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11.15: A.12.4- Section 12.4 Answers

  • Page ID
    121472
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    1. \(\displaystyle u(r,\theta )=\alpha_{0}\,\frac{\ln r/\rho }{\ln \rho_{0}/\rho}+ \sum_{n=1}^{\infty}\frac{r^{n}\rho^{-n}-\rho^{n}r^{-n}}{\rho_{0}^{n}\rho^{-n}-\rho^{n}\rho_{0}^{-n}}(\alpha_{n}\cos n\theta +\beta_{n}\sin n\theta);\) \(\quad\) \(\displaystyle \alpha_{0}=\frac{1}{2\pi }\int_{-\pi }^{\pi }f(\theta )d\theta,\quad \alpha_{n}=\frac{1}{\pi }\int_{-\pi }^{\pi }f(\theta )\cos n\theta d\theta,\quad \beta_{n}=\frac{1}{\pi }\int_{-\pi }^{\pi}f(\theta )\sin n\theta d\theta,\quad n=1,2,3,\ldots.\)

    2. \(\displaystyle u(r,\theta )= \sum_{n=1}^{\infty}\alpha_{n}\frac{\rho_{0}^{-n\pi /\gamma} r^{n\pi /\gamma}-\rho_{0}^{n\pi /\gamma}r^{-n\pi /\gamma}}{\rho_{0}^{-n\pi /\gamma}\rho^{n\pi /\gamma}-\rho_{0}^{n\pi /\gamma}\rho^{-n\pi /\gamma}}\sin\frac{n\pi\theta}{\gamma};\)\(\quad\)\(\displaystyle\alpha_{n}=\frac{1}{\gamma}\int_{0}^{\gamma}f(\theta )\sin\frac{n\pi\theta}{\gamma}d\theta ,\quad n=1,2,3,\ldots \)

    3. \(\displaystyle u(r,\theta )=\rho\alpha_{0}\ln\frac{r}{\rho_{0}}+\frac{\rho\gamma}{\pi} \sum_{n=1}^{\infty}\frac{\alpha_{n}}{n}\frac{\rho_{0}^{-n\pi /\gamma} r^{n\pi /\gamma}-\rho_{0}^{n\pi /\gamma}r^{-n\pi /\gamma}}{\rho_{0}^{-n\pi /\gamma}\rho^{n\pi /\gamma}-\rho_{0}^{n\pi /\gamma}\rho^{-n\pi /\gamma}}\cos\frac{n\pi\theta}{\gamma};\quad\alpha_{0}=\frac{1}{\gamma}\int_{0}^{\gamma}f(\theta )d\theta,\quad \alpha_{n}=\frac{2}{\gamma }\int_{0}^{\gamma}f(\theta )\cos\frac{n\pi\theta }{\gamma}d\theta,\quad n=1,2,3,\ldots\)

    4. \(\displaystyle u(r,\theta )=\sum_{n=1}^{\infty}\alpha_{n}\frac{r^{(2n-1)\pi /2\gamma}}{\rho^{(2n-1)\pi /2\gamma}}\cos\frac{(2n-1)\pi\theta}{2\gamma};\quad\alpha_{n}=\frac{2}{\gamma}\int_{0}^{\gamma}f(\theta )\cos\frac{(2n-1)\pi\theta}{2\gamma}d\theta,\ n=1,2,3,\ldots\)

    5. \(\displaystyle u(4,\theta )=\frac{2\gamma\rho_{0}}{\pi}\sum_{n=1}^{\infty}\frac{\alpha_{n}}{2n-1}\frac{\rho ^{-(2n-1)\pi /2\gamma}r^{(2n-1)\pi /2\gamma}+\rho^{(2n-1)\pi /2\gamma}r^{-(2n-1)\pi /2\gamma} }{\rho^{-(2n-1)\pi /2\gamma}\rho_{0}^{(2n-1)\pi /2\gamma}-\rho^{(2n-1)\pi /2\gamma}\rho_{0}^{-(2n-1)\pi /2\gamma} }\sin\frac{(2n-1)\pi\theta}{2\gamma}; \quad\alpha_{n}=\frac{2}{\gamma}g(\theta )\sin\frac{(2n-1)\pi\theta }{2\gamma}d\theta ,\quad n=1,2,3,\ldots \)

    6. \(\displaystyle u(r,\theta )=\alpha_{0}+\sum_{n=1}^{\infty}\alpha_{n}\frac{r^{n\pi /\gamma}}{\rho^{n\pi /\gamma}}\cos\frac{n\pi\theta }{\gamma };\quad\alpha_{0}=\frac{1}{\gamma}\int_{0}^{\gamma}f(\theta )d\theta,\quad \alpha_{n}=\frac{2}{\gamma }\int_{0}^{\gamma }f(\theta )\cos\frac{n\pi\theta }{\gamma}\,d\theta,\ n=1,2,3,\ldots\)

    7. \(\displaystyle u(r,\theta )=c+\sum_{n=1}^{\infty}\frac{r^{n}}{n\rho^{n-1}}(\alpha _{n}\cos n\theta +\beta_{n}\sin n\theta );\) \(\quad\) \(c\in\mathbb{R}\) arbitrary, \(\displaystyle\alpha_{n}=\frac{1}{\pi}\int_{-\pi }^{\pi}f(\theta )\cos n\theta d\theta\), \(\displaystyle\beta_{n}=\frac{1}{\pi }\int_{-\pi }^{\pi }f(\theta )\sin n\theta d\theta,\) \(n=1,2,3,\ldots \)


    This page titled 11.15: A.12.4- Section 12.4 Answers is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by William F. Trench.

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