Question

Consider the problem $$ X^{\prime \prime}+\lambda X=0, \quad X^{\prime}(0)=0, \quad X^{\prime}(L)=0 $$ Let \(\lambda=\mu^{2},\) where \(\mu=v+i \sigma\) with \(v\) and \(\sigma\) real. Show that if \(\sigma \neq 0,\) then the only solution of Eqs. (i) is the trivial solution \(X(x)=0 .\) Hint: Use an argument similar to that in Problem 17 of Section \(10.1 .\)

Short answer

#tag_title#Conclusion#tag_content#In conclusion, we showed that for the given differential equation, under certain conditions on the parameter λ, the only possible solution is the trivial solution \(X(x) = 0\). We found the real and imaginary parts of the equation, applied the given boundary conditions, integrated both parts of the equation, and used integration by parts. Through this process, we determined that the only way for the equation to hold is if σ is zero, which ultimately implies the trivial solution.

Step-by-step solution

Rewriting the Differential Equation with λ

Using the given substitution for \(\lambda\), i.e. \(\lambda = \mu^2\) with \(\mu=v+i\sigma\), rewrite the differential equation as: $$ X^{\prime \prime}+ (v+i\sigma)^2 X=0 $$ Step 2: Find the real and imaginary parts of the differential equation.

Real and Imaginary Parts

To separate the real and imaginary parts, let \(X(x) = \operatorname{Re}[X(x)] + i\cdot \operatorname{Im}[X(x)] = R(x) + iI(x)\), where \(R(x)\) and \(I(x)\) are real valued functions. Now substitute \(X(x)\) in the rewritten differential equation and equate the real and imaginary parts. The system of real differential equations becomes: $$ \begin{cases} R''(x) - (2vi\sigma + \sigma^2)R(x) = 0 \\ I''(x) - (2vi\sigma + \sigma^2)I(x) = 0 \end{cases} $$ Step 3: Apply the given boundary conditions.

Boundary Conditions

The given boundary conditions are \(X'(0) = 0\) and \(X'(L) =0\). Therefore, we have: $$ \begin{cases} R'(0) = 0 \\ I'(0) = 0 \\ R'(L) = 0 \\ I'(L) = 0 \end{cases} $$ Step 4: Integrate the real and imaginary differential equations.

Integrating Real and Imaginary Differential Equations

Now, integrate both differential equations on the interval \((0, L)\) for \(R(x)\) and \(I(x)\). This will give us two expressions: $$ \int_{0}^{L} R''(x) - (2vi\sigma + \sigma^2)R(x) dx = 0 $$ and $$ \int_{0}^{L} I''(x) - (2vi\sigma + \sigma^2)I(x) dx = 0 $$ Step 5: Apply integration by parts

Integration by Parts and Using Boundary Conditions

Now apply integration by parts to both expressions. Take \(u=R\) and \(dv = -(2vi\sigma + \sigma^2) R dx\) for the first expression, and \(u=I\) and \(dv = -(2vi\sigma + \sigma^2) I dx\) for the second one. Use boundary conditions to simplify the expressions. We obtain: $$ \int_{0}^{L} (2vi\sigma R^{2}(x) + \sigma^{2}R^{2}(x)) dx = 0 $$ and $$ \int_{0}^{L} (2vi\sigma I^{2}(x) + \sigma^{2}I^{2}(x)) dx = 0 $$ Step 6: Show that σ must be zero.

Argue that σ cannot be nonzero

Notice that, since \(R^2(x)\) and \(I^2(x)\) are positive or zero, and \(\sigma^{2}>0\), the only way to satisfy the above integrals is to require that \(2vi\sigma = 0\), which implies that, if \(\sigma \neq 0\), then \(X(x) = R(x) + iI(x) \equiv 0\).

Key concepts

Differential Equations

A differential equation is a mathematical equation that relates a function with its derivatives. These equations play a crucial role in describing various phenomena in engineering, physics, and other sciences. When solving differential equations, we are often tasked with finding functions that satisfy given conditions or represent real-world processes.

In the exercise provided, we encounter a second-order differential equation: \[X'' + \lambda X = 0\]The term \(X''\) represents the second derivative of the function \(X\), which describes how the rate of change of the rate of change of \(X\) behaves. Here, \(\lambda\) is a parameter that can alter the behavior of the solution, acting as a scaling factor. The solution to differential equations like these often involves understanding the behavior of how functions grow or decay depending on the parameter \(\lambda\).

Studying such equations requires methods like substitution, finding characteristic equations, and applying initial or boundary conditions. Solutions can vary greatly, demonstrating oscillatory, exponential, or other dynamic behavior. Being able to solve such differential equations is fundamental in modeling many natural and engineering systems.

Complex Numbers

Complex numbers are an extension of the real numbers and are used extensively in physics and engineering. They consist of a real part and an imaginary part, formulated as \( a + bi \), where \( i \) is the imaginary unit defined as \( \sqrt{-1} \). This means that \( i^2 = -1 \).

In the exercise, we introduce complex numbers through the parameter \( \lambda = \mu^2 \), where \( \mu = v + i\sigma \). This formulation shows how complex numbers can be applied to transform real parameters into scenarios where dynamics of systems can be more comprehensively explored.

By incorporating both real and imaginary components, complex numbers allow us to tackle a wider range of problems, such as oscillations, damped systems, and wave phenomena. The imaginary part introduces rotational elements to the analysis, often connecting the real and imaginary through Euler's formula \( e^{ix} = \cos x + i \sin x \). This plays a vital role in simplifying and solving differential equations, especially in scenarios involving trigonometric functions and exponential growth or decay.

Trivial Solution

A trivial solution in the context of differential equations refers to a solution that applies either universally or under particular conditions but doesn't depict any dynamic or meaningful change. This usually means that the solution is the zero solution.

In the exercise, the task is to prove that the only solution possible under the given conditions is the trivial solution \( X(x) = 0 \) if \( \sigma eq 0 \). This means the system described by the differential equation doesn't deviate from zero, and thus exhibits no motion or change. This might seem unexciting at first, but it is a vital aspect as it suggests stability or possibly decaying oscillations in real-world terms.

Determining when only the trivial solution exists often involves analyzing the boundary or initial conditions, as well as the parameters embedded in the differential equation. Such solutions tell us something about the constraints of the system being studied, influencing how real-world systems can be controlled or understood.

Boundary Conditions

Boundary conditions in differential equations specify the values that a solution must satisfy at the edges or boundaries of the domain in which the solution is valid. These conditions help determine the specific solution to a differential equation from a family of potential solutions.

In the given problem, the boundary conditions are \( X'(0) = 0 \) and \( X'(L) = 0 \). These conditions require the derivative of the function \( X \) to be zero at both ends of the interval \( [0, L] \).

Boundary conditions severely limit the solutions of a differential equation and can often lead to the conclusion that only the trivial solution exists, especially when the system in question doesn't allow for non-zero solutions to match the given constraints. This is because satisfying such conditions may only be possible if the solution function is entirely undistorted by the dynamics represented in the differential equation.

• Dirichlet Boundary Conditions: Specified values of the function at the boundaries.
• Neumann Boundary Conditions: Specified values of the first derivative at the boundaries.

These conditions are crucial because they integrate practical constraints into the solving process, guiding us to realistic solutions that can meaningfully describe physical phenomena.