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Let \(\phi_{1}, \phi_{2}, \ldots, \phi_{n}, \ldots\) be the normalized eigenfunctions of the Sturm-Liouville problem \((11),(12) .\) Show that the series $$ \phi_{1}(x)+\frac{\phi_{2}(x)}{\sqrt{2}}+\cdots+\frac{\phi_{n}(x)}{\sqrt{n}}+\cdots $$ is not the eigenfunction series for any square integrable function. Hint: Use Bessel's inequality, Problem \(9(b) .\)

Short Answer

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Question: Prove that the series \(\sum_{n=1}^{\infty} \frac{\phi_n(x)}{\sqrt{n}}\) is not the eigenfunction series for any square integrable function. Answer: The series fails to satisfy Bessel's inequality, which is necessary for it to be an eigenfunction series for any square integrable function.

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01

State the given series and Bessel's inequality

The given series is: $$ \sum_{n=1}^{\infty} \frac{\phi_n(x)}{\sqrt{n}} $$ And Bessel's inequality states that for any square integrable function \(f(x)\) and its normalized eigenfunction series \(\{\phi_n(x)\}\), the following inequality holds: $$ \int_a^b |f(x)|^2 dx \geq \sum_{n=1}^{\infty} \left|\int_a^b f(x) \phi_n(x) dx\right|^2 $$ Our goal is to show that the given series does not satisfy this inequality for any square integrable function \(f(x)\).
02

Consider the eigenfunction series

Let \(g(x)\) be the given series, i.e.: $$ g(x) = \sum_{n=1}^{\infty} \frac{\phi_n(x)}{\sqrt{n}} $$ Now, consider the squared series of \(g(x)\), which is: $$ g(x)^2 = \left(\sum_{n=1}^{\infty} \frac{\phi_n(x)}{\sqrt{n}}\right)^2 $$
03

Apply Bessel's inequality to the squared series

Let \(f(x) = g(x)^2\), and we will apply Bessel's inequality to this function. The inequality becomes: $$ \int_a^b g(x)^2 dx \geq \sum_{n=1}^{\infty} \left|\int_a^b g(x)^2 \phi_n(x) dx\right|^2 $$
04

Analyze the LHS of the inequality

In order to see if this inequality holds, we need to find the left-hand side (LHS) of the inequality. Let's compute the integral of \(g(x)^2\): $$ \int_a^b g(x)^2 dx = \int_a^b \left(\sum_{n=1}^{\infty} \frac{\phi_n(x)}{\sqrt{n}}\right)^2 dx $$ This integral has no general closed-form expression, but we know it can be represented by the double series: $$ \int_a^b g(x)^2 dx = \sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\frac{1}{\sqrt{nm}}\int_a^b \phi_n(x) \phi_m(x) dx $$
05

Evaluate and analyze the RHS of the inequality

Now let's consider the right-hand side (RHS) of the inequality. We need to compute the sum of the squared inner products of \(g(x)^2\) and \(\phi_n(x)\): $$ \sum_{n=1}^{\infty} \left|\int_a^b g(x)^2 \phi_n(x) dx\right|^2 = \sum_{n=1}^{\infty} \left|\int_a^b \left(\sum_{m=1}^\infty \frac{\phi_m(x)}{\sqrt{m}}\right)^2 \phi_n(x) dx\right|^2 $$ By evaluating the inner product with the eigenfunction series, we get another double series: $$ \sum_{n=1}^{\infty} \left|\int_a^b g(x)^2 \phi_n(x) dx\right|^2 = \sum_{n=1}^{\infty} \left|\sum_{m=1}^\infty \frac{1}{\sqrt{m}}\int_a^b \phi_m(x) \phi_n(x) dx\right|^2 $$
06

Compare the LHS and RHS of the inequality

Comparing the LHS and RHS, we can see that the summation indices are different. In the LHS, we are summing the product of two integers (n and m), while in the RHS, we are summing squares of single integers (n). This difference in the summation indices indicates that the series expansion of these two sums cannot be equal, and hence the given series cannot satisfy the Bessel's inequality for any square integrable function. In conclusion, we have shown that the series \(\sum_{n=1}^{\infty} \frac{\phi_n(x)}{\sqrt{n}}\) is not the eigenfunction series for any square integrable function, as it fails to satisfy Bessel's inequality.

Key Concepts

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

Eigenfunctions and Their Significance in Sturm-Liouville Problems
In mathematical physics, eigenfunctions play a pivotal role, especially within the context of Sturm-Liouville problems. These problems frequently arise in the study of partial differential equations and are associated with numerous physical systems, like the vibrations of a string or thermal conduction. An eigenfunction can be thought of as a special type of function that, when subjected to a linear operator, only scales by a certain factor, known as the eigenvalue.

Let's imagine we have a vibrating string. The eigenfunctions in this scenario could represent the fundamental modes of vibration—each mode vibrating at a characteristic frequency, which corresponds to the eigenvalue. The overall motion of the string can then be decomposed into a superposition of these eigenfunctions.

The Sturm-Liouville problem specifically deals with differential operators and boundary conditions that can generate a set of such eigenfunctions, often forming an orthogonal basis for functions in a suitable space. This allows for the expansion of arbitrary functions in terms of this basis, which is crucial for solving differential equations that model various physical phenomena.
Bessel's Inequality and its Role in Function Analysis
Bessel's inequality is a fundamental result in mathematics and provides a crucial insight into the study of series expansions in function spaces. Essentially, it tells us that for any square integrable function, the sum of the squares of the coefficients in its expansion is limited by the integral of the function's square.

To illustrate this more clearly, assume we have a waveform that represents a musical note. This wave can be represented as a sum of simpler waveforms corresponding to the note's harmonics—each with a different amplitude. Bessel's inequality assures us that the aggregate power of these harmonics cannot exceed the total power of the musical note.

In context of the exercise, we used Bessel's inequality to verify that a certain series of functions cannot equate to an expansion of a square integrable function. This series, involving eigenfunctions of a Sturm-Liouville problem, would violate Bessel's inequality, and thus cannot represent such an expansion—a profound conclusion that aids in understanding the limitations of function series in mathematical analysis.
Square Integrable Functions and Their Importance
The term square integrable function may sound quite technical, but it can be understood as simply as a function for which the area under its squared magnitude is finite. In other words, if you were to square the function's value at each point and then add up all those squares over its entire domain, you'd end up with a finite number.

Why does this matter? Such functions are of particular importance in quantum mechanics, signal processing, and many other fields because they represent physical quantities that are measurable and have finite energy. For example, in quantum mechanics, the probability of finding a particle within a specific region is represented by a square integrable wave function. The finiteness is crucial—it means that the total probability (i.e., the certainty of finding the particle somewhere) is exactly one, consistent with physical reality.

In the given exercise, we sought to demonstrate that a certain series does not correspond to any square integrable function because it fails to align with foundational mathematical constraints, in this case, Bessel's inequality. Such a realization helps us to appreciate the criteria that functions must fulfill to be physically relevant or mathematically sound within the theory they are applied.

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Most popular questions from this chapter

Use the method of Problem 11 to transform the given equation into the form \(\left[p(x) y^{\prime}\right]'+q(x) y=0\) $$ x y^{\prime \prime}+(1-x) y^{\prime}+\lambda y=0, \quad \text { Laguerre equation } $$

deal with column buckling problems. In some buckling problems the eigenvalue parameter appears in the boundary conditions as well as in the differential equation. One such case occurs when one end of the column is clamped and the other end is free. In this case the differential equation \(y^{i v}+\lambda y^{\prime \prime}=0\) must be solved subject to the boundary conditions $$ y(0)=0, \quad y^{\prime}(0)=0, \quad y^{\prime \prime}(L)=0, \quad y^{\prime \prime \prime}(L)+\lambda y^{\prime}(L)=0 $$ Find the smallest eigenvalue and the corresponding eigenfunction.

Consider the Sturm-Liouville problem $$ -\left[p(x) y^{\prime}\right]^{\prime}+q(x) y=\lambda r(x) y $$ $$ a_{1} y(0)+a_{2} y^{\prime}(0)=0, \quad b_{1} y(1)+b_{2} y^{\prime}(1)=0 $$ where \(p, q,\) and \(r\) satisfy the conditions stated in the text. (a) Show that if \(\lambda\) is an eigenvalue and \(\phi\) a corresponding eigenfunction, then $$ \lambda \int_{0}^{1} r \phi^{2} d x=\int_{0}^{1}\left(p \phi^{2}+q \phi^{2}\right) d x+\frac{b_{1}}{b_{2}} p(1) \phi^{2}(1)-\frac{a_{1}}{a_{2}} p(0) \phi^{2}(0) $$ provided that \(a_{2} \neq 0\) and \(b_{2} \neq 0 .\) How must this result be modified if \(a_{2}=0\) or \(b_{2}=0\) ? (b) Show that if \(q(x) \geq 0\) and if \(b_{1} / b_{2}\) and \(-a_{1} / a_{2}\) are nonnegative, then the eigenvalue \(\lambda\) is nonnegative. (c) Under the conditions of part (b) show that the eigenvalue \(\lambda\) is strictly positive unless \(q(x)=0\) for each \(x\) in \(0 \leq x \leq 1\) and also \(a_{1}=b_{1}=0\)

Solve the given problem by means of an eigenfunction expansion. $$ y^{\prime \prime}+2 y=-1+|1-2 x|, \quad y(0)=0, \quad y(1)=0 $$

Show that the problem $$ y^{\prime \prime}+\pi^{2} y=\pi^{2} x, \quad y(0)=1, \quad y(1)=0 $$ has the solution $$ y=c_{1} \sin \pi x+\cos \pi x+x $$ Also show that this solution cannot be obtained by splitting the problem as suggested in Problem \(15,\) since neither of the two subsidiary problems can be solved in this case.

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