Problem 29
Question
$$ \int_{0}^{x} r J_{0}(r) d r=x J_{1}(x) $$
Step-by-Step Solution
Verified Answer
The integral is equal to \( x J_{1}(x) \).
1Step 1: Understanding the Problem
We need to evaluate the definite integral \( \int_{0}^{x} r J_{0}(r) \, dr \) and show that it is equal to \( x J_{1}(x) \), where \( J_{0} \) and \( J_{1} \) are Bessel functions of the first kind of order 0 and 1, respectively.
2Step 2: Recall Bessel Function Properties
The Bessel functions \( J_{0}(x) \) and \( J_{1}(x) \) are solutions to Bessel's differential equation. An important identity for the Bessel functions of the first kind is: \( \frac{d}{dx}[xJ_{1}(x)] = xJ_{0}(x) \).
3Step 3: Differentiate the Right Side
Differentiate the right-hand expression \( x J_{1}(x) \) with respect to \( x \):\[ \frac{d}{dx}(xJ_{1}(x)) = J_{1}(x) + xJ_{1}'(x) \].Using the identity \( J_{1}'(x) = -J_{0}(x) \), this simplifies to \( J_{1}(x) + x(-J_{0}(x)) = J_{1}(x) - xJ_{0}(x) \).
4Step 4: Apply Fundamental Theorem of Calculus
To solve the integral \( \int_{0}^{x} r J_{0}(r) \, dr \), recognize it as a result of the anti-derivative. From Step 3, since \( \frac{d}{dx}(xJ_{1}(x)) = xJ_{0}(x) \), by the Fundamental Theorem of Calculus:\[ \int_{0}^{x} r J_{0}(r) \, dr = xJ_{1}(x) - 0 \cdot J_{1}(0) = xJ_{1}(x) \].
5Step 5: Verify the Boundary Conditions
Evaluate the boundary conditions of this integral. At \( x = 0 \), both sides are zero since \( x J_{1}(x) \) equals zero and the definite integral from 0 to 0 equals zero. Thus, the equality holds for all \( x \geq 0 \).
Key Concepts
Definite IntegralFundamental Theorem of CalculusBessel's Differential Equation
Definite Integral
A definite integral is a mathematical expression that represents the area under the curve of a function, between two specific limits. In our original exercise, the definite integral is \[ \int_{0}^{x} r J_{0}(r) \, dr \] Here, the function we are integrating is \( r J_{0}(r) \), where \( J_{0} \) is a Bessel function of the first kind of order 0. This type of integral provides valuable information about how functions behave over intervals, particularly when they involve special functions like Bessel functions which arise frequently in physics and engineering.
- The lower limit of integration is 0, and the upper limit is \( x \), a variable which can take any non-negative value.
- The result of a definite integral is usually a number, which in this case, depends on \( x \).
Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus bridges the concept of differentiation with integration, two main concepts in calculus. It tells us that if a function has an anti-derivative, then the value of its definite integral can be calculated using this anti-derivative.
In simple terms, it states:
In simple terms, it states:
- The definite integral of a function from \( a \) to \( b \) is the difference in the values of the anti-derivative at \( b \) and at \( a \).
Bessel's Differential Equation
Bessel's differential equation arises in problems with cylindrical symmetry, common in physics and engineering fields like acoustics, electromagnetism, and fluid dynamics. The equation is \[x^2 y'' + x y' + (x^2 - n^2) y = 0\] where \( n \) is a constant. Bessel functions, including \( J_{0}(x) \) and \( J_{1}(x) \), are solutions to this differential equation.
Key aspects of Bessel functions include:
Key aspects of Bessel functions include:
- They are oscillatory functions that generalize trigonometric functions but are suited for radial components.
- \( J_{0}(x) \) represents the zeroth-order Bessel function, while \( J_{1}(x) \) represents the first order. Both are routinely encountered in wave propagation and heat conduction problems.
Other exercises in this chapter
Problem 29
Use the power series method to solve the given initial-value problem. $$ (x-1) y^{\prime \prime}-x y^{\prime}+y=0, y(0)=-2, y^{\prime}(0)=6 $$
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\(x=0\) is a regular singular point of the given differential equation. Show that the indicial roots of the singularity differ by an integer. Use the method of
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Use the power series method to solve the given initial-value problem. $$ (x+1) y^{\prime \prime}-(2-x) y^{\prime}+y=0, y(0)=2, y^{\prime}(0)=-1 $$
View solution