Question

Show that the solution of the initial value problem $$ m u^{\prime \prime}+\gamma u^{\prime}+k u=0, \quad u\left(t_{0}\right)=u_{0}, \quad u^{\prime}\left(t_{0}\right)=u_{0}^{\prime} $$ can be expressed as the sum \(u=v+w,\) where \(v\) satisfies the initial conditions \(v\left(t_{0}\right)=\) \(u_{0}, v^{\prime}\left(t_{0}\right)=0, w\) satisfies the initial conditions \(w\left(t_{0}\right)=0, w^{\prime}\left(t_{0}\right)=u_{0}^{\prime},\) and both \(v\) and \(w\) satisfy the same differential equation as \(u\). This is another instance of superposing solutions of simpler problems to obtain the solution of a more general problem.

Short answer

Answer: The solution of the initial value problem can be expressed as the sum of two functions \(v(t)\) and \(w(t)\), both of which satisfy the same differential equation as \(u(t)\) and the given initial conditions. This sum is given by: $$ u(t) = v(t) + w(t) = (A_1 e^{r_1 t} + A_2 e^{r_2 t}) + (B_1 e^{r_1 t} + B_2 e^{r_2 t}) $$ where \(A_1\), \(A_2\), \(B_1\), and \(B_2\) are constants determined by the initial conditions and \(r_1\) and \(r_2\) are the solutions of the characteristic equation.

Step-by-step solution

Find the general solution of the given differential equation

To find the general solution of the given differential equation \( m u''+\gamma u'+ku=0\), first, we need to find the characteristic equation, which is given by: $$ m r^2 + \gamma r + k = 0 $$ Now, solve this quadratic equation for \(r\). The solutions for the characteristic equation will give us the general solution for the given differential equation in the form of: $$ u(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t} $$ where \(C_1\) and \(C_2\) are constants and \(r_1\) and \(r_2\) are the solutions of the characteristic equation.

Determine the specific solutions \(v\) and \(w\)

To find the specific solutions \(v\) and \(w\) that satisfy the given initial conditions, we first need to find their corresponding general solutions. Both \(v\) and \(w\) satisfy the same differential equation as \(u\), so their general solutions will also be in the form of: $$ v(t) = A_1 e^{r_1 t} + A_2 e^{r_2 t} $$ $$ w(t) = B_1 e^{r_1 t} + B_2 e^{r_2 t} $$ where \(A_1\), \(A_2\), \(B_1\), and \(B_2\) are constants. Now, we must satisfy the given initial conditions: 1. For \(v(t)\): - \(v(t_0) = u_0 \Rightarrow A_1 e^{r_1 t_0} + A_2 e^{r_2 t_0} = u_0\) - \(v'(t_0) = 0 \Rightarrow r_1 A_1 e^{r_1 t_0} + r_2 A_2 e^{r_2 t_0} = 0\) 2. For \(w(t)\): - \(w(t_0) = 0 \Rightarrow B_1 e^{r_1 t_0} + B_2 e^{r_2 t_0} = 0\) - \(w'(t_0) = u_0' \Rightarrow r_1 B_1 e^{r_1 t_0} + r_2 B_2 e^{r_2 t_0} = u_0'\) Solve this system of linear equations for \(A_1\), \(A_2\), \(B_1\), and \(B_2\) to find the specific solutions \(v\) and \(w\).

Show that \(u=v+w\)

Now that we have the specific solutions \(v\) and \(w\), we can show that their sum satisfies the original initial value problem by proving that \(u=v+w\) and that both \(u\) and \(v+w\) satisfy the initial conditions. The sum of the specific solutions \(v\) and \(w\) is given by: $$ u(t) = v(t) + w(t) = (A_1 e^{r_1 t} + A_2 e^{r_2 t}) + (B_1 e^{r_1 t} + B_2 e^{r_2 t}) $$ $$ u(t) = (A_1 + B_1) e^{r_1 t} + (A_2 + B_2) e^{r_2 t} $$ This has the same form as the general solution of the given differential equation, so \(u\) is indeed the sum of \(v\) and \(w\). In addition, since \(v\) and \(w\) have been chosen to satisfy their respective initial conditions, their sum \(u\) also satisfies the given initial conditions for \(u(t_0)=u_0\) and \(u'(t_0)=u_0'\). Therefore, the solution of the initial value problem can be expressed as the sum \(u=v+w\) with \(v\) and \(w\) satisfying the same differential equation as \(u\) and the given initial conditions.

Key concepts

Initial Value Problem

In differential equations, an initial value problem is a type of problem that not only involves solving the differential equation but also finding a specific solution that satisfies given initial conditions. These conditions are typically provided for the function and its derivatives at a certain point.
Consider the second-order linear differential equation:

• \( m u'' + \gamma u' + k u = 0 \)

Here, the initial value problem specifies values for \(u(t_0) = u_0\) and \(u'(t_0) = u_0'\) at the initial time \(t_0\). The aim is to find a unique function \(u(t)\) that not only solves the differential equation but also meets these initial conditions.
Initial value problems are crucial in applications involving time-dependent processes, such as physics problems where initial positions and velocities are often specified. By rigorously applying the initial conditions, we can determine the constants in the general solution, leading to a specific and unique solution.

Characteristic Equation

The characteristic equation is a tool used to solve linear differential equations with constant coefficients. It arises from substituting a trial solution of the form \( u(t) = e^{rt} \) into the differential equation.
For the given differential equation \( m u'' + \gamma u' + k u = 0 \), substituting the trial solution yields:

• \( m r^2 + \gamma r + k = 0 \)

This equation is a quadratic equation in terms of \(r\), and its roots, \(r_1\) and \(r_2\), help form the general solution to the differential equation.
Depending on the nature of the roots (real and distinct, real and repeated, or complex conjugates), different forms of the general solution arise:

• Real and distinct roots: \( u(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t} \)
• Repeated roots: \( u(t) = (C_1 + C_2 t) e^{rt} \)
• Complex roots: \( u(t) = e^{\alpha t}(C_1 \cos \beta t + C_2 \sin \beta t) \)

These solutions form the backbone of solving linear differential equations and enable us to tailor the solution to meet specific initial conditions.

Superposition Principle

The superposition principle is a fundamental concept in linear algebra and differential equations, particularly useful for solving linear systems. It states that if \(u_1(t)\) and \(u_2(t)\) are solutions to a homogeneous linear differential equation, then any linear combination of these solutions, say, \(a u_1(t) + b u_2(t)\), is also a solution.
This principle allows us to construct new solutions from existing ones, which can be particularly useful in solving complex initial value problems. In practice, we often break down a complicated problem into simpler parts, solve each part individually, and then combine them.
In the context of the initial value problem, we can express the solution as \(u = v + w\), where:

• \(v(t)\) satisfies the initial condition \(v(t_0) = u_0\) and \(v'(t_0) = 0\)
• \(w(t)\) satisfies the initial condition \(w(t_0) = 0\) and \(w'(t_0) = u_0'\)

Both \(v\) and \(w\) solve the same differential equation, allowing their sum to form the solution \(u(t)\). This technique demonstrates the utility of the superposition principle in obtaining the complete solution from pieces satisfying simpler initial conditions.