Problem 31 - Advanced Engineering Mathematics

Question

Assuming \(u(x, y)=X(x) Y(y)\) and substituting into \(\partial^{2} u / \partial x^{2}-u=0\) we get \(X^{\prime \prime} Y-X Y=0\) or \(Y\left(X^{\prime \prime}-X\right)=0\) This implies \(X(x)=c_{1} e^{x}\) or \(X(x)=c_{2} e^{-x}\). For these choices of \(X, Y\) can be any function of \(y .\) Two solutions of the partial differential cquation are then $$u_{1}(x, y)=A(y) e^{x} \quad \text { and } \quad u_{2}(x, y)=H(y) e^{x}.$$ since the partial differential equation is linear and homogencous the superposition principle indicates that another solution is $$u(x, y)=u_{1}(x, y)+u_{2}(x, y)=A(y) e^{x}+B(y) e^{-x}.$$

Step-by-Step Solution

Answer

The solution to the PDE is \( u(x, y) = A(y)e^x + B(y)e^{-x} \).

Step 1: Understand the Problem

We start by identifying the given partial differential equation (PDE): \( \frac{\partial^{2} u}{\partial x^{2}} - u = 0 \). Here, \( u(x,y) = X(x) Y(y) \) is an assumed solution which means we're exploring solutions that can be separated into functions of \( x \) and \( y \).

Step 2: Substitute Assumed Solution

Substituting \( u(x, y) = X(x)Y(y) \) into the PDE, we have \( X''(x)Y(y) - X(x)Y(y) = 0 \). Factor out \( Y(y) \) to get \( Y(y)(X''(x) - X(x)) = 0 \). Since \( Y(y) eq 0 \), this implies \( X''(x) - X(x) = 0 \).

Step 3: Solve for X(x)

The equation \( X''(x) - X(x) = 0 \) is a linear homogeneous differential equation with constant coefficients. Its characteristic equation is \( r^2 - 1 = 0 \), which factors to \( (r-1)(r+1) = 0 \). Thus, \( r = 1 \) or \( r = -1 \), giving solutions \( X(x) = c_1 e^x + c_2 e^{-x} \).

Step 4: Choose Particular Solutions

To find particular solutions based on the results from Step 3, choose \( X(x) = c_1 e^x \) and \( X(x) = c_2 e^{-x} \). For these choices of \( X(x) \), no condition is imposed on \( Y(y) \), so \( Y(y) \) can be any function of \( y \).

Step 5: Formulate Solutions u1 and u2

Two particular solutions of the PDE are \( u_1(x, y) = A(y)e^x \) and \( u_2(x, y) = B(y)e^{-x} \), where \( A(y) \) and \( B(y) \) are arbitrary functions that only depend on \( y \).

Step 6: Use Superposition Principle

Since the given PDE is linear and homogeneous, the superposition principle allows us to combine \( u_1(x, y) \) and \( u_2(x, y) \) to form a general solution: \( u(x, y) = A(y)e^x + B(y)e^{-x} \).

Key Concepts

Partial Differential Equations

Partial Differential Equations (PDEs) are equations that involve partial derivatives of a multivariable function. In essence, these equations serve as a fundamental tool in describing physical phenomena, such as heat diffusion and wave propagation, where the system's behavior depends on multiple variables.
When dealing with PDEs, we often seek to find solutions that demonstrate how these variables interact over a specific domain.
Key elements of PDEs include:

• Order: The order is determined by the highest derivative present in the equation. For instance, our example has a second-order PDE since it involves a second-order derivative with respect to x.
• Linearity: Linear PDEs, like the one given, show solutions that can be directly added together if they solve the equation individually.

In the exercise, the PDE is \( \frac{\partial^{2} u}{\partial x^{2}} - u = 0 \), suggesting that the behavior of the function \( u(x,y) \) is dependent on how its second derivative with respect to x relates to its original form.

Homogeneous Differential Equation

A Homogeneous Differential Equation is a type of differential equation where terms can be grouped so that each term is proportional to the solution function itself.
These equations often arise with constant coefficients, whereby the sum of individual solutions coincidentally provides a uniform solution under certain conditions.
Homogeneous equations, like in our example where \( X''(x) - X(x) = 0 \), portray the concept of zero forcing function, meaning no external input is modifying the equation's inherent solution behavior.
Key aspects include:

• Linear Combination: Solutions of homogeneous differential equations can be added or scaled, resulting in other valid solutions.
• Constant Coefficients: This feature enables the use of characteristic equations to derive solutions, due to the unchanging nature across different derivatives.

The problem at hand illustrates the homogeneous nature by simplifying into a form solved by an exponential solution approach, yielding forms like \( c_1 e^x \), allowing other components to freely adapt.

Superposition Principle

The Superposition Principle is a fascinating mathematical property that applies to linear systems, particularly significant in solving homogeneous linear differential equations like the one given.
It states that if a function can be expressed as a sum of simpler solutions, then these simpler solutions are also valid for the equation. In simpler terms, when dealing with linear equations, you can "superpose" or "add" solutions together.
Implementation aspects include:

• Linearity: The principle is valid only for linear systems and implies that the system's behavior can be attributed to the aggregate effect of its components.
• Solution Flexibility: It provides the opportunity to construct a general solution from particular solutions discovered independently.

In the context of the problem, recognizing \( u_1(x, y) = A(y)e^x \) and \( u_2(x, y) = B(y)e^{-x} \) as solutions allows the formation of a general solution \( u(x, y) = A(y)e^x + B(y)e^{-x} \), underlining the principle's practical utility.

Characteristic Equation

The Characteristic Equation is a vital tool used in solving linear homogeneous differential equations with constant coefficients by transforming them into algebraic equations.
It stems from the assumption that solutions might resemble exponential functions, turning derivatives into a polynomial of a characteristic variable, often denoted as \( r \).
Core features to understand:

• Transformation: The process involves seeking exponential solutions \( X(x) = e^{rx} \), which simplifies differential equations into characteristic polynomials.
• Root Analysis: Solving the characteristic equation \( r^2 - 1 = 0 \) leads to roots \( r = 1 \) and \( r = -1 \), which in turn dictates the form of the solution in terms of \( e^x \) and \( e^{-x} \).

In the exercise at hand, the characteristic equation guides toward simple and clear exponential solutions, leveraging the homogeneous and linear nature of the differential equation to establish general solution forms efficiently.