Question

determine whether the given boundary value problem is self-adjoint. $$ y^{\prime \prime}+y^{\prime}+2 y=0, \quad y(0)=0, \quad y(1)=0 $$

Short answer

No, the given boundary value problem is not self-adjoint because the given operator, \(L\), and its adjoint, \(L^*\), are not equal.

Step-by-step solution

Find the adjoint of the linear operator

To find the adjoint operator, we can use the Green's formula: $$ \int_{0}^{1} (Ly)u dx = \int_{0}^{1} y(L^*u) dx + [uv^{\prime} - u^{\prime}v]_0^1 $$ Here, \(L\) is the given differential operator, and \(L^*\) is its adjoint. Let \(Ly = y^{\prime \prime} + y^{\prime} + 2y\), and substitute it into the formula. Then, integrate by parts to find the expression for \(L^*u\).

Integrate by parts

Let's integrate by parts: $$ \int_{0}^{1} (y^{\prime \prime} + y^{\prime} + 2y)u dx =\int_{0}^{1} y(-u^{\prime \prime} + u^{\prime} + 2u) dx + [uv^{\prime} - u^{\prime}v]_0^1 $$ Now, we will integrate by parts on the first two terms of the integrals. First term: $$ \int yu^{\prime \prime} dx = yu^{\prime}\big|_{0}^{1} - \int y^{\prime}u^{\prime} dx $$ Second term: $$ \int y^{\prime}u dx = y^{\prime}u\big|_{0}^{1} - \int y^{\prime \prime}u dx $$ We will substitute these into the Green's formula.

Substitute the results in Green's formula and simplify

Substitute the integration by parts back into the Green's formula: $$ yu^{\prime}\big|_{0}^{1} - \int y^{\prime}u^{\prime} dx + y^{\prime}u\big|_{0}^{1} - \int y^{\prime \prime}u dx + \int 2yu dx = \int y(-u^{\prime \prime} + u^{\prime} + 2u) dx + [uv^{\prime} - u^{\prime}v]_0^1 $$ Notice that the terms with y' and y'' cancel out. Now we can simplify the equation: $$ 2 \int yu dx = \int y(2u - u^{\prime \prime} + u^{\prime}) dx + [uv^{\prime} - u^{\prime}v]_0^1 $$ We can rewrite the expression as: $$ \int y(2u - u^{\prime \prime} + u^{\prime}) dx = 2 \int yu dx - [uv^{\prime} - u^{\prime}v]_0^1 $$ Now we compare this with the formula for L, and we find the expression of the adjoint operator, \(L^*\): $$ L^*u = -u^{\prime \prime} + u^{\prime} + 2u $$

Compare the given operator and its adjoint

Now let's compare the given operator, \(L\), and its adjoint, \(L^*\): $$ L = \frac{d^2}{dt^2} + \frac{d}{dt} + 2 \\ L^* = -\frac{d^2}{dt^2} + \frac{d}{dt} + 2 $$ Since \(L\) and \(L^*\) are not equal, the given boundary value problem is not self-adjoint.

Key concepts

Adjoint Operator

An adjoint operator is an essential idea in the context of differential equations and boundary value problems. It can be thought of as a 'partner' to a given operator. To find the adjoint of a linear differential operator, we often use an important mathematical tool called Green's formula. The process involves manipulating the original operator to reveal another operator, which is its adjoint. For example, if we have a differential operator defined as \( L = \frac{d^2}{dt^2} + \frac{d}{dt} + 2 \), the goal is to find the operator \( L^* \), such that - The Green's formula holds - Certain boundary terms vanish or maintain symmetryThe adjoint operator helps in determining properties like the self-adjoint nature of boundary value problems, which is a condition involving equality of the differential operator and its adjoint under prescribed boundary conditions.

Green's Formula

Green's formula is an incredibly useful mathematical expression for working with differential equations, particularly when dealing with operators. Essentially, it provides a way to relate each side of an integration equation, and it's instrumental in finding adjoint operators.The formula is expressed as: \[\int_{0}^{1} (Ly)u \, dx = \int_{0}^{1} y(L^*u) \, dx + [uv^{\prime} - u^{\prime}v]_0^1\]In this formula: - \( Ly \) represents the original differential operator acting on a function \( y \)- \( L^*u \) represents the adjoint operator acting upon another function \( u \)- The terms \([uv^{\prime} - u^{\prime}v]_0^1\) are boundary terms, which can often simplify or cancel under specific conditions.Green's formula serves as the backbone of many calculations in this field, allowing us to explore relationships between functions and their derivatives over a specific interval.

Linear Differential Operator

Linear differential operators are foundational to the study of differential equations. They are operators that involve derivatives of a function and include linear terms. Linear in this context means that the operator \( L \) satisfies the property: \[ L(c_1y_1 + c_2y_2) = c_1Ly_1 + c_2Ly_2 \]for any functions \( y_1 \) and \( y_2 \), and constants \( c_1 \) and \( c_2 \).In the given boundary value problem, the operator was - \( L = \frac{d^2}{dt^2} + \frac{d}{dt} + 2 \).These operators are used to model a variety of physical systems and equations, often popping up in contexts like wave equations, heat equations, and more.Understanding how linear differential operators work is key to solving many types of differential equations, as it leads to identifying the nature of solutions and whether or not certain properties like self-adjointness are present.