Solving the wave equation with piecewise initial conditions

[songoku]

Homework Statement

Given that u(x,t) is displacement of string vibration, find the solution of the following wave equation based on the initial and boundary conditions:
$u_{tt}=u_{xx}, 0<x<2$
$u(0)=u(2)=0$
$u(x,0)= \begin{cases} 0.1x & \text{if } 0<x<1 \\ 0.1(2-x) & \text{if } 1\leq x<2 \end{cases}$
$u_{t}(x,0)=0$

Relevant Equations

D’Alembert’s Formula

I have solved the question using formula:
$$u(x,t)=\Sigma_{n=1}^{\infty} \sin \left(\frac{n \pi x}{2}\right) \left[a_n \frac{2}{n \pi} \sin \left(\frac{n \pi t}{2}\right)+b_n \cos \left(\frac{n \pi t}{2}\right)\right]$$

and I get:
$$u(x,t)=\Sigma_{n=1}^{\infty} \frac{0.8}{n^2 \pi^2}\sin \left(\frac{n \pi}{2}\right) \cos \left(\frac{n \pi t}{2}\right) \sin \left(\frac{n \pi x}{2}\right)$$

as the solution.

I want to try using d’Alembert’s Formula: $u(x,t)=A(x-t)+B(x+t)$ where
$$A(y)=\frac{1}{2} \left(F(y)-\int_{0}^{y} G(x) dx \right)$$
$$B(y)=\frac{1}{2} \left(F(y)+\int_{0}^{y} G(x) dx \right)$$
and $F$ and $G$ are and odd periodic extensions of $f(x)$ and $g(x)$ ($f(x)=u(x,0)$ and $g(x)=u_{t}(x,0)$)

For this question, $g(x)=0$ so $G(x)$ is also 0. For $F(y)$:
$$F(y)=
\begin{cases}
-0.1(2+y) &, -2<y<-1 \\
0.1y & , -1<y<1 \\
0.1(2-y) & , 1\leq y<2
\end{cases}$$

$$A(x-t)=\frac{1}{2} F(x-t)=
\begin{cases}
-0.1(2+x-t) &, -2<x<-1 \\
0.1(x-t) & , -1<x<1 \\
0.1(2-x+t) & , 1\leq x<2
\end{cases}
$$

$$B(x+t)=\frac{1}{2} F(x+t)=
\begin{cases}
-0.1(2+x+t) &, -2<x<-1 \\
0.1(x+t) & , -1<x<1 \\
0.1(2-x-t) & , 1\leq x<2
\end{cases}
$$

So:
$$u(x,t)=A(x-t)+B(x+t)$$
$$=
\begin{cases}
-0.2(2+x) &, -2<x<-1 \\
0.2x & , -1<x<1 \\
0.2(2-x) & , 1\leq x<2
\end{cases}
$$

Is this even correct? Is it possible to get the same solution in sigma form using D’Alembert’s Formula?

Thanks

 

[Orodruin]

That doesn't look like a proper periodic extension of $f$ at all. Also, $F$ has conditions on its argument, which is $x - t$ or $x+t$, not $x$.

It is easier to just describe $F$ on one full period of $F$ and then just refer to it as periodic with that period.

 

[songoku]

That doesn't look like a proper periodic extension of $f$ at all. Also, $F$ has conditions on its argument, which is $x - t$ or $x+t$, not $x$.

It is easier to just describe $F$ on one full period of $F$ and then just refer to it as periodic with that period.

I am sorry I don't understand the hint.

This is what I thought: F should be periodic in interval (-2, 2). The initial graph of $f(x)$ has the shape of triangle and the reflection about origin will produce inverted triangle for interval (-2, 0). This is the graph with the corresponding equation for each interval:

Is that not the odd periodic extension of $f(x)$?

Thanks

 

[Orodruin]

I am sorry I don't understand the hint.

This is what I thought: F should be periodic in interval (-2, 2). The initial graph of $f(x)$ has the shape of triangle and the reflection about origin will produce inverted triangle for interval (-2, 0). This is the graph with the corresponding equation for each interval:

View attachment 359404

Is that not the odd periodic extension of $f(x)$?

Thanks

A periodic function repeats once every period forever. Your function is only defined on (-2,2).

 

[songoku]

A periodic function repeats once every period forever. Your function is only defined on (-2,2).

Do you mean I need to use Fourier sine series to represent the periodic function to cover every period forever?

 

[Orodruin]

Do you mean I need to use Fourier sine series to represent the periodic function to cover every period forever?

No. You need to somehow specify that your function has a period of 4.

 

[songoku]

No. You need to somehow specify that your function has a period of 4.

$$F(x)=
\begin{cases}
-0.1(2+x) &, -2+4t<x\leq -1+4t \\
0.1x & , -1+4t<x\leq 1+4t \\
0.1(2-x) & , 1+4t< x\leq 2+4t
\end{cases}$$

where $t \in \mathbb Z$

Is this what you mean?

Thanks

 

[pasmith]

$$F(x)=
\begin{cases}
-0.1(2+x) &, -2+4t<x\leq -1+4t \\
0.1x & , -1+4t<x\leq 1+4t \\
0.1(2-x) & , 1+4t< x\leq 2+4t
\end{cases}$$

where $t \in \mathbb Z$

Is this what you mean?

Thanks

That doesn't work; adding 4 to x adds -0.4 to F(x) in the first and third cases and 0.4 in the second case. The function is therefore not periodic.

The idea is to say that if 4n \leq x &lt; 4n+1 (don't use t for this purpose; the question has already defined that to mean something else) then F(x) = F_0(x - 4n) where <br /> F_0 : [0,4) \to \mathbb{R} : x \mapsto \begin{cases}<br /> 0.1x &amp; 0 \leq x &lt; 1 \\<br /> 0.1(2 - x) &amp; 0 \leq x &lt; 3 \\<br /> 0.1(x-4) &amp; 3 \leq x &lt; 4 \end{cases} or perhaps that if 4n - 1 \leq x &lt; 4n+3 then F(x) = F_1(x - 4n) where <br /> F_1 : [-1,3) \to \mathbb{R}: x \mapsto \begin{cases}<br /> 0.1x &amp; -1 \leq x &lt; 1 \\<br /> 0.1(2-x) &amp; 1 \leq x &lt; 3 \end{cases}

 

[songoku]

That doesn't work; adding 4 to x adds -0.4 to F(x) in the first and third cases and 0.4 in the second case. The function is therefore not periodic.

For a function to be periodic, each case of the piecewise function should increase by same amount if we add the period to the function?

or perhaps that if 4n - 1 \leq x &lt; 4n+3 then F(x) = F_1(x - 4n) where <br /> F_1 : [-1,3) \to \mathbb{R}: x \mapsto \begin{cases}<br /> 0.1x &amp; -1 \leq x &lt; 1 \\<br /> 0.1(2-x) &amp; 1 \leq x &lt; 3 \end{cases}

This is $F_1 (x)$ so:
$$F_1(x - 4n)=
\begin{cases}
0.1(x-4n) & -1 \leq x-4n < 1 \\
0.1(2-(x-4n)) & 1 \leq x-4n < 3 \end{cases}
$$

Do I interpret that correctly? Thanks

 

[WWGD]

For a function to be periodic, each case of the piecewise function should increase by same amount if we add the period to the function?

This is $F_1 (x)$ so:
$$F_1(x - 4n)=
\begin{cases}
0.1(x-4n) & -1 \leq x-4n < 1 \\
0.1(2-(x-4n)) & 1 \leq x-4n < 3 \end{cases}
$$

Do I interpret that correctly? Thanks

No, f(x) is periodic with period p if for k Integer, f(x)=f(x+kp).

 

[Gavran]

The Fourier solution in the original post is correct and it is a periodic function with a period of $4$.
The d’Alembert solution is correct too, but only for $t=0$ and the multiplication by $1/2$ should be included.
Intervals in the expression for $A(x-t)$ should be changed into $\begin{cases}...-2\lt x-t\le -1\\...-1\lt x-t\lt 1\\...1\le x-t\lt 2\end{cases}$ and the multiplication by $1/2$ should be included and intervals in the expression for $B(x+t)$ should be changed into $\begin{cases}...-2\lt x+t\le -1\\...-1\lt x+t\lt 1\\...1\le x+t\lt 2\end{cases}$ and the multiplication by $1/2$ should be included.

There is a connection between two solutions. See https://www.math.purdue.edu/~eremenko/dvi/dalembert2.pdf.

$$F(y)=
\begin{cases}
-0.1(2+y) &, -2<y<-1 \\
0.1y & , -1<y<1 \\
0.1(2-y) & , 1\leq y<2
\end{cases}$$

You can add that $F(x+4n)=F(x)$ where $n$ is an integer to make $F(x)$ a periodic function with a period of $4$. The same holds for $A(x)$, $B(x)$ and the final d’Alembert solution in the original post.

 

[songoku]

I am really sorry for late reply

The Fourier solution in the original post is correct and it is a periodic function with a period of $4$.
The d’Alembert solution is correct too, but only for $t=0$ and the multiplication by $1/2$ should be included.
Intervals in the expression for $A(x-t)$ should be changed into $\begin{cases}...-2\lt x-t\le -1\\...-1\lt x-t\lt 1\\...1\le x-t\lt 2\end{cases}$ and the multiplication by $1/2$ should be included and intervals in the expression for $B(x+t)$ should be changed into $\begin{cases}...-2\lt x+t\le -1\\...-1\lt x+t\lt 1\\...1\le x+t\lt 2\end{cases}$ and the multiplication by $1/2$ should be included.

There is a connection between two solutions. See https://www.math.purdue.edu/~eremenko/dvi/dalembert2.pdf.


You can add that $F(x+4n)=F(x)$ where $n$ is an integer to make $F(x)$ a periodic function with a period of $4$. The same holds for $A(x)$, $B(x)$ and the final d’Alembert solution in the original post.

I think I get this.

Thank you very much for all the help and explanation Orodruin, pasmith, WWGD, Gavran