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

Question

Let $A$ be a square matrix. If $A^2 = A$, then the matrix $A$ is called:

Answer 1

The initial displacement here is a bump of height $1$ sitting between $x=-1$ and $x=1$, starting from rest (zero initial velocity). D'Alembert's solution splits this bump into two waves, each of half the original height, one moving right at speed $c$ and one moving left at speed $c$:

\[ u(x,t) = \frac{1}{2} f(x-ct) + \frac{1}{2} f(x+ct) \]

where $f$ is the indicator function that is $1$ on $(-1,1)$ and $0$ elsewhere.

The first piece, $\frac{1}{2} f(x-ct)$, is the original bump shrunk to half height and shifted right by $ct$. The second piece, $\frac{1}{2} f(x+ct)$, is the same half height bump shifted left by $ct$. At $t=0$ these two half height bumps sit exactly on top of each other over $(-1,1)$, adding up to the full height $1$, matching the given initial shape.

Now watch the point $x=0$ as $t$ increases from $0$. The right moving half bump has support at $x \in (ct-1, ct+1)$, and the left moving half bump has support at $x \in (-ct-1,-ct+1)$. As long as $0$ lies inside both supports, both terms contribute $\frac{1}{2}$ each at $x=0$, giving $u(0,t) = \frac{1}{2}+\frac{1}{2}=1$.

Check when $0$ is inside the right moving support $(ct-1,ct+1)$: this needs $ct-1<0<ct+1$, that is $-1<ct<1$, and since $ct>0$ this means $ct<1$, i.e. $t<1/c$. By symmetry the left moving bump gives the same condition. So for every $t$ with $0<t<1/c$, both half bumps still cover $x=0$, and $u(0,t)=1$ exactly. Once $t \geq 1/c$, both half bumps have moved past $x=0$ and $u(0,t)$ drops to $0$.

Let's summarize:

D'Alembert's solution splits a zero velocity bump into two half height copies moving apart at speed $c$.
At $x=0$, both half height copies overlap for $0<t<1/c$, adding to the full value $1$.
The value never exceeds $1$ at any time, so $1$ is the maximum.

So the maximum value of $u(0,t)$ for $t>0$ is $1$.

\[\boxed{u(0,t)_{max} = 1}\]

Answer 2

The initial displacement here is a bump of height $1$ sitting between $x=-1$ and $x=1$, starting from rest (zero initial velocity). D'Alembert's solution splits this bump into two waves, each of half the original height, one moving right at speed $c$ and one moving left at speed $c$:

\[ u(x,t) = \frac{1}{2} f(x-ct) + \frac{1}{2} f(x+ct) \]

where $f$ is the indicator function that is $1$ on $(-1,1)$ and $0$ elsewhere.

The first piece, $\frac{1}{2} f(x-ct)$, is the original bump shrunk to half height and shifted right by $ct$. The second piece, $\frac{1}{2} f(x+ct)$, is the same half height bump shifted left by $ct$. At $t=0$ these two half height bumps sit exactly on top of each other over $(-1,1)$, adding up to the full height $1$, matching the given initial shape.

Now watch the point $x=0$ as $t$ increases from $0$. The right moving half bump has support at $x \in (ct-1, ct+1)$, and the left moving half bump has support at $x \in (-ct-1,-ct+1)$. As long as $0$ lies inside both supports, both terms contribute $\frac{1}{2}$ each at $x=0$, giving $u(0,t) = \frac{1}{2}+\frac{1}{2}=1$.

Check when $0$ is inside the right moving support $(ct-1,ct+1)$: this needs $ct-1<0<ct+1$, that is $-1<ct<1$, and since $ct>0$ this means $ct<1$, i.e. $t<1/c$. By symmetry the left moving bump gives the same condition. So for every $t$ with $0<t<1/c$, both half bumps still cover $x=0$, and $u(0,t)=1$ exactly. Once $t \geq 1/c$, both half bumps have moved past $x=0$ and $u(0,t)$ drops to $0$.

Let's summarize:

D'Alembert's solution splits a zero velocity bump into two half height copies moving apart at speed $c$.
At $x=0$, both half height copies overlap for $0<t<1/c$, adding to the full value $1$.
The value never exceeds $1$ at any time, so $1$ is the maximum.

So the maximum value of $u(0,t)$ for $t>0$ is $1$.

\[\boxed{u(0,t)_{max} = 1}\]

Show Hint

The Correct Option is A

Solution and Explanation

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