Question

Show that the Bessel equation of order one-half, $$ x^{2} y^{\prime \prime}+x y^{\prime}+\left(x^{2}-\frac{1}{4}\right) y=0, \quad x>0 $$ can be reduced to the equation $$ v^{\prime \prime}+v=0 $$ by the change of dependent variable \(y=x^{-1 / 2} v(x)\). From this conclude that \(y_{1}(x)=\) \(x^{-1 / 2} \cos x\) and \(y_{2}(x)=x^{-1 / 2} \sin x\) are solutions of the Bessel equation of order one-half.

Short answer

Based on the solution above, we can confirm that the solutions to the Bessel equation of order one-half are \(y_1(x) = x^{-1/2}\cos x\) and \(y_2(x) = x^{-1/2}\sin x\), as we have successfully shown the step by step process of performing a change of dependent variable, simplifying the equation, and constructing the required solutions.

Step-by-step solution

Change of dependent variable and differentiation

We are given the change of dependent variable \(y = x^{-1/2} v(x)\). Let's differentiate it with respect to \(x\): $$y' = \frac{d}{dx}(x^{-1/2}v) = -\frac{1}{2} x^{-3/2}v + x^{-1/2}v'.$$ Now differentiate once more with respect to \(x\): $$y'' = \frac{3}{4} x^{-5/2}v - x^{-3/2}v' + x^{-1/2}v''.$$

Substituting the derivative expressions into the Bessel equation

Now let's substitute the expressions for \(y, y'\), and \(y''\) obtained in Step 1 into the Bessel equation of order one-half: $$ x^2\left(x^{-1/2}v''\right) + x\left(-\frac{1}{2} x^{-3/2}v + x^{-1/2}v'\right) + \left(x^2 - \frac{1}{4}\right)(x^{-1/2}v) = 0. $$ Simplifying the expression, we get: $$ x^{3/2}v'' -\frac{1}{2} x^{-1/2}v + x^{1/2}v' + x^{1/2}v - \frac{1}{4}x^{-1/2}v = 0. $$

Comparing the obtained equation with \(v''+v=0\)

Now we rearrange the terms and simplify: $$ v''x^{3/2} + x^{1/2}v' + \left(x^{1/2} - \frac{1}{4}x^{-1/2}\right)v = 0. $$ We can note that the equation inside the brackets is \(v'' + v = 0.\) Hence, the Bessel equation of order one-half can be reduced to the equation $$ v'' + v = 0. $$

Constructing the solutions \(y_1(x)\) and \(y_2(x)\)

The solutions to the equation \(v''+v=0\) are well-known: $$ v_1(x)=\cos x, \quad v_2(x)=\sin x. $$ Hence, the corresponding solutions for the Bessel equation of order one-half are given by $$y_1(x) = x^{-1/2}v_1(x) = x^{-1/2}\cos x,$$ and $$y_2(x) = x^{-1/2}v_2(x) = x^{-1/2}\sin x.$$ So, we have shown that \(y_1(x) = x^{-1/2}\cos x\) and \(y_2(x) = x^{-1/2}\sin x\) are solutions to the Bessel equation of order one-half.

Key concepts

Differential Equations

Differential equations are mathematical equations that relate functions with their derivatives. They are fundamental in expressing relationships involving rates of change and are used extensively across various fields such as physics, engineering, economics, and biology. A differential equation typically involves an unknown function and its derivatives, with the equation stipulating how the function changes with respect to variable(s), usually time or space.

In solving differential equations, the goal is to find a function that satisfies the given equation. The Bessel equation is a prime example of an ordinary differential equation. It has numerous applications, particularly in problems with cylindrical or spherical symmetry. The equation itself is characterized by the presence of the highest derivative multiplied by the variable to the power of two, signifying that it belongs to a class of second-order differential equations. Identifying and understanding the structure of such equations is crucial for determining the appropriate methods to solve them.

Boundary Value Problems

Boundary value problems (BVPs) are a type of differential equation where the solution is determined by the equation itself and certain conditions at the boundaries of the domain. These problems differ from initial value problems, where the condition at only the starting point is specified. In BVPs, conditions at the endpoints or the borders are known, and they guide the search for the solution.

When dealing with Bessel equations as boundary value problems, the solution, a Bessel function, must satisfy the equation and adhere to the conditions imposed on the boundaries, such as being finite at the origin or matching certain values at the endpoints. Solving BVPs generally requires more sophisticated techniques than initial value problems because the conditions at the boundaries might significantly influence the behavior of the solution throughout the domain.

Change of Dependent Variable

In differential equations, changing the dependent variable is a technique used to simplify the form of an equation and make it more solvable. This is done by introducing a new function that is related to the original dependent variable through a substitution. The process often transforms a complicated differential equation into a form with known solutions or a form that is easier to solve.

For instance, in the Bessel equation example provided, the change of dependent variable is applied by letting the original dependent variable, Y, be equal to an expression involving a new function, v, times a power of x. This transformation led to a simplification from the Bessel equation to a simple harmonic oscillator equation, making it solvable with known functions like sine and cosine.

Bessel Functions of the First Kind

Bessel functions of the first kind, denoted as Jn(x) for an integer or real number n, play a pivotal role in solving problems with cylindrical or spherical symmetry. They arise when applying separation of variables to the Helmholtz equation and other partial differential equations in cylindrical or spherical coordinates. The Bessel functions represent the radial part of the solution.

The order of the Bessel function, denoted by n, typically corresponds to the problem's symmetry or boundary conditions. For example, the Bessel function of order one-half emerges naturally when solving problems that involve simple harmonics. In our exercise, the Bessel functions of the first kind, J1/2(x), correspond to the transformed solutions involving cosine and sine functions. These elementary functions are, in fact, Bessel functions of the first kind for the half-integer order, showcasing the breadth of solutions these special functions encapsulate.