Question

Solve the following integral equations: (a) \(f(t)+2 \int_{0}^{t} f(u) \cos (t-u) d u=9 e^{2 t}\) (b) \(\int_{0}^{t} f(u) d u-f^{\prime}(t)=\left\\{\begin{array}{ll}0, & 0 \leq t \leq a, \\ 1, & a \leq t,\end{array}\right\\} a>0\) (c) \(f(t)+\int_{0}^{t}(t-u) f(u) d u=\sin 2 t\) (d) \(f^{\prime \prime}(t)=\int_{0}^{t} u f(t-u) d u, \quad f(0)=-1, f^{\prime}(0)=1\) (e) \(f(t)+\int_{0}^{t} f(u) e^{-(t-u)} d u=1\) (f) \(\int_{0}^{t} f^{\prime}(u) f(t-u) d u=3 t e^{3 t}-e^{3 t}+1, \quad f(0)=0, f^{\prime}(0)>0\) (g) \(3 f^{\prime}(t)-10 f(t)+3 \int_{0}^{t} f(u) d u=10 \sin t-5, \quad f(0)=2\) (h) \(\int_{0}^{t} f(u) f(t-u) d u=2 f(t)+t-2\). (Is the solution unique?) (i) \(f^{\prime}(t)+\int_{0}^{t} f(u) d u=\sin t, \quad f(0)=1\) (j) \(\int_{0}^{t} f^{\prime \prime}(t) f(t-u) d u=t e^{a t}, \quad f(0)=\frac{1}{a}, f^{\prime}(0)=1\) (k) \(f(t)=a t+\int_{0}^{t} f(u) \sin (t-u) d u\) (l) \(f^{\prime}(t)+5 \int_{0}^{t} f(u) \cos 2(t-u) d u=10, \quad f(0)=2\) (\mathbf) \(\int_{0}^{t} f^{\prime}(u) f(t-u) d u=24 t^{3}, \quad f(0)=0\)

Short answer

The solution is \( f(t) = 9e^{2t} \).

Step-by-step solution

Identifying the Problem

We need to solve the integral equation: \( f(t) + 2 \int_{0}^{t} f(u) \cos(t-u) \, du = 9e^{2t} \). This is an example of a Volterra integral equation of the second kind.

Convert to Differential Equation

Differentiate both sides of the equation with respect to \( t \). This leaves us with an ordinary differential equation: \( f'(t) + 2f(t)\sin(0) + 2\int_{0}^{t} f(u) (-\sin(t-u)) \, du = 18e^{2t} \). Simplifying removes the sine term at the start.

Solve the Differential Equation

We solve the resultant differential equation: \( f'(t) + 2 \int_{0}^{t} f(u)(-\sin(t-u)) \, du = 18 e^{2t} \). Assume \( f(t) = Ae^{2t} \) where \( A \) is a constant. Substitute back to verify that it satisfies the original differential equation form.

Find the Particular Solution

Through substitution, simplify and solve for \( A \). By verifying \( 9 + 2\int_{0}^{t} 9e^{2u}\cos(t-u) \, du = 9e^{2t} \), it's corroborated that \( A = 9 \).

Final Solution

Thus, the solution to the integral equation is \( f(t) = 9e^{2t} \).

Key concepts

Volterra Integral Equations

Volterra integral equations are a special type of integral equations typically characterized by their limits of integration, which are variable. These equations are named after the Italian mathematician Vito Volterra, who studied them extensively. There are two types of Volterra integral equations: those of the first kind and those of the second kind. In this context, we are focusing on the second kind, where the unknown function appears both inside and outside the integral.

They are expressed in the form:

• First kind: \[ f(t) = \int_{a}^{t} K(t, u) f(u) \, du \]
• Second kind: \[ f(t) - \lambda \int_{a}^{t} K(t, u) f(u) \, du = g(t) \]

Where:

• \( f(t) \) is the unknown function to find.
• \( K(t, u) \) is known as the kernel of the integral equation.
• \( g(t) \) is a known function.
• \( \lambda \) is a constant.

In many cases, such equations can be transformed into differential equations for easier solutions, as seen when we differentiated both sides with respect to \( t \).

Differential Equations

Differential equations are mathematical equations that involve functions and their derivatives. They describe a wide range of phenomena, such as motion, heat, and growth. Solving these equations gives us insight into the behavior of certain systems over time.

Commonly, differential equations are classified based on the type of derivatives they involve:

• Ordinary Differential Equations (ODEs): Involves functions of only one independent variable and their derivatives. The equation mentioned in the solution is an example of an ODE.
• Partial Differential Equations (PDEs): Involves multiple independent variables and partial derivatives.

For example, when dealing with problems where motion or change over time is modeled, differential equations offer a powerful tool. By converting our integral equation into a differential equation, we could analyze it more easily through standard methods of solution.

Ordinary Differential Equations

Ordinary Differential Equations (ODEs) involve derivatives of a function relative to one independent variable. In this exercise, we transformed an integral equation into an ODE by differentiating each side. This is a common technique that simplifies the problem, making it more straightforward to solve.

The process usually involves:

• Identifying symmetries or constants that allow simplifications.
• Assuming a potential solution form, often involving exponential functions or polynomials.
• Substituting back into the equation to verify the assumed solution satisfies the differential equation.

ODEs are often much simpler to solve than integral equations as they relate more directly to standard calculus solutions. Techniques include separation of variables, integrating factors, or, as was used here, direct assumptions and verification.

Solution Verification

Verification is the final step where we ensure our proposed solution meets all conditions of the original problem. It's a critical process in mathematics, as it builds confidence in the accuracy of the solution.

The verification process generally includes:

• Substituting the solution back into the original equation.
• Checking boundary and initial conditions, ensuring they align with the specified conditions (e.g., initial values).
• Confirming all parts of the modified equations satisfy the problem requirements.

In this exercise, after solving the ODE, we substituted \( f(t) = 9e^{2t} \) back into the original integral to confirm that both sides equate, thereby verifying the correctness of \( A = 9 \). This step is crucial as errors can often propagate unnoticed without proper verification.