r^2} + \frac{1}{r} \frac{\partial u}{\partial r} + \frac{1}{r^2} \frac{\partial^2 u}{\partial \theta^2} = 0\]in polar coordinates, which appears in this problem.

Laplace's equation is significant in describing steady-state distributions in systems, meaning they do not change over time. It assumes conditions where there are no internal sources or sinks, making it based solely on spatial variables.

• Its solutions are known as harmonic functions, which possess unique properties such as the inability to achieve local maxima or minima within their domain.
• The equation is crucial in defining the equilibrium states in various physical phenomena.

Understanding the nature of Laplace's equation helps in grasping the steady-state behavior of potential fields.

Bessel's Equation

Bessel's equation arises naturally when solving PDEs with cylindrical or spherical symmetry through separation of variables. For this context, the equation obtained involves solutions of Laplace's PDE in polar coordinates, resulting in differential equations similar to Bessel's equation.

Bessel's equation typically takes the form:\[ x^2 y'' + x y' + (x^2 - n^2)y = 0\]Solutions to Bessel’s equation are known as Bessel functions, denoted as \( J_n(x) \) and \( Y_n(x) \). These functions are crucial in problems involving radial symmetry.

• Bessel functions model wave-like solutions in various cylindrical systems, such as electromagnetic waves or heat conduction in a cylindrical object.
• They are used extensively where radial dependence appears alongside oscillatory behaviors.

Neglecting or misunderstanding Bessel functions can significantly affect the analysis of situations involving rotational symmetries.