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Chapter 10: Problem 11

Problems deal with the regular Sturm-Liouville problem $$ \begin{aligned} y^{\prime \prime}+\lambda y &=0 & &(00 .\) Note that Theorem 1 does not exclude the possibility of negative eigenvalues. Show that \(\lambda_{0}=0\) is an eigenvalue if and only if \(h L=1\), in which case the associated eigenfunction is \(y_{0}(x)=x\).

Short Answer

Expert verified
\( \lambda_0 = 0 \) is an eigenvalue if and only if \( hL = 1 \), and the eigenfunction is \( y_0(x) = x \).

Step by step solution

01

Apply the Boundary Conditions

To determine if \( \lambda_0 = 0 \) is an eigenvalue, substitute \( \lambda_0 = 0 \) in the differential equation \[ y'' = 0. \] The general solution to this differential equation is \[ y(x) = C_1 x + C_2, \]where \( C_1 \) and \( C_2 \) are constants.
02

Apply the First Boundary Condition

From the boundary condition \( y(0) = 0 \), substitute \( x=0 \) into the general solution:\[ y(0) = C_1 \times 0 + C_2 = 0 \Rightarrow C_2 = 0. \] Thus, the solution simplifies to\[ y(x) = C_1 x. \]
03

Apply the Second Boundary Condition

Apply the second boundary condition \( hy(L) - y'(L) = 0 \), substitute for \( y(x) = C_1 x \):1. Compute \( y'(x) = C_1 \).2. Substitute \( x=L \) into \( y(x) \): \[ y(L) = C_1 L. \]3. The second boundary condition becomes \[ h C_1 L - C_1 = 0. \]
04

Simplify the Boundary Condition

Factor out \( C_1 \) from the equation \[ h C_1 L - C_1 = 0: \]\[ C_1(hL - 1) = 0. \]Since we want a nontrivial solution for \( y(x) \), \( C_1 eq 0 \). Thus,\[ hL - 1 = 0 \Rightarrow hL = 1. \]
05

Conclusion on Eigenvalues

We have shown that for \( \lambda_0 = 0 \) to be an eigenvalue, the condition \( hL = 1 \) must be met. In this case, the associated eigenfunction is\[ y_0(x) = C_1 x \] which can be simplified to \( y_0(x) = x \) when normalizing \( C_1 = 1 \).

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Key Concepts

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

Eigenvalues
Eigenvalues are key to understanding many mathematical and physical systems, especially when dealing with the Sturm-Liouville problem. In mathematical terms, eigenvalues are special scalars associated with a particular linear transformation characterized by a given differential equation. They emerge when a non-zero scalar multiple of a vector, known as an eigenvector, meets the conditions of the transformation. In the context of our problem, we consider the eigenvalue \( \lambda = 0 \). This situation occurs under specific conditions, where the differential equation and boundary conditions are satisfied simultaneously.

In the Sturm-Liouville problem, eigenvalues arise from the solutions of a differential equation subject to given boundary conditions. For an eigenvalue \( \lambda_0 = 0 \) to appear, it must ensure that the entire system remains balanced under these conditions. This leads us to the main result in our exercise, \( hL = 1 \), indicating that the specific relationship between the length \( L \) and parameter \( h \) defines when \( \lambda_0 = 0 \) is indeed an eigenvalue.
  • Eigenvalues determine the stability and behavior of physical systems.
  • They are pivotal in oscillation and wave problems.
  • Each eigenvalue corresponds to a specific eigenfunction that satisfies the problem requirements.
Boundary Conditions
Boundary conditions are crucial in solving differential equations, particularly in the Sturm-Liouville problem. They specify values or expressions that solutions must satisfy at the endpoints of the domain, affecting which solutions are viable. In this problem, the boundary conditions are:
  • \( y(0) = 0 \)
  • \( hy(L) - y'(L) = 0 \)
Applying boundary conditions is essential to determine specific solutions from the general solution of a differential equation. By substituting these conditions into the general form, we refine the potential solutions to only those that satisfy both boundary requirements. It helps in verifying under what situations the given eigenvalue is true. Here, the boundary conditions played a pivotal role in proving that \( \lambda_0 = 0 \) can be an eigenvalue if the relationship \( hL = 1 \) holds.

  • Boundary conditions can be initial or final; here, they define solution behavior at the endpoints.
  • They filter the possible solutions from the general differential equation solution.
  • Correct application aligns solution behavior with physical or theoretical constraints of the problem context.
Differential Equations
Differential equations are equations that contain an unknown function and its derivatives. They describe a wide array of phenomena such as sound, heat, electrodynamics, and mechanics. In the context of the Sturm-Liouville problem, we are working with a second-order linear differential equation:
\[y'' + \lambda y = 0\]

This particular form is significant because it represents systems that oscillate, and solving these equations allows us to understand wave patterns and resonant frequencies. The key to solving such equations typically involves finding solutions that satisfy additional criteria like boundary conditions. In our problem, the implication that \( \lambda = 0 \) simplifies the differential equation substantially to \( y'' = 0 \). This change reveals that the potential solutions must be linear in form, such as \( y(x) = C_1 x + C_2 \).

  • Differential equations model dynamic systems and changing quantities.
  • Solutions can often take on an infinite number of forms; boundary conditions help refine them.
  • In our problem, simplifying the differential equation involved setting specific values (like \( \lambda = 0 \)) to find manageable solutions.

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Most popular questions from this chapter

A uniform bar of length \(L\) is made of material with density \(\delta\) and Young's modulus \(E .\) Substitute \(u(x, t)=X(x) \cos \omega t\) in \(\delta u_{t t}=E u_{x x}\) to find the natural frequencies of longitudinal vibration of the bar with the two given conditions at its ends \(x=0\) and \(x=L\). Both ends are fixed.

Problems deal with the regular Sturm-Liouville problem $$ \begin{aligned} y^{\prime \prime}+\lambda y &=0 & &(00 .\) Note that Theorem 1 does not exclude the possibility of negative eigenvalues. Suppose that \(h L=1\) in (33) and that \(f(x)\) is piecewise smooth. Show that $$ f(x)=c_{0} x+\sum_{n=1}^{\infty} c_{n} \sin \frac{\beta_{n} x}{L} $$ where \(\left\\{\beta_{n}\right\\}_{1}^{\infty}\) are the positive roots of \(\tan x=x\), and $$ \begin{aligned} c_{0} &=\frac{3}{L^{3}} \int_{0}^{L} x f(x) d x \\ c_{n} &=\frac{2}{L \sin ^{2} \beta_{n}} \int_{0}^{L} f(x) \sin \frac{\beta_{n} x}{L} d x \end{aligned} $$

A string with fixed ends is acted on by a periodic force \(F(x, t)=F(x) \sin \omega t\) per unit mass, so $$ y_{t t}=a^{2} y_{x x}+F(x) \sin \omega t. $$ Substitute $$ y(x, t)=\sum_{n=1}^{\infty} c_{n} \sin \frac{n \pi x}{L} \sin \omega t $$ and $$ F(x)=\sum_{n=1}^{\infty} F_{n} \sin \frac{n \pi x}{L}, $$ to derive the steady periodic solution $$ y(x, t)=\sum_{n=1}^{\infty} \frac{F_{n} \sin (n \pi x / a) \sin \omega t}{\omega_{n}^{2}-\omega^{2}} $$ where \(\omega_{n}=n \pi a / L\). Hence resonance does not result if \(\omega=\omega_{n}\) but \(F_{n}=0\).

The telephone equation for the voltage \(e(x, t)\) in a long transmission line at the point \(x \geqq 0\) at time \(t\) is $$ \frac{\partial^{2} e}{\partial x^{2}}=L C \frac{\partial^{2} e}{\partial t^{2}}+(R C+L G) \frac{\partial e}{\partial t}+R G e, $$ where \(R, L, G\), and \(C\) denote resistance, inductance, conductance, and capacitance (all per unit length of line), respectively. The condition \(e(0, t)=E_{0} \cos \omega t\) represents a periodic signal voltage at the origin of transmission at \(x=0\). Assume that \(e(x, t)\) is bounded as \(x \rightarrow+\infty\). Substitute \(\widetilde{e}(x, t)=E(x) e^{i \omega t}\) to derive the steady periodic solution $$ e(x, t)=E_{0} e^{-\alpha x} \cos (\omega t-\beta x), $$ where \(\alpha\) and \(\beta\) are the real and imaginary parts, respectively, of the complex number $$ \left[\left(R G-L C \omega^{2}\right)+i \omega(R C+L G)\right]^{1 / 2} . $$

Suppose that \(w=b x\) in (34). Derive the deflection function $$ y(x)=\frac{2 b L^{5}}{E I \pi^{5}} \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^{5}} \sin \frac{n \pi x}{L} . $$ The method used in Problems 17 and 18 succeeds because the functions \(\sin (n \pi x / L)\) satisfy the hinged /hinged conditions \(y(0)=y^{\prime \prime}(0)=y(L)=y^{\prime \prime}(L)=0\), so that \(y(x)\) does also. If, instead, both ends of the beam are fixed, in place of the sine functions we can use the eigenfunctions of the problem $$ \begin{aligned} y^{(4)}-\lambda y &=0 & &(0
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