# 10.5 There and back again - an exploration of the liouville  (Page 4/10)

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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

$1={\rho }^{1/2}\left(x,\left(,t,\right)\right){x}^{\text{'}}\left(t\right),$

which can be rearranged to give

${x}^{\text{'}}\left(t\right)={\rho }^{-1/2}\left(x,\left(,t,\right)\right).$

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 $\sigma \left(t\right)=\left(\rho \circ x\right)\left(t\right)=\rho \left(x,\left(,t,\right)\right)$ . By the chain rule,

${\sigma }^{\text{'}}\left(t\right)={\rho }^{\text{'}}\left(x,\left(,t,\right)\right)·{x}^{\text{'}}\left(t\right).$

Since ${x}^{\text{'}}\left(t\right)={\sigma }^{-1/2}\left(t\right)$ , we obtain

${\sigma }^{\text{'}}\left(t\right)=\frac{{\rho }^{\text{'}}\left(x,\left(,t,\right)\right)}{\sqrt{\sigma \left(t\right)}}.$

Applying the quotient rule produces

$\begin{array}{ccc}\hfill {\sigma }^{\text{'}\text{'}}\left(t\right)& =& \frac{\sqrt{\sigma \left(t\right)}{\rho }^{\text{'}\text{'}}\left(x,\left(,t,\right)\right){x}^{\text{'}}\left(t\right)-\frac{1}{2}{\rho }^{\text{'}}\left(x,\left(,t,\right)\right){\sigma }^{-1/2}\left(t\right){\sigma }^{\text{'}}\left(t\right)}{\sigma \left(t\right)}\hfill \\ & =& \frac{\sqrt{\sigma \left(t\right)}{\rho }^{\text{'}\text{'}}\left(x,\left(,t,\right)\right){\sigma }^{-1/2}\left(t\right)-\frac{1}{2}{\left({\sigma }^{\text{'}},\left(t\right)\right)}^{2}}{\sigma \left(t\right)}\hfill \\ & =& \frac{{\rho }^{\text{'}\text{'}}\left(x,\left(,t,\right)\right)}{\sigma \left(t\right)}-\frac{{\left({\sigma }^{\text{'}},\left(t\right)\right)}^{2}}{2\sigma \left(t\right)}.\hfill \end{array}$

Hence,

${\rho }^{\text{'}\text{'}}\left(x,\left(,t,\right)\right)=\sigma \left(t\right){\sigma }^{\text{'}\text{'}}\left(t\right)+\frac{1}{2}{\left({\sigma }^{\text{'}},\left(t\right)\right)}^{2}.$

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

$q\left(t\right)=\frac{\sigma \left(t\right){\sigma }^{\text{'}\text{'}}\left(t\right)+\frac{1}{2}{\left({\sigma }^{\text{'}},\left(t\right)\right)}^{2}}{4{\left(\sigma ,\left(,t,\right)\right)}^{2}}-\frac{5{\left({\sigma }^{\text{'}},\left(t\right)\right)}^{2}\sigma \left(t\right)}{16{\left(\sigma ,\left(,t,\right)\right)}^{3}},$

which may be simplified to

$q\left(t\right)=\frac{{\sigma }^{\text{'}\text{'}}\left(t\right)}{4\sigma \left(t\right)}-\frac{3}{16}{\left(\frac{{\sigma }^{\text{'}}\left(t\right)}{\sigma \left(t\right)}\right)}^{2}.$

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, $\rho \left(x\right)$ , $x\in \left[0,L\right]$ , in the wave equation to a potential function, $q\left(t\right)$ , $t\in \left[0,1\right]$ , 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.

## Numerical results

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 $\lambda =\left\{{\pi }^{2},4{\pi }^{2},9{\pi }^{2},16{\pi }^{2},...\right\}$ . 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 $\lambda \left(q\right)=\left\{{\mu }_{1},4{\pi }^{2},9{\pi }^{2},16{\pi }^{2},...\right\}$ specify unique, even $q\left(t\right)$ 's given by

$q\left(t\right)=-2\frac{{d}^{2}}{d{t}^{2}}\mathrm{log}\left[\frac{\mathrm{sin}\sqrt{{\mu }_{1}}\left(1-t\right)-\mathrm{sin}\sqrt{{\mu }_{1}}t}{\mathrm{sin}\sqrt{{\mu }_{1}}},,,\mathrm{sin},\pi ,t\right]$

where $\left[f,g\right]=f{g}^{\text{'}}-{f}^{\text{'}}g$ , the Wronskian. In [link] , a function $q$ is even if $q\left(1-t\right)=q\left(t\right)$ for $t\in \left[0,1\right]$ .

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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