Question

Indicate a way of solving nonhomogeneous boundary value problems that is analogous to using the inverse matrix for a system of linear algebraic equations. The Green's function plays a part similar to the inverse of the matrix of coefficients. This method leads to solutions expressed as definite integrals rather than as infinite series. Except in Problem 35 we will assume that \(\mu=0\) for simplicity. (a) Show by the method of variation of parameters that the general solution of the differential equation $$ -y^{\prime \prime}=f(x) $$ can be written in the form $$ y=\phi(x)=c_{1}+c_{2} x-\int_{0}^{x}(x-s) f(s) d s $$ where \(c_{1}\) and \(c_{2}\) are arbitrary constants. (b) Let \(y=\phi(x)\) also be required to satisfy the boundary conditions \(y(0)=0, y(1)=0\) Show that in this case $$ c_{1}=0, \quad c_{2}=\int_{0}^{1}(1-s) f(s) d s $$ (c) Show that, under the conditions of parts (a) and (b), \(\phi(x)\) can be written in the form $$ \phi(x)=\int_{0}^{x} s(1-x) f(s) d s+\int_{x}^{1} x(1-s) f(s) d s $$ (d) Defining $$ G(x, s)=\left\\{\begin{array}{ll}{s(1-x),} & {0 \leq s \leq x} \\ {x(1-s),} & {x \leq s \leq 1}\end{array}\right. $$ show that the solution takes the form $$ \phi(x)=\int_{0}^{1} G(x, s) f(s) d s $$ The function \(G(x, s)\) appearing under the integral sign is a Green's function. The usefulness of a Green's function solution rests on the fact that the Green's function is independent of the nonhomogencous term in the differential equation. Thus, once the Green's function is determined, the solution of the boundary value problem for any nonhomogeneous term \(f(x)\) is obtained by a single integration. Note further that no determination of arbitrary constants is required, since \(\phi(x)\) as given by the Green's function integral formula automatically satisfies the boundary conditions.

Short answer

Question: Define Green's function and use it to express the solution of the nonhomogeneous boundary value problem with the given differential equation and boundary conditions. Differential equation: \( -y^{\prime \prime} = f(x) \) Boundary conditions: \( y(0) = y(1) = 0 \) Solution: Green's function is defined as: $$ G(x,s) = \left\{\begin{array}{ll} s(1-x), & {0 \leq s \leq x} \\ x(1-s), & {x \leq s \leq 1} \end{array}\right. $$ The solution of the nonhomogeneous boundary value problem can be expressed using Green's function as: $$ \phi(x) = \int_{0}^{1} G(x, s) f(s) ds $$

Step-by-step solution

(a) Variation of parameters

To find a general solution for the given differential equation, we should use the method of variation of parameters. The given equation is: $$ -y^{\prime \prime} = f(x) $$ First, we need to find the complementary function. In this case it is a simple equation with constant coefficients. The complementary function is a general solution to the homogeneous version of this equation: $$ -y_c^{\prime \prime} = 0 $$ The solutions of the above equation are linear functions, such as: $$ y_c = c_1 + c_2x $$ Now, let's find the particular solution of the given differential equation. Using variation of parameters, substitute the complementary function into the original equation, replacing \(c_1\) and \(c_2\) with functions of \(x\) that we have yet to determine: $$ -\frac{d^2}{dx^2}(u_1(x) + u_2(x)x) = f(x) $$ Since the first derivative of the complementary function is zero, the second derivative will also be zero. Thus, the equation becomes: $$ -\frac{d^2}{dx^2}(u_2(x)x) = f(x) $$ Now we will integrate both sides: $$ -\int(2 u_{2}^{\prime}(x) + x u_{2}^{\prime \prime}(x)) dx = \int f(x) dx $$ Substituting back into the form of variation of parameters, we get the general solution: $$ y = \phi(x) = c_1 + c_2x - \int_{0}^{x}(x-s) f(s) ds $$

(b) Applying the boundary conditions

Now, we apply the boundary conditions to this general solution: 1. \(y(0) = \phi(0) = c_1 + c_2(0) - \int_{0}^{0}(0-s)f(s) ds = 0\) From this, we can deduce that \(c_1 = 0\). 2. \(y(1) = \phi(1) = c_2(1) - \int_{0}^{1}(1-s)f(s) ds = 0\) From this, we can deduce that \(c_2 = \int_{0}^{1}(1-s)f(s) ds\).

(c) Expressing the function \(\phi(x)\) in a new form

Now, we will express the function \(\phi(x)\) found above in the form stated in the problem: $$ \phi(x) = \int_{0}^{x} s(1-x) f(s) ds + \int_{x}^{1} x(1-s) f(s) ds $$ First, let's rewrite the expression given in part (b) by inserting the values of \(c_1\) and \(c_2\): $$ \phi(x) = x \int_{0}^{1} (1-s) f(s) ds - \int_{0}^{x}(x-s) f(s) ds $$ Now, split the second term on the right side into two integrals: $$ \phi(x) = x \int_{0}^{1} (1-s) f(s) ds - \int_{0}^{x} x f(s) ds + \int_{0}^{x} s f(s) ds $$ Rearrange the terms and combine the integrals: $$ \phi(x) = \int_{0}^{x} s(1-x) f(s) ds + \int_{x}^{1} x(1-s) f(s) ds $$

(d) Defining the Green's function and expressing the solution

To define the Green's function, let's express \(G(x,s)\) as: $$ G(x,s) = \left\{\begin{array}{ll} s(1-x), & {0 \leq s \leq x} \\ x(1-s), & {x \leq s \leq 1} \end{array}\right. $$ Now, let's prove that the solution takes the form: $$ \phi(x) = \int_{0}^{1} G(x, s) f(s) ds $$ From part (c), we already have the function \(\phi(x)\) expressed as a sum of two integrals. The first integral has the limits from \(0\) to \(x\), while the second integral has the limits from \(x\) to \(1\). When \(0 \leq s \leq x\), the first part of \({G(x, s)}\) is \(s(1-x)\) and when \(x \leq s \leq 1\), the second part is \(x(1-s)\). Hence, the solution can be written as: $$ \phi(x) = \int_{0}^{1} G(x, s) f(s) ds $$ The function \(G(x, s)\) is called Green's function, and it is independent of the nonhomogeneous term \(f(x)\). The advantage of this solution is that once the Green's function is established, the solution of the boundary value problem for any nonhomogeneous term \(f(x)\) is achieved by a single integration.

Key concepts

Differential Equations

A differential equation is a mathematical equation involving derivatives of a function. These can represent a variety of phenomena, including motion, heat, and other physical processes. In the simplest terms, they describe how things change.

Nonhomogeneous differential equations have an added function, say \( f(x) \), that makes them different from their homogeneous counterparts. For example, the equation \( -y'' = f(x) \) indicates a nonhomogeneous differential equation because of the presence of \( f(x) \).

Solving differential equations can be quite complex, often involving multiple steps to find a general solution. This general solution usually combines complementary and particular solutions. The complementary solution solves the homogeneous part, while the particular solution addresses the entire equation with added terms.

Green's Functions

The Green's function is a valuable tool for solving linear nonhomogeneous differential equations with specified boundary conditions. Suppose we denote it by \( G(x, s) \). It acts like an inverse operator for differential equations, similar to how an inverse matrix works for systems of algebraic equations. This function is specially designed to handle the boundary constraints, making it independent of the nonhomogeneous term \( f(x) \).

By employing the Green's function, we can transform the problem of solving a differential equation into a single definite integral. This approach simplifies the process, as the Green function only needs to be determined once for the given boundary value problem. The specific solution for any \( f(x) \) is then obtained using:

• \(\phi(x) = \int_{0}^{1} G(x, s) f(s) ds\)

This integral gives us the function \(\phi(x)\), directly accommodating the boundary conditions.

Variation of Parameters

The method of variation of parameters is a technique used to find particular solutions of nonhomogeneous differential equations. Instead of guessing the form of the particular solution, this method involves varying the constants in the complementary solution to functions.

For example, in the equation \( -y'' = f(x) \), we start with the complementary solution, \( y_c = c_1 + c_2x \), where \(c_1\) and \(c_2\) are constants derived from the homogeneous part of the equation. The method of variation of parameters replaces these constants by functions, leading us to a particular solution that satisfies the entire equation.

Through integration and solving, we transform the solution into a form that accounts for both the differential equation and the boundary conditions.

Boundary Conditions

Boundary conditions are constraints necessary to define a unique solution to a differential equation. They specify the behavior of the solution at specific points, often at the endpoints of an interval. For boundary value problems, these conditions are critical in determining the function's values precisely at these boundaries.

For instance, given the boundary conditions \( y(0) = 0 \) and \( y(1) = 0 \), we ensure that the solution \( \phi(x) \) satisfies these conditions at \( x = 0 \) and \( x = 1 \).

The method of using Green's functions inherently incorporates these boundary conditions into the solution. Therefore, once you have the Green's function for a given differential equation, applying it ensures the solution meets all specified boundary constraints without further modification.