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

Expert verified
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

01

Set up the wave equation

First, we have to write the wave equation for this problem: $$\frac{\partial^2 w}{\partial t^2}= v^2\frac{\partial^2 w}{\partial x^2}$$
02

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 \(c_0\).
03

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. \(X''(x) = \frac{k}{v^2} X(x)\)
04

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: $$w(x,t) = XY = (A\cos(\sqrt{k}t) + B\sin(\sqrt{k}t))(C\cos(\sqrt{\frac{k}{v^2}}x) + D\sin(\sqrt{\frac{k}{v^2}}x))$$
05

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.

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