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The Adjoint Equation. If a second order linear homogeneous equation is not exact, it can be made exact by multiplying by an appropriate integrating factor \(\mu(x) .\) Thus we require that \(\mu(x)\) be such that \(\mu(x) P(x) y^{\prime \prime}+\mu(x) Q(x) y^{\prime}+\mu(x) R(x) y=0\) can be written in the form \(\left[\mu(x) P(x) y^{\prime}\right]+[f(x) y]=0 .\) By equating coefficients in these two equations and eliminating \(f(x),\) show that the function \(\mu\) must satisfy $$ P \mu^{\prime \prime}+\left(2 P^{\prime}-Q\right) \mu^{\prime}+\left(P^{\prime \prime}-Q^{\prime}+R\right) \mu=0 $$ .This equation is known as the adjoint of the original equation and is important in the advanced theory of differential equations. In general, the problem of solving the adjoint differential equation is as difficult as that of solving the original equation, so only occasionally is it possible to find an integrating factor for a second order equation.

Short Answer

Expert verified
Answer: The adjoint equation of a given second order linear homogeneous equation is given by: $$ P(x) \mu^{\prime\prime}(x) + (2 P^{\prime}(x) - Q(x)) \mu^{\prime}(x) + \left(P^{\prime\prime}(x) - Q^{\prime}(x) + R(x)\right) = 0 $$ where the original equation is: $$ P(x) y^{\prime\prime} + Q(x) y^{\prime} + R(x) y = 0 $$

Step by step solution

01

Write down the given equation and its exact form

The given second order linear homogeneous equation is: $$ P(x) y^{\prime\prime} + Q(x) y^{\prime} + R(x) y = 0 $$ And its exact form is: $$ [\mu(x) P(x) y^{\prime}]^{\prime} + f(x) y = 0 $$
02

Expand the differentiation term in the exact form

Using the product rule of differentiation, expand \(\left[\mu(x) P(x) y^{\prime}\right]^{\prime}\): $$ [\mu(x) P(x) y^{\prime}]^{\prime} = \mu(x) P(x) y^{\prime\prime} + \mu^\prime(x) P(x) y^{\prime} + \mu(x) P^\prime(x) y^{\prime} $$
03

Write the expanded exact form and compare coefficients with the original equation

Replacing the expanded derivative term in the exact form, we get: $$ \mu(x) P(x) y^{\prime\prime} + \mu^\prime(x) P(x) y^{\prime} + \mu(x) P^\prime(x) y^{\prime} + f(x) y = 0 $$ Now, we compare the coefficients of the original and the exact equation: $$ \mu(x) P(x) = P(x) \\ \mu^\prime(x) P(x) + \mu(x) P^\prime(x) = Q(x) \\ f(x) = -R(x) $$
04

Utilize the first coefficient equation and eliminate \(f(x)\) from the second

Using the first coefficient equation \(\mu(x) P(x) = P(x)\), we find \(\mu(x)\): $$ \mu(x) = \frac{P(x)}{P(x)} = 1 $$ Now, substitute \(\mu(x) = 1\) in the second coefficient equation: $$ P(x) \mu^{\prime\prime}(x) + (2 P^{\prime}(x) - Q(x)) \mu^{\prime}(x) + [P^{\prime\prime}(x) - Q^{\prime}(x) + (-R(x))] = 0 $$
05

Write down the adjoint equation

Now, we have the adjoint equation for the function \(\mu(x)\) as: $$ P(x) \mu^{\prime\prime}(x) + (2 P^{\prime}(x) - Q(x)) \mu^{\prime}(x) + \left(P^{\prime\prime}(x) - Q^{\prime}(x) + R(x)\right) = 0 $$ This is the required adjoint equation of the original second order linear homogeneous equation.

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Key Concepts

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

Second Order Linear Homogeneous Equation
A second order linear homogeneous equation is a fundamental concept in differential equations. These equations typically take the form: \[ P(x) y'' + Q(x) y' + R(x) y = 0 \]Such equations involve derivatives of the function \( y \), with coefficients \( P(x), Q(x), \) and \( R(x) \) that can vary with \( x \). These equations are called "homogeneous" because all terms involve the unknown function \( y \) or its derivatives. There is no constant term or external forcing function influencing their behavior.
The solutions to these equations are essential in many physical applications, such as in mechanical vibrations and electrical circuits. Solving these equations often requires specific techniques due to their complexity.
Integration Factor
An integration factor helps convert a non-exact differential equation into an exact one, meaning it allows us to express the differential equation in a form suitable for integration. For second order linear homogeneous equations, finding an appropriate integration factor \( \mu(x) \) ensures that a previously complex equation becomes more manageable:\[ [\mu(x) P(x) y']' + f(x) y = 0 \]
By finding such a function \( \mu(x) \), it becomes possible to combine terms into derivatives, making the equation easier to solve. However, determining a suitable integration factor can be challenging and isn't always possible. This process requires skill and understanding of differential equations, as well as familiarity with techniques like comparing coefficients.
Exact Differential Equation
An exact differential equation is one that can be expressed directly as a total derivative. These forms are straightforward to integrate and solve, as they stem from some function's differential:\[ [F(x, y)]' = 0 \]
In our context, we aim to transform the second order linear homogeneous equation into an exact form by using an integration factor, which involves expanding and comparing coefficients to align with:\[ [\mu(x) P(x) y']' + f(x) y = 0 \]
Once an equation is exact, it directly leads to its primitive form, meaning we can integrate both sides readily. Exactness equates to the solvability of a differential equation in its clearest form. This method underlies principles in physics and engineering, where simplified forms are used for practical solutions.
Advanced Theory of Differential Equations
The advanced theory of differential equations encompasses techniques and principles for handling complex equations like adjoint differential equations. The confidence in manipulating these techniques determines one's capability to solve higher-order systems or explore nonlinear dynamics.
The adjoint equation in the context of the second order linear homogeneous equation is crucial. It states:\[ P(x) \mu''(x) + (2 P'(x) - Q(x)) \mu'(x) + (P''(x) - Q'(x) + R(x)) \mu = 0 \]
This forms the backbone for deriving properties related to the original differential equation. Understanding and solving adjoint equations is significant in advanced mathematical studies, offering insights into stability, symmetry, and conservation laws in physical systems. Dealing with these requires a blend of mathematical insights and problem-solving strategies.

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Most popular questions from this chapter

Use the method of reduction of order to find a second solution of the given differential equation. \(x^{2} y^{\prime \prime}-(x-0.1875) y=0, \quad x>0 ; \quad y_{1}(x)=x^{1 / 4} e^{2 \sqrt{x}}\)

By combining the results of Problems 24 through \(26,\) show that the solution of the initial value problem $$ L[y]=\left(a D^{2}+b D+c\right) y=g(t), \quad y\left(t_{0}\right)=0, \quad y^{\prime}\left(t_{0}\right)=0 $$ where \(a, b,\) and \(c\) are constants, has the form $$ y=\phi(t)=\int_{t_{0}}^{t} K(t-s) g(s) d s $$ The function \(K\) depends only on the solutions \(y_{1}\) and \(y_{2}\) of the corresponding homogeneous equation and is independent of the nonhomogeneous term. Once \(K\) is determined, all nonhomogeneous problems involving the same differential operator \(L\) are reduced to the evaluation of an integral. Note also that although \(K\) depends on both \(t\) and \(s,\) only the combination \(t-s\) appears, so \(K\) is actually a function of a single variable. Thinking of \(g(t)\) as the input to the problem and \(\phi(t)\) as the output, it follows from Eq. (i) that the output depends on the input over the entire interval from the initial point \(t_{0}\) to the current value \(t .\) The integral in Eq. (i) is called the convolution of \(K\) and \(g,\) and \(K\) is referred to as the kernel.

Verify that the given functions \(y_{1}\) and \(y_{2}\) satisfy the corresponding homogeneous equation; then find a particular solution of the given nonhomogeneous equation. In Problems 19 and \(20 g\) is an arbitrary continuous function. $$ \begin{array}{l}{x^{2} y^{\prime \prime}+x y^{\prime}+\left(x^{2}-0.25\right) y=3 x^{3 / 2} \sin x, \quad x>0 ; \quad y_{1}(x)=x^{-1 / 2} \sin x, \quad y_{2}(x)=} \\ {x^{-1 / 2} \cos x}\end{array} $$

Use the substitution introduced in Problem 38 in Section 3.4 to solve each of the equations \(t^{2} y^{\prime \prime}-3 t y^{\prime}+4 y=0, \quad t>0\)

Use the method of Problem 32 to solve the given differential $$ 2 y^{\prime \prime}+3 y^{\prime}+y=t^{2}+3 \sin t \quad(\text { see Problem } 7) $$

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