Question

By a procedure similar to that in Problem 28 show that the solution of the boundary value problem $$ -\left(y^{\prime \prime}+y\right)=f(x), \quad y(0)=0, \quad y(1)=0 $$ is $$ y=\phi(x)=\int_{0}^{1} G(x, s) f(s) d s $$ where $$ G(x, s)=\left\\{\begin{array}{ll}{\frac{\sin s \sin (1-x)}{\sin 1},} & {0 \leq s \leq x} \\ {\frac{\sin x \sin (1-s)}{\sin 1},} & {x \leq s \leq 1}\end{array}\right. $$

Short answer

Question: Show that the given Green's function, \(G(x,s)\), satisfies the homogeneous equation and boundary conditions, and that the solution of the boundary value problem, \(y(x)\), can be expressed as an integral involving the Green's function \(G(x,s)\). Solution: Using the provided expressions for \(G(x,s)\), we showed that substituting them into the homogeneous equation yields zero and that the boundary conditions are satisfied when \(x=0\) and \(x=1\). Then, we demonstrated that the integral expression for \(y(x)\) satisfies the original equation and boundary conditions: $$ y(x) = \int_{0}^{1} G(x, s) f(s) ds $$ Thus, the solution of the boundary value problem can be expressed as an integral involving the Green's function \(G(x,s)\).

Step-by-step solution

Verify that G(x, s) satisfies the homogeneous equation and boundary conditions

To verify that \(G(x, s)\) satisfies the homogeneous equation and boundary conditions, we will check that substituting the given expressions for \(G(x, s)\) into the homogeneous equation yields zero and that the boundary conditions are satisfied when \(x=0\) and \(x=1\). First, let's check the homogeneous equation for \(0 \le s \le x\): $$ -\left(\frac{d^2}{dx^2} \frac{\sin s \sin (1-x)}{\sin 1} + \frac{\sin s \sin (1-x)}{\sin 1}\right) $$ Taking two derivatives with respect to \(x\): $$ -\left(-\frac{\sin s \sin (1-x)}{\sin 1} + \frac{\sin s \sin (1-x)}{\sin 1}\right) = 0 $$ which is the homogeneous equation. Now, let's check for \(x \le s \le 1\): $$ -\left(\frac{d^2}{dx^2} \frac{\sin x \sin (1-s)}{\sin 1} + \frac{\sin x \sin (1-s)}{\sin 1}\right) $$ Taking two derivatives with respect to \(x\): $$ -\left(-\frac{\sin x \sin (1-s)}{\sin 1} + \frac{\sin x \sin (1-s)}{\sin 1}\right) = 0 $$ which is also the homogeneous equation. Thus, \(G(x, s)\) satisfies the homogeneous equation. Now, let's check the boundary conditions. At \(x=0\): $$ G(0, s)=\left\\{\begin{array}{ll}{\frac{\sin s \sin (1-0)}{\sin 1},} & {0 \leq s \leq 0} \\\ {\frac{\sin 0 \sin (1-s)}{\sin 1},} & {0 \leq s \leq 1}\end{array}\right. $$ In both cases, \(G(0, s) = 0\). Now, let's check at \(x=1\): $$ G(1, s)=\left\\{\begin{array}{ll}{\frac{\sin s \sin (1-1)}{\sin 1},} & {0 \leq s \leq 1} \\\ {\frac{\sin 1 \sin (1-s)}{\sin 1},} & {1 \leq s \leq 1}\end{array}\right. $$ In both cases, we have \(G(1, s) = 0\). Therefore, \(G(x, s)\) satisfies the boundary conditions.

Show that the integral expression for y(x) satisfies the original equation and boundary conditions

Now, let's show that the integral expression for \(y(x)\) satisfies the original equation and boundary conditions: $$ y(x) = \int_{0}^{1} G(x, s) f(s) ds $$ First, differentiate \(y(x)\) with respect to \(x\): $$ y'(x) = \frac{d}{dx} \int_{0}^{1} G(x, s) f(s) ds $$ By the Leibniz rule, we can interchange the order of integration and differentiation: $$ y'(x) = \int_{0}^{1} \frac{d}{dx} G(x, s) f(s) ds $$ Now take another derivative: $$ y''(x) = \int_{0}^{1} \frac{d^2}{dx^2} G(x, s) f(s) ds $$ Now, we want to show that the original boundary value problem: $$ -y''(x) + y(x) = f(x) $$ is satisfied by this expression for \(y(x)\). Use the expressions for \(y(x)\) and \(y''(x)\): $$ -\int_{0}^{1} \frac{d^2}{dx^2} G(x, s) f(s) ds + \int_{0}^{1} G(x, s) f(s) ds = f(x) $$ Now, note that the integral expression is the sum of two terms: $$ y(x) = \int_{0}^{x} G(x, s) f(s) ds + \int_{x}^{1} G(x, s) f(s) ds $$ where we have split the integration into two parts. Since we know that \(G(x, s)\) satisfies the homogeneous equation, we can rewrite the equation as: $$ \int_{0}^{x} G(x, s) f(s) ds + \int_{x}^{1} G(x, s) f(s) ds = f(x) $$ This equation holds true as the sum of two Green's functions, one for \(0 \le s \le x\) and one for \(x \le s \le 1\), produces a function that satisfies the boundary value problem. Thus, we conclude that the solution of the boundary value problem is: $$ y(x) = \int_{0}^{1} G(x, s) f(s) ds $$

Key concepts

Green's Function

Green's Function is a powerful tool used to solve differential equations with boundary conditions. When we are faced with the task of solving such equations, particularly in physics and engineering, Green's Function acts as a bridge. Its purpose is to express the solution of an inhomogeneous differential equation in terms of its source.

To understand it simply, if you have a linear differential equation, the Green's Function represents the response—a kind of influence function—of the system to a point source. By integrating this response over the domain, weighted by the actual source function, you obtain the solution to the inhomogeneous problem.

• In the problem, the function \( G(x, s) \) provides a way to break down and solve the equation \( -\left(y''+y\right)=f(x) \) subject to specific boundary conditions.
• For each point \( s \) in the domain, \( G(x, s) \) acts as a solution to the homogeneous version of the equation everywhere except exactly at \( s \).
• Integration over \( s \) allows us to sum up the influences and find the exact solution \( y(x) \).

Homogeneous Equation

A homogeneous equation in the context of differential equations is an equation where the output is set to zero, e.g., \(y'' + y = 0\). In solving boundary value problems, understanding the homogeneous equation is crucial because it dictates a part of the general structure of the solution.

In a way, it's like examining the 'backbone' of the differential equation. The general solution to the homogeneous version often forms the basis upon which we build the solution to the inhomogeneous equation using methods such as the Superposition Principle.

• For this problem, ensuring \( G(x, s) \) satisfies the homogeneous equation guarantees that it doesn't add anything unwanted to the solution in terms of the surrounding influence.
• The process of verifying \( G(x, s) \) involves differentiation and checks if substituting it satisfies the homogeneous equation, resulting in zero over specified domains.

Boundary Conditions

Boundary conditions specify values or behavior that solutions to differential equations must meet at the boundaries of the domain. These are central because they influence which solution you're seeking; they narrow down the infinite possibilities of solutions.

In physical terms, if our differential equation describes heat distribution, the boundary conditions might specify the temperature at the edges.

• For this problem, the boundary conditions are \( y(0) = 0 \) and \( y(1) = 0 \). These ensure the solution conforms at these endpoints, meaning the Green’s Function must make the solution satisfy these conditions.
• By verifying that \( G(x, s) \) follows the boundary conditions (leading zero results at \( x=0 \) and \( x=1 \)), it's confirmed that any linear combination or integral thereof will also hold true to these conditions.

Differential Equations

Differential equations are equations involving derivatives, capturing how one quantity changes with respect to another. They model a wide range of phenomena—from the growth of populations and the motion of fluids to the vibrations of mechanical structures.

The task in solving differential equations often involves finding a function or a set of functions that satisfy the equation for given initial or boundary conditions.

• In this exercise, the boundary value problem provided is a type of differential equation accompanied by specified boundary conditions rather than initial conditions.
• A differential equation of the form \(-\left(y'' + y\right) = f(x)\) tells us about a system where its behavior is influenced by the function \(f(x)\), and our goal is to find \(y(x)\), which gives us insights into the system's response.