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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

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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.

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

This problem illustrates that the eigenvalue parameter sometimes appears in the boundary conditions as well as in the differential equation. Consider the longitudinal vibrations of a uniform straight elastic bar of length \(L .\) It can be shown that the axial displacement \(u(x, t)\) satisfies the partial differential equation $$ (E / \rho) u_{x x}=u_{u^{i}} \quad 00 $$ $$ \begin{array}{l}{\text { where } E \text { is Young's modulus and } \rho \text { is the mass per unit volume. If the end } x=0 \text { is fixed, }} \\\ {\text { then the boundary condition there is }}\end{array} $$ $$ u(0, t)=0, \quad t>0 $$ $$ \begin{array}{l}{\text { Suppose that the end } x=L \text { is rigidly attached to a mass } m \text { but is otherwise unrestrained. }} \\ {\text { We can obtain the boundary condition here by writing Newton's law for the mass. From }} \\ {\text { the theory of elasticity it can be shown that the force exerted by the bar on the mass is given }} \\ {\text { by }-E A u_{x}(L, t) \text { . Hence the boundary condition is }}\end{array} $$ $$ E A u_{x}(L, t)+m u_{u}(L, t)=0, \quad t>0 $$ $$ \begin{array}{l}{\text { (a) A ssume that } u(x, t)=X(x) T(t), \text { and show that } X(x) \text { and } T(t) \text { satisfy the differential }} \\\ {\text { equations }}\end{array} $$ $$ \begin{array}{c}{X^{\prime \prime}+\lambda X=0} \\ {T^{\prime \prime}+\lambda(E / \rho) T=0}\end{array} $$ $$ \text { (b) Show that the boundary conditions are } $$ $$ X(0)=0, \quad X^{\prime}(L)-\gamma \lambda L X(L)=0 $$ $$ \begin{array}{l}{\text { where } y=m / \rho A L \text { is a dimensionless parameter that gives the ratio of the end mass to the }} \\ {\text { mass of the rod }} \\ {\text { Hile the differentitil equation for } T(t) \text { in simplify ing the boundary conditionat } x=L \text { . }} \\ {\text { (c) Detchine the form of the eigenfunctions the equation satisficaby the real cigen- }} \\ {\text { values of Eqs. (iv) and (vi). Find the first two eigenvalues } \lambda_{1} \text { and } \lambda_{2} \text { if } \gamma=0.5 .}\end{array} $$

In this problem we consider a higher order eigenvalue problem. In the study of transverse vibrations of a uniform elastic bar one is led to the differential equation $$ y^{\mathrm{w}}-\lambda y=0 $$ $$ \begin{array}{l}{\text { where } y \text { is the transverse displacement and } \lambda=m \omega^{2} / E I ; m \text { is the mass per unit length of }} \\\ {\text { the rod, } E \text { is Young's modulus, } I \text { is the moment of inertia of the cross section about an }} \\ {\text { axis through the centroid perpendicular to the plane of vibration, and } \omega \text { is the frequency of }} \\ {\text { vibration. Thus for a bar whose material and geometric properties are given, the eigenvalues }} \\ {\text { determine the natural frequencies of vibration. Boundary conditions at each end are usually }} \\ {\text { one of the following types: }}\end{array} $$ $$ \begin{aligned} y=y^{\prime} &=0, \quad \text { clamped end } \\ y=y^{\prime \prime} &=0, \quad \text { simply supported or hinged end, } \\ y^{\prime \prime}=y^{\prime \prime \prime} &=0, \quad \text { free end } \end{aligned} $$ $$ \begin{array}{l}{\text { For each of the following three cases find the form of the eigenfunctions and the equation }} \\ {\text { satisfied by the eigenvalues of this fourth order boundary value problem. Determine } \lambda_{1} \text { and }} \\ {\lambda_{2}, \text { the two eigenvalues of smallest magnitude. Assume that the eigenvalues are real and }} \\ {\text { positive. }}\end{array} $$ $$ \begin{array}{ll}{\text { (a) } y(0)=y^{\prime \prime}(0)=0,} & {y(L)=y^{\prime \prime}(L)=0} \\ {\text { (b) } y(0)=y^{\prime \prime}(0)=0,} & {y(L)=y^{\prime \prime}(L)=0} \\ {\text { (c) } y(0)=y^{\prime}(0)=0,} & {y^{\prime \prime}(L)=y^{\prime \prime \prime}(L)=0 \quad \text { (cantilevered bar) }}\end{array} $$

In each of Problems 7 through 10 determine the form of the eigenfunctions and the determinantal equation satisfied by the nonzero eigenvalues. Determine whether \(\lambda=0\) is an eigenvalue, and find approximate values for \(\lambda_{1}\) and \(\lambda_{2},\) the nonzero eigenvalues of smallest absolute value. Estimate \(\lambda_{n}\) for large values of \(n\). Assume that all eigenvalues are real. $$ \begin{array}{l}{y^{\prime \prime}+\lambda y=0} \\ {y(0)=0, \quad y(\pi)+y^{\prime}(\pi)=0}\end{array} $$

Solve the given problem by means of an eigenfunction expansion. $$ y^{\prime \prime}+2 y=-x, \quad y(0)=0, \quad y^{\prime}(1)=0 ; \quad \text { see Section } 11.2, \text { Problem } 7 $$

determine the normalized eigenfunctions of the given problem. $$ y^{\prime \prime}+\lambda y=0, \quad y^{\prime}(0)=0, \quad y^{\prime}(1)+y(1)=0 $$
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