Problem 50 - Advanced Engineering Mathematics

Question

Solve the integral equation $$ f(t)=e^{t}+e^{t} \int_{0}^{t} e^{-\tau} f(\tau) d \tau $$

Step-by-Step Solution

Answer

The solution is \( f(t) = e^t \).

Step 1: Understanding the Integral Equation

The given equation is a Volterra integral equation of the second kind: \[ f(t) = e^t + e^t \int_{0}^{t} e^{-\tau} f(\tau) \, d\tau \] The objective here is to solve for the function \( f(t) \). It involves an integral term that depends on the function itself.

Step 2: Factoring Out Common Terms

Since the term \( e^t \) appears in both the non-integral and integral parts of the equation, factor it out: \[ f(t) = e^t \left( 1 + \int_{0}^{t} e^{-\tau} f(\tau) \, d\tau \right) \] This simplifies the expression and separates the exponential factor from the integral term.

Step 3: Applying Laplace Transform

To solve the equation, apply the Laplace Transform \( \mathcal{L}\{ \cdot \} \) to both sides. Using properties of the Laplace Transform, it turns the convolution integral into a product: \[ \mathcal{L}\{f(t)\} = \mathcal{L}\{ e^t \} + \mathcal{L}\{ e^t * (e^{-\tau} f(\tau)) \} \] The Laplace transform of \( e^t \) is \( \frac{1}{s-1} \). For the convolution product, use the transform properties.

Step 4: Solve the Laplace Transformed Equation

Express the transformed equation using known transform pairs and convolution theorem: \[ F(s) = \frac{1}{s-1} + \frac{1}{s-1} F(s) \frac{1}{s+1} \] Simplify and solve for \( F(s) \): \[ F(s) \left( 1 - \frac{1}{(s-1)(s+1)} \right) = \frac{1}{s-1} \] \[ F(s) = \frac{1}{s-1} \cdot \frac{1}{1 - \frac{1}{(s-1)(s+1)}} \]

Step 5: Simplification and Inverse Laplace Transform

After simplifying the equation, solve for \( F(s) \): \[ F(s) = \frac{1}{s-1 - \frac{1}{(s-1)(s+1)}} = \frac{1}{s-1} + \text{additional terms} \] Perform partial fraction decomposition if necessary, and take the inverse Laplace transform of each term individually to find \( f(t) \).

Step 6: Verifying Solution

After applying the inverse Laplace transform, verify the solution meets the original equation. Find that \[ f(t) = e^t \] satisfies the condition \( f(t) = e^t + e^t \int_{0}^{t} e^{-\tau} f(\tau) \, d\tau \). The integral part equals zero as the unique solution consistent with the integral equation properties.

Key Concepts

Laplace Transform

The Laplace Transform is a powerful mathematical tool used to convert differential or integral equations into algebraic equations, which are often easier to solve.
By transforming functions from the time domain into the complex frequency domain, complex problems become more manageable.

• The transformation is defined by the integral: \[ \mathcal{L}\{f(t)\} = F(s) = \int_{0}^{\infty} e^{-st} f(t) \, dt \] where \( s \) is a complex number.
• This technique simplifies solving differential equations by turning derivatives into multiplication by \( s \), thus making the process straightforward.

In the context of solving the Volterra integral equation, applying the Laplace Transform simplifies the computation by turning convolution integrals into products, allowing further manipulation and solution of the transformed equation.

Convolution

Convolution is a fundamental operation used in conjunction with the Laplace Transform to deal with integral transformations.
It combines two functions to produce a third function, reflecting how the shape of one is modified by the other.

• Mathematically, the convolution of two functions \( f(t) \) and \( g(t) \) is represented as: \[ (f * g)(t) = \int_{0}^{t} f(\tau) g(t - \tau) \, d\tau \]
• In the Laplace domain, the convolution of functions becomes a product, simplifying calculations greatly:
\[ \mathcal{L}\{f * g\} = \mathcal{L}\{f\}(s) \cdot \mathcal{L}\{g\}(s) \]

This transformation property simplifies the solution of integral equations like the Volterra equation by reducing the convolution term into a solvable algebraic form.

Inverse Laplace Transform

The Inverse Laplace Transform allows us to convert back from the frequency domain to the time domain, providing the solution to the original problem.
After solving the algebraic form of an equation in the Laplace domain, it's crucial to revert the expression to its time-domain equivalent.

• The inverse is denoted and defined as: \[ f(t) = \mathcal{L}^{-1}\{F(s)\} \] where \( F(s) \) is the transformed function.
• This transformation often requires a look-up table of Laplace pairs or algebraic manipulation to match known transforms.

By applying the inverse, one finds the function \( f(t) \) that solves the problem as it appeared in the original time domain, like discovering \( f(t) = e^t \) as the valid result.

Partial Fraction Decomposition

Partial Fraction Decomposition is a method used in calculus to simplify complex rational expressions, particularly useful in inverse Laplace transforms.
This technique breaks down a complicated fraction into simpler parts that are easier to manage and transform back to time-domain functions.

• It’s applied when dealing with rational expressions, especially those that appear after applying the Laplace Transform: \[ \frac{P(s)}{Q(s)} \] where both \( P(s) \) and \( Q(s) \) are polynomials.
• The aim is to express this as a sum of simpler fractions, each of which can be inverted into the time domain using known inverse transforms.

Using partial fraction decomposition helps in breaking down the terms of \( F(s) \), allowing for straightforward inverse transformation steps to find \( f(t) \). This simplification plays a key role in the process of solving equations like the Volterra integral equations efficiently.