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(193)

where

${B}_{a}=\frac{1-{\nu }_{a}^{2}}{{E}_{a}{t}_{a}}\frac{1}{{\mathrm{\Gamma }}^{2}}C$ (194)

(194)

Proof: substitution of Eq. (193) into Eq. (189) yields

${B}_{a}\left(\frac{{d}^{2}{I}_{1}\left(\mathrm{\Gamma }r\right)}{d{r}^{2}}+\frac{1}{r}\frac{d{I}_{1}\left(\mathrm{\Gamma }r\right)}{dr}-\frac{{I}_{1}\left(\mathrm{\Gamma }r\right)}{{r}^{2}}\right)=\frac{1-{\nu }_{a}^{2}}{{E}_{a}}\frac{1}{{t}_{a}}C{I}_{1}\left(\mathrm{\Gamma }r\right)$ (195)

(195)

But ${I}_{1}\left(\mathrm{\Gamma }r\right)$ satisfies Eq. (180), i.e.,

$\frac{{d}^{2}{I}_{1}\left(\mathrm{\Gamma }r\right)}{d{r}^{2}}+\frac{1}{r}\frac{d{I}_{1}\left(\mathrm{\Gamma }r\right)}{dr}-\frac{{I}_{1}\left(\mathrm{\Gamma }r\right)}{{r}^{2}}={\mathrm{\Gamma }}^{2}{I}_{1}\left(\mathrm{\Gamma }r\right)$ (196)

(196)

Substitution of Eq. (196) into Eq. (195) gives Eq. (194). QED.