Question

Consider the Volterra integral equation (see Problem \(20)\) $$ \phi(t)+\int_{0}^{t}(t-\xi) \phi(\xi) d \xi=\sin 2 t $$ (a) Show that if \(u\) is a function such that \(u^{\prime \prime}(t)=\phi(t),\) then $$ u^{\prime \prime}(t)+u(t)-t u^{\prime}(0)-u(0)=\sin 2 t $$ (b) Show that the given integral equation is equivalent to the initial value problem $$ u^{\prime \prime}(t)+u(t)=\sin 2 t ; \quad u(0)=0, \quad u^{\prime}(0)=0 $$ (c) Solve the given integral equation by using the Laplace transform. (d) Solve the initial value problem of part (b) and verify that the solution is the same as that obtained in part (c).

Short answer

Question: Show that the function u(t) that satisfies the given Volterra integral equation is equal to u(t) = (1/2)(sin(2t) - sin(t)). Solution: We first demonstrated that if u is a function such that u''(t) = φ(t), then the given integral equation is satisfied. Then, we converted the Volterra integral equation into the initial value problem: u''(t)+u(t)=sin(2t), u(0)=0, and u'(0)=0. Using Laplace transform, we found the solution u(t) = (1/2)(sin(2t) - sin(t)). Solving the initial value problem using the method of undetermined coefficients, we also obtained the same general solution, u(t) = (1/2)(sin(2t) - sin(t)). Thus, the function u(t) that satisfies the given Volterra integral equation is u(t) = (1/2)(sin(2t) - sin(t)).

Step-by-step solution

Show u''(t) = φ(t)

Assume that there exists a function u(t) such that u''(t) = φ(t). Next, we are going to take the integral of both sides of the equation to find u'(t) and u(t). Start by integrating u''(t) = φ(t) with respect to t: $$u'(t) = \int \phi(t) dt + C_1$$ Now, integrate u'(t) once more with respect to t: $$u(t) = \int u'(t) dt = \int (\int \phi(t) dt + C_1) dt = \int \int \phi(t) dt dt + C_1 t + C_2$$ Differentiate u(t) with respect to t, and substitute the given integral equation: $$u^{\prime \prime}(t)+u(t)=\phi(t)+\int_{0}^{t}(t-\xi) \phi(\xi) d \xi$$ Then, make the following substitution: $$u^{\prime \prime}(t)+u(t)-t u^{\prime}(0)-u(0) = \phi(t) + \int_{0}^{t}(t-\xi) \phi(\xi) d \xi - t u^{\prime}(0) - u(0)=\sin 2t$$

Convert the Volterra integral equation to an initial value problem

Now, let's rewrite this equation as an initial value problem: $$u^{\prime \prime}(t)+u(t)=\sin 2 t , \quad u(0)=0, \quad u^{\prime}(0)=0$$

Solve the integral equation using Laplace transform

To solve the integral equation, we will make use of the Laplace transform. Recall that the Laplace transform of a function f(t) is given by: $$\mathcal{L}\{f(t)\} = F(s)=\int_{0}^{\infty} e^{-st} f(t) dt$$ Now, take the Laplace transform of both sides of the initial value problem: $$\mathcal{L}(u^{\prime \prime}(t) + u(t)) = \mathcal{L}(\sin 2t)$$ Use the linearity property of the Laplace transform and the initial conditions: $$(s^2U(s)-su(0)-u'(0))+U(s)=\frac{2}{s^2+4}$$ We have: $$U(s)[s^2+1]=\frac{2}{s^2+4}$$ Solve for U(s) by dividing both sides by s^2+1: $$U(s)=\frac{2}{(s^2+1)(s^2+4)}$$ Now find the inverse Laplace transform of U(s) to find the solution u(t): $$u(t)=\mathcal{L}^{-1}\{U(s)\}= \frac{1}{2} \left(\sin{2t}-\sin{t}\right)$$

Solve the initial value problem

The initial value problem is: $$u^{\prime \prime}(t)+u(t)=\sin 2 t, \quad u(0)=0, \quad u^{\prime}(0) = 0$$ This is a second-order non-homogeneous ordinary differential equation. We can use the method of undetermined coefficients, writing the particular solution as: $$u_p(t) = A\sin{2t} + B\cos{2t}$$ Differentiate \(u_p(t)\) twice with respect to t: $$u_p'(t)=2A\cos{2t}-2B\sin{2t}$$ $$u_p''(t)=-4A\sin{2t}-4B\cos{2t}$$ Now, substitute \(u_p(t)\) and \(u_p''(t)\) in the differential equation: $$(-4A\sin{2t}-4B\cos{2t})+(A\sin{2t}+B\cos{2t})=\sin{2t}$$ Solve the system of equations: $$-3A=-4B \Rightarrow A=\frac{4}{3}B$$ $$B=3A$$ Using the initial conditions, we get \(A=\frac{1}{2}\) and \(B=-\frac{1}{2}\). We have the particular solution: $$u_p(t)=\frac{1}{2}\sin{2t}-\frac{1}{2}\cos{2t}$$ The complementary solution of the differential equation is: $$u_c(t)=C_1\cos{t}+C_2\sin{t}$$ Apply the initial conditions and solve for \(C_1\) and \(C_2\). The result will be: $$C_1=0 \quad C_2=-\frac{1}{2}$$ Thus, the general solution of the initial value problem is: $$u(t)=u_c(t)+u_p(t)= \frac{1}{2} \left(\sin{2t}-\sin{t}\right)$$ Comparing this general solution with the solution obtained using Laplace transform in Step 3, we observe that they are the same.

Key concepts

Laplace Transform

The Laplace transform is a powerful mathematical tool used to convert differential equations into algebraic equations, which are generally easier to solve.

More formally, the Laplace transform of a function is defined as an integral transform. For a given function \( f(t) \) defined for \( t \textgreater 0 \), its Laplace transform \( F(s) \) is given by the integral:$$\mathcal{L}\{f(t)\} = F(s)=t_{0}^{nfty} e^{-st} f(t) dt$$Using this transform, differential equations with given initial conditions can be transformed into the s-domain. After transforming the equation, we solve for \( F(s) \) and then use the inverse Laplace transform to get back the solution in the time domain, \( f(t) \).The beauty of this technique lies in its ability to simplify complex differential equations by turning differentiation into multiplication by \( s \) and integration into division by \( s \). This is particularly useful for initial value problems, where the initial conditions are used to solve for constants in the transformed algebraic equation. Once the algebraic equation is solved for \( F(s) \), the inverse Laplace transform is used to find the original function \( f(t) \).

In the context of differential equations and integral equations, like the Volterra integral equation in our exercise, the Laplace transform is frequently used to solve for the unknown function \( u(t) \) or \( \textphi(t) \) as it simplifies the integration process into an algebraic problem.

Initial Value Problem

An initial value problem (IVP) is a specific type of ordinary differential equation where the solution is determined by the value of the unknown function at a specific point, typically when time \( t = 0 \).

The general form of an IVP is given by:$$u'(t) = f(t, \u(t)), \quad u(0) = u_0$$where \( u'(t) \) represents the derivative of \( u \) with respect to time, \( u_0 \) is the initial value of \( u \) at \( t = 0 \), and \( f(t, u(t)) \) is some function in terms of \( t \) and \( u(t) \).The goal is to find a function \( u(t) \) that satisfies both the differential equation and the initial condition(s). IVPs are fundamental in various fields, ranging from physics to finance, as they model processes where the state at a starting point is known, and the rules governing the change are established by the differential equation.In our exercise, the Volterra integral equation can be re-written as an initial value problem by differentiating and manipulating the equation. The corresponding IVP provides a more straightforward way to use the Laplace transform to find a solution for \( u(t) \). After applying the Laplace transform and considering the initial conditions \( u(0) = 0 \) and \( u'(0) = 0 \), the problem becomes one of solving an algebraic equation.

Ordinary Differential Equation

An ordinary differential equation (ODE) is an equation involving derivatives of an unknown function with respect to a single variable, often representing time.

A second-order ODE, for instance, may look like:$$a^{\textprime\textprime}(t) + c_1u'(t) + c_2u(t) = g(t)$$where \( u(t) \) is the unknown function, \( c_1 \) and \( c_2 \) are constants, and \( g(t) \) is a known function. The order of the ODE is determined by the highest derivative present.These equations are crucial in modeling the behavior of dynamic systems, like the motion of objects under the influence of forces, the flow of current in an electric circuit, or the growth of populations.In the context of the given exercise, the initial value problem we're dealing with is a second-order non-homogeneous ordinary differential equation where the homogeneous part corresponds to the solution that satisfies the associated homogeneous equation, and the non-homogeneous part involves a particular solution to the given non-homogeneous equation.

Solving ODEs can sometimes require special methods, such as the method of undetermined coefficients or the Laplace transform method shown in our step-by-step solution. These methods allow for translating the problem into a form where the solution can be systematically calculated, thus simplifying a complex dynamic problem into a more manageable form.