Question

Solve problem \(10.30\) if the string is released from the given configuration with the velocity \(v_{0}(x)=c_{0} w_{0}(x)\), where \(c_{0}\) is a constant.

Short answer

Answer: The method used to solve the wave equation in this problem is called separation of variables. It involves breaking down the original partial differential equation (PDE) into ordinary differential equations (ODEs), which can then be solved for the general solution. The initial conditions help to find the particular solution by applying these conditions to the general solution, narrowing it down to a specific form that satisfies the given problem.

Step-by-step solution

Set up the wave equation

First, we have to write the wave equation for this problem: $$\frac{\partial^2 w}{\partial t^2}= v^2\frac{\partial^2 w}{\partial x^2}$$

Apply the initial conditions

We are given two initial conditions: 1. \(w(x,0) = w_0(x)\): The initial displacement of the string is given by the function \(w_0(x)\). 2. \(v_0(x) = c_0w_0(x)\): The initial velocity of each point on the string is proportional to its displacement, with a constant of proportionality \(c_0\).

Solve the wave equation using separation of variables

In order to solve the wave equation, we can use the method of separation of variables. Assume that the solution can be written in the form: $$w(x,t) = X(x)T(t)$$ Plugging this form into the wave equation, we get: $$X(x)T''(t) = v^2 X''(x)T(t)$$ Divide both sides by \(X(x)T(t)\) to separate the variables: $$\frac{T''(t)}{T(t)} = v^2\frac{X''(x)}{X(x)}$$ Since each side of the equation depends on a different variable (one on \(t\) and the other on \(x\)), this expression must be equal to a constant, say \(k\): $$\frac{T''(t)}{T(t)} = v^2\frac{X''(x)}{X(x)}=k$$ Now, we have two ordinary differential equations (ODEs): 1. \(T''(t) = k T(t)\) 2. \(X''(x) = \frac{k}{v^2} X(x)\)

Solve the ODEs

Now, solve the ODEs for \(T(t)\) and \(X(x)\), respectively. For \(T(t)\), we know that the general solution of the ODE is: $$T(t) = A\cos(\sqrt{k}t) + B\sin(\sqrt{k}t)$$ For \(X(x)\), the general solution of the ODE is: $$X(x) = C\cos(\sqrt{\frac{k}{v^2}}x) + D\sin(\sqrt{\frac{k}{v^2}}x)$$ Now, we can write the general solution for \(w(x,t)\) as: $$w(x,t) = XY = (A\cos(\sqrt{k}t) + B\sin(\sqrt{k}t))(C\cos(\sqrt{\frac{k}{v^2}}x) + D\sin(\sqrt{\frac{k}{v^2}}x))$$

Apply the initial conditions to find the particular solution

Given that we have the initial conditions \(w(x,0) = w_0(x)\) and \(v_0(x) = c_0w_0(x)\), we can now find the particular solution by applying these conditions to the general solution: 1. For initial displacement: $$w(x,0) = w_0(x) = AC\cos(\sqrt{\frac{k}{v^2}}x)$$ 2. For initial velocity: $$v_0(x) = c_0w_0(x) = \frac{\partial}{\partial t}(w(x,t))\Big|_{t=0} = \sqrt{k}BD\sin(\sqrt{\frac{k}{v^2}}x)$$ From here, we can find the expressions for the constants \(A\), \(B\), \(C\), and \(D\) in terms of \(w_0(x)\), \(c_0\), \(k\) and \(v^2\). However, the expressions of these constants will depend on the specific form of \(w_0(x)\), which is not given in the problem statement. In conclusion, the solution of the problem involves finding the particular solution of the wave equation under given initial conditions. The method of separation of variables helps in breaking down the original PDE into ODEs, which can then be solved for the general solution. This general solution can be narrowed down to a particular solution once the given initial conditions are applied. Without knowing the specific form of \(w_0(x)\), we cannot give a definitive expression for the constants or the solution itself.

Key concepts

Separation of Variables

The method of separation of variables is a powerful technique used to solve partial differential equations, like the wave equation in this problem. The idea is to assume that the solution can be expressed as a product of functions, each depending only on one of the variables. For our wave equation, we assume that the solution can be written as \( w(x,t) = X(x)T(t) \).
This assumption turns the original partial differential equation (PDE) into ordinary differential equations (ODEs) that are easier to handle. By plugging \( w(x,t) = X(x)T(t) \) into the wave equation, we find that the PDE can be separated such that:

• \( X(x)T''(t) = v^2 X''(x)T(t)\)
• \( \frac{T''(t)}{T(t)} = v^2\frac{X''(x)}{X(x)} = k \)

Here, \( k \) is a separation constant, and each side of the equation depends on different variables, making it possible to equate them to a constant.
The outcome is two simpler ODEs that can be solved individually. This makes separation of variables a useful tool for finding solutions in problems with consistent boundary and initial conditions.

Initial Conditions

Initial conditions play a crucial role in solving differential equations as they allow us to find a unique solution to a problem. In the context of this wave equation problem, two initial conditions are specified:

• Initial displacement: \( w(x,0) = w_0(x) \)
• Initial velocity: \( v_0(x) = c_0 w_0(x) \)

The first condition describes the initial shape of the string, while the second gives its initial velocity, which is proportional to the displacement function \( w_0(x) \) by the constant \( c_0 \).
Applying these conditions to the general solution helps locate the specific solution that satisfies these premises. It involves substituting \( t=0 \) into the general expression and matching it with the given conditions.
By doing this, we can determine the values of the constants within the general solution equation, making it specific to the problem at hand. Initial conditions ensure that the solution is not just theoretically valid but also practically applicable to the described physical scenario.

Ordinary Differential Equations

Ordinary differential equations (ODEs) arise in this problem after using the separation of variables method on the wave equation. We end up with two key ODEs:

• \( T''(t) = k T(t) \)
• \( X''(x) = \frac{k}{v^2} X(x) \)

These equations are simpler than the original PDE because they only involve one variable each, instead of two. Solving an ODE typically involves finding a function that satisfies the differential equation for all values of the variable.
The general solutions to these ODEs for this problem are:

• For \( T(t) \): \( T(t) = A\cos(\sqrt{k}t) + B\sin(\sqrt{k}t) \)
• For \( X(x) \): \( X(x) = C\cos(\sqrt{\frac{k}{v^2}}x) + D\sin(\sqrt{\frac{k}{v^2}}x) \)

The constants \( A, B, C, \) and \( D \) need to be determined through boundary and initial conditions. Solving ODEs like these helps simplify the process of finding the wave function \( w(x,t) \) in terms of the given initial conditions.

Vibrating String

The physical problem we're solving involves a vibrating string, a classic example of a wave problem. The wave equation models how waves propagate along the string, and understanding it helps explain phenomena like musical instruments. Originating from different disturbances or initial conditions, the string's movement can be quite complex to analyze.
For a string fixed at both ends, there are characteristic waves that can travel along it. These are described mathematically by the wave function \( w(x,t) \), where \( x \) represents the position along the string and \( t \) the time past.
The solutions to the wave equation describe these vibrations naturally, including how they change over time. Due to boundary conditions, like fixed ends, only certain frequencies (or harmonics) are allowed. This is why a violin string, for example, can produce distinct musical notes.

• The given initial conditions define how the string was displaced and the velocity each point had initially.
• As the system evolves, it transitions through states described by the wave function, softly decaying in real-world situations due to resistance.

Understanding vibrating strings combines the beauty of physics and mathematics, providing insights into not only practical applications but also natural phenomena.