Question:
Let \(u(x,t)\) satisfy the wave equation
\[ \frac{\partial^2 u}{\partial t^2} = c^2 \frac{\partial^2 u}{\partial x^2}, \qquad -\infty < x < \infty, \; t > 0, \; c > 0 \] with
\[ u(x,0) = \begin{cases} 1 & \text{if } |x| < 1 \\ 0 & \text{otherwise} \end{cases} \] and
\[ \frac{\partial u}{\partial t}(x,0) = 0. \] By using D'Alembert's formula, the maximum value of \(u(0,t)\) for \(t > 0\) is
Let \(u(x,t)\) satisfy the wave equation
\[ \frac{\partial^2 u}{\partial t^2} = c^2 \frac{\partial^2 u}{\partial x^2}, \qquad -\infty < x < \infty, \; t > 0, \; c > 0 \] with
\[ u(x,0) = \begin{cases} 1 & \text{if } |x| < 1 \\ 0 & \text{otherwise} \end{cases} \] and
\[ \frac{\partial u}{\partial t}(x,0) = 0. \] By using D'Alembert's formula, the maximum value of \(u(0,t)\) for \(t > 0\) is
\[ \frac{\partial^2 u}{\partial t^2} = c^2 \frac{\partial^2 u}{\partial x^2}, \qquad -\infty < x < \infty, \; t > 0, \; c > 0 \] with
\[ u(x,0) = \begin{cases} 1 & \text{if } |x| < 1 \\ 0 & \text{otherwise} \end{cases} \] and
\[ \frac{\partial u}{\partial t}(x,0) = 0. \] By using D'Alembert's formula, the maximum value of \(u(0,t)\) for \(t > 0\) is
Show Hint
Use D'Alembert's formula with zero initial velocity: u(0,t) = f(ct) since f is even, and f is 1 whenever ct is less than 1.
Updated On: Jul 21, 2026
- \(1\)
- \(\dfrac{1}{2}\)
- \(\dfrac{1}{4}\)
- \(0\)
Show Solution
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The Correct Option is A
Solution and Explanation
Step 1: Write down D'Alembert's formula.
For the 1D wave equation \(u_{tt} = c^2 u_{xx}\) on the whole real line with initial position \(u(x,0) = f(x)\) and initial velocity \(u_t(x,0) = g(x)\), D'Alembert's formula gives
\[ u(x,t) = \frac{1}{2}\big[f(x-ct) + f(x+ct)\big] + \frac{1}{2c}\int_{x-ct}^{x+ct} g(s)\, ds. \]
Here \(f(x) = 1\) if \(|x| < 1\) and \(0\) otherwise, and \(g(x) = 0\) for every \(x\).
Step 2: Drop the velocity term.
Since \(g \equiv 0\), the integral term vanishes, so
\[ u(x,t) = \frac{1}{2}\big[f(x-ct) + f(x+ct)\big]. \]
Step 3: Plug in x = 0.
At \(x=0\),
\[ u(0,t) = \frac{1}{2}\big[f(-ct) + f(ct)\big]. \]
\(f\) only depends on \(|x|\) (it is \(1\) exactly when the argument is strictly between \(-1\) and \(1\)), so \(f(-ct) = f(ct)\) for every \(t\), and
\[ u(0,t) = \frac{1}{2}\big[f(ct)+f(ct)\big] = f(ct). \]
A common slip is to assume one of \(f(ct)\) or \(f(-ct)\) could be \(1\) while the other is \(0\), giving an average like \(1/2\); but since both terms are always equal here, that never happens at \(x=0\).
Step 4: Evaluate f(ct).
Since \(c > 0\) and \(t > 0\), \(ct > 0\), so \(f(ct) = 1\) exactly when \(ct < 1\), that is when \(0 < t < 1/c\), and \(f(ct) = 0\) once \(t \geq 1/c\). So
\[ u(0,t) = \begin{cases} 1 & 0 < t < 1/c \\ 0 & t \geq 1/c. \end{cases} \]
Final Answer:
For every \(t\) between \(0\) and \(1/c\), \(u(0,t)\) equals \(1\), and it never exceeds that, so the maximum value of \(u(0,t)\) for \(t>0\) is \(1\). \[ \boxed{1} \]
For the 1D wave equation \(u_{tt} = c^2 u_{xx}\) on the whole real line with initial position \(u(x,0) = f(x)\) and initial velocity \(u_t(x,0) = g(x)\), D'Alembert's formula gives
\[ u(x,t) = \frac{1}{2}\big[f(x-ct) + f(x+ct)\big] + \frac{1}{2c}\int_{x-ct}^{x+ct} g(s)\, ds. \]
Here \(f(x) = 1\) if \(|x| < 1\) and \(0\) otherwise, and \(g(x) = 0\) for every \(x\).
Step 2: Drop the velocity term.
Since \(g \equiv 0\), the integral term vanishes, so
\[ u(x,t) = \frac{1}{2}\big[f(x-ct) + f(x+ct)\big]. \]
Step 3: Plug in x = 0.
At \(x=0\),
\[ u(0,t) = \frac{1}{2}\big[f(-ct) + f(ct)\big]. \]
\(f\) only depends on \(|x|\) (it is \(1\) exactly when the argument is strictly between \(-1\) and \(1\)), so \(f(-ct) = f(ct)\) for every \(t\), and
\[ u(0,t) = \frac{1}{2}\big[f(ct)+f(ct)\big] = f(ct). \]
A common slip is to assume one of \(f(ct)\) or \(f(-ct)\) could be \(1\) while the other is \(0\), giving an average like \(1/2\); but since both terms are always equal here, that never happens at \(x=0\).
Step 4: Evaluate f(ct).
Since \(c > 0\) and \(t > 0\), \(ct > 0\), so \(f(ct) = 1\) exactly when \(ct < 1\), that is when \(0 < t < 1/c\), and \(f(ct) = 0\) once \(t \geq 1/c\). So
\[ u(0,t) = \begin{cases} 1 & 0 < t < 1/c \\ 0 & t \geq 1/c. \end{cases} \]
Final Answer:
For every \(t\) between \(0\) and \(1/c\), \(u(0,t)\) equals \(1\), and it never exceeds that, so the maximum value of \(u(0,t)\) for \(t>0\) is \(1\). \[ \boxed{1} \]
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