Question

Either solve the given boundary value problem or else show that it has no solution. \(y^{\prime \prime}+2 y=0, \quad y^{\prime}(0)=1, \quad y^{\prime}(\pi)=0\)

Short answer

Answer: Yes, the given boundary value problem has a solution, which is \(y(x) = \frac{1}{\sqrt{2}}\sin(\sqrt{2}x)\).

Step-by-step solution

Solve the ODE

The given ODE is \(y^{\prime \prime} + 2y = 0\). To solve this second-order linear homogeneous ODE, we first determine the characteristic equation: \(m^2 + 2 = 0\). Solving this quadratic equation, we get the complex roots \(m = 0 \pm \sqrt{-2} = 0 \pm i\sqrt{2}\). Therefore, the general solution of the ODE will have the form \(y(x) = C_1 \cos(\sqrt{2}x) + C_2 \sin(\sqrt{2}x)\), where \(C_1\) and \(C_2\) are constants.

Apply the boundary conditions

Now we apply the given boundary conditions: 1. \(y^{\prime}(0) = 1\): Differentiate \(y(x)\) to find \(y'(x) = -C_1 \sqrt{2} \sin(\sqrt{2}x) + C_2 \sqrt{2} \cos(\sqrt{2}x)\). At \(x = 0\), we get \(y'(0) = C_2\sqrt{2} = 1 \Rightarrow C_2 = \frac{1}{\sqrt{2}}\). 2. \(y^{\prime}(\pi) = 0\): At \(x = \pi\), we get \(y'(\pi) = -\frac{1}{\sqrt{2}} \sin(2\pi) + C_1 \sqrt{2} \cos(2\pi)\). Since \(\sin(2\pi) = 0\) and \(\cos(2\pi) = 1\), this simplifies to \(y'(\pi) = C_1 \sqrt{2} = 0\). Thus, \(C_1 = 0\).

Determine if the solution satisfies the boundary value problem

Since we found values for \(C_1\) and \(C_2\) that satisfy both boundary conditions, the given boundary value problem has a solution. The solution is \(y(x) = 0\cdot \cos(\sqrt{2}x) + \frac{1}{\sqrt{2}}\cdot \sin(\sqrt{2}x) = \frac{1}{\sqrt{2}}\sin(\sqrt{2}x)\).

Key concepts

Linear Homogeneous ODE

In the mathematical realm, the problem addressed relates to an ODE—short for Ordinary Differential Equation—which represents a profound field within calculus. Specifically, the equation tackled here is classified as linear and homogeneous. To break this down, an ODE is characterized as linear when its solution can be expressed as a superposition of its homogeneous solutions. What does this mean? Essentially, the principle of superposition allows us to add separate solutions together to obtain new solutions. A homogeneous ODE, on the other hand, has properties that make it even more special; it is an equation in which the output is zero whenever the input is. This allows mathematicians to craft an always applicable, standard form for these types of problems, which often simplifies the process of finding a solution.

In our textbook exercise, the homogeneous linear ODE given is \(y^{\prime \prime} + 2y = 0\). Observe that it features only the function \(y\) and its derivatives, without any additional functions of \(x\) multiplied by \(y\) or its derivatives. This unique characteristic implies that solutions to this equation will involve constants and functions naturally occurring from the characteristic equation—a pivotal element that will be discussed more in the next section.

Characteristic Equation

Advancing to the core of our resolution process, we encounter the concept of the characteristic equation—a cornerstone in solving linear homogeneous ODEs. This equation surfaces from an algebraic approach to differential equations, where we substitute trial solutions into the ODE to identify potential patterns or forms that could solve it. In essence, the characteristic equation is the key that can unlock the general solution to the ODE.

For the exercise at hand, by assuming a solution of \(y = e^{mx}\), we transform the original ODE into a quadratic equation of the form \(m^2 + 2 = 0\), devoid of differential terms. Solving this yields the values of \(m\) and subsequently the form of our solution. The roots here, being complex numbers, lead us directly to the intriguing world of sines and cosines due to Euler's formula, which connects complex exponentials to trigonometric functions. As a result, our general solution transforms into \(y(x) = C_1 \cos(\sqrt{2}x) + C_2 \sin(\sqrt{2}x)\), with \(C_1\) and \(C_2\) as constants that must be determined. To provide context, such trigonometric solutions are indicative of oscillatory behavior, commonly observed in physical phenomena such as waves or harmonic motion.

Boundary Conditions

The final, determinative stage of our problem-solving exercise is the application of boundary conditions. Boundary conditions are the constraints that a solution to a differential equation must satisfy at specific points or over particular intervals. Think of them as the 'rules' that guide the possible solutions of a differential equation down a more defined path. In many respects, boundary conditions root a potentially abstract mathematical solution in the physical realm, reflecting the constraints of the real-life system modeled by the equation.

In our case, the two boundary conditions are \(y^{\prime}(0) = 1\) and \(y^{\prime}(\pi) = 0\). These conditions stipulate that the derivative of \(y\) at \(x = 0\) and \(x = \pi\) must match the given values. By applying these conditions, we determine our unknown constants \(C_1\) and \(C_2\), cementing our specific solution to the differential equation. It’s crucial to ensure that your solution is not just a broad expression but one that toes the line of these mathematical 'rules'—without adherence to the imposed boundary conditions, the solution would be insufficient for the specific scenario. Hence, meticulously applying these conditions entails taking derivatives, evaluating at the given points, and solving for the constants. The completeness of the process emanates from the seamless fusion of theoretical mathematics with practical limitations—the beauty of boundary value problems.