Question

Express the solution of the given initial value problem in terms of a convolution integral. \(y^{\prime \prime}+\omega^{2} y=g(t) ; \quad y(0)=0, \quad y^{\prime}(0)=1\)

Short answer

Question: Determine the complete solution for the initial value problem (IVP) involving a second-order linear inhomogeneous ordinary differential equation (ODE) with constant coefficients and given initial conditions: \(y^{\prime \prime}+\omega^2 y = g(t)\), with \(y(0)=0\) and \(y'(0)=1\). Answer: The complete solution for the initial value problem is given by: \(y(t) = \frac{1}{\omega}\sin(\omega t) + \frac{1}{\omega} \int_{0}^{t} \sin(\omega (t-s)) g(s) ds\)

Step-by-step solution

Find the complementary solution

For the homogeneous equation \(y^{\prime \prime}+\omega^2 y = 0\), we can find the characteristic equation: \(r^2 + \omega^2 = 0\). Solving for r, we get two complex roots, \(r_{1} = i\omega\) and \(r_{2} = -i\omega\). Therefore, the complementary solution of the homogeneous equation is \(y_c(t) = C_1\cos(\omega t)+C_2\sin(\omega t)\), where \(C_1\) and \(C_2\) are constants.

Find the GreenFunction

We will now find the Green's function, which will help us find the particular solution in terms of the convolution integral. The Green's function \(G(t)\) satisfies the following equation: \(G''(t)+\omega^2 G(t)=\delta(t)\), with the conditions \(G(0) = 0\) and \(G'(0^+)-G'(0^-) = G'(0)=1\). The homogeneous solution of \(G''(t)+\omega^2 G(t)=0\) is the same as the complementary solution we found in step 1. Let \(G(t)\) be of the form \(G(t)=a(t)\cos(\omega t) + b(t)\sin(\omega t)\). Then \(G'(t)=a'(t)\cos(\omega t)-a(t)\omega\sin(\omega t)+b'(t)\sin(\omega t)+b(t)\omega\cos(\omega t)\) and \(G''(t)=-a''(t)\sin(\omega t)-a'(t)\omega\cos(\omega t)+b''(t)\cos(\omega t)-b'(t)\omega\sin(\omega t)-a(t)\omega^2\cos(\omega t)-b(t)\omega^2\sin(\omega t)\). Plugging back into the GreenFunction equation, we get: \(-a''(t)\sin(\omega t)-a'(t)\omega\cos(\omega t)+b''(t)\cos(\omega t)-b'(t)\omega\sin(\omega t)=\delta(t)\). Comparing corresponding coefficients of \(\cos(\omega t)\) and \(\sin(\omega t)\), we get two differential equations: - \(-\omega a'(t)+b''(t)=0;\) , \(a(0)=-\frac{1}{\omega}\) - \(a''(t)+\omega b'(t)=0;\) , \(b(0)=0\). Solving these integral equations, we find that \(G(t)=\frac{1}{\omega}\sin(\omega t)u(t)\), where \(u(t)\) is Heaviside unit step function.

Determine the particular solution using the convolution integral

Now we can express the particular solution \(y_p(t)\) using the convolution integral: \(y_p(t)=\int_{0}^{t} G(t-s) g(s) ds\) = \(\frac{1}{\omega} \int_{0}^{t} \sin(\omega (t-s)) g(s) ds\)

Apply the initial conditions to find the constants in the complementary solution

We have the initial conditions \(y(0)=0\) and \(y'(0)=1\). Applying the initial condition to the complementary solution \(y_c(t)\): \(y_c(0) = C_1\cos(0)+C_2\sin(0) = C_1 = 0\). So, we found that \(C_1 = 0\). Now, differentiate \(y_c(t)\) to find \(y'_c(t)\): \(y'_c(t) = -C_1\omega\sin(\omega t)+C_2\omega\cos(\omega t)\) Using the second initial condition with \(y'_c(0) = 1\): \(y'_c(0)=C_2\omega = 1 \Rightarrow C_2=\frac{1}{\omega}\).

Write the complete solution for y(t)

Finally, we can write the complete solution for y(t) by combining the complementary solution and the particular solution: \(y(t) = y_c(t) + y_p(t)\) \(y(t) = \frac{1}{\omega}\sin(\omega t) + \frac{1}{\omega} \int_{0}^{t} \sin(\omega (t-s)) g(s) ds\) This is the solution of the initial value problem expressed in terms of the convolution integral.

Key concepts

Differential Equations

Differential equations are mathematical expressions involving functions and their derivatives. They are used to describe various physical phenomena such as motion, growth, and decay. The order of a differential equation is determined by the highest derivative present. For example, if a function involves a second derivative, it is called a second-order differential equation. Differential equations can be categorized as either ordinary or partial. Ordinary differential equations (ODEs) involve functions of a single variable and their derivatives. Partial differential equations involve multiple variables.

• Linear vs. Non-linear: Linear differential equations have solutions that can be added together and multiplied by constants to produce another solution. Non-linear equations do not have this property.
• Homogeneous vs. Non-homogeneous: A homogeneous differential equation has zero on one side of the equation. Non-homogeneous includes a non-zero term.

In our exercise, we worked with a second-order linear non-homogeneous differential equation: \(y'' + \omega^2 y = g(t)\). Understanding these categories is essential for solving and analyzing the behavior of solutions to differential equations.

Initial Value Problem

An initial value problem is a type of differential equation accompanied by specific conditions known as initial conditions. These conditions specify the values of the function and its derivatives at the starting point. Solving an initial value problem involves finding a function that satisfies both the differential equation and the given initial conditions.

• Purpose of Initial Conditions: They determine the particular solution out of the infinite possible solutions of a differential equation.
• Typical Form: An initial value problem can be generally stated as \( y(t_0) = y_0, \) and \( y'(t_0) = y'_0 \).

In our example, the initial value problem given was \(y'' + \omega^2 y = g(t); \quad y(0)=0, \quad y'(0)=1\). The initial conditions help determine the constants in the complementary solution, ensuring that the final solution aligns with the problem's starting conditions.

Green's Function

Green's Function is a powerful tool used to solve inhomogeneous differential equations. It allows the construction of a particular solution from a source function using an integral representation. For linear differential operators, Green's functions provide a way of expressing the solution in terms of an integral that involves the Green's function itself and the source term.

• Definition: A Green's function, \( G(t) \), satisfies \( G''(t) + \omega^2 G(t) = \delta(t) \), where \( \delta(t) \) is the Dirac delta function.
• Properties: It satisfies certain conditions at a boundary or initial point, which, in our exercise, were \( G(0) = 0 \) and \( G'(0^+) - G'(0^-) = 1 \).

For our task, the Green's function was found to be \( G(t) = \frac{1}{\omega} \sin(\omega t)u(t) \), where \( u(t) \) is the Heaviside step function. This function helps transform the differential equation solution into a manageable convolution integral.

Complementary Solution

The complementary solution refers to the part of the solution of a differential equation that relates to the homogeneous equation component. To derive it, solve the associated homogeneous equation, which excludes any external forces or inputs (i.e., \( g(t) = 0 \)).

• Homogeneous Equation: The complementary solution is found from \( y'' + \omega^2 y = 0 \).
• Form: For second-order linear homogeneous differential equations, solutions are typically expressed in terms of exponential, sine, or cosine functions depending on the nature of the characteristic equation's roots.

In the exercise, the complementary solution is \( y_c(t) = C_1 \cos(\omega t) + C_2 \sin(\omega t) \). By applying initial conditions, the constants \( C_1 \) and \( C_2 \) were determined, leading us to a complete physical interpretation of the system's free response behavior.