Question

Find the general solution of the wave equation $$ \frac{\partial^{2} u}{\partial t^{2}}-a^{2} \frac{\partial^{2} u}{\partial x^{2}}=0, \quad 00 . $$ such that \(u(0, t)=u(2 \pi, t), u_{x}(0, t)=u_{x}(2 \pi, t)\).

Short answer

Based on the given wave equation and boundary conditions, we used the separation of variables technique and Fourier series expansions to find the general solution. After assuming a separable solution, substituting it into the given PDE, and solving the resulting ODEs, we were able to derive the general solution: $$ u(x, t) = \sum_{n=-\infty}^{\infty}\left[ \left(A_n\cos(nx) + B_n\sin(nx) \right) \left( C_n\cos(an t) + D_n\sin(an t) \right) \right], $$ where \(A_n, B_n, C_n,\) and \(D_n\) are constants to be determined by the initial conditions.

Step-by-step solution

Assume a Separable Solution

Assume a solution of the form \(u(x, t) = X(x)T(t)\). This is where we separate the variables \(x\) and \(t\), which will later help us simplify the given wave equation.

Substitute the Separable Solution into the Wave Equation

Substitute \(u(x, t) = X(x)T(t)\) into the given PDE: $$ \frac{\partial^2}{\partial t^2}(XT) - a^2 \frac{\partial^2}{\partial x^2}(XT) = 0 \Rightarrow X(x) \frac{d^2T(t)}{dt^2} = a^2 T(t) \frac{d^2X(x)}{dx^2} \Rightarrow \frac{\frac{d^2T(t)}{dt^2}}{a^2T(t)} = \frac{\frac{d^2X(x)}{dx^2}}{X(x)} = -\lambda^2, $$ where \(\lambda^2\) is an arbitrary constant.

Solve the ODEs for X(x) and T(t)

We now have two ordinary differential equations: $$ \frac{d^2X}{dx^2} + \lambda^2X = 0, $$ and $$ \frac{d^2T}{dt^2} + a^2\lambda^2T = 0. $$ Solve these ODEs for \(X(x)\) and \(T(t)\). For the spatial ODE, the solution is given by: $$ X(x) = A\cos(\lambda x) + B\sin(\lambda x), $$ where \(A\) and \(B\) are constants. For the temporal ODE, the solution is given by: $$ T(t) = C\cos(a\lambda t) + D\sin(a\lambda t), $$ where \(C\) and \(D\) are constants.

Apply the Boundary Conditions

Apply the boundary conditions given in the problem to determine the constants: $$ u(0, t) = u(2\pi, t) \Rightarrow X(0)T(t) = X(2\pi)T(t) \Rightarrow X(0) = X(2\pi), $$ and $$ u_x(0, t) = u_x(2\pi, t) \Rightarrow X'(0)T(t) = X'(2\pi)T(t) \Rightarrow X'(0) = X'(2\pi). $$ For \(X(x)\), since \(X(0) = X(2\pi)\): $$ A\cos(0) + B\sin(0) = A\cos(2\pi\lambda) + B\sin(2\pi\lambda) \Rightarrow A = A\cos(2\pi\lambda) $$ This means that either \(A=0\) or \(\lambda = n\), where \(n\) is an integer. For \(X'(x)\), since \(X'(0) = X'(2\pi)\): $$ -\lambda A\sin(0) + \lambda B\cos(0) = -\lambda A\sin(2\pi\lambda) + \lambda B\cos(2\pi\lambda) \Rightarrow B = B\cos(2\pi\lambda) $$ This means that either \(B=0\) or \(\lambda = n\), where \(n\) is an integer.

Combine the Solutions for X(x) and T(t)

Since \(u(x, t)=X(x)T(t)\) and considering that \(\lambda = n\): $$ u(x, t) = \sum_{n=-\infty}^{\infty}\left[ \left(A_n\cos(nx) + B_n\sin(nx) \right) \left( C_n\cos(an t) + D_n\sin(an t) \right) \right], $$ where \(A_n, B_n, C_n,\) and \(D_n\) are constants to be determined by the initial conditions. This is the general solution to the given wave equation with the given boundary conditions.

Key concepts

Partial Differential Equations (PDEs)

A Partial Differential Equation (PDE) involves multiple independent variables and partial derivatives of an unknown function. The wave equation presented in the exercise is a classic PDE that models phenomena like sound and light waves. In this wave equation, we have two key variables, time \(t\) and spatial dimension \(x\). The equation is expressed as:

• \( \frac{\partial^2 u}{\partial t^2} - a^2 \frac{\partial^2 u}{\partial x^2} = 0 \)

This reflects how changes over time and space influence each other. Solving such an equation requires specialized methods because we're dealing with more dimensions than ordinary differential equations, complicating the situation. To simplify, we might convert it into simpler forms of equations by breaking it down using various techniques including separation of variables which deals with one variable at a time.

Boundary Conditions

Boundary conditions are crucial for finding specific solutions to PDEs. They provide the necessary constraints that a solution must satisfy. In this problem, we have two boundary conditions:

• \( u(0, t) = u(2\pi, t) \): This means that the function \(u(x,t)\) should be the same at the spatial boundaries \(x=0\) and \(x=2\pi\).
• \( u_x(0, t) = u_x(2\pi, t) \): The spatial derivative of \(u\), denoted as \(u_x\), must also match at these boundaries.

These conditions imply that the wave is continuous and smooth across the spatial boundary. By integrating these conditions into our solution, such as in the step where we set \(X(0) = X(2\pi)\) and \(X'(0) = X'(2\pi)\), we can ensure that the general solution respects these constraints, leading to meaningful physical interpretations.

Separation of Variables

Separation of variables is a neat technique to tackle PDEs, especially when faced with problems like the wave equation. This approach assumes that the solution can be written as a product of single-variable functions. For the wave equation, we tried:

• \( u(x, t) = X(x)T(t) \)

By doing this, we divide the complex PDE into simpler ordinary differential equations (ODEs): one for time \(T(t)\) and one for space \(X(x)\). This simplification is feasible due to the structure of the wave equation, where derivatives with respect to different variables appear separately.The step of multiplying both sides by \(\frac{1}{a^2 X(x) T(t)}\) helped us achieve independent equations for \(T(t)\) and \(X(x)\). This method shows the power of exploring solutions in segmented parts, enabling simplified calculations and more manageable solutions.

Ordinary Differential Equations (ODEs)

Once we separate variables in the PDE, we derive ordinary differential equations which deal with functions of a single variable. In this exercise, two ODEs come from separation:

• The spatial ODE: \( \frac{d^2X(x)}{dx^2} + \lambda^2X(x) = 0 \)
• The temporal ODE: \( \frac{d^2T(t)}{dt^2} + a^2\lambda^2T(t) = 0 \)

Both ODEs can be solved using standard techniques, yielding sinusoidal solutions because of their forms. The spatial ODE results give solutions like:

• \( X(x) = A\cos(\lambda x) + B\sin(\lambda x) \)

For the time component, we find:

• \( T(t) = C\cos(a\lambda t) + D\sin(a\lambda t) \)

These solutions blend constants \(A, B, C,\) and \(D\), which we adjust using boundary and initial conditions, painting a full picture of behavior across both time and space dimensions.