Problem 30 - Differential Equations and Linear Algebra

Question

Use the Convolution Theorem and the table of Laplace transforms to show that $$ \int_{0}^{x}(x-w)^{a} w^{b} d w=\frac{a ! b !}{(a+b+1) !} x^{a+b+1} $$ \(a > -1, b > -1, x > 0\)

Step-by-Step Solution

Answer

To prove the given integral equation using the Convolution Theorem and the table of Laplace transforms, \(f(x-w) = (x-w)^{a}\) and \(g(w) = w^{b}\) are identified. Then, Laplace transforms for both f(t) and g(t) are found, and \(F(s) = \frac{a!}{s^{a+1}}\), \(G(s) = \frac{b!}{s^{b+1}}\). By applying the Convolution Theorem, we find the inverse Laplace transform of \(F(s)G(s)\), which gives us the convolution of the functions as \(f(t) * g(t) = \frac{a!b!}{(a+b+1)!} t^{a+b+1}\). Finally, we show that the given integral is equal to the convolution of f and g, yielding the result \(\int_{0}^{x}(x-w)^{a} w^{b} d w = \frac{a!b!}{(a+b+1)!} x^{a+b+1}\).

Step 1: Identify the functions to apply convolution

Given integral is, $$ \int_{0}^{x}(x-w)^{a} w^{b} d w $$ Compare this to convolution formula, $$ f(t) * g(t) = \int_{0}^{x}f(x - w)g(w) dw $$ Here, $$ f(x-w) = (x-w)^{a}, g(w) = w^{b} $$ Now, we need to find the Laplace transforms of these functions.

Step 2: Find Laplace transforms

Using the Laplace transform table, we can find the Laplace transforms for f(t) and g(t) as follows: Laplace transform of \(f(t) = t^a\) is: $$ L\{t^a\} = \frac{a!}{s^{a+1}} $$ Laplace transform of \(g(t) = t^b\) is: $$ L\{t^b\} = \frac{b!}{s^{b+1}} $$ So, we have: $$ F(s) = \frac{a!}{s^{a+1}}, G(s) = \frac{b!}{s^{b+1}} $$

Step 3: Apply Convolution Theorem

Convolution theorem states that, $$ L^{-1}\{F(s)G(s)\} = f(t) * g(t) $$ First, find F(s)G(s): $$ F(s)G(s) = \frac{a!b!}{s^{a+1}s^{b+1}} $$ We now need to find the inverse Laplace transform of F(s)G(s). So, $$ L^{-1}\{F(s)G(s)\} = f(t)*g(t) $$

Step 4: Find the inverse Laplace transform

Using the Laplace transform table, the inverse Laplace transform of \(p! / s^{p+1}\) is \(t^p\). So the inverse Laplace transform of F(s)G(s) is: $$ L^{-1}\left\{\frac{a!b!}{s^{a+1}s^{b+1}}\right\} = \frac{a!b!}{(a+b+1)!} t^{a+b+1} $$ Therefore, the convolution of the functions is: $$ f(t) * g(t) = \frac{a!b!}{(a+b+1)!} t^{a+b+1} $$ Now remember the given integral is the convolution of f and g. So, we can replace the convolution with the given integral: $$ \int_{0}^{x}(x-w)^{a} w^{b} d w = \frac{a!b!}{(a+b+1)!} x^{a+b+1} $$ There you have it. The integral has been proven to be equal to the given expression, using the convolution theorem and the table of Laplace transforms.

Key Concepts

Laplace Transform

The Laplace Transform is a powerful tool for solving differential equations and evaluating integrals. This mathematical technique converts functions of time (often denoted as \( t \)) into functions of complex frequency (denoted as \( s \)). Here's why this tool is favored:

• Transforms differential equations into algebraic equations, making them easier to handle.
• Converts convolutions in the time domain into simple multiplications in the frequency domain.
• Allows for easy application of initial and boundary conditions.

The Laplace Transform of a function \( f(t) \) is defined as:\[L\{f(t)\} = \int_{0}^{\infty} e^{-st} f(t) \, dt\]In the problem, we use the fact that the Laplace Transform of \( t^a \) is \( \frac{a!}{s^{a+1}} \), illustrating how simple polynomials are transformed. This formula connects complex operations on time-domain functions with arithmetic in the frequency domain.

Inverse Laplace Transform

To go back from the frequency domain to the time domain, we need the Inverse Laplace Transform. This function undoes the Laplace Transform by turning algebraic equations back into functions of time.

• Essential for interpreting results in the original problem's terms.
• Recovers original functions so that real-world scenarios are understandable.

The inverse is expressed generally as:\[L^{-1}\{F(s)\} = f(t)\],where the primary goal is to find an expression for \( f(t) \) in terms of \( t \). In the exercise, we determined the inverse of \( \frac{a!b!}{s^{a+1}s^{b+1}} \) by recognizing it matches the form \( \frac{p!}{s^{p+1}} \), giving the result \( t^{a+b+1} \). This result helped illustrate how the convolution integral was equivalent to the given expression for the integral.

Convolution Integral

The Convolution Integral is central in the analysis of linear systems. It describes the amount of overlap of one function as it is shifted over another function. Often, convolutions appear as integrals:\[(f * g)(t) = \int_{0}^{t} f(u) g(t-u) \, du\]For solving problems using Laplace Transforms, the Convolution Theorem is invaluable. This theorem states that the Laplace Transform of a convolution is simply the product of their individual Laplace Transforms:\[L\{f * g\} = L\{f\} \cdot L\{g\}\]In the problem, we applied it to handle an integral involving two functions to find the Laplace Transforms, multiply them, and then apply the inverse Laplace Transform. This streamlined the computation of the integral given in the exercise.

Gamma Function

The Gamma Function, denoted \( \Gamma(n) \), is an extension of the factorial function to complex numbers. It helps in evaluating integrals and appears frequently in probability and statistics.

• Defined for any complex number except negative integers.
• For positive integers \( n \), \( \Gamma(n) = (n-1)! \).
• Used to generalize the notion of factorials.

The relation to factorials is given through the formula:\[\Gamma(n) = \int_{0}^{\infty} t^{n-1} e^{-t} \, dt\]In the exercise, the Gamma Function is implied as it relates to factorial calculations used in the transforms. This ensures that solutions are valid even when dealing with real or complex numbers, not just integers. Understanding the Gamma Function aids in working with transforms and integrals involving factorials, as seen in our step-by-step solution.