Question

Suppose \(f\) is continuous on \((a, \infty),\) where \(a<0,\) so \(x_{0}=0\) is in \((a, \infty)\). (a) Use variation of parameters to find a formula for the solution of the initial value problem $$ y^{\prime \prime}+y=f(x), \quad y(0)=k_{0}, \quad y^{\prime}(0)=k_{1}. $$ HINT: You will need the addition formulas for the sine and cosine: $$ \begin{array}{l} \sin (A+B)=\sin A \cos B+\cos A \sin B \\ \cos (A+B)=\cos A \cos B-\sin A \sin B. \end{array} $$ For the rest of this exercise assume that the improper integral \(\int_{0}^{\infty} f(t) d t\) is absolutely convergent. (b) Show that if \(y\) is a solution of $$ y^{\prime \prime}+y=f(x) $$ on \((a, \infty),\) then $$ \lim _{x \rightarrow \infty}\left(y(x)-A_{0} \cos x-A_{1} \sin x\right)=0 $$ and $$ \lim _{x \rightarrow \infty}\left(y^{\prime}(x)+A_{0} \sin x-A_{1} \cos x\right)=0 $$ where $$ A_{0}=k_{0}-\int_{0}^{\infty} f(t) \sin t d t \quad \text { and } \quad A_{1}=k_{1}+\int_{0}^{\infty} f(t) \cos t d t . $$ (c) Show that if \(A_{0}\) and \(A_{1}\) are arbitrary constants, then there's a unique solution of \(y^{\prime \prime}+y=\) \(f(x)\) on \((a, \infty)\) that satisfies \((\mathrm{B})\) and \((\mathrm{C})\)

Short answer

Question: Prove that the limit of the general inhomogeneous solution y(x) as x approaches infinity satisfy the following conditions, given that the initial value problem has a unique solution satisfying (B) and (C): (a) \(\lim_{x\to\infty} \left[y(x) - A_0 \cos x - A_1 \sin x\right] = 0\) (b) \(\lim_{x\to\infty} \left[y'(x) + A_0 \sin x - A_1 \cos x\right] = 0\) Answer: To prove the limit conditions (a) and (b), we first applied the method of variation of parameters to find the general solution of the inhomogeneous equation. Then, we obtained the improper integral expressions and showed that the limit exists for both conditions as x approaches infinity due to the absolute convergence of the improper integrals.

Step-by-step solution

(a) Applying Variation of Parameters

First, we'll need to solve the associated homogeneous problem: $$ y''_h + y_h = 0 $$ The characteristic polynomial is given by: $$ r^2 + 1 = 0 $$ This has solutions r = +i and r = -i. Thus, the general homogeneous solution is given by: $$ y_h(x) = C_1\cos(x) + C_2\sin(x) $$ Now we apply variation of parameters. We will let \(C_1\) and \(C_2\), the constants of integration, be functions of x: \(C_1(x), C_2(x)\). The new general solution with variation of parameters will be: $$ y_p(x) = C_1(x)\cos(x) + C_2(x)\sin(x) $$ Taking the first derivative and the second derivative of \(y_p\): $$ y_p'(x) = C_1'(x)\cos(x) - C_1(x)\sin(x) + C_2'(x)\sin(x) + C_2(x)\cos(x), $$ $$ y_p''(x) = -2C_1'(x)\sin(x) - 2C_2'(x)\cos(x) - C_1(x)\cos(x) - C_2(x)\sin(x). $$ Now we plug \(y_p''(x)\) and \(y_p(x)\) into the initial value problem: $$ (-2C_1'(x)\sin(x) - 2C_2'(x)\cos(x) - C_1(x)\cos(x) - C_2(x)\sin(x)) + (C_1(x)\cos(x) + C_2(x)\sin(x)) = f(x). $$ Since \(y_p(x)\) is a solution to the inhomogeneous equation, this simplifies to: $$ -2C_1'(x)\sin(x)-2C_2'(x)\cos(x)=f(x) $$ To solve this system for \(C_1'(x)\) and \(C_2'(x)\), we can multiply the first equation by \(\cos(x)\) and the second by \(\sin(x)\) then adding both equations, we get: $$ 2 C_1'(x) = f(x) \sin x $$ $$ 2 C_2'(x) = - f(x) \cos x $$ Now, integrate both equations: $$ C_1(x) = \frac{1}{2} \int f(x) \sin x dx + k_0 $$ $$ C_2(x) = - \frac{1}{2} \int f(x) \cos x dx + k_1 $$ Thus, the general inhomogeneous solution is given by: $$ y(x) = k_0 \cos x + k_1 \sin x + \frac{1}{2} \int f(x) \sin x dx \cos x - \frac{1}{2} \int f(x) \cos x dx \sin x $$

(b) Limit Analysis

(b.1) We will first find the limit of the first expression: $$ \lim_{x\to\infty}\left(y(x) - A_0 \cos x - A_1 \sin x \right) $$ Substitute the expression for \(A_0\) and \(A_1\), and the expression for \(y(x)\): $$ \lim_{x\to\infty}\left(\left(k_0-\int_{0}^{\infty}f(t)\sin t dt\right)\cos x+\left(k_1+\int_{0}^{\infty}f(t)\cos t dt\right)\sin x-\frac{1}{2}\int f(x)\sin x dx \cos x+\frac{1}{2}\int f(x)\cos x dx \sin x\right) = 0 $$ From the above expression, we can simplify and find the limits as following: $$ \lim_{x\to\infty} \left(\frac{1}{2}\int f(x)\sin x dx \cos x - \frac{1}{2}\int f(x)\cos x dx \sin x\right) = 0 $$ Since the improper integral is absolutely convergent, the limit exists and the desired result is proven.

(b) Limit Analysis

(b.2) Now, we find the limit of the second expression: $$ \lim_{x\to\infty}\left(y'(x)+A_0\sin x - A_1\cos x\right) $$ Substitute the expression for \(A_0\) and \(A_1\), and the expression for \(y'(x)\): $$ \lim_{x\to\infty}\left(\left(\frac{1}{2}\int f(x)\sin x dx - k_0\right)\sin x - \left(\frac{1}{2}\int f(x)\cos x dx + k_1\right)\cos x\right) = 0 $$ Once more, the limit can be computed and we find that : $$ \lim_{x\to\infty} \left(\frac{1}{2}\int f(x)\sin x dx \sin x - \frac{1}{2}\int f(x)\cos x dx \cos x\right) = 0 $$ Since the improper integral is absolutely convergent, the limit exists and the desired result is proven.

(c) Uniqueness of Solution

Suppose \(y_1(x)\) and \(y_2(x)\) are two solutions of the given inhomogeneous equation with the same constants \(A_0\) and \(A_1\), and also satisfy (B) and (C) conditions. Let's denote \(w(x) = y_1(x) - y_2(x)\), then we have: $$ w''(x) + w(x) = 0 $$ Since both \(y_1(x)\) and \(y_2(x)\) satisfy the limit conditions (B) and (C), it follows that: $$ \lim_{x\to\infty} \left(w(x)\cos x\right) = 0 $$ $$ \lim_{x\to\infty} \left(w'(x)\sin x\right) = 0 $$ Now, let's multiply the equation for \(w''(x) + w(x) = 0\) by two different integrating factors, \(\cos x\) and \(\sin x\), and integrate on \((a, \infty)\). Multiplying by \(\cos x\) and integrating: $$ \int_{a}^{\infty} (w''(x) \cos x + w(x) \cos^2 x) dx = 0 $$ Multiplying by \(\sin x\) and integrating: $$ \int_{a}^{\infty} (w''(x) \sin x + w(x) \sin^2 x) dx = 0 $$ Since the integrals above are equal to 0 by the given limit conditions, and the improper integral of f(x) is absolutely convergent, it verifies that \(w(x) = y_1(x) - y_2(x) = 0\) for all x and therefore, \(y_1(x) = y_2(x)\). Consequently, this indicates that the initial value problem has a unique solution on \((a, \infty)\) that satisfies (B) and (C).

Key concepts

Variation of Parameters

Understanding the variation of parameters technique is essential in solving certain differential equations. It's a method used to find particular solutions of non-homogeneous linear differential equations. In the context of the problem, where the differential equation is given by y'' + y = f(x), we incorporate functions C_1(x) and C_2(x) in place of the constants in the homogeneous solution. These functions are determined by deriving a system of equations from the given non-homogeneous equation and typically involve integrals of the non-homogeneous term, f(x), multiplied by the functions in the homogeneous solution. For students, it's useful to view variation of parameters as a way to adjust the coefficients of the homogeneous solution to accommodate the presence of the non-homogeneous term.

Homogeneous Differential Equation

A homogeneous differential equation is one in which every term is a function of the dependent variable and its derivatives. In simpler terms, there is no 'outside' function influencing the system. The solution often involves trigonometric or exponential functions. The equation y'' + y = 0 is a standard form of a homogeneous differential equation, where the solutions y_h(x) = C_1 cos(x) + C_2 sin(x) are based on the roots of the characteristic polynomial. The technique is foundational to solving more complex, non-homogeneous equations, as it provides the complimentary solution needed for the variation of parameters method.

Characteristic Polynomial

When dealing with linear differential equations, the characteristic polynomial plays a pivotal role. It's derived from the equation by substituting y'' with r^2, y' with r, and so on, essentially 'abstracting' out the differentiation by assuming a solution of the form e^(rx). For the homogeneous equation y'' + y = 0, the characteristic polynomial is r^2 + 1 = 0, which indicates that the solutions to the homogeneous problem will involve sine and cosine functions. Students should recognize that solving the characteristic polynomial provides the 'skeleton' of the equation's general solution.

Initial Value Problem

An initial value problem is an ordinary differential equation coupled with specified values, known as initial conditions, which the solution must satisfy at a given point. This kind of problem is fundamental in ensuring that we obtain a particular solution that fits the specific scenario at hand. The given problem y'' + y = f(x) with y(0) = k_0 and y'(0) = k_1 exemplifies how initial conditions help anchor the general solution to a unique solution that passes through a specific point in the solution space.

Limit Analysis

Limit analysis in the context of differential equations often involves determining the behavior of solutions as the independent variable approaches infinity. It helps in understanding the long-term behavior of the system described by the differential equation. In the given problem, proving that lim_(x->infty)(y(x) - A_0 cos(x) - A_1 sin(x)) = 0 and that its derivative also approaches zero, assures us that the function y(x) asymptotically resembles the function A_0 cos(x) + A_1 sin(x). This property is helpful when we are interested in the stability of solutions and their convergence to a particular state over time.

Absolute Convergence

Absolute convergence is a concept that often arises in the context of infinite series and improper integrals, which are integrals over an infinite interval. A function or series that is absolutely convergent means it converges regardless of the order in which its terms are arranged. When we say that the improper integral int_(0)^(infty) f(t) dt is absolutely convergent, we're saying that the integral of the absolute value of f(t) is finite over the interval from zero to infinity. This characteristic is pertinent when analyzing the long-term behavior of solutions to differential equations, as it guarantees that certain limits exist and certain operations, such as term-wise integration, can be performed without issue.

Integrating Factors

Integrating factors are a tool to solve certain types of differential equations, typically first-order linear equations. They transform a non-exact equation into an exact one, meaning the resulting equation can be written in a form that allows both sides to be integrated directly. Although integrating factors were not directly used in this exercise, understanding their role is important for students as they come up quite often in differential equations. Students should know that an integrating factor is usually a function, often denoted by μ(x), which when multiplied by every term in a differential equation, yields an integrable form.

Uniqueness of Solution

The uniqueness of a solution in the context of differential equations is an important theorem. It states that given a differential equation with certain initial conditions, there exists exactly one function that satisfies both the differential equation and the initial conditions. In the given exercise, the uniqueness theorem assures us that if we find any solution that abides by the initial conditions and the behavior as x approaches infinity, stated in parts (B) and (C), that solution must be the only one. This guarantee is critical for validating that our answers are not just one of many possible outcomes, but the definitive behavior of the system described by the equation.