Question

By combining the results of Problems 24 through \(26,\) show that the solution of the initial value problem $$ L[y]=\left(a D^{2}+b D+c\right) y=g(t), \quad y\left(t_{0}\right)=0, \quad y^{\prime}\left(t_{0}\right)=0 $$ where \(a, b,\) and \(c\) are constants, has the form $$ y=\phi(t)=\int_{t_{0}}^{t} K(t-s) g(s) d s $$ The function \(K\) depends only on the solutions \(y_{1}\) and \(y_{2}\) of the corresponding homogeneous equation and is independent of the nonhomogeneous term. Once \(K\) is determined, all nonhomogeneous problems involving the same differential operator \(L\) are reduced to the evaluation of an integral. Note also that although \(K\) depends on both \(t\) and \(s,\) only the combination \(t-s\) appears, so \(K\) is actually a function of a single variable. Thinking of \(g(t)\) as the input to the problem and \(\phi(t)\) as the output, it follows from Eq. (i) that the output depends on the input over the entire interval from the initial point \(t_{0}\) to the current value \(t .\) The integral in Eq. (i) is called the convolution of \(K\) and \(g,\) and \(K\) is referred to as the kernel.

Short answer

Answer: The solution y(t) can be expressed using an integral representation with a convolution kernel K that depends only on the solutions of the corresponding homogeneous equation, as follows: $$y(t) = \int_{t_0}^{t} K(t-s) g(s) ds$$

Step-by-step solution

Homogeneous Equation and Its Solutions

The given nonhomogeneous second-order differential equation is: $$(aD^2 + bD + c)y(t) = g(t)$$ with the initial conditions \(y(t_0) = 0\) and \(y'(t_0)=0\). The corresponding homogeneous equation is: $$(aD^2 + bD + c)y(t) = 0$$ Let's consider that \(y_1(t)\) and \(y_2(t)\) be the two linearly independent solutions to the homogeneous equation.

Laplace Transform of the Differential Equation

Apply the Laplace transform to the given nonhomogeneous differential equation: $$a\mathcal{L}[y''(t)] + b\mathcal{L}[y'(t)] + c\mathcal{L}[y(t)] = \mathcal{L}[g(t)]$$

Solving for the Laplace Transform of y(t)

We will now solve for the Laplace transform of y(t), denoted by \(Y(s)\), using the relationships for the Laplace transform of the derivatives: $$a(s^2Y(s) - sy(t_0) - y'(t_0)) + b(sY(s) - y(t_0)) + cY(s) = G(s)$$ Substitute the initial conditions \(y(t_0) = 0\) and \(y'(t_0)=0\) to get: $$a(s^2Y(s)) + b(sY(s)) + cY(s) = G(s)$$ The solution for the Laplace transform of y(t) is then given by: $$Y(s) = \frac{1}{as^2 + bs + c}G(s)$$

Applying Inverse Laplace Transform and Convolution

Now, we apply the inverse Laplace transform to find y(t): $$y(t) = \mathcal{L}^{-1}[Y(s)](t) = \mathcal{L}^{-1}\left[\frac{1}{as^2 + bs + c}G(s)\right](t)$$ Using the convolution theorem, we can rewrite y(t) as the convolution of K(t-s) and g(s), where K(t-s) is the inverse Laplace transform of the function \(\frac{1}{as^2 + bs + c}\): $$y(t) = \mathcal{L}^{-1}\left[\frac{1}{as^2 + bs + c}\right](t-s)\ast g(s)$$

Expressing the Solution y(t) with Integral Representation

Now, we express the solution y(t) using the integral representation, where K(t-s) only depends on the solutions \(y_1(t)\) and \(y_2(t)\) of the homogeneous equation: $$y(t) = \int_{t_0}^{t} K(t-s) g(s) ds$$ This is the desired result. Now, all nonhomogeneous problems involving the same differential operator L are reduced to the evaluation of an integral, with the kernel K depending only on the solutions of the corresponding homogeneous equation.

Key concepts

Laplace Transform

The Laplace Transform is a powerful tool for solving differential equations. It's a technique that changes functions from the time domain, often denoted as a function of t, to the frequency domain, where functions are noted as a function of s.

This transformation makes it easier to solve complex differential equations and it brings great advantage since it converts differentiation into algebraic operations.

• The Laplace Transform of a function \(f(t)\) is given by: \[\mathcal{L}[f(t)] = \int_{0}^{\infty} e^{-st} f(t) \, dt\]
• Once in the s-domain, we use algebraic methods to solve the equation.
• Finally, we apply the inverse Laplace Transform to switch back to the time domain.

In solving the differential equation from the original exercise, the Laplace Transform helps simplify the equation by eliminating derivatives. It allows us to solve for \(Y(s)\), the Laplace Transform of our solution function \(y(t)\).

Once solved, we convert \(Y(s)\) back to \(y(t)\) using the inverse transform, often involving convolution when the equation is nonhomogeneous.

Homogeneous Differential Equation

A homogeneous differential equation is one where the function is set equal to zero, signifying no external influence or inputs.
These equations define systems in equilibrium without external forces or sources.

• The general form for linear homogeneous differential equations is given by: \[(aD^2 + bD + c)y(t) = 0\]
• Solutions to this type of equation, known also as complementary solutions, involve finding the characteristic equation and solving for its roots.
• The solutions are typically denoted by \(y_1(t)\) and \(y_2(t)\), indicating two linearly independent solutions.

In the problem at hand, the homogeneous version sets the stage for solving the corresponding nonhomogeneous equation.
These solutions, \(y_1(t)\) and \(y_2(t)\), are essential because they form the kernel function \(K(t-s)\) used in the convolution for nonhomogeneous equations.
Thus, homogeneous solutions provide the structure that underpins solving more complex linked problems.

Kernel Function

The Kernel Function in the context of differential equations serves as a bridge linking input functions to their resulting effects or outputs.
It essentially governs how an input modifies a system over time. This concept emerges naturally in the process of solving nonhomogeneous differential equations using convolution.

• The Kernel Function, represented as \(K(t-s)\), depends solely on the solutions \(y_1(t)\) and \(y_2(t)\) of the corresponding homogeneous equation.
• The kernel \(K\) simplifies the problem significantly since it abstracts away specifics of the nonhomogeneous function \(g(t)\).
• Once \(K(t-s)\) is found, we use convolution to determine the solution \(y(t)\), by integrating the product of \(K(t-s)\) and \(g(s)\).

The kernel effectively acts as a signature or fingerprint of how the system would behave under any input modifications, facilitating solving for the solution in an otherwise complex system.

Nonhomogeneous Differential Equation

Nonhomogeneous differential equations include an external force or source, unlike their homogeneous counterparts. This addition, in equations, is the term \(g(t)\), representing external input or forcing function.
These types of equations are generally more complex but are more reflective of real-world scenarios.

• Such an equation is represented as: \[(aD^2 + bD + c)y(t) = g(t)\]
• To tackle these, solutions often involve first solving the corresponding homogeneous equation to derive the Kernel Function \(K(t-s)\).
• We then use this kernel in conjunction with \(g(t)\) to find the system's response through an integral known as convolution.

In our example, this approach simplifies to evaluating an integral for \(y(t)\), using the kernel derived from the homogeneous portion. This method elegantly transforms solving complex differential equations into manageable integration tasks.
Thus, while nonhomogeneous equations present additional challenges, leveraging techniques like convolution through kernel functions streamlines finding their solutions.