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A vibrating string of length 1 in a resistant medium with fixed ends, linear initial displacement, and zero initial velocity is modeled by the following problem

$$\left\{ \begin{array}{l l} u_{tt} - c^2u_{xx} + ru_t = 0 & \quad \mbox{$0<x<1, t>0$} \\ \quad u(x,0) = \begin{cases} x & \textrm{ if $0\le x\le 1/2$} \\ 1-x & \textrm{ if $1/2\le x\le 1,$} \end{cases} \\ \quad u_t(x,0) = 0, \\ u(0,t) = u(1,t) = 0, \\ \end{array} \right. $$

where $r$ is a constant, and $0<r<2\pi c$. Use separation of variable to find a series solution.

I have left the condition for $u(x,0)$ blank because I'm not sure how to code into the problem another brace for the two initial conditions that it has which are $x$ if $0 \leq x \leq 1/2$ and $1-x$ if $1/2 \leq x \leq 1$.

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  • $\begingroup$ What is your initial condition for $u(x,0)$? The formula says zero, the text describes something else. $\endgroup$ – Joonas Ilmavirta Mar 26 '18 at 15:03
  • $\begingroup$ @JoonasIlmavirta Oh, it is not zero. I have corrected it. It is what is described in text. Will this change the explanation that you have given me? $\endgroup$ – KBG Mar 26 '18 at 18:47
  • $\begingroup$ My answer is still valid. The initial conditions for the PDE change the initial conditions for the ODEs. That's all. $\endgroup$ – Joonas Ilmavirta Mar 26 '18 at 18:57
  • $\begingroup$ You could try (copying and pasting) this for formatting the initial condition: $$u(x,0)=\begin{cases}x,&\text{ when }x<1/2\\1-x,&\text{ when }x\geq1/2\end{cases}$$ $\endgroup$ – Joonas Ilmavirta Mar 26 '18 at 19:03
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Since you are looking at a string between $x=0$ and $x=1$, it is convenient to write it as a Fourier series in $x$. The endpoints are attached, so we only get sine terms — if you want, you can also include cosine terms and later figure out why their coefficients vanish. The coefficients depend on $t$.

So, we write $$ u(x,t) = \sum_{k=1}^\infty a_k(t)\sin(k\pi x). $$ In each term the time- and space-dependency are separate; this is your separation of variables. The PDE becomes $$ \sum_{k=1}^\infty [a_k''(t)+(ck\pi)^2a_k(t)+ra_k'(t)]\sin(k\pi x)=0. $$ Since this vanishes for all $x$ and $t$, we get for each $k$ the ODE $$ a_k''(t)+(ck\pi)^2a_k(t)+ra_k'(t)=0. $$ This can be solved with elementary methods, once you know $a_k(0)$ and $a_k'(0)$. These come from your initial conditions for $u$ at $t=0$.

Using the series representation, we have $$ u(x,0) = \sum_{k=1}^\infty a_k(0)\sin(k\pi x) $$ and $$ \partial_tu(x,0) = \sum_{k=1}^\infty a_k'(0)\sin(k\pi x), $$ so the initial conditions for the ODEs are the Fourier coefficients of the initial coefficients for $u$. You have $\partial_tu(x,0)=0$, so $a_k'(0)=0$. To find $a_k(0)$, you need to compute the Fourier series of your function $u(x,0)$. The Fourier series of that function can be found in tables, including the one in Wikipedia.

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  • $\begingroup$ How do I solve for $a_k(0)$ and $a'_k(0)$ ? $\endgroup$ – KBG Mar 29 '18 at 13:44
  • $\begingroup$ @KBG By computing the Fourier series of your $u(x,0)$. See the updated answer. $\endgroup$ – Joonas Ilmavirta Mar 29 '18 at 14:10

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