Problem 5 - Advanced Engineering Mathematics

Question

Use separation of variables to find, if possible, product solutions for the given partial differential equation. $$ x \frac{\partial u}{\partial x}=y \frac{\partial u}{\partial y} $$

Step-by-Step Solution

Answer

The solutions are of the form \( u(x, y) = C(xy)^{-\lambda} \).

Step 1: Assume a Product Solution

Begin by assuming a solution of the form \( u(x, y) = X(x)Y(y) \), where \( X \) is a function of \( x \) only, and \( Y \) is a function of \( y \) only.

Step 2: Substitute into the PDE

Substitute \( u(x, y) = X(x)Y(y) \) into the given partial differential equation: \( x \frac{d}{dx}(X(x)Y(y)) = y \frac{d}{dy}(X(x)Y(y)) \). This expands to \( xY(y)X'(x) = yX(x)Y'(y) \), where primes denote derivatives with respect to \( x \) and \( y \) respectively.

Step 3: Separate Variables

Rearrange the equation to separate variables: \( \frac{X'(x)}{X(x)} = \frac{y}{x} \frac{Y'(y)}{Y(y)} \). This implies that both sides must be equal to a constant \( -\lambda \).

Step 4: Solve the X Equation

From \( \frac{X'(x)}{X(x)} = -\lambda \), integrate to obtain \( \ln|X(x)| = -\lambda \ln|x| + C \). Exponentiate to find \( X(x) = Cx^{-\lambda} \).

Step 5: Solve the Y Equation

From \( \frac{y}{x} \frac{Y'(y)}{Y(y)} = -\lambda \), rewrite as \( yY'(y) = -\lambda xY(y) \). Integrate with respect to \( y \) to get \( \ln|Y(y)| = -\lambda \ln|y| + C \), and exponentiate to find \( Y(y) = Cy^{-\lambda} \).

Step 6: Form the General Solution

Combine the solutions for \( X(x) \) and \( Y(y) \) to form the general solution: \( u(x, y) = Cx^{-\lambda}y^{-\lambda} = C(xy)^{-\lambda} \).

Step 7: Conclusion

The product solutions for the given partial differential equation are of the form \( u(x, y) = C(xy)^{-\lambda} \), where \( C \) and \( \lambda \) are constants.

Key Concepts

Partial Differential Equation

A partial differential equation (PDE) is an equation that involves multiple independent variables and their partial derivatives. Unlike ordinary differential equations, which have derivatives with respect to a single variable, PDEs can involve multiple unknown functions and several corresponding variables. They are crucial in various scientific fields like physics, engineering, and finance because they model phenomena with more than one independent factor, such as temperature in a heated room over time.

To solve a PDE, one typically seeks a function that satisfies the given differential equation. One example from our exercise is:

• \( x \frac{\partial u}{\partial x} = y \frac{\partial u}{\partial y} \)

This equation showcases how the partial derivatives of a function \( u(x, y) \) relate the independent variables \( x \) and \( y \) through a proportionate relationship. Understanding how to find solutions to such equations is where techniques like separation of variables become valuable.

Product Solutions

When dealing with PDEs, one method to find solutions is by assuming that the solution can be expressed as a product of functions. This technique is called product solutions, where we break down a complex problem into simpler parts. The idea is to express the overall function \( u(x, y) \) as a product of two separate functions:

• \( u(x, y) = X(x)Y(y) \)

Here, \( X(x) \) is a function dependent only on \( x \) and \( Y(y) \) depends on \( y \). The assumption of product solutions helps simplify a PDE because it allows us to handle each variable's effect independently, dividing a complex differential equation into smaller, more manageable pieces. These pieces can then be tackled individually, often yielding straightforward results after application of methods like separation of variables.

Integration

Integration is applied in solving differential equations to find the function that satisfies a given rate of change. In our exercise, once we have separated the variables in the PDE, both sides of the equation can be tackled using integration to find functions \( X(x) \) and \( Y(y) \).

For example, once we achieve separation:

• \( \frac{X'(x)}{X(x)} = -\lambda \)

We integrate both sides to find \( X(x) \). This involves integrating:

• \( \int \frac{dX(x)}{X(x)} = \int -\lambda \, dx \)

Logarithmic integration leads to \( \ln|X(x)| \), resulting in a power-law function upon exponentiation. Similarly, this process applies to find \( Y(y) \). Integrating across a separation of variables framework enables us to derive functions that satisfy each variable's independent part of the PDE, simplifying the path to a general solution.

Solution Method

The method we use in solving a PDE involving product solutions is separation of variables. This approach breaks a PDE into parts that can be individually solved. Here's a step-by-step overview:

• Assume a product solution form \( u(x, y) = X(x)Y(y) \).
• Substitute this form into the PDE to break it down into simpler components.
• Separate the variables so that each side of the equation depends on only one independent variable.
• Set each side equal to a constant \( -\lambda \) because they are equal across all values, leading to simpler ordinary differential equations (ODEs).
• Integrate each ODE separately to find \( X(x) \) and \( Y(y) \).
• Combine them back to form the general solution of the original PDE.

This method is powerful because it leverages symmetry or simplicity inherent in many physical problems, helping solve complex situations in a straightforward manner. It’s particularly effective in linear PDEs with boundary and initial conditions, providing function solutions that can be adapted to specific scenarios.