4.1方程的齐次化方法: Duhamel 原理习题参考解答

zhangzujin

# 方程的齐次化方法: Duhamel 原理习题参考解答 --- 1、 如果已知下述常微分方程的特定初值问题 \begin\{aligned\} \left\\{\begin\{array\}\{ll\} -y''+y=0, x > 0,\\\\ y(0)=0,\quad y'(0)=1 \end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 的解为 $\displaystyle y=Y(x)$, 试通过它写出一般初值问题 \begin\{aligned\} \left\\{\begin\{array\}\{ll\} -y''+y=f(x), x > 0,\\\\ y(0)=a,\quad y'(0)=b \end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 的解的表达式. [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) / 由齐次化原理知解为 \begin\{aligned\} y(x)=aY'(x)+b Y(x)-\int\_0^x f(t)Y(x-t)\mathrm\{ d\} t. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 事实上, \begin\{aligned\} y'(x)=&aY''(x)+bY'(x)-\int\_0^x f(t)Y'(x-t)\mathrm\{ d\} t\\\\ =&aY(x)+bY'(x)-\int\_0^x f(t)Y'(x-t)\mathrm\{ d\} t,\\\\ y''(x)=&aY'(x)+bY''(x)-\left\[\int\_0^x f(t)Y''(x-t)\mathrm\{ d\} t+f(x)Y'(0)\right\]\\\\ =&aY'(x)+bY(x)-\int\_0^x f(t)Y(x-t)\mathrm\{ d\} t-f(x) =y(x)-f(x), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 且 \begin\{aligned\} y(0)=aY'(0)+bY(0)=a, y'(0)=aY(0)+bY'(0)=b. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 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、 如果已知以下初值问题 \begin\{aligned\} \left\\{\begin\{array\}\{ll\} y^\{(k)\}+a\_1y^\{(k-1)\}+\cdots+a\_ky=0,&x > 0,\\\\ y(0)=y'(0)=\cdots=y^\{(k-2)\}(0)=0, y^\{(k-1)\}(0)=1,&k\geq 2 \end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 的解为 $\displaystyle y=Y(x)$, 其中 $\displaystyle a\_1,\cdots,a\_k$ 皆为常数. 试通过它写出一般初值问题 \begin\{aligned\} \left\\{\begin\{array\}\{ll\} y^\{(k)\}+a\_1y^\{(k-1)\}+\cdots+a\_ky=f(x),&x > 0,\\\\ y(0)=\alpha\_0, \cdots, y^\{(k-1)\}(0)=\alpha\_\{k-1\} \end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 的解的表达式. [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) / \begin\{aligned\} \sum\_\{i=0\}^\{k-1\} \alpha\_iY^\{(k-1-i)\}(x)+\int\_0^x f(t)Y(x-t)\mathrm\{ d\} t. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 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、 证明定理 4.3. [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) / 显然, \begin\{aligned\} v\_2(0,t)=v\_2(l,t)=0,\quad v\_2(x,0)=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 另外, 也有 \begin\{aligned\} \frac\{\partial v\_2\}\{\partial t\} &= w(x,t,t)+\int\_0^t \frac\{\partial\}\{\partial t\}w(t,x,\tau)\mathrm\{ d\} \tau =\int\_0^t \frac\{\partial \}\{\partial t\}w(t,x\tau)\mathrm\{ d\} \tau\Rightarrow \frac\{\partial v\_2\}\{\partial t\}(x,0)=0,\\\\ \frac\{\partial^2v\_2\}\{\partial t^2\} &=\frac\{\partial w\}\{\partial t\}(x,t,t) +\int\_0^t \frac\{\partial^2\}\{\partial t^2\}w(x,t,\tau)\mathrm\{ d\}\tau =f\_1(x,t)+\int\_0^t a^2\frac\{\partial^2\}\{\partial x^2\}w(x,t,\tau)\mathrm\{ d\} \tau\\\\ &=f\_1(x,t)+a^2\frac\{\partial^2v\_2\}\{\partial x^2\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 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)/ --- 4、 已知 Cauchy 问题 \begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\}w\_\{tt\}-w\_\{xx\}=0,&x\in\mathbb\{R\}, t > \tau,\\\\ w|\_\{t=\tau\}=0, w\_t|\_\{t=\tau\}=2, &x\in\mathbb\{R\}\end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 的解为 $\displaystyle w(x,t;\tau)=2(t-\tau)$, 试写出 Cauchy 问题 \begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\} u\_\{tt\}-u\_\{xx\}=2,&x\in\mathbb\{R\}, t > 0,\\\\ u|\_\{t=0\}=0, u\_t|\_\{t=0\}=0,&x\in\mathbb\{R\}\end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 的解. [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) / 由定理 4.1 知 \begin\{aligned\} u(x,t)=\int\_0^t w(x,t;\tau)\mathrm\{ d\} \tau=\int\_0^t 2(t-\tau)\mathrm\{ d\} \tau=t^2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 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)/