3.1分离变量法的理论基础习题参考解答

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# 分离变量法的理论基础习题参考解答 --- 1、 求下列特征值问题的特征值和特征函数: (1)、 $\displaystyle \left\\{\begin\{array\}\{llllllllllll\}X''(x)+\lambda X(x)=0, 0 < x < l,\\\\ X(0)=X'(l)=0;\end\{array\}\right.$ [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1-1)、 当 $\displaystyle \lambda < 0$ 时, $\displaystyle X(x)=c\_1\mathrm\{e\}^\{\sqrt\{-\lambda\}x\}+c\_2\mathrm\{e\}^\{-\sqrt\{-\lambda\}x\}$. 联合边界条件知 $\displaystyle c\_1=c\_2=0$, 而 $\displaystyle X\equiv 0$. (1-2)、 当 $\displaystyle \lambda=0$ 时, $\displaystyle X(x)=c\_1+c\_2x$. 联合边界条件知 $\displaystyle c\_1=c\_2=0$, 而 $\displaystyle X\equiv 0$. (1-3)、 当 $\displaystyle \lambda > 0$ 时, $\displaystyle X(x)=c\_1\cos \sqrt\{\lambda\}x+c\_2\sin \sqrt\{\lambda x\}$. 联合边界条件知 \begin\{aligned\} \cos (\sqrt\{\lambda\}l)=0\Rightarrow 0 < \sqrt\{\lambda l\}=k\pi+\frac\{\pi\}\{2\} \Leftrightarrow \lambda=\left\[\frac\{\left(k+\frac\{1\}\{2\}\right)\pi\}\{l\}\right\]^2, k\geq 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 此时它有特征函数 $\displaystyle X\_k(x)=\sin\frac\{\left(k+\frac\{1\}\{2\}\right)\pi\}\{l\}x$. 综上, 原问题有特征值 $\displaystyle \left\[\frac\{\left(k+\frac\{1\}\{2\}\right)\pi\}\{l\}\right\]^2, k\geq 0$, 对应的特征函数为 $\displaystyle X\_k(x)=\sin\frac\{\left(k+\frac\{1\}\{2\}\right)\pi\}\{l\}x$. 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ (2)、 $\displaystyle \left\\{\begin\{array\}\{llllllllllll\}X''(x)+\lambda X(x)=0, &0 < x < l,\\\\ X'(0)=X(l)=0;\end\{array\}\right.$ [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (2-1)、 当 $\displaystyle \lambda < 0$ 时, $\displaystyle X(x)=c\_1\mathrm\{e\}^\{\sqrt\{-\lambda\}x\}+c\_2\mathrm\{e\}^\{-\sqrt\{-\lambda\}x\}$. 联合边界条件知 $\displaystyle c\_1=c\_2=0$, 而 $\displaystyle X\equiv 0$. (2-2)、 当 $\displaystyle \lambda=0$ 时, $\displaystyle X(x)=c\_1+c\_2x$. 联合边界条件知 $\displaystyle c\_1=c\_2=0$, 而 $\displaystyle X\equiv 0$. (2-3)、 当 $\displaystyle \lambda > 0$ 时, $\displaystyle X(x)=c\_1\cos \sqrt\{\lambda\}x+c\_2\sin \sqrt\{\lambda x\}$. 联合边界条件知 \begin\{aligned\} \cos (\sqrt\{\lambda\}l)=0\Rightarrow 0 < \sqrt\{\lambda l\}=k\pi+\frac\{\pi\}\{2\} \Leftrightarrow \lambda=\left\[\frac\{\left(k+\frac\{1\}\{2\}\right)\pi\}\{l\}\right\]^2, k\geq 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 此时它有特征函数 $\displaystyle X\_k(x)=\cos\frac\{\left(k+\frac\{1\}\{2\}\right)\pi\}\{l\}x$. 综上, 原问题有特征值 $\displaystyle \left\[\frac\{\left(k+\frac\{1\}\{2\}\right)\pi\}\{l\}\right\]^2, k\geq 0$, 对应的特征函数为 $\displaystyle X\_k(x)=\cos\frac\{\left(k+\frac\{1\}\{2\}\right)\pi\}\{l\}x$. 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ (3)、 $\displaystyle \left\\{\begin\{array\}\{llllllllllll\}X''(x)+\lambda X(x)=0, 0 < x < l,\\\\ X(0)=X'(l)+hX(l)=0\left(\mbox\{$h$ 为正常数\}\right);\end\{array\}\right.$ [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (3-1)、 当 $\displaystyle \lambda < 0$ 时, $\displaystyle X(x)=c\_1\mathrm\{e\}^\{\sqrt\{-\lambda\}x\}+c\_2\mathrm\{e\}^\{-\sqrt\{-\lambda\}x\}$. 联合边界条件知 $\displaystyle c\_1=c\_2=0$, 而 $\displaystyle X\equiv 0$. (3-2)、 当 $\displaystyle \lambda=0$ 时, $\displaystyle X(x)=c\_1+c\_2x$. 联合边界条件知 $\displaystyle c\_1=c\_2=0$, 而 $\displaystyle X\equiv 0$. (3-3)、 当 $\displaystyle \lambda > 0$ 时, $\displaystyle X(x)=c\_1\cos \sqrt\{\lambda\}x+c\_2\sin \sqrt\{\lambda x\}$. 联合边界条件知 \begin\{aligned\} \sqrt\{\lambda\}\cos\left(\sqrt\{\lambda\}l\right)+h \sin \left(\sqrt\{\lambda\}l\right)=0 \Rightarrow \tan \left(\sqrt\{\lambda\}l\right)=-\frac\{\sqrt\{\lambda\}\}\{h\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 设 $\displaystyle \lambda\_k$ 是上述方程的第 $\displaystyle k(\geq 1)$ 个正根, 则对应的特征函数为 $\displaystyle X\_k(x)=\sin \sqrt\{\lambda\_k\}x$. 综上, 原问题有特征值 $\displaystyle \lambda\_k$, 它是 $\displaystyle \tan \left(\sqrt\{\lambda\}l\right)=-\frac\{\sqrt\{\lambda\}\}\{h\}$ 第 $\displaystyle k$ 个正根, 对应的特征函数为 $\displaystyle X\_k(x)=\sin \sqrt\{\lambda\_k\}x$. 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/