考研数学:算子法
于二階常系數非齊次線性微分方程:
y′′+py′+qy=f(x)y^{\prime \prime}+p y^{\prime}+q y=f(x) y′′+py′+qy=f(x)
定義算子
P(D)=D2+pD+qP(D)=D^{2}+p D+q P(D)=D2+pD+q
其中微分算子DDD滿足Dy=y′,D2y=y′′D y=y^{\prime}, D^{2} y=y^{\prime \prime}Dy=y′,D2y=y′′,則二階常系數非齊次線性微分方程等價于
P(D)y=f(x)P(D) y=f(x) P(D)y=f(x)
其特解可形式地記為
y?=1P(D)f(x)y^{*}=\frac{1}{P(D)} f(x) y?=P(D)1?f(x)
算子P(D)P(D)P(D)具有如下性質:
(1)設函數y1,y2y_1,y_2y1?,y2?具有二階導數,c1,c2c_1,c_2c1?,c2?為常數,則有P(D)(c1y1+c2y2)=c1P(D)y1+c2P(D)y2P(D)\left(c_{1} y_{1}+c_{2} y_{2}\right)=c_{1} P(D) y_{1}+c_{2} P(D) y_{2}P(D)(c1?y1?+c2?y2?)=c1?P(D)y1?+c2?P(D)y2?
(2)P(D)eλt=eλtP(λ)P(D) e^{\lambda t}=e^{\lambda t} P(\lambda)P(D)eλt=eλtP(λ)
(3)P(D2)cos?at=cos?atP(?a2)P\left(D^{2}\right) \cos a t=\cos a t P\left(-a^{2}\right)P(D2)cosat=cosatP(?a2)
(4)P(D2)sin?at=sin?atP(?a2)P\left(D^{2}\right) \sin a t=\sin a t P\left(-a^{2}\right)P(D2)sinat=sinatP(?a2)
(5)P(D)eλtv(t)=eλtP(D+λ)v(t)P(D) e^{\lambda t} v(t)=e^{\lambda t} P(D+\lambda) v(t)P(D)eλtv(t)=eλtP(D+λ)v(t)
證明:
(1)簡單,不證
(2)
P(D)eλt=(Dn+a1Dn?1+?+an)eλt==eλt(λn+a1λn?1+?+αn)=eλtP(λ)\begin{aligned} P(D) e^{\lambda t} &=\left(D^{n}+a_{1} D^{n-1}+\cdots+a_{n}\right) e^{\lambda t}=\\ &=e^{\lambda t}\left(\lambda^{n}+a_{1} \lambda^{n-1}+\cdots+\alpha_{n}\right)=e^{\lambda t} P(\lambda) \end{aligned} P(D)eλt?=(Dn+a1?Dn?1+?+an?)eλt==eλt(λn+a1?λn?1+?+αn?)=eλtP(λ)?
(3)因為
eiat=cos?at+isin?ate?iat=cos?at?isin?at\begin{array}{l} e^{i a t}=\cos a t+i \sin a t \\ e^{-i a t}=\cos a t-i \sin a t \end{array} eiat=cosat+isinate?iat=cosat?isinat?
故
cos?at=eiat+e?iat2,sin?at=eiat?e?iat2i\cos at=\frac{e^{i a t}+e^{-i a t}}{2}, \quad \sin a t=\frac{e^{i a t}-e^{-i a t}}{2 i} cosat=2eiat+e?iat?,sinat=2ieiat?e?iat?
于是
P(D2)cos?at=P(D2)(eiαt+e?iat2)==12P(D2)eiat+12P(D2)e?iat==12P((ia)2)eiat+12P((?ia)2)e?iat==12(eiat+e?iat)P(?a2)=cos?atP(?a2)\begin{aligned} P\left(D^{2}\right) \cos a t &=P\left(D^{2}\right)\left(\frac{e^{i \alpha t}+e^{-i a t}}{2}\right)=\\ &=\frac{1}{2} P\left(D^{2}\right) e^{i a t}+\frac{1}{2} P\left(D^{2}\right) e^{-i a t}=\\ &=\frac{1}{2} P\left((i a)^{2}\right) e^{i a t}+\frac{1}{2} P\left((-i a)^{2}\right) e^{-i a t}=\\ &=\frac{1}{2}\left(e^{i a t}+e^{-i a t}\right) P\left(-a^{2}\right)=\cos a t P\left(-a^{2}\right) \end{aligned} P(D2)cosat?=P(D2)(2eiαt+e?iat?)==21?P(D2)eiat+21?P(D2)e?iat==21?P((ia)2)eiat+21?P((?ia)2)e?iat==21?(eiat+e?iat)P(?a2)=cosatP(?a2)?
(4)
P(D2)sin?at=P(D2)(eiat?e?iat2i)==12iP(D2)eiat?12iP(D2)e?iat==12iP((ia)2)eiat?12iP((?ia)2)e?iat==12i(eiat?e?iat)P(?a2)=sin?atP(?a2)\begin{aligned} P\left(D^{2}\right) \sin a t &=P\left(D^{2}\right)\left(\frac{e^{i a t}-e^{-i a t}}{2 i}\right)=\\ &=\frac{1}{2 i} P\left(D^{2}\right) e^{i a t}-\frac{1}{2 i} P\left(D^{2}\right) e^{-i a t}=\\ &=\frac{1}{2 i} P\left((i a)^{2}\right) e^{i a t}-\frac{1}{2 i} P\left((-i a)^{2}\right) e^{-i a t}=\\ &=\frac{1}{2 i}\left(e^{i a t}-e^{-i a t}\right) P\left(-a^{2}\right)=\sin a t P\left(-a^{2}\right) \end{aligned} P(D2)sinat?=P(D2)(2ieiat?e?iat?)==2i1?P(D2)eiat?2i1?P(D2)e?iat==2i1?P((ia)2)eiat?2i1?P((?ia)2)e?iat==2i1?(eiat?e?iat)P(?a2)=sinatP(?a2)?
(5)
Dmeλtv(t)=∑k=0mCkmDkeλt?Dm?kv(t)==∑k=0mCkmλkeλtDm?kv(t)==eλt(∑k=0mChmλkDm?k)v(t)==eλt(D+λ)mv(t)\begin{aligned} D^{m} e^{\lambda t} v(t) &=\sum_{k=0}^{m} C_{k}^{m} D^{k} e^{\lambda t} \cdot D^{m-k} v(t)=\\ &=\sum_{k=0}^{m} C_{k}^{m} \lambda^{k} e^{\lambda t} D^{m-k} v(t)=\\ &=e^{\lambda t}\left(\sum_{k=0}^{m} C_{h}^{m} \lambda^{k} D^{m-k}\right) v(t)=\\ &=e^{\lambda t}(D+\lambda)^{m} v(t) \end{aligned} Dmeλtv(t)?=k=0∑m?Ckm?Dkeλt?Dm?kv(t)==k=0∑m?Ckm?λkeλtDm?kv(t)==eλt(k=0∑m?Chm?λkDm?k)v(t)==eλt(D+λ)mv(t)?
此處mmm是任意的非負整數,由此可以得到公式4
下面討論非齊次項f(x)f(x)f(x)與特解y?(x)y^{*}(x)y?(x)的關系
- f(x)=ekxf(x)=\mathrm{e}^{k x}f(x)=ekx
此時y?(x)y^{*}(x)y?(x)的形式:y?(x)=1F(D)ekx=1F(k)ekxy^{*}(x)=\frac{1}{F(D)} \mathrm{e}^{k x}=\frac{1}{F(k)} \mathrm{e}^{k x}y?(x)=F(D)1?ekx=F(k)1?ekx,其中F(k)≠0F(k) \neq 0F(k)?=0,F(k)F(k)F(k)為F(D)F(D)F(D)中的DDD用kkk代替所得值
注:若F(k)=0F(k)=0F(k)=0,不妨設kkk為F(k)F(k)F(k)的mmm重根,則1F(D)ekx=xm1F(m)(D)ekx=xm1F(m)(k)ekx\frac{1}{F(D)} \mathrm{e}^{k x}=x^{m}\frac{1}{F^{(m)}(D)} \mathrm{e}^{k x}=x^{m} \frac{1}{F^{(m)}(k)} \mathrm{e}^{k x}F(D)1?ekx=xmF(m)(D)1?ekx=xmF(m)(k)1?ekx,其中F(m)(D)F^{(m)}(D)F(m)(D)表示F(D)F(D)F(D)對DDD的mmm階導數。
- f(x)=sin?axf(x)=\sin a xf(x)=sinax或cos?ax\cos a xcosax
y?(x)=1F(D2)sin?ax=sin?axF(?a2)y^{*}(x)=\frac{1}{F\left(D^{2}\right)} \sin a x=\frac{\sin a x}{F\left(-a^{2}\right)}y?(x)=F(D2)1?sinax=F(?a2)sinax?,或y?(x)=1F(D2)cos?ax=cos?axF(?a2)y^{*}(x)=\frac{1}{F\left(D^{2}\right)} \cos a x=\frac{\cos a x}{F\left(-a^{2}\right)}y?(x)=F(D2)1?cosax=F(?a2)cosax?,其中F(?a2)≠0F\left(-a^{2}\right) \neq 0F(?a2)?=0
注:若F(?a2)=0F\left(-a^{2}\right)=0F(?a2)=0,不妨設(?a2)\left(-a^{2}\right)(?a2)為F(?a2)=0F\left(-a^{2}\right)=0F(?a2)=0的mmm重根,則
1F(D2)sin?ax=xm?1F(m)(D2)sin?ax\frac{1}{F\left(D^{2}\right)} \sin a x=x^{m} \cdot \frac{1}{F^{(m)}\left(D^{2}\right)} \sin a x F(D2)1?sinax=xm?F(m)(D2)1?sinax
1F(D2)cos?ax=xm?1F(m)(D2)cos?ax\frac{1}{F\left(D^{2}\right)} \cos a x=x^{m} \cdot \frac{1}{F^{(m)}\left(D^{2}\right)} \cos a x F(D2)1?cosax=xm?F(m)(D2)1?cosax
- f(x)=ekxv(x)f(x)=e^{k x} v(x)f(x)=ekxv(x)
y?(x)=1F(D)ekxv(x)=ekx1F(D+k)v(x)y^{*}(x)=\frac{1}{F(D)} \mathrm{e}^{k x} v(x)=e^{k x} \frac{1}{F(D+k)} v(x) y?(x)=F(D)1?ekxv(x)=ekxF(D+k)1?v(x)
- f(x)f(x)f(x)為kkk此多項式,當P(0)≠0P(0) \neq 0P(0)?=0時,由多項式除法得:
1P(D)=(b0+b1D+?+bkDk)+(bk+1Dk+1+bk+2Dk+2+?)\begin{array}{l} \frac{1}{P(D)}=\left(b_{0}+b_{1} D+\cdots+b_{k} D^{k}\right)+\left(b_{k+1} D^{k+1}+\right. \\ \left.b_{k+2} D^{k+2}+\cdots\right) \end{array} P(D)1?=(b0?+b1?D+?+bk?Dk)+(bk+1?Dk+1+bk+2?Dk+2+?)?
由于kkk次多項式超過kkk階的導數全為000,故由上式得特解:
y?=1P(D)f(x)=(b0+b1D+?+bkDk)f(x)y^{*}=\frac{1}{P(D)} f(x)=\left(b_{0}+b_{1} D+\cdots+b_{k} D^{k}\right) f(x) y?=P(D)1?f(x)=(b0?+b1?D+?+bk?Dk)f(x)
幾個注解:
DDD表示微分,則1Dcos?x=∫cos?xdx=sin?x+C\frac{1}{D} \cos x=\int \cos x d x=\sin x+CD1?cosx=∫cosxdx=sinx+C,積分常數CCC不寫。
總結
- 上一篇: php fopen函数 返回值,php中
- 下一篇: Java入门书籍