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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

In this problem we show that pointwise convergence of a sequence \(S_{n}(x)\) does not imply mean convergence, and conversely. (a) Let \(S_{n}(x)=n \sqrt{x} e^{-n x^{2} / 2}, 0 \leq x \leq 1 .\) Show that \(S_{n}(x) \rightarrow 0\) as \(n \rightarrow \infty\) for each \(x\) in \(0 \leq x \leq 1 .\) Show also that $$ R_{n}=\int_{0}^{1}\left[0-S_{n}(x)\right]^{2} d x=\frac{n}{2}\left(1-e^{-n}\right) $$ and hence that \(R_{n} \rightarrow \infty\) as \(n \rightarrow \infty .\) Thus pointwise convergence does not imply mean convergence. (b) Let \(S_{n}(x)=x^{n}\) for \(0 \leq x \leq 1\) and let \(f(x)=0\) for \(0 \leq x \leq 1 .\) Show that $$ R_{n}=\int_{0}^{1}\left[f(x)-S_{n}(x)\right]^{2} d x=\frac{1}{2 n+1} $$ and hence \(S_{n}(x)\) converges to \(f(x)\) in the mean. Also show that \(S_{n}(x)\) does not converge to \(f(x)\) pointwise throughout \(0 \leq x \leq 1 .\) Thus mean convergence does not imply pointwise convergence.

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

Determine a formal eigenfunction series expansion for the solution of the given problem. Assume that \(f\) satisfies the conditions of Theorem \(11.3 .1 .\) State the values of \(\mu\) for which the solution exists. $$ y^{\prime \prime}+\mu y=-f(x), \quad y^{\prime}(0)=0, \quad y^{\prime}(1)+y(1)=0 $$

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

Consider the problem $$ y^{\prime \prime}+p(x) y^{\prime}+q(x) y=0, \quad y(0)=a, \quad y(1)=b $$ Let \(y=u+v,\) where \(v\) is any twice differentiable function satisfying the boundary conditions (but not necessarily the differential equation). Show that \(u\) is a solution of the problem $$ u^{\prime \prime}+p(x) u^{\prime}+q(x) u=g(x), \quad u(0)=0, \quad u(1)=0 $$ where \(g(x)=-\left[v^{\prime \prime}+p(x) v^{\prime}+q(x) v\right],\) and is known once \(v\) is chosen. Thus nonhomogeneities can be transferred from the boundary conditions to the differential equation. Find a function \(v\) for this problem.
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