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Problem of the Week #65 - June 24th, 2013

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Chris L T521

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Jan 26, 2012
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Thanks again to those who participated in last week's POTW! Here's this week's problem!

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Background Info: From PDEs, d'Alembert's solution to the boundary value problem of the vibrating string
\[\left\{\begin{array}{l} \frac{\partial^2 u}{\partial t^2}=c^2\frac{\partial^2u}{\partial x^2},\qquad 0<x<L,\,\, t>0\\ u(0,t)=0\text{ and }u(L,t)=0\text{ for all $t>0$}\\ u(x,0)=f(x)\text{ and }u_t(x,0)=g(x)\text{ for $0<x<L$}\end{array}\right.\]
is given by
\[u(x,t)=\frac{1}{2}\left[f^{\ast}(x-ct)+f^{\ast}(x+ct)\right]+\frac{1}{2c}\int_{x-ct}^{x+ct} g^{\ast}(s)\,ds,\]
where $f^{\ast}$ and $g^{\ast}$ are the odd extensions of $f$ and $g$ respectively.

Problem: Suppose that both $f$ and $g$ are symmetric about $x=\frac{L}{2}$; that is, $f(L-x)=f(x)$ and $g(L-x)=g(x)$. Show that
\[u\left(x,t+\frac{L}{c}\right)=-u(x,t)\]
for all $0<x<L$ and $t>0$.

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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Chris L T521

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Jan 26, 2012
995
No one answered this week's question. You can find my solution below.


Proof: We note here that $f^{\ast}$ and $g^{\ast}$ are both odd $2L$-periodic functions and also satisfy the symmetry relations $f^{\ast}(L-x)=f^{\ast}(x)$ and $g^{\ast}(L-x)=g^{\ast}(x)$ (which they inherit from $f$ and $g$). With this, we see that
\[\begin{aligned} u\left(x,t+\frac{L}{c}\right) &= \frac{1}{2}\left[f^{\ast} \left(x-c\left(t+\frac{L}{c}\right)\right) + f^{\ast}\left(x+c\left(t+\frac{L}{c}\right)\right)\right] + \frac{1}{2c}\int_{x-c\left(t+\frac{L}{c}\right)}^{ x+c\left(t+\frac{L}{c}\right)} g^{\ast}(s)\,ds\\ &= \frac{1}{2}\left[ f^{\ast}(-(L-(x-ct))) + f^{\ast}(L-(-x-ct))\right] + \frac{1}{2c}\int_{x-ct-L}^{x+ct+L} g^{\ast}(s)\,ds\\ &= \frac{1}{2} \left[ -f^{\ast}(L-(x-ct)) + f^{\ast}(L - (-x - ct)) \right] + \frac{1}{2c} \int_{x-ct-L}^{x+ct+L} g^{\ast}(s)\,ds\\ &= \frac{1}{2} \left[ -f^{\ast}(x-ct) + f^{\ast}(-x-ct)\right] + \frac{1}{2c}\int_{x-ct-L}^{x+ct+L} g^{\ast}(s)\,ds\\
&= \frac{1}{2}\left[ -f^{\ast}(x-ct)- f^{\ast}(x+ct)\right] + \frac{1}{2c}\int_{x-ct}^{x+ct+2L} g^{\ast}(s+L)\,ds\\ &= -\frac{1}{2}\left[ f^{\ast}(x-ct) + f^{\ast}(x+ct)\right] +\frac{1}{2c} \int_{x-ct}^{x+ct} -g^{\ast}(-s-L)\,ds\\ &= -\frac{1}{2} \left[f^{\ast}(x+ct)+ f^{\ast}(x-ct)\right] - \frac{1}{2c}\int_{x-ct}^{x+ct} g^{\ast}(-s-L+2L)\,ds\\ &= -\frac{1}{2}\left[f^{\ast}(x+ct)+f^{\ast} (x-ct) \right] - \frac{1}{2c}\int_{x-ct}^{x+ct} g^{\ast}(L-s)\,ds\\ &= -\left(\frac{1}{2}\left[ f^{\ast}(x+ct)+ f^{\ast}(x-ct)\right] + \frac{1}{2c}\int_{x-ct}^{x+ct} g^{\ast}(s)\,ds\right) \\ &= -u(x,t),\end{aligned}\]
Thus, $u\left(x,t+\dfrac{L}{c}\right)= -u(x,t)$.$\hspace{.25in}\blacksquare$
 
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