Chapter 7: Problem 8
Find the expansion of
$$
f(x)=x, \quad 0
Short Answer
Expert verified
Question: Determine the expansion of the function f(x) = x in a series of eigenfunctions of the Sturm-Liouville system: \(y'' + \lambda y = 0\), \(y'(0) = 0\), and \(y'(m) = 0\).
Answer: The expansion of the function f(x) = x in the series of eigenfunctions of the given Sturm-Liouville system can be written as:
$$
f(x) \approx \sum_{n=1}^{\infty} c_n A_n\cos\left(\frac{n\pi x}{m}\right),
$$
where \(c_n\) and \(A_n\) are the coefficients determined by the orthogonality property of eigenfunctions for each \(n = 1, 2, 3, ...\).
Step by step solution
01
Solve the Sturm-Liouville System for Eigenvalues λ and Eigenfunctions y_n(x)
Begin by solving the Sturm-Liouville system:
$$
\begin{aligned}
y^{\prime\prime}+\lambda y &= 0, \\
y^{\prime}(0) &= 0, \\
y^{\prime}(m) &= 0.
\end{aligned}
$$
Let's assume a solution of the form \(y(x) = A \cos(\sqrt{\lambda} x) + B \sin(\sqrt{\lambda}x)\). Differentiating y(x) with respect to x, we get:
$$
y^{\prime}(x)=-A\sqrt{\lambda} \sin(\sqrt{\lambda}x) + B\sqrt{\lambda} \cos(\sqrt{\lambda}x)
$$
Applying the boundary conditions:
$$
y^{\prime}(0) = B\sqrt{\lambda}=0 \Rightarrow B=0
$$
and
$$
y^{\prime}(m)=-A\sqrt{\lambda}\sin(\sqrt{\lambda}m) = 0
$$
Since A cannot be zero, otherwise our eigenfunction would be null, \(\sin(\sqrt{\lambda}m)= 0\). Therefore,
$$
\sqrt{\lambda}m=n\pi, \quad n=1,2,3, \cdots
$$
Eigenvalues and eigenfunctions can be expressed as:
$$
\lambda_n =\frac{n^2\pi^2}{m^2}, \quad y_n(x)=A_n\cos\left(\frac{n\pi x}{m}\right)
$$
02
Calculate Coefficients for the Expansion
We approximate f(x) with a series of eigenfunctions as follows:
$$
f(x) \approx \sum_{n=1}^{\infty} c_n y_n(x)
$$
To find the coefficients \(c_n\), we will use the orthogonality property of eigenfunctions. Multiply both sides of the equation by \(y_k(x)\), with \(k \neq n\), and integrate over the interval (0, m):
$$
\int_{0}^{m} f(x) y_k(x) dx \approx \sum_{n=1}^{\infty} c_n \int_{0}^{m} y_n(x) y_k(x) dx
$$
Using the orthogonality property, the integral on the right-hand side will be zero for \(n \neq k\) and non-zero for \(n=k\). Therefore, we get:
$$
c_k = \frac{\int_{0}^{m} f(x) y_k(x) dx}{\int_{0}^{m} y_k^2(x) dx}
$$
Since \(f(x) = x\) and \(y_k(x) = A_k \cos\left(\frac{k\pi x}{m}\right)\), we can simplify this expression to find \(c_k\):
$$
c_k = \frac{\int_{0}^{m} x A_k \cos\left(\frac{k\pi x}{m}\right) dx}{\int_{0}^{m} A_k^2 \cos^2\left(\frac{k\pi x}{m}\right) dx}
$$
03
Write the Expansion Series
Now that we have the coefficients \(c_k\), we can write the expansion of f(x) in the series of eigenfunctions:
$$
f(x) \approx \sum_{n=1}^{\infty} c_n A_n\cos\left(\frac{n\pi x}{m}\right)
$$
Thus, we have found the expansion of \(f(x) = x\) in a series of eigenfunctions of the given Sturm-Liouville system.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Eigenvalues
Eigenvalues are a fundamental part of solving differential equations in Sturm-Liouville theory. They act like special numbers that allow differential equations to have non-trivial solutions. In the context of the exercise, we explore eigenvalues for the Sturm-Liouville problem
- Given by the equation: \( y^{\prime\prime} + \lambda y = 0 \)
- Subject to boundary conditions: \( y^{\prime}(0) = 0 \) and \( y^{\prime}(m) = 0 \)
- The condition \( y^{\prime}(0) = 0 \) simplifies to \( B \sqrt{\lambda} = 0 \), implying \( B = 0 \).
- The condition \( y^{\prime}(m) = 0 \) gives \( -A \sqrt{\lambda} \sin(\sqrt{\lambda}m) = 0 \).
Eigenfunctions
Eigenfunctions represent the backbone solutions to the differential equations in the Sturm-Liouville problem once the eigenvalues are known. For each eigenvalue \( \lambda_n \), there corresponds an eigenfunction \( y_n(x) \). These eigenfunctions develop as:
- With the general form: \( y(x) = A \cos(\sqrt{\lambda}x) + B \sin(\sqrt{\lambda}x) \)
- With \( B = 0 \) due to boundary conditions, leading to \( y(x) = A \cos(\sqrt{\lambda}x) \)
Orthogonality of Eigenfunctions
The orthogonality of eigenfunctions is a crucial property in Sturm-Liouville theory. It means that different eigenfunctions, corresponding to different eigenvalues, are mathematically orthogonal over a given interval (like a dot product being zero). For our eigenfunctions \( y_n(x) \) and \( y_k(x) \) associated with different eigenvalues \( \lambda_n eq \lambda_k \), they satisfy:
- The integral of their product over the interval \((0, m)\) is zero, \( \int_0^m y_n(x) y_k(x) \, dx = 0 \).
Series Expansion
Series expansion in the context of Sturm-Liouville problems enables us to express a function as a sum of scaled eigenfunctions over the defined interval. Given a function \( f(x) = x \) and the system’s eigenfunctions \( y_n(x) \), we approximate the function:
- As \( f(x) \approx \sum_{n=1}^{\infty} c_n y_n(x) \)
- Integrate the product of \( f(x) \) and each eigenfunction \( y_k(x) \) over the interval \((0, m)\)