Question

Consider the equation $$ \phi(t)+\int_{0}^{t} k(t-\xi) \phi(\xi) d \xi=f(t) $$ in which \(f\) and \(k\) are known functions, and \(\phi\) is to be determined. Since the unknown function \(\phi\) appears under an integral sign, the given equation is called an integral equation; in particular, it belongs to a class of integral equations known as Voltera integral equations. Take the Laplace transform of the given integral equation and obtain an expression for \(\mathcal{L}\\{\phi(t)\\}\) in terms of the transforms \(\mathcal{L}\\{f(t)\\}\) and \(\mathcal{L}\\{k(t)\\}\) of the given functions \(f\) and \(k .\) The inverse transform of \(\mathcal{L}\\{\phi(t)\\}\) is the solution of the original integral equation.

Short answer

Answer: The solution for the unknown function \(\phi(t)\) is given by the inverse Laplace transform of the following expression: \(\phi(t) = \mathcal{L}^{-1}\\{\frac{\mathcal{L}\\{f(t)\\}}{1 + \mathcal{L}\\{k(t)\\}}\\}\).

Step-by-step solution

Apply Laplace Transform to the Integral Equation

Take the Laplace transform of the given integral equation: $$\mathcal{L}\\{\phi(t) + \int_{0}^{t} k(t-\xi) \phi(\xi) d \xi\\} = \mathcal{L}\\{f(t)\\}$$

Apply Laplace Transform Properties to the Integral Term

We know that the Laplace transform of the convolution of two functions is the product of their Laplace transforms, i.e., \(\mathcal{L}\\{(f*g)(t)\\} = \mathcal{L}\\{f(t)\\} \cdot \mathcal{L}\\{g(t)\\}\). Here, the integral is the convolution of \(\phi(t)\) and \(k(t)\): $$\int_{0}^{t} k(t-\xi) \phi(\xi) d \xi = (k * \phi)(t)$$ Applying the convolution theorem, we get: $$\mathcal{L}\\{(k * \phi)(t)\\} = \mathcal{L}\\{k(t)\\} \cdot \mathcal{L}\\{\phi(t)\\}$$

Substitute the Laplace Transforms into the Integral Equation

Now, substitute the Laplace transforms from Step 1 and Step 2 into the integral equation: $$\mathcal{L}\\{\phi(t)\\} + \mathcal{L}\\{k(t)\\} \cdot \mathcal{L}\\{\phi(t)\\} = \mathcal{L}\\{f(t)\\}$$

Solve for the Laplace Transform of the Unknown Function \(\phi\)

Factor out \(\mathcal{L}\\{\phi(t)\\}\) from the left side of the equation: $$(1 + \mathcal{L}\\{k(t)\\}) \cdot \mathcal{L}\\{\phi(t)\\} = \mathcal{L}\\{f(t)\\}$$ Now, divide by \((1 + \mathcal{L}\\{k(t)\\})\) to isolate \(\mathcal{L}\\{\phi(t)\\}\): $$\mathcal{L}\\{\phi(t)\\} = \frac{\mathcal{L}\\{f(t)\\}}{1 + \mathcal{L}\\{k(t)\\}}$$

Obtain the Unknown Function \(\phi\) using the Inverse Laplace Transform

Take the inverse Laplace transform of both sides of the equation to find the unknown function \(\phi(t)\): $$\phi(t) = \mathcal{L}^{-1}\\{\frac{\mathcal{L}\\{f(t)\\}}{1 + \mathcal{L}\\{k(t)\\}}\\}$$ Now, the inverse Laplace transform \(\mathcal{L}^{-1}\\{\frac{\mathcal{L}\\{f(t)\\}}{1 + \mathcal{L}\\{k(t)\\}}\\}\) is the solution of the original integral equation.

Key concepts

Integral Equation

Integral equations are mathematical equations in which an unknown function appears under an integral sign. They are an essential part of mathematical analysis and arise naturally in various fields, such as physics and engineering. In our specific problem, we are dealing with a Volterra integral equation. This particular class of equations is characterized by integrals with limits that are functions of the upper bound (in this case, 't').
This means we have an integral starting from a fixed point (zero) up to a variable point 't'.

• Behavior: Volterra integral equations are typically used to model processes that are accumulative over time, such as population growth or the spread of disease.
• Linear vs Nonlinear: An integral equation can be linear if the unknown function is only multiplied by a constant or another function.

Understanding integral equations is crucial because they often provide solutions that cannot be tackled by ordinary differential equations alone.

Laplace Transform

The Laplace Transform is a powerful tool for simplifying complex equations, especially when dealing with differential and integral equations. It transforms a time-domain function into a frequency-domain function, making complex mathematical operations more manageable. This transformation can be thought of as a way to "untangle" the components of a function.
Here’s why it's useful:

• Linearity: It can turn integral and differential equations into algebraic equations, which are easier to solve.
• Operational rules: The Laplace Transform has well-established rules that simplify handling of convolutions and derivatives.

In the case of our problem, applying the Laplace Transform to each term in the given integral equation helps us express the unknown function's transform, making the original equation algebraically solvable.

Convolution Theorem

The Convolution Theorem is an essential tool for solving problems involving integral equations using the Laplace Transform. The theorem states that the Laplace transform of the convolution of two functions is the product of their individual Laplace transforms. Convolution in mathematics is a kind of blending or merging process.
This is useful because:

• Integration Simplified: It allows the conversion of an integral involving an unknown function into a simple multiplication of transforms.
• Ease of Calculations: Once multiplexed by their Laplace Transforms, algebraic manipulation becomes straightforward.

This theorem is particularly crucial in our equation, where the integral term signifies the convolution of two functions. Without this theorem, handling the integral part of the equation symbolically would be considerably more challenging.

Inverse Laplace Transform

The Inverse Laplace Transform is the process of converting a function from its frequency domain representation back to its original time domain. This involves finding the original time-domain function from its transformation, which is typically denoted by an inverse operator, \( \mathcal{L}^{-1} \).
The utility of this transform lies in:

• Reconstruction: It helps reconstruct the unknown function from its Laplace-transformed form once a solution in the frequency domain is found.
• Solution Finalization: In the given problem, after simplifying and rearranging the transformed equation algebraically, the inverse Laplace Transform provides us with the solution to the original integral equation, thus determining the function \( \phi(t) \). This allows us to use transforms to approach a solution iteratively.

This step is crucial when an explicit time-domain solution is required, completing the cycle of transformation and simplification initiated by the Laplace Transform.