Question

use the result of Problem 32 to find the adjoint of the given differential equation. $$ x^{2} y^{\prime \prime}+x y^{\prime}+\left(x^{2}-v^{2}\right) y=0, \quad \text { Bessel's equation } $$

Short answer

#Answer# The adjoint of the given Bessel's differential equation is: $$ x^2(u''(x)y + 2u'(x)y'+ u(x)y'') + x(u'(x)y' + u(x)y') - xu'(x)y'+ (x^2 - v^2)(u(x)y) = 0. $$

Step-by-step solution

Identify the coefficients

Here, we are given the equation $$ x^2y'' + xy' + (x^2 - v^2)y = 0 .$$ The coefficients are \(x^2\) for the second derivative term ( \(y''\) ), \(x\) for the first derivative term ( \(y'\) ), and \((x^2 - v^2)\) for the \(y\) term.

Multiply by a weighting function

Now, we will multiply each term by a weighting function, which is often represented by the symbol \(u(x)\), to obtain the following expression: $$u(x)x^2y'' + u(x)xy' + u(x)(x^2 - v^2)y = 0.$$

Reverse the order of the derivatives

Next, we reverse the order of the derivatives in each term: $$x^2(u(x)y)'' + x((u(x)y)')' + (x^2 - v^2)(u(x)y) = 0.$$

Simplify the resulting equation

Now, simplify the equation to obtain the adjoint: $$ \begin{aligned} x^2(u(x)y)'' + x(u(x)y)' - xu'(x)y' +(x^2 - v^2)(u(x)y) &= 0 \\ x^2\left(u''(x)y + 2u'(x)y'+ u(x)y''\right) + x\left(u'(x)y' + u(x)y'\right) - xu'(x)y' +(x^2 - v^2)(u(x)y) &= 0. \end{aligned} $$ Thus, the adjoint of the given Bessel's differential equation is: $$ x^2(u''(x)y + 2u'(x)y'+ u(x)y'') + x(u'(x)y' + u(x)y') - xu'(x)y'+ (x^2 - v^2)(u(x)y) = 0. $$