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Observe that the ODE makes sense only over the domain of definition of the $t$ variable, while the function $\rho $ makes sense only over the domain of definition of the $x$ variable. Our goal was to learn about the $q$ in Chapter 6 of [link] , shown in [link] . Since the function $q$ is known, we have knowledge of the domain of $t$ . In principle, we can obtain the corresponding domain of $x$ by inverting the integral [link] , but this requires knowledge of $\rho $ as a function of $x$ , which is what we are trying to find in the first place. We need a way to obtain this domain informationfrom other quantities we can compute.
Our answer to this problem is the following. Treating [link] as an equation that defines $x$ implicitly as a function of $t$ , we differentiate both sides with respect to $t$ to obtain
which can be rearranged to give
Thus, we obtain a differential equation for $x\left(t\right)$ . Furthermore, we have an initial condition: $x\left(0\right)=0$ , since $x=0$ and $t=0$ correspond to each other by [link] . We can therefore obtain the $x$ -values that correspond to given $t$ -values, provided that we can compute ${\rho}^{-1/2}\left(x,\left(,t,\right)\right)$ for arbitrary $t$ .
We accomplish this by changing the equation [link] into an ODE for $\sigma \left(t\right)=(\rho \circ x)\left(t\right)=\rho \left(x,\left(,t,\right)\right)$ . By the chain rule,
Since ${x}^{\text{'}}\left(t\right)={\sigma}^{-1/2}\left(t\right)$ , we obtain
Applying the quotient rule produces
Hence,
We can substitute these expressions for ${\rho}^{\text{'}}\left(x,\left(,t,\right)\right)$ and ${\rho}^{\text{'}\text{'}}\left(x,\left(,t,\right)\right)$ in equation [link] to obtain
which may be simplified to
This is a differential equation that we may solve for $\sigma \left(t\right)$ over a grid that coincides with the domain of $q$ . We can then use the ODE for ${x}^{\text{'}}\left(t\right)$ , [link] , to obtain the values of $x$ at which $\rho $ takes on the values of $\sigma \left(t\right)$ .
We have shown that it is possible to convert a mass density function, $\rho \left(x\right)$ , $x\in [0,L]$ , in the wave equation to a potential function, $q\left(t\right)$ , $t\in [0,1]$ , in the Sturm-Liouville equation and discussed our method to numerically convert from $q$ to $\rho $ . In the next section, the results from numerically reverting this change of variables, arriving at a mass density function for a string of unit length from a specific Sturm-Liouville potential, are described.
The simplest vibrating string problem is when the string has uniform mass density, $\rho \left(x\right)=1$ . The wave equation with uniform mass density corresponds to the Sturm-Liouville equation with potential function $q\left(t\right)=0$ . For the resulting eigenvalue problem, the eigenvalues are $\lambda =\{{\pi}^{2},4{\pi}^{2},9{\pi}^{2},16{\pi}^{2},...\}$ . What if just one eigenvalue is changed? In chapter 6 of [link] it is shown for ${\mu}_{1}<4{\pi}^{2}$ the Dirichlet spectra with variable first eigenvalue $\lambda \left(q\right)=\{{\mu}_{1},4{\pi}^{2},9{\pi}^{2},16{\pi}^{2},...\}$ specify unique, even $q\left(t\right)$ 's given by
where $[f,g]=f{g}^{\text{'}}-{f}^{\text{'}}g$ , the Wronskian. In [link] , a function $q$ is even if $q(1-t)=q\left(t\right)$ for $t\in [0,1]$ .
To illustrate our results in this section, we have set this first eigenvalue, ${\mu}_{1}$ , to 1, 9, and 15. [link] displays the Sturm-Liouville potentials which correspond to these three values of ${\mu}_{1}$ being substituted into [link] .
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