Problem 4 - Advanced Engineering Mathematics

Question

Substituting \(u(x, y)=X(x) Y(y)\) into the partial differential equation yields \(X^{\prime} Y=X Y^{\prime}+X Y .\) Separating variables and using the separation constant \(-\lambda\) we obtain $$\frac{X^{\prime}}{X}=\frac{Y+Y^{\prime}}{Y}=-\lambda.$$ Then $$X^{\prime}+\lambda X=0 \quad \text { and } \quad y^{\prime}+(1+\lambda) Y=0$$ so that $$\begin{aligned} &X=c_{1} e^{-\lambda x} \quad \text { and }\\\ &Y=c_{2} e^{-(1+\lambda) y}=0. \end{aligned}$$ A particular product solution of the partial differential equation is $$u=X Y=c_{3} e^{-y-\lambda(x+y)}.$$

Step-by-Step Solution

Answer

The product solution is \(u = c_3 e^{-y - \lambda(x+y)}\).

Step 1: Substitute Variables into the PDE

We start with the substitution of the form \( u(x, y) = X(x) Y(y) \) into the given partial differential equation (PDE). This process allows us to express the PDE in terms of separate functions of \(x\) and \(y\). Hence, the equation becomes \( X' Y = X Y' + X Y \).

Step 2: Separate Variables

To separate variables, we rearrange the equation to obtain two equal expressions each dependent only on one variable. We get \( \frac{X'}{X} = \frac{Y + Y'}{Y} = -\lambda \), where \(-\lambda\) is the separation constant. This step separates the equation into two ordinary differential equations (ODEs).

Step 3: Solve the ODE for X(x)

First, solve the ODE \( \frac{X'}{X} = -\lambda \). Integrating both sides, we find \( X' = -\lambda X \). The solution to this ODE is \( X = c_1 e^{-\lambda x} \), where \( c_1 \) is a constant determined by boundary conditions.

Step 4: Solve the ODE for Y(y)

Next, solve the ODE \( \frac{Y' + Y}{Y} = -\lambda \). Simplifying to obtain \( Y' + (1 + \lambda)Y = 0 \), which is a first-order linear homogeneous ODE. Solving, we get \( Y = c_2 e^{-(1+\lambda)y} \), where \( c_2 \) is another constant.

Step 5: Form the Product Solution

The particular solution to the PDE is given by the product \( u = X \cdot Y \). Substituting the expressions for \(X\) and \(Y\) gives \( u = c_1c_2 e^{-\lambda x} e^{-(1+\lambda) y} \). Simplifying, we have \( u = c_3 e^{-\lambda(x+y)} e^{-y} \) where \( c_3 = c_1c_2 \). Further simplification yields \( u = c_3 e^{-y - \lambda(x+y)} \).

Key Concepts

Separation of Variables

Separation of variables is a fundamental technique in solving partial differential equations (PDEs). It involves rewriting a PDE so that each side of the equation depends on different variables. This helps break down complex equations into simpler, more manageable ones. The core idea is to assume that the solution can be expressed as a product of functions, each dependent on a single variable, say, \( u(x, y) = X(x)Y(y) \).
By substituting \( u(x, y) = X(x)Y(y) \) into the PDE, we can separate the variables. This involves manipulating the equation to isolate expressions of single variables on either side, often leading to simpler ordinary differential equations (ODEs).
The magical part of this method is that it turns a complicated PDE into a set of easier-to-solve ordinary differential equations, substantially simplifying the problem. Understanding and practicing this technique is crucial because it provides a pathway to solve boundary value problems in physics and engineering.

Ordinary Differential Equations

Ordinary Differential Equations (ODEs) are equations containing functions of one independent variable and their derivatives. In the context of separation of variables for a PDE, these equations are typically obtained after separating variables.
Let's take the given ODEs from our exercise: \( X' + \lambda X = 0 \) and \( Y' + (1 + \lambda)Y = 0 \). These are first-order, linear, homogeneous ODEs. Here, \( \lambda \) is the separation constant which arises when we equate the separated parts to a constant value. Solving these requires integration or applying known methods for linear ODEs.The solutions, \( X = c_1 e^{-\lambda x} \) and \( Y = c_2 e^{-(1+\lambda)y} \), demonstrate the classic behavior of exponential solutions to linear homogeneous ODEs. Mastering these cannot be overstated as they are building blocks for more complex solutions.

Boundary Conditions

Boundary conditions are essential in defining particular solutions for differential equations, especially for PDEs derived from physical systems. They offer constraints that help determine the specific solution that corresponds to the physical scenario being modeled.
In practice, boundary conditions are extra information specified in the problem, often taking forms like \( f(a) = b \), where \( a \) and \( b \) are known values at the boundaries. These conditions are used to find the constants present in the solutions of ODEs, such as \( c_1 \) and \( c_2 \) in our example.
They ensure uniqueness and proper alignment with the physical context or restrictions of the problem. Without boundary conditions, many mathematical solutions would remain ambiguous. Thus, they are pivotal in obtaining fully determined, applicable solutions.

Homogeneous Equations

Homogeneous equations are a special type of differential equations where all terms are a function of the dependent variable or its derivatives. In simpler terms, these equations are 'balanced' with nothing on the right-hand side except zero (e.g., \( X' + \lambda X = 0 \)).
In the process of solving PDEs using separation of variables, you often arrive at homogeneous ODEs like in this exercise. These equations typically characterize situations where no external forces are acting (everything comes from within the system).
Solving them usually results in exponential functions, thanks to the nature of their coefficients being constants or functions of the independent variable alone. Understanding this category of equations is crucial as they often appear across various domains, including physics, engineering, and mathematics.