9.6. One-Dimensional Wave Equation

Example 1

Solve the wave equation

subject to boundary conditions

and initial conditions

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Solution:

Step 1: Expand the initial condition

Use the identity

Hence,

Step 2: General solution form

Step 3: Apply initial velocity condition

Step 4: Match initial displacement

Comparing coefficients gives:

Final solution:

Example 2: Sudden Impact on a String

A guitar string lies across the back of a pickup truck. At time \(t = 0\), the truck suddenly slams into a brick wall with speed \(v_0\), producing an instantaneous uniform vertical velocity along the entire string. Determine the resulting displacement \(y(x,t)\) of the string for \(0 < x < L,\; t > 0\), assuming both ends are fixed.

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Solution:

The wave equation is

with boundary conditions

and initial conditions

Using separation of variables, we seek a Fourier sine series solution of the form

Differentiating with respect to \(t\) and applying the initial velocity condition gives

Hence the Fourier sine coefficients are

Evaluating the integral gives

Final Answer:

Example 3

Solve the following boundary value problem:

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Solution:

Step 1: The wave equation is \(y_{tt} = a^2 y_{xx}\). Given \(16y_{tt} = y_{xx}\), we rewrite it as:

Step 2: The general solution for a string of length \(L = \pi\) fixed at both ends is:

Step 3: Apply the initial position condition \(y(x,0) = \dfrac{1}{3}\sin x\). By substituting \(t=0\), the sum collapses to a single term:

Step 4: Apply the initial velocity condition \(y_t(x,0) = \dfrac{1}{3}\sin x\). Differentiating the general solution with respect to \(t\) and setting \(t=0\):

Final Answer:

Example 4

Solve the following boundary value problem:

Final Answer:

Example 5

Solve the following boundary value problem:

Final Answer:

Example 6

Solve the following boundary value problem:

Final Answer:

Example 7

The solution to the wave equation

is given by

for constants \(c_n\) and \(d_n\), \(n \geq 1\). Find \(c_n\).

1. \[ \boxed{\displaystyle \frac{2}{n\pi}\int_0^5 (3-x) \sin \left( \frac{n\pi x}{5} \right)\,dx} \]
2. \[ \frac{2}{5}\int_0^5 (2x) \sin \left( \frac{n\pi x}{5} \right)\,dx \]
3. \[ \frac{2}{n\pi}\int_0^5 (2x) \sin \left( \frac{n\pi x}{5} \right)\,dx \]
4. \[ \frac{2}{5}\int_0^5 (3-x) \sin \left( \frac{n\pi x}{5} \right)\,dx \]
5. \[ \frac{2}{n\pi}\int_0^5 (3-x) \cos \left( \frac{n\pi x}{5} \right)\,dx \]

Example 8: Wave Equation with Mixed Boundary Conditions

Consider a stretched string of length \(L\), initially at rest. The displacement \(y(x,t)\) satisfies the wave equation with one fixed end and one free (sliding) end:

The boundary conditions are:

and the initial conditions are:

Use the method of separation of variables to derive the solution

We solve this using separation of variables.

1. Separation of variables

Assume a product solution \(y(x,t)=X(x)T(t)\). Substituting into the PDE gives:

Divide by \(a^2XT\):

This leads to two ODEs:

2. Spatial eigenvalue problem

We consider different cases for \(\lambda\):

Case 1: \(\lambda = 0\)

Apply boundary conditions: \(X(0)=0 \Rightarrow A=0\), and \(X'(L)=0 \Rightarrow B=0\). Hence only the trivial solution \(X \equiv 0\).

Case 2: \(\lambda < 0\)

Let \(\lambda = -\mu^2\), where \(\mu > 0\). Then:

Applying the boundary conditions again forces \(A=B=0\), giving only the trivial solution.

Case 3: \(\lambda > 0\)

Let \(\lambda=k^2\). Then:

General solution:

Apply boundary conditions:

At \(x=0\):

So:

At \(x=L\):

For nontrivial solutions:

Hence the eigenfunctions are:

3. Time equation

For each \(n\), the time equation becomes:

The general solution is:

Using \(y_t(x,0)=0\), we get \(D_n=0\).

4. Full solution

Hence each solution (building block) is:

By superposition, the general solution is obtained by summing over all odd integers \(n\):

5. Coefficients

From \(y(x,0)=f(x)\), we obtain:

This is a Fourier sine series expansion of \(f(x)\) on \((0,L)\). Hence

Example 9:

A string of length \(\pi\) is fixed at the left end and the right end is allowed to move vertically but not horizontally, so

The tension and density are such that \(a^2=1\), giving the wave equation:

The string has initial displacement \(f(x)=x\), \(0 \le x \le \pi\), and no initial velocity. The solution can be written as:

where \(c_n\) are constants determined by the initial condition. Find \(T_n(t)\).

1. \[ \boxed{T_n(t)=\cos\left(\frac{2n-1}{2}t\right)} \]
2. \[ T_n(t)=\sinh\left(\frac{2n-1}{2}t\right) \]
3. \[ T_n(t)=\sin\left(\frac{2n-1}{2}t\right) \]
4. \[ T_n(t)=\cos(nt) \]
5. \[ T_n(t)=\sin(nt) \]

Example 10: The d’Alembert solution (an alternate method)

Part 1: Suppose that the function \(F\) is twice differentiable for all \(x\). Use the chain rule to verify that \(F(x+at)\) and \(F(x-at)\) are solutions to the wave equation

Part 2: Let \(F\) be the odd periodic extension of the initial displacement \(f(x)\) with period \(2L\). Verify that the solution

satisfies the boundary conditions the initial conditions

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Solution:

Part 1: Verification via Chain Rule

Let \(u = x+at\). By the chain rule:

Substituting into the wave equation gives \(a^2F''(u) = a^2(F''(u))\), confirming it is a solution. The verification for \(F(x-at)\) follows the same logic.

Part 2: Verification of Conditions

Given

• Boundary Conditions: Due to odd symmetry
  \[ y(0,t) = \frac{1}{2}[F(-at) + F(at)] = 0 \]
  and similarly for \(x=L\).
• Initial Displacement:
  \[ y(x,0) = \frac{1}{2}[F(x) + F(x)] = F(x) = f(x), \quad \forall x \in (0,L). \]
• Initial Velocity:
  \[y_t(x,t) = \frac{1}{2}[-aF'(x-at) + aF'(x+at)\]
  At \(t=0\),
  \[y_t(x,0) = \frac{1}{2}[-aF'(x) + aF'(x)] = 0\]

Example 11: The d’Alembert solution

A vibrating string of length \(L=\pi\) is fixed at both ends and released from rest at time \(t=0\). Its initial displacement is given by

The motion of the string satisfies the wave equation

with boundary conditions \(y(0,t)=y(\pi,t)=0\), and zero initial velocity: \(y_t(x,0)=0\).

1. Find the odd periodic extension \(F(x)\) of \(f(x)\) with period \(2\pi\).
2. Use the d’Alembert’s formula to determine the displacement \(y(x,t)\) of the string.
3. Evaluate and graph the displacement \(y(x,t)\) at \(t=\frac{\pi}{4}\).

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Solution:

1. Odd Periodic Extension \(F(x)\)

To satisfy the boundary conditions \(y(0,t)=y(\pi,t)=0\), we extend \(f(x)=1-\cos(2x)\) as an odd function, then periodically with period \(2\pi\). For \(x \in [-\pi, 0]\), we use the odd property \(F(x) = -f(-x)\).

2. The d’Alembert’s Formula

With wave speed \(a=1\) and zero initial velocity \(y_t(x,0)=0\), the general solution simplifies to the superposition of two traveling waves:

3. Evaluation and Graphing at \(t=\frac{\pi}{4}\)

Substituting \(t=\frac{\pi}{4}\) into the formula gives:

The first term, \(\frac{1}{2}F(x-\frac{\pi}{4})\), shifts the wave to the right, and the second term, \(\frac{1}{2}F(x+\frac{\pi}{4})\), shifts it to the left. The total displacement is the sum of these two components.

Practice Problems

Solve the following boundary value problems:

1. \[ y_{tt} = 4y_{xx}, \quad 0 < x < \pi, \quad t > 0; \quad y(0,t) = y(\pi,t) = 0, \quad y(x,0) = \frac{1}{10} \sin 2x, \quad y_t(x,0) = 0 \]
2. \[ y_{tt} = y_{xx}, \quad 0 < x < 1, \quad t > 0; \quad y(0,t) = y(1,t) = 0, \quad y(x,0) = \frac{1}{10} \sin \pi x - \frac{1}{20} \sin 3\pi x, \quad y_t(x,0) = 0 \]
3. \[ 4y_{tt} = y_{xx}, \quad 0 < x < \pi, \quad t > 0; \quad y(0,t) = y(\pi,t) = 0, \quad y(x,0) = y_t(x,0) = \frac{1}{10} \sin x \]
4. \[ 4y_{tt} = y_{xx}, \quad 0 < x < 2, \quad t > 0; \quad y(0,t) = y(2,t) = 0, \quad y(x,0) = \frac{1}{5} \sin \pi x \cos \pi x, \quad y_t(x,0) = 0 \]
5. \[ y_{tt} = 25y_{xx}, \quad 0 < x < 3, \quad t > 0; \quad y(0,t) = y(3,t) = 0, \quad y(x,0) = \frac{1}{4} \sin \pi x, \quad y_t(x,0) = 10 \sin 2\pi x \]
6. \[ y_{tt} = 100y_{xx}, \quad 0 < x < \pi, \quad t > 0; \quad y(0,t) = y(\pi,t) = 0, \quad y(x,0) = x(\pi - x), \quad y_t(x,0) = 0 \]
7. \[ y_{tt} = 100y_{xx}, \quad 0 < x < 1, \quad t > 0; \quad y(0,t) = y(1,t) = 0, \quad y(x,0) = 0, \quad y_t(x,0) = x \]
8. \[ y_{tt} = 4y_{xx}, \quad 0 < x < 1, \quad t > 0; \quad y(0,t) = y(1,t) = 0, \quad y(x,0) = 0, \quad y_t(x,0) = x(1 - x) \]