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In Exercises \(16-19\) find all values of \(\omega\) such that boundary problem has a unique solution, and find the solution by the method used to prove Theorem 13.1.3. For other values of \(\omega\), find conditions on \(F\) such that the problem has a solution, and find all solutions by the method used to prove Theorem \(13.1 .4 .\) $$ y^{\prime \prime}+\omega^{2} y=F(x), \quad y(0)=0, \quad y^{\prime}(\pi)=0 $$

Short Answer

Expert verified
**Short Answer** For the boundary value problem given, we have a unique solution y(x) = Bsin((2n+1)x/2) if ω = (2n+1)/2, where n = 0, 1, 2,... For other values of ω, the problem has a solution only if F(x) = 0 for all x, or if the conditions on F found by the Green's function method are satisfied.

Step by step solution

01

General Solution

First, we need to find the general solution y(x) of the given homogeneous differential equation: $$ y^{\prime \prime}+\omega^{2} y=0 $$ This is the equation of a simple harmonic oscillator, and its general solution have the form: $$ y(x) = Acos(\omega x) + Bsin(\omega x) $$
02

Applying the boundary condition y(0) = 0

Now our task is to apply the boundary conditions. First, let's apply the boundary condition y(0) = 0: $$ y(0) = Acos(0) + Bsin(0) = A $$ So, we have A = 0, which simplifies our general solution to: $$ y(x) = Bsin(\omega x) $$
03

Applying the boundary condition y'(π) = 0

Next, let's apply the other boundary condition y'(π) = 0. First, we need to differentiate y(x) with respect to x: $$ y'(x) = \omega Bcos(\omega x) $$ Now, we can apply the boundary condition y'(π) = 0: $$ 0 = \omega Bcos(\omega \pi) $$
04

Finding values of ω for unique solution and conditions on F

The only way for the above equation to hold true is if cos(ωπ) = 0 or if ωB = 0. For a unique solution, we must assume that B is nonzero, which means cos(ωπ) must be 0. The function cos(ωπ) has zeroes at the points ωπ = (2n+1)π/2 for n = 0, 1, 2, ... So, for ω = (2n+1)/2, we have a unique solution: $$ y(x) = Bsin\left(\frac{(2n+1)x}{2}\right) $$ For other values of ω, we have two possibilities. Either ω is not equal to (2n+1)/2, or ωB = 0. If ωB = 0 and ω is nonzero, B must be zero, so the solution is y(x) = 0, which is valid only if F(x) = 0 for all x. On the other hand, if ω is not equal to (2n+1)/2, we can apply the method used to prove Theorem 13.1.4 to find conditions on F to have a solution. This involves finding the Green's function G(x, ξ) and checking the integrability conditions.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Unique Solution Conditions
To determine the unique solution conditions for the boundary value problem presented in this exercise, we need to ensure that the solution exists and is singularly defined. This involves solving the homogeneous differential equation and applying specific boundary conditions. For the differential equation: \[ y'' + \omega^2 y = 0 \]The general solution is: \[ y(x) = A \cos(\omega x) + B \sin(\omega x) \]Applying the first boundary condition, \( y(0) = 0 \), we find that \( A = 0 \), which simplifies our general solution to: \[ y(x) = B \sin(\omega x) \]Next, by applying the condition \( y'(\pi) = 0 \), we develop the equation: \[ 0 = \omega B \cos(\omega \pi) \]Here, uniqueness requires that \( B eq 0 \) and thus \( \cos(\omega \pi) = 0 \), which translates into specific values of \( \omega \) such as \( \omega = \frac{(2n+1)}{2} \) for integers \( n \). These values ensure that the solution is not only existent but unique.
Harmonic Oscillator
The harmonic oscillator is a foundational element in physics and engineering, representing systems that experience restoring force proportional to their displacement. The behavior is captured by the differential equation: \[ y'' + \omega^2 y = 0 \]This equation models oscillatory systems such as springs and pendulums when friction is negligible. The term \( \omega \) represents the angular frequency, a key parameter that dictates the speed of oscillations.
  • In its solutions, you will find sinusoidal functions: sine and cosine, which signify regular periodic motion.
  • The general solution is \( y(x) = A \cos(\omega x) + B \sin(\omega x) \), where \( A \) and \( B \) are constants determined by initial conditions.
The harmonic oscillator’s importance extends beyond simple mechanical systems to electrical circuits, and even quantum systems, making it a versatile and crucial concept in understanding oscillatory behavior.
Green's Function
In the context of inhomogeneous differential equations, such as the boundary value problem explored here, Green's function plays a vital role. It acts as an integral kernel that transforms a given inhomogeneous equation into a solvable integral form. For our problem:\[ y'' + \omega^2 y = F(x) \]The solution technique involves finding a relevant Green's function \( G(x, \xi) \) that satisfies boundary conditions and transforms the problem into:\[ y(x) = \int G(x, \xi)F(\xi) \, d\xi \]
  • Green's function simplifies solving complex differential equations by converting them into integral equations.
  • The method ensures that the solution accounts for initial and boundary conditions effectively.
By using the Green's function, one can handle cases where \( F(x) \) is a non-zero function, thereby finding all possible solutions when \( \omega eq \frac{(2n+1)}{2} \).
Differential Equation Solutions
Differential equations are mathematical expressions that define relationships involving derivatives of unknown functions. Solutions to differential equations can be specific or general, depending on the presence of arbitrary constants determined via initial or boundary conditions. In solving a differential equation:
  • The general solution includes arbitrary constants, capturing all possible behaviors of the system.
  • Boundary conditions help pinpoint a unique solution from the family described by the general solution.
For example, the differential equation:\[ y'' + \omega^2 y = F(x) \]required treating the homogeneous part separately before tackling the inhomogeneous form. Understanding how to find these solutions forms the backbone of effectively dealing with boundary value problems, crucial for mathematical modeling in various scientific fields.

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Most popular questions from this chapter

In Exercises \(16-19\) find all values of \(\omega\) such that boundary problem has a unique solution, and find the solution by the method used to prove Theorem 13.1.3. For other values of \(\omega\), find conditions on \(F\) such that the problem has a solution, and find all solutions by the method used to prove Theorem \(13.1 .4 .\) $$ y^{\prime \prime}+\omega^{2} y=F(x), \quad y^{\prime}(0)=0, \quad y(\pi)=0 $$

Rewrite the equation in Sturm-Liouville form (with \(\lambda=0) .\) Assume that \(b, c, \alpha,\) and \(\nu\) are contants. $$ \left(1-x^{2}\right) y^{\prime \prime}-2 x y^{\prime}+\alpha(\alpha+1) y=0 $$

(a) Determine whether \(\lambda=0\) is an eigenvalue. If it is, find an associated eigenfunction. (b) Compute the negative eigenvalues with errors not greater than \(5 \times 10^{-8}\). State the form of the associated eigenfunctions. (c) Compute the first four positive eigenvalues with errors not greater than \(5 \times 10^{-8}\). State the form of the associated eigenfunctions. $$ y^{\prime \prime}+\lambda y=0, \quad y(0)+y^{\prime}(0)=0, \quad y(2)+2 y^{\prime}(2)=0 $$

In Example 13.2 .4 we found that the eigenvalue problem $$ x^{2} y^{\prime \prime}+x y^{\prime}+\lambda y=0, \quad y(1)=0, \quad y(2)=0 $$ is equivalent to the Sturm-Liouville problem $$ \left(x y^{\prime}\right)^{\prime}+\frac{\lambda}{x} y=0, \quad y(1)=0, \quad y(2)=0 $$ Multiply the differential equation in (B) by \(y\) and integrate to show that $$ \lambda \int_{1}^{2} \frac{y^{2}(x)}{x} d x=\int_{1}^{2} x\left(y^{\prime}(x)\right)^{2} d x $$ Conclude from this that the eigenvalues of (A) are all positive.

Find the Green's function for the boundary value problem $$ x^{2} y^{\prime \prime}+x y^{\prime}+\left(x^{2}-1 / 4\right) y=F(x), \quad y(\pi / 2)=0, \quad y(\pi)=0 $$ given that $$ y_{1}(x)=\frac{\cos x}{\sqrt{x}} \text { and } y_{2}(x)=\frac{\sin x}{\sqrt{x}} $$ are solutions of the complementary equation. Then use the Green's function to solve (A) with (a) \(F(x)=x^{3 / 2}\) and (b) \(F(x)=x^{5 / 2}\)

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