Question

The following equations are called integral equations because the unknown dependent variable appears within an integral. When the equation also contains derivatives of the dependent variable, it is referred to as an integro- differential equation. In each exercise, the given equation is defined for \(t \geq 0\). Use Laplace transforms to obtain the solution. \(\int_{0}^{t} y(t-\lambda) y(\lambda) d \lambda=6 t^{3}\). Is the solution \(y(t)\) unique? If not, find all possible solutions.

Short answer

Based on the given integro-differential equation, after applying Laplace Transforms, inverse Laplace Transform, and utilizing the convolution theorem, we found that the solution of y(t) is $C\frac{t^4}{24}$ for any constant C in the real numbers. Therefore, the solution is not unique as it has different values for C. The possible solutions for y(t) are given by $y(t) = C\frac{t^4}{24}$, for all C in ℝ.

Step-by-step solution

Apply Laplace Transform on Both Sides

First, we will apply the Laplace Transform on both sides of the integro-differential equation. Let's denote the Laplace Transform of \(y(t)\) as \(Y(s)\). Using the convolution theorem, which states that $$\mathcal{L}\{\int_{0}^{t} f(t-\lambda)g(\lambda) d\lambda\}=F(s)G(s)$$ We then have: $$Y(s)Y(s) = \mathcal{L}\{6t^3\}$$

Simplify and Solve for Y(s)

Now we can proceed to solve the equation for \(Y(s)\): $$Y(s)^2 = \frac{6}{s^4}$$ $$Y(s) = \sqrt{\frac{6}{s^4}}$$

Get the Inverse Laplace Transform for Y(s)

Next, we need to find the inverse Laplace Transform of \(Y(s)\), which will give us \(y(t)\): $$y(t) = \mathcal{L}^{-1}\{\sqrt{\frac{6}{s^4}}\}$$ At this point, we realize that it's difficult to find the inverse Laplace Transform directly. We can, however, use the convolution theorem, which was mentioned in the first step, again in the following way: $$y(t) = \mathcal{L}^{-1}\{\sqrt{\frac{6}{s^4}}\} = \mathcal{L}^{-1}\{\frac{\sqrt{6}}{s^4} * 1\}$$ $$y(t) = \mathcal{L}^{-1}\{\frac{\sqrt{6}}{s^4}\} * \mathcal{L}^{-1}\{1\}$$ We know that \(\mathcal{L}^{-1}\{1\} = \delta(t)\), where \(\delta(t)\) is the Dirac delta function. So, $$y(t) = \mathcal{L}^{-1}\{\frac{\sqrt{6}}{s^4}\} * \delta(t)$$

Convolve the Two Functions

Now we can convolve the two functions: $$y(t) = \int_{0}^{t} \mathcal{L}^{-1}\{\frac{\sqrt{6}}{s^4}\}(t-\lambda)\delta(\lambda) d\lambda$$ As the Dirac delta function "picks" the value of the function at \(\lambda = 0\): $$y(t) = \mathcal{L}^{-1}\{\frac{\sqrt{6}}{s^4}\}(t)$$ Now we see that it is possible to find the inverse Laplace Transform of the remaining expression. By using the Laplace Transform table, we can find that $$\mathcal{L}^{-1}\{\frac{\sqrt{6}}{s^5}\} = C\frac{t^4}{24},$$ where \(C\) is a constant. So, the solution for \(y(t)\) is: $$y(t) = C\frac{t^4}{24}$$

Check for Unique Solution

In the final step, we need to check if the solution for \(y(t)\) is unique. Observe that \(y(t)\) can have different values for \(C\). Hence, the solution for \(y(t)\) is not unique. We can then say that all possible solutions for \(y(t)\) are given by $$y(t) = C\frac{t^4}{24}, \space \forall C \in \mathbb{R}$$

Key concepts

Laplace Transforms

Laplace transforms are a powerful mathematical tool used to convert complex differential equations into simpler algebraic forms. By transforming the time-domain functions, usually denoted as functions of the variable t, into the frequency-domain representations, functions of the complex variable s, solving calculus problems becomes more manageable.

Laplace transforms are particularly useful because they can handle initial conditions directly and turn differentiation and integration operations into algebraic multiplication and division. The Laplace transform of a function f(t) is typically given by the following integral:
\[ \mathcal{L}\{f(t)\} = F(s) = \int_{0}^{\infty} e^{-st} f(t) dt. \]
Another benefit is that common functions and their transforms are tabulated, making it easier for students to work with them. When facing a complex integro-differential equation, as in the original exercise, using Laplace transforms simplifies the process of finding solutions by converting it into a form that might be more readily solvable.

Convolution Theorem

An integral part of the efficiency of Laplace transforms lies with the convolution theorem. It allows the multiplication of two Laplace transforms to represent the operation of convolution in the time domain.

The theorem states that if you have two functions, f(t) and g(t), with respective Laplace transforms F(s) and G(s), then the Laplace transform of their convolution is the product of their individual transforms:
\[ \mathcal{L}\{f(t) * g(t)\} = F(s) \cdot G(s). \]
Conversely, this property also applies when taking the inverse Laplace Transform, which means: \[ \mathcal{L}^{-1}\{F(s)G(s)\} = f(t) * g(t). \]

• The convolution of two functions is defined as \[ (f * g)(t) = \int_{0}^{t} f(t-\lambda)g(\lambda) d\text{\textlambda}. \]
Using convolution provides a method to incorporate integral equations and invert them back into the time domain, which was crucial in the step-by-step solution provided.

Unique Solution

In the context of differential equations, a unique solution refers to a single solution that satisfies the equation under a given set of initial conditions. If a differential equation has a unique solution, it means that there is only one function or curve that meets the requirements of the equation for the specified initial or boundary conditions.

However, not all differential equations have unique solutions. Some may have multiple solutions or even no solution at all, depending on factors such as the nature of the equation and the conditions applied. In the example problem with the integro-differential equation, after going through the Laplace transformation and inverse processes, it's shown that the solution is not unique. This is because the constant C can take any real value (\( C \in \mathbb{R} \)), which means there is an infinite number of possible solutions, each corresponding to a different value of C.

Inverse Laplace Transform

The inverse of the Laplace transform is a method used to revert a frequency-domain function back into its original time-domain form. Just as the Laplace transform is integral in simplifying differential equations, the inverse Laplace transform (\( \mathcal{L}^{-1} \)) is key for returning to the necessary time-dependent solution after solving an equation algebraically in the frequency domain.

To find the inverse Laplace Transform, often a table of known Laplace Transforms is consulted, making it possible to match a given F(s) to its corresponding f(t). In cases where direct matching isn't possible, as in the example problem, techniques such as partial fractions, complex inversion formulas, or utilizing theorems like the convolution theorem become valuable.

In our example, the inverse transform was found using the convolution theorem, demonstrating a practical application of the theorem and the inverse Laplace Transform's role in determining the solution y(t) of the original integro-differential equation.