Solving 1-D Wave Equation w/ Fixed Boundaries

R and S are arbitrary functions.In summary, the problem is to solve the one-dimensional wave equation with isotropic, homogeneous conditions and fixed boundary conditions. The solution can be written as a superposition of sine and cosine functions, but the goal is to find a functional form using d'Alembert's solution. To do this, a bijective coordinate change is made, and after some calculations, it is shown that the solution can be written as u(x,t) = S(x+ct) + R(x-ct), where R and S are arbitrary functions.
  • #1
Mindscrape
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[SOLVED] 1-D Wave Eqn

Alright, so this problem is giving me troubles, and I must just be missing the trick. The equation to solve is the one dimensional wave equation with isotropic, homogeneous, etc. (i.e. wave in a vacuum). Which means the PDE is

[tex]\frac{\partial^2 u}{\partial t^2} = c^2 \frac{\partial^2 u}{\partial x^2}[/tex]

with the boundary conditions that each end is fixed at zero, [itex]u(0,t) = 0[/itex] and [itex]u(L, t)[/itex]. I also know the initial conditions for the initial position and initial velocity are functions of x [itex] u(x,0) = f(x)[/itex] and [itex]\frac{\partial u}{\partial t}(x, 0) = g(x) [/itex]. My job is to show that the solution follows d'Alembert's solution (which I can prove), that the wave equation can be written as [itex]u(x,t) = R(x-ct) + S(x+ct)[/itex] where R and S are some functions.

The solution is easily found

[tex]u(x,t) = \sum_{n=1}^\infty \left(A_n sin\frac{n \pi x}{L} cos\frac{n \pi c t}{L} + B_n sin\frac{n \pi x}{L} sin\frac{n \pi c t}{L} \right)[/tex]

Mostly what I am having trouble with is getting rid of the summations. The first initial condition is helpful because

[tex] \sum_{n=1}^\infty A_n sin \frac{n \pi x}{L} = f(x)[/tex]

which can be directly substituted in (though I don't know if I want or need to). The other IC is not so helpful

[tex] \sum_{n=1}^\infty \frac{n \pi c}{L} B_n sin\frac{n \pi x}{L} = g(x)[/tex]

and I cannot really do much other. The other thing I know I have to do is use trig identities, which will rewrite the solution as

[tex]u(x,t) = \sum \left(\frac{A_n}{2} [sin \frac{n \pi}{L} (x + ct) + sin \frac{n \pi}{L} (x - ct)] + \frac{B_n}{2}[cos \frac{n \pi}{L} (x - ct) - cos\frac{n \pi}{L} (x + ct)] \right)[/tex]

This starts to towards making some functions, and I converted this all into exponentials to get (after some grouping)

[tex] u(x,t) = \sum (\frac{A'_n}{2i} - \frac{B'_n}{2}) Exp(i\frac{n\pi}{L}(x+ct)) + (-\frac{A'_n}{2i} - \frac{B'_n}{2}) Exp(-i\frac{n\pi}{L}(x+ct)) + (\frac{A'_n}{2i} + \frac{B'_n}{2}) Exp(i\frac{n\pi}{L}(x-ct)) + (-\frac{A'_n}{2i} + \frac{B'_n}{2}) Exp(-i\frac{n\pi}{L}(x-ct))[/tex]

Where [itex]A'_n = A_n/2[/itex] and [itex]B'_n = B_n/2[/itex].

Unfortunately, I can't rewrite these back into sines and cosines as two functions. Does someone know the trick, specifically to this problem, or in general, that gets from the superposed answer to the functional answer?
 
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  • #2
Ok

make a bijective coordinate change

[tex]
\begin{gathered}
\xi \left( {x,t} \right) \hfill \\
\psi (x,t) \hfill \\
\end{gathered}
[/tex]

Using the chain rule for differentiation you get

[tex]
\begin{gathered}
\frac{{\partial ^2 u}}
{{\partial t^2 }} = \frac{{\partial ^2 u}}
{{\partial \psi ^2 }}\left( {\frac{{\partial \psi }}
{{\partial t}}} \right)^2 + 2\frac{{\partial ^2 u}}
{{\partial \xi \partial \psi }}\frac{{\partial \xi }}
{{\partial t}}\frac{{\partial \psi }}
{{\partial t}} + \frac{{\partial u}}
{{\partial \psi }}\frac{{\partial ^2 \psi }}
{{\partial t^2 }} \hfill \\
+ \frac{{\partial ^2 u}}
{{\partial \xi ^2 }}\left( {\frac{{\partial \xi }}
{{\partial t}}} \right)^2 + \frac{{\partial u}}
{{\partial \xi }}\frac{{\partial ^2 \xi }}
{{\partial t^2 }} \hfill \\
\end{gathered}
[/tex]

do the same thing for

[tex]
\frac{{\partial ^2 u}}
{{\partial x^2 }}
[/tex]

plug all this crap into the wave equation, and bring everything to the same side, and you get.

[tex]
\begin{gathered}
\left( {\left( {\frac{{\partial \psi }}
{{\partial t}}} \right)^2 - c^2 \left( {\frac{{\partial \psi }}
{{\partial x}}} \right)^2 } \right)\frac{{\partial ^2 u}}
{{\partial \psi ^2 }} + 2\frac{{\partial ^2 u}}
{{\partial \xi \partial \psi }}\left( {\frac{{\partial \xi }}
{{\partial t}}\frac{{\partial \psi }}
{{\partial t}} - c^2 \frac{{\partial \xi }}
{{\partial x}}\frac{{\partial \psi }}
{{\partial x}}} \right) \hfill \\
\left( {\left( {\frac{{\partial \xi }}
{{\partial t}}} \right)^2 - c^2 \left( {\frac{{\partial \xi }}
{{\partial x}}} \right)^2 } \right)\frac{{\partial ^2 u}}
{{\partial \xi ^2 }} + \left( {\frac{{\partial ^2 \psi }}
{{\partial t^2 }} - c^2 \frac{{\partial ^2 \psi }}
{{\partial x^2 }}} \right)\frac{{\partial u}}
{{\partial \psi }} + \left( {\frac{{\partial ^2 \xi }}
{{\partial t^2 }} - c^2 \frac{{\partial ^2 \xi }}
{{\partial x^2 }}} \right)\frac{{\partial u}}
{{\partial \xi }} = 0 \hfill \\
\end{gathered}
[/tex]

Now, you can pick [tex]\xi ,\psi[/tex] to be whatever you want as long as it's a bijective map. Let's pick:

[tex]
\begin{gathered}
\psi = x + ct \hfill \\
\xi = x - ct \hfill \\
\end{gathered}
[/tex]

This simplifies everything to [tex]
- 4c^2 \frac{{\partial ^2 u}}
{{\partial \psi \partial \xi }} = 0
[/tex]

so

[tex]
\frac{{\partial ^2 u}}
{{\partial \psi \partial \xi }} = 0
[/tex]

This is easy enough to integrate

namely:

[tex]
\begin{gathered}
\frac{{\partial u}}
{{\partial \psi }} = \frac{{\partial S(\psi )}}
{{\partial \psi }} \hfill \\
u\left( {\xi ,\psi } \right) = S\left( \psi \right) + R\left( \xi \right) \hfill \\
\therefore u\left( {x,t} \right) = S\left( {x + ct} \right) + R\left( {x - ct} \right) \hfill \\
\end{gathered}
[/tex]
 

1. What is the 1-D wave equation with fixed boundaries?

The 1-D wave equation with fixed boundaries is a mathematical model used to describe the behavior of a wave in one dimension. It takes into account factors such as time, displacement, and the speed of the wave.

2. How is the 1-D wave equation with fixed boundaries solved?

The equation can be solved using various methods, such as separation of variables, Fourier series, or the method of characteristics. These methods involve breaking down the equation into simpler equations and then finding solutions for each part.

3. What are the fixed boundaries in the 1-D wave equation?

The fixed boundaries refer to the points at which the wave is constrained to stay within. These boundaries can be physical, such as a string or a rod, or they can be mathematical, such as a finite interval on a graph.

4. What are some real-life applications of the 1-D wave equation with fixed boundaries?

The 1-D wave equation is used in various fields, such as physics, engineering, and acoustics, to model and understand the behavior of waves in different systems. Some examples include analyzing vibrations in buildings and bridges, studying sound waves in musical instruments, and predicting the behavior of seismic waves during an earthquake.

5. What are some limitations of the 1-D wave equation with fixed boundaries?

While the 1-D wave equation is a useful tool for understanding wave behavior, it has some limitations. It assumes that the wave is traveling in one direction and that the medium through which the wave is traveling is homogeneous and isotropic. In real-life situations, these assumptions may not always hold, leading to discrepancies between the predicted and actual behavior of waves.

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