Question

Lösen Sie die Integralgleichung $$ \int_{0}^{t} \cos (t-\tau) f(\tau) d \tau=t \sin t $$ unter Nutzung der Rechenregeln für Faltung von Funktionen und deren LAPLACE- Transformierten.

Short answer

The solution is \( f(t) = 2 \sin(t) \).

Step-by-step solution

Understand the Problem Statement

We are given an integral equation involving a convolution integral and asked to solve for the unknown function \( f(\tau) \). The equation is \( \int_{0}^{t} \cos(t - \tau) f(\tau) \, d\tau = t \sin t \). This can benefit from using Laplace transforms, which simplify convolution integrals.

Identify Known Functions

In the equation, \( \, \cos(t-\tau) \, \) and \( \, f(\tau) \, \) are the functions involved in the convolution. The right-hand side of the equation is \( \, t \sin t \, \). We aim to find \( f(t) \) by using these functions' Laplace transforms.

Apply the Laplace Transform to Both Sides

Taking the Laplace transform of both sides of the integral equation, the convolution theorem states: \( \mathcal{L}\{\cos(t-\tau) \ast f(\tau)\} = \mathcal{L}\{\cos(t)\} \cdot \mathcal{L}\{f(t)\} \). Thus, \( F(s) = \mathcal{L}\{f(t)\} \text{ and } T(s) = \mathcal{L}\{t \sin t\} \).

Find Laplace Transforms of Known Functions

The Laplace transform of \( \cos(t) \) is \( \frac{s}{s^2+1} \), and the transform of \( t \sin t \) is \( \frac{2s}{(s^2+1)^2} \). Using these, the equation in the Laplace domain is: \( \frac{s}{s^2+1} \cdot F(s) = \frac{2s}{(s^2+1)^2} \).

Solve for F(s) in the Laplace Domain

Rearrange the equation to solve for \( F(s) \): \[ F(s) = \frac{\frac{2s}{(s^2+1)^2}}{\frac{s}{s^2+1}} = \frac{2}{(s^2+1)} \].

Inverse Laplace Transform to Find f(t)

The inverse Laplace transform of \( \frac{2}{s^2+1} \) is \( 2 \sin(t) \). Therefore, \( f(t) = 2 \sin(t) \).

Verify the Solution

Substitute \( f(t) = 2 \sin(t) \) back into the original equation to verify: \( \int_{0}^{t} \cos(t-\tau) \cdot 2\sin(\tau) \, d\tau \). Simplifying, this equals \( t \sin t \), confirming \( f(t) \) is correct.

Key concepts

Convolution theorem

The convolution theorem is a fundamental concept used to simplify the process of solving integral equations involving convolution integrals. In this context, a convolution integral is an expression of the form \( \int_{0}^{t} \phi(t-\tau) \psi(\tau) \, d\tau \). The theorem states that the Laplace transform of a convolution of two functions is the product of their individual Laplace transforms.

This powerful result helps us handle convolutions in the frequency domain rather than the time domain, making calculations significantly easier.

• Transforms convolution integrals from time to frequency domain.
• Allows multiplication of transformed functions instead of convolution.
• Simplifies solving linear systems and differential equations.

Laplace Transform

The Laplace transform is an integral transform used to convert a function from its original domain (often time) into the frequency domain. It is given by: \( \mathcal{L}\{f(t)\} = \int_{0}^{\infty} e^{-st} f(t) \, dt \). This transformation is particularly useful for solving linear ordinary differential equations and integral equations where it turns differentiation and integration into multiplication and division.

When applied to the problem at hand, the Laplace transform converts each term into an algebraic form, making the equation much more manageable. Key properties include:

• Linearity: \( \mathcal{L}\{a f(t) + b g(t)\} = aF(s) + bG(s) \).
• Transforms derivatives into algebraic equations: \( \mathcal{L}\{f'(t)\} = sF(s) - f(0) \).
• Enables use of convolution theorem.

Inverse Laplace Transform

The inverse Laplace transform is the process of transforming a function expressed in the frequency domain back into the time domain. If \( \mathcal{L}\{f(t)\} = F(s) \), then the inverse transform is denoted as \( \mathcal{L}^{-1}\{F(s)\} = f(t) \).

Finding the inverse Laplace transform is crucial, as it allows us to obtain the original time-domain function from its transformed-frequency form.

• Often involves using tables for common transform pairs.
• Utilizes partial fractional decomposition to simplify inversion.
• Essential for finding solutions to differential and integral equations in original domain.

In the exercise, applying the inverse Laplace transform to \( \frac{2}{s^2+1} \) yields \( f(t) = 2 \sin(t) \).

Function solving

Solving for an unknown function within an integral equation involves several systematic steps. Initially, applying transformations like the Laplace transform can convert complex integral expressions into simpler algebraic forms.

In this particular exercise, solving for \( F(s) \) in the Laplace domain gave \( \frac{2}{s^2+1} \). Through inverse transformation, we determined \( f(t) = 2\sin(t) \). Here are some general steps involved in such problems:

• Identify known functions and setup the equation.
• Apply Laplace transform to convert to frequency domain.
• Solve algebraic equations for transform of unknown function.
• Apply inverse Laplace transform for solution in time domain.

These steps transform the problem from solving an integral equation to handling a simpler algebraic equation.

Mathematical verification

Verification of solutions is a crucial step to ensure correctness in mathematical problem-solving. It involves substituting the calculated solution back into the original problem to confirm its validity.

For the given integral equation, post calculation and Laplace transformations, it's important to substitute \( f(t) = 2 \sin(t) \) back into the convolution integral \( \int_{0}^{t} \cos(t-\tau) \cdot 2 \sin(\tau) \, d\tau \). Carried out correctly, this should establish the original condition \( t \sin t \). Steps for a sound verification include:

• Substitute the solution into the original equation.
• Use known identities or simplifications to check the result.
• Ensure the calculated result fulfills the original conditions.

Successfully verifying the solution confirms both the method and solution are correct.