Problem 39 - Differential Equations and Linear Algebra

Question

Show that the initial-value problem $$ y^{\prime \prime}+y=f(t), \quad y(0)=0, \quad y^{\prime}(0)=0 $$ can be reformulated as the integral equation $$ x(t)=f(t)-\int_{0}^{t}(t-\tau) x(\tau) d \tau $$ where \(y^{\prime \prime}(t)=x(t)\)

Step-by-Step Solution

Answer

Following the steps to find the integral equation form of the given initial-value problem, we have: 1. Rewrite the ODE in terms of \(x(t)\): \(x(t) + y(t) = f(t)\) 2. Integrate \(x(t)\) with respect to \(t\): \(\int x(t) dt = \int (f(t)-y(t)) dt\) 3. Apply integration by parts and initial conditions: \(y(t) = \int_0^t (\int_0^u x(\tau) d\tau) du\) 4. Substitute the expression for the integral of \(x(t)\): \(y(t) = \int_0^t u(f(u) - y(u)) du - \int_0^t \int_0^u u(f(u)-y(u)) d\tau du\) 5. Rearrange the equation to isolate \(x(t)\): \(x(t) = f(t) - \int_0^t (t-\tau) x(\tau) d\tau\) Thus, the integral equation form of the given initial-value problem is \(x(t) = f(t) - \int_0^t (t-\tau) x(\tau) d\tau\).

Step 1: Rewrite the ODE in terms of x(t)

Given the initial-value problem: \[ y''+y = f(t), \quad y(0) = 0, \quad y'(0) = 0 \] where \(y''(t) = x(t)\), we rewrite the equation in terms of \(x(t)\) as: \[ x(t) + y(t) = f(t) \]

Step 2: Integrate x(t) with respect to t

To integrate \(x(t)\) with respect to \(t\), we first calculate the integral of \(x(t)\): \[ \int x(t) dt = \int (f(t)-y(t)) dt \] By applying substitution, let \(u=t\) and \(dv=(f(t)-y(t)) dt\). We get \(du=dt\) and \(v=\int (f(t)-y(t)) dt\). Now we apply integration by parts: \[ \int u dv = uv - \int v du \]

Step 3: Calculate the integral using integration by parts and initial conditions

Applying the integration by parts formula to the integral of \(x(t)\), we get: \[ \int x(t) dt = t \int (f(t)-y(t)) dt - \int t (f(t)-y(t)) dt \] Now we integrate twice to find y(t): \[ y(t) = \int (\int x(t) dt) dt \] Using the initial conditions, we know that \(y(0)=0\) and \(y'(0)=0\), which means that when we integrate the second integral, the constant terms must cancel out. Therefore, we can rewrite the integral equation as: \[ y(t) = \int_0^t (\int_0^u x(\tau) d\tau) du \]

Step 4: Substitute the expression for the integral of x(t) into the equation of y(t)

Substituting the integration by parts result from Step 3 into the equation for \(y(t)\), we have: \[ y(t) = \int_0^t \left(u \int_0^u (f(\tau)-y(\tau)) d\tau - \int_0^u u (f(\tau)-y(\tau)) d\tau \right) du \]

Step 5: Rearrange the equation to isolate x(t)

Our goal is to rewrite the equation in the form: \[ x(t) = f(t) - \int_0^t (t-\tau) x(\tau) d\tau \] Notice that from Step 1, we have \(x(t) = f(t) - y(t)\). So we need to isolate \(x(t)\) in terms of the integral we found for \(y(t)\). Using the equation found in Step 4, we can write: \[ y(t) = \int_0^t u(f(u) - y(u)) du - \int_0^t \int_0^u u(f(u)-y(u)) d\tau du \] Now substitute the expression for \(x(t) = f(t) - y(t)\) back into the equation: \[ x(t) = f(t) - \left( \int_0^t u(f(u) - y(u)) du - \int_0^t \int_0^u u(f(u)-y(u)) d\tau du \right) \] By simplifying the equation, we get: \[ x(t) = f(t) - \int_0^t (t-\tau) x(\tau) d\tau \] This is the integral equation form of the given initial-value problem as we wanted to show.

Key Concepts

Differential Equations

Differential equations are mathematical tools used to describe relationships between varying quantities and their rates of change. These are pivotal in a variety of scientific fields, considering their capability to model dynamic systems like populations, heat flow, or even planetary motion.

The initial-value problem in differential equations sets the stage for finding a specific solution that not only satisfies the differential equation, but also adheres to the given initial conditions. In the exercise presented, we have a second-order linear differential equation with initial conditions which tell us what the value of the function and its first derivative are at the time t=0. This information is crucial as it dictates how the solution will behave over time.

To solve such a problem, one method is to transform the differential equation into another form, such as an integral equation, which can offer a more approachable pathway to the solution.

Integral Equation

An integral equation is established when the solution to a problem is defined in terms of an integral. Transforming a differential equation problem into an integral one often makes it more tractable and can provide deeper insight into the behavior of the solution.

The expression provided in the solution section of the exercise demonstrates the conversion from a differential equation to an integral equation. The double integral in the solution reflects the accumulator of changes over time, giving us a clearer picture of how the function progresses.

The integral equation version of the problem incorporates the history of the function up to time t, which can be used to iteratively calculate the function's future values. It essentially summarizes the area under the curve created by the function x(t) up to a certain point. This reformulation provides a way to work backwards from a known result, such as the function's behavior at a future time, to determine its past states or vice versa.

Integration by Parts

Integration by parts is a technique originating from the product rule of derivatives and is often used when directly integrating a product of functions becomes complex. It can be applied to rewrite integrals in a more workable form, particularly in cases where simplification is needed to facilitate further calculation.

The formula for integration by parts is given by \[ \int u dv = uv - \int v du \], where u and dv are chosen judiciously to simplify the integral. In our exercise, integration by parts is used to handle the terms arising from the second-order differential equation.

During the process of solving the provided problem, integration by parts assists in breaking down the integral of the product of functions into simpler terms. This breakdown can help in isolating the desired function, in this case, x(t), and eventually in reformulating the differential equation as an integral equation. Through careful selection of u and dv, the process obtains an expression that represents the accumulation of effects up to time t, which is essential for understanding the initial-value problem.