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

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.

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 $\lambda =\{{\pi }^2,4{\pi }^2,9{\pi }^2,16{\pi }^2,...\}$ . 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] .