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Chapter 11: Problem 4

State whether the given boundary value problem is homogeneous or non homogeneous. $$ -y^{\prime \prime}+x^{2} y=\lambda y, \quad y^{\prime}(0)-y(0)=0, \quad y^{\prime}(1)+y(1)=0 $$

Short Answer

Expert verified
Differential equation: \(- y^{\prime \prime} + x^{2}y = \lambda y\) Boundary conditions: \(y^{\prime}(0) - y(0)=0\) and \(y^{\prime}(1) + y(1)=0\) Answer: The given boundary value problem is homogeneous.

Step by step solution

01

Analyze the differential equation

The given differential equation is: $$ - y^{\prime \prime} + x^{2}y = \lambda y $$ This is a linear differential equation. To determine if it's homogeneous or not, we need to examine the right-hand side of the equation, which is \(\lambda y\). Since \(\lambda y\) does not involve any function other than \(y\), this equation is homogeneous.
02

Analyze the boundary conditions

Now let's examine the boundary conditions: 1. \(y^{\prime}(0) - y(0)=0\) 2. \(y^{\prime}(1) + y(1)=0\) For boundary condition 1: $$ y^{\prime}(0) - y(0) = 0 \implies y^{\prime}(0) = y(0) $$ Since this equation is true when \(y^{\prime}(0)=0\) and \(y(0)=0\), the first boundary condition is homogeneous. For boundary condition 2: $$ y^{\prime}(1) + y(1) = 0 \implies y^{\prime}(1) = -y(1) $$ Similar to the first boundary condition, this equation is true when \(y^{\prime}(1)=0\) and \(y(1)=0\). Therefore, the second boundary condition is also homogeneous.
03

Determine if the boundary value problem is homogeneous or non-homogeneous

Since both the differential equation and the boundary conditions are homogeneous, the given boundary value problem is homogeneous.

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Key Concepts

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

Homogeneous Differential Equation
A homogeneous differential equation is one where, generally, each term is a function of the dependent variable and its derivatives. This means every term involves the function or its derivatives in a linear manner without any external additive terms.

In the exercise provided, the differential equation is \(- y'' + x^2 y = \lambda y\). We look at the right-hand side, \(\lambda y\), which indeed involves only the function \(y\) multiplied by \(\lambda\), a parameter or constant.

This indicates that there are no terms involving other functions or non-constant terms that do not include \(y\). Thus, the equation is homogeneous.

Essential characteristics of homogeneous differential equations include:
  • No constant or specific external forces act on the system within the equation.
  • Solutions can often be expressed as combinations of functions, known as superpositions.
  • The zero function (\(y = 0\) everywhere) is always a solution.
Linear Differential Equation
Linear differential equations are a type of equation where the dependent variable \(y\) and all its derivatives appear to the first degree and are not multiplied by one another. This makes them "linear" in nature, similar to linear algebra where variables are raised only to the first power.

In our exercise, the linearity is evident as \(- y'' + x^2 y = \lambda y\); every term involving \(y\) and its derivative appears linearly.

These equations are very useful due to:
  • Simplicity - They are easier to solve compared to non-linear equations.
  • Superposition - The principle of superposition applies, allowing solutions to be added together to form new solutions.
  • Predictability - Their behavior is well-understood and predictable.
The trait of being linear is crucial for deciding solution methods and understanding the system being described.
Boundary Conditions
Boundary conditions are constraints necessary for solving differential equations that depict real-world situations. They define how the solution behaves at the edges of the domain and ensure that the solution suits the physical boundaries of the problem.

In the provided exercise, we have two boundary conditions:
1. \(y'(0) - y(0) = 0\).
2. \(y'(1) + y(1) = 0\).

For these conditions to be homogeneous, they need to hold when \(y\) is zero over its entire domain.

These conditions assist in:
  • Ensuring uniqueness of solutions, meaning the same boundary value problem does not have multiple solutions.
  • Reflecting the physical constraints or symmetries of the system being studied.
  • Allowing for the identification of only solutions that are physically feasible or meaningful in context.
Essentially, boundary conditions are crucial for fully understanding and resolving differential equations in applicable scenarios.

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Most popular questions from this chapter

Consider the problem $$ y^{\prime \prime}+\lambda y=0, \quad \alpha y(0)+y^{\prime}(0)=0, \quad y(1)=0 $$ $$ \begin{array}{l}{\text { where } \alpha \text { is a given constant. }} \\\ {\text { (a) Show that for all values of } \alpha \text { there is an infinite sequence of positive eigenvalues. }} \\ {\text { (b) If } \alpha<1, \text { show that all (real) eigenvalues are positive. Show the smallest eigenvalue }} \\\ {\text { approaches zero as } \alpha \text { approaches } 1 \text { from below. }} \\ {\text { (c) Show that } \lambda=0 \text { is an eigenvalue only if } \alpha=1} \\ {\text { (d) If } \alpha>1 \text { , show that there is exactly one negative eigenvalue and that this eigenvalue }} \\ {\text { decreases as } \alpha \text { increases. }}\end{array} $$

The method of eigenfunction expansions is often useful for nonhomogeneous problems related to the wave equation or its generalizations. Consider the problem $$ r(x) u_{u}=\left[p(x) u_{x}\right]_{x}-q(x) u+F(x, t) $$ $$ \begin{aligned} u_{x}(0, t)-h_{1} u(0, t)=0, & u_{x}(1, t)+h_{2} u(1, t)=0 \\\ u(x, 0)=f(x), & u_{t}(x, 0)=g(x) \end{aligned} $$ This problem can arise in connection with generalizations of the telegraph equation (Problem 16 in Section 11.1 ) or the longitudinal vibrations of an elastic bar (Problem 25 in Section \(11.1) .\) (a) Let \(u(x, t)=X(x) T(t)\) in the homogeneous equation corresponding to Eq. (i) and show that \(X(x)\) satisfies Eqs. ( 28) and ( 29) of the text. Let \(\lambda_{n}\) and \(\phi_{n}(x)\) denote the eigenvalues and normalized eigenfunctions of this problem. (b) Assume that \(u(x, t)=\sum_{n=1}^{\infty} b_{n}(t) \phi_{n}(x),\) and show that \(b_{n}(t)\) must satisfy the initial value problem $$ b_{n}^{\prime \prime}(t)+\lambda_{n} b_{n}(t)=\gamma_{n}(t), \quad b_{n}(0)=\alpha_{n}, \quad b_{n}^{\prime}(0)=\beta_{n} $$ where \(\alpha_{n}, \beta_{n},\) and \(\gamma_{n}(t)\) are the expansion coefficients for \(f(x), g(x),\) and \(F(x, t) / r(x)\) in terms of the eigenfunctions \(\phi_{1}(x), \ldots, \phi_{n}(x), \ldots\)

Using the method of Problem 17 , transform the problem $$ y^{\prime \prime}+2 y=2-4 x, \quad y(0)=1, \quad y(1)+y^{\prime}(1)=-2 $$ into a new problem in which the boundary conditions are homogeneous. Solve the latter problem by reference to Example 1 of the text.

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 \(a_{1}=b_{1}=0\)

In this problem we explore a little further the analogy between Sturm- Liouville boundary value problems and Hermitian matrices. Let \(\mathbf{A}\) be an \(n \times n\) Hermitian matrix with eigenvalues \(\lambda_{1}, \ldots, \lambda_{n}\) and corresponding orthogonal eigenvectors \(\xi^{(1)}, \ldots, \xi^{(n)} .\) Consider the nonhomogeneous system of equations $$ \mathbf{A x}-\mu \mathbf{x}=\mathbf{b} $$ where \(\mu\) is a given real number and \(\mathbf{b}\) is a given vector. We will point out a way of solving Eq. ( \(i\) ) that is analogous to the method presented in the text for solving Eqs. (1) and (2). (a) Show that \(\mathbf{b}=\sum_{i=1}^{n} b_{i} \xi^{(i)},\) where \(b_{i}=\left(\mathbf{b}, \boldsymbol{\xi}^{(i)}\right)\) (b) Assume that \(\mathbf{x}=\sum_{i=1}^{n} a_{i} \xi^{(i)}\) and show that for Eq. (i) to be satisficd, it is necessary that \(a_{i}=b_{i} /\left(\lambda_{i}-\mu\right) .\) Thus $$ \mathbf{x}=\sum_{i=1}^{n} \frac{\left(\mathbf{b}, \boldsymbol{\xi}^{(i)}\right)}{\lambda_{i}-\mu} \boldsymbol{\xi}^{(i)} $$ provided that \(\mu\) is not one of the eigenvalues of \(\mathbf{A}, \mu \neq \lambda_{i}\) for \(i=1, \ldots, n .\) Compare this result with \(\mathrm{Eq.}(13) .\)
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