### （五）分數階微分方程的解法及其適定性問題介紹

a ) 為此介紹一些常見的變換及其性質
Laplace變換的定義為
$$\mathscr{L} \{f(t)\}=\int_{0}^{\infty}f(t)e^{-st}dt$$
Laplace反演變換公式為
$$\mathscr{L}^{-1}F(s)=\int_{0}^{\infty}F(s)e^{st}ds$$

$$f(t)\ast g(t)=\int_{0}^{t}f(t-\tau)g(\tau)d\tau=\int_{0}^{t}f(t)g(t-\tau)d\tau=g(t) \ast f(t)$$

$$\mathscr{L}\{f(t)\ast g(t)\}=F(s)G(s)$$
$$\mathscr{L}^{-1}\{F(s)G(s)\}=f(t)\ast g(t)$$
$n$階導數的$Laplace$變換公式
$$\mathscr{L}\{D^{n}f(x)\}=s^{n}F(s)-\sum_{k=0}^{n-1}s^{k}D^{n-k-1}f(0)$$

$$\mathscr{L}\{D^{n}f(x)\}=s^{n}\mathscr{L}\{f(x)\}$$

$$\mathscr{L}\{_{0}^{RL}D_{t}^{\alpha}f(x)\}=s^{\alpha}F(s)-\sum_{k=0}^{n-1}s^{k}D^{\alpha-k-1}f(0)$$

$$D^{\alpha}f(x)=D^{n}D^{-(n-\alpha)}f(x)=D^{n}g(x)$$

$$g(x)=D^{-(n-a)}f(x)=\frac{x^{n-\alpha-1}}{\Gamma(n-\alpha)}\ast f(x)$$

$$\mathscr{L}\{_{0}^{RL}D_{t}^{\alpha}\}=s^{n}G(s)-\sum_{k=0}^{n-1}s^{k}[D^{n-k-1}g(s)|_{s \to 0^{+}}]$$

$$G(s)=\mathscr{L}\{\frac{x^{n-\alpha-1}}{\Gamma(n-\alpha)}\}F(s)=s^{\alpha-n}F(s)$$

$$\mathscr{L}\{_{0}^{RL}D_{t}^{\alpha}\}=s^{\alpha}F(s)-D^{\alpha-1}f(0)$$

$$\mathscr{L}\{_{0}^{RL}D_{t}^{\alpha}\}=s^{\alpha}F(s)-sD^{\alpha-2}f(0)-D^{\alpha-1}f(0)$$

$$\mathscr{L}\{D^{-\alpha}f(x)\}=\mathscr{L}\{\frac{x^{\alpha-1}}{\Gamma(\alpha)}\}F(s)=s^{-\alpha}F(s)$$

$Caputo$型的分數階導數為
$$\mathscr{L}\{_{0}^{C}D_{t}^{\alpha}f(x)\}=s^{\alpha}F(s)-\sum_{k=0}^{n-1}s^{a-k-1}D^{k}f(0)$$

$$\mathscr{L}\{{}_{0}^{C}D_{t}^{\alpha}f(x)\}=\mathscr{L}\{D^{-(n-\alpha)}D^{n}f(x\}=s^{a-n}[s^{n}F(s)-\sum_{k=0}^{n-1}s^{n-k-1}D^{k}f(0)]$$
Fourier變換的定義為
$$\mathscr{F}\{f(x)\}=\int_{-\infty}^{+\infty}e^{-i\omega x}f(x)dx$$
$$\mathscr{F}^{-1}\{F(\omega)\}=\frac{1}{2\pi}\int_{-\infty}^{+\infty}F(\omega)e^{i \omega x}dk$$
$n$階導數的Fourier變換為
$$\mathscr{F}\{D^{n}f(x)\}=(-i\omega)^{n}F(\omega)$$

$$\mathscr{F}\{_{-\infty}^{RL}D_{t}^{\alpha}f(x)\}=$$
$$\mathscr{F}\{_{-\infty}^{C}D_{t}^{\alpha}f(x)\}=$$

$$\mathscr{M}\{f(t)\}=\int_{0}^{+\infty}t^{s-1}f(t)dt$$
Millin逆變換的定義為
$$f(t)=\frac{1}{2\pi i}\int_{a-i\infty}^{a+i\infty}t^{-s}F(s)ds$$

$$\mathscr{M}\{_{0}^{RL}D_{t}^{\alpha}\}=$$
$$\mathscr{M}\{_{0}^{C}D_{t}^{\alpha}\}=$$

c). M-L函數、Wright函數、Fox函數

M-L函數:

$$e^{x}=\sum_{n=0}^{\infty}\frac{x^{n}}{n!}=\sum_{n=0}^{\infty}\frac{x^{n}}{\Gamma(n+1)},x \in R$$
M-L函數正是指數函數的推廣，這是由數學家 Mittag-Leffler 引進的函數，正如指數函數在整數階微分方程中的地位一樣，M-L函數在分數階微分方程中也起到很重要的作用，我們也可以稱其為廣義的指數函數.

$$E_{\alpha}(z)=\sum_{k=0}^{\infty}\frac{z^{k}}{\Gamma(k\alpha+1)}$$

$$E_{\alpha,\beta}(z)=\sum_{k=0}^{\infty}\frac{z^{k}}{\Gamma(\alpha k+\beta)}$$

$$\overline{\lim_{k \to \infty}}\sqrt[k]{a_{k}}=\overline{\lim_{k \to \infty}}(\alpha {k}+\beta-1)^{-\frac{\alpha k+\beta}{2k}}e^{\alpha}=0$$

$$\Gamma(1+x)=\sqrt{2\pi x}x^{x+\frac{1}{2}}e^{-x+\frac{\theta}{12x}},(0<\theta<1)$$

$$E_{1,1}(z)=\sum_{k=0}^{\infty}\frac{z^{k}}{\Gamma(k+1)}=\sum_{k=0}^{\infty}\frac{z^{k}}{k!}=e^{z}$$
$$E_{1,2}(z)=\sum_{k=0}^{\infty}\frac{z^{k}}{\Gamma(k+2)}=\sum_{k=0}^{\infty}\frac{z^{k}}{(k+1)!}=\frac{1}{z}\sum_{k=0}^{\infty}\frac{z^{k+1}}{(k+1)!} =\frac{e^{z}-1}{z}$$
$$E_{1,3}(z)=\sum_{k=0}^{\infty}\frac{z^{k}}{\Gamma(k+3)}=\sum_{k=0}^{\infty}\frac{z^{k}}{(k+2)!}=\frac{1}{z^{2}}\sum_{k=0}^{\infty}\frac{z^{k+2}}{(k+2)!} =\frac{e^{z}-1-z}{z^{2}}$$
$$E_{1,m}(z)=\sum_{k=0}^{\infty}\frac{z^{k}}{\Gamma(k+m)}=\sum_{k=0}^{\infty}\frac{z^{k}}{(k+m-1)!}=\frac{1}{z^{m-1}}\sum_{k=0}^{\infty}\frac{z^{k+m-1}}{(k+m-1)!} =\frac{e^{z}-\sum_{k=0}^{m-2}\frac{z^{k}}{k!}}{z^{m-1}}$$

$$E_{2,1}(z^{2})=\sum_{k=0}^{\infty}\frac{z^{2k}}{\Gamma(2k+1)}=\sum_{k=0}^{\infty}\frac{z^{2k}}{(2k)!}=cosh(z)$$
$$E_{2,2}(z^{2})=\sum_{k=0}^{\infty}\frac{z^{2k}}{\Gamma(2k+2)}=\sum_{k=0}^{\infty}\frac{z^{2k}}{(2k+1)!}=\frac{sinh(z)}{z}$$

$$\cosh(z)=\frac{e^{z}+e^{-z}}{2}$$
$$\sinh(z)=\frac{e^{z}-e^{-z}}{2}$$

$$E_{\frac{1}{2},1}(z)=\sum_{k=0}^{\infty}\frac{z^{k}}{\Gamma(\frac{k}{2}+1)}=e^{z^2} erfc(-z)$$

$$erf(z)=\frac{2}{\sqrt{\pi}}\int_{0}^{z}e^{-t^{2}}dt$$
$$erfc(z)=1-erf(z)=\frac{2}{\sqrt{\pi}}\int_{z}^{\infty}e^{-t^{2}}dt$$

\begin{eqnarray*}
g'(z)&=&\sum_{n=0}^{\infty}\frac{(-1)^{n}n z^{n-1}}{\Gamma(1+\frac{n}{2})}\\
&=& -\frac{2}{\sqrt{\pi}}+\sum_{n=2}^{\infty}\frac{(-1)^{n}n z^{n-1}}{\Gamma(1+\frac{n}{2})}\\
&=&- \frac{2}{\sqrt{\pi}}+2z\sum_{m=0}^{\infty}\frac{(-1)^{m}z^{m}}{\Gamma(1+\frac{m}{2})},(Let \ \ n=m+2)\\
&=&- \frac{2}{\sqrt{\pi}}+2zg(z)
\end{eqnarray*}

$$g(z)=e^{z^{2}}erfc(z)$$

$$\frac{d}{dz}E_{\frac{1}{k},1}(z)=kz^{k-1}E_{\frac{1}{k},1}(z)+k\sum_{n=1}^{k-1}\frac{z^{n-1}}{\Gamma(n/k)}$$

$$E_{\frac{1}{k},1}(z)=e^{z^{k}}+e^{z^{k}}\int_{0}^{z}ke^{-t^{k}}\sum_{n=1}^{k-1}\frac{t^{n-1}}{\Gamma(n/k)}dt$$

$$E_{\alpha}(z)=\frac{1}{2\pi i}\int_{Ha}\frac{\xi^{\alpha-1}}{\xi^{\alpha}-z}\ e^{\xi}d\xi,\ \ \ \ \alpha>0,z\in \mathbb{C}$$

$$\frac{1}{\Gamma(z)}=\frac{1}{2\pi i}\int_{Ha}\ e^{\xi}\xi^{-z}d\xi$$
Hankel圍道圖形如下:

$$E_{\beta}(-t^{\beta})=\sum_{k=0}^{\infty}\frac{(-t^{\beta})^{k}}{\Gamma(\beta k+1)} \sim \frac{sin(\beta \pi)}{\pi}\ \frac{\Gamma(\beta)}{t^{\beta}}$$
M-L函數的Laplace變換在解分數階偏微分方程中起到核心作用:
$$\mathscr{L}\{t^{n\alpha+\beta-1}E_{\alpha,\beta}^{(n)}(t^{\alpha})\}=\frac{n!s^{\alpha-\beta}}{(s^{\alpha}-1)^{n+1}}$$

$$\mathscr{L}\{t^{\beta-1}E_{\alpha,\beta}(t^{\alpha})\}=\frac{s^{\alpha-\beta}}{s^{\alpha}-1}$$

$$E_{\alpha,\beta}^{(n)}(z)=\frac{d^{n}}{dz^{n}}\sum_{k=0}^{\infty}\frac{z^{k}}{\Gamma(k\alpha+\beta)}=\sum_{k=0}^{\infty}\frac{k(k-1)(k-2)\cdots (k-n+1)z^{k-n}}{\Gamma(k\alpha+\beta)}$$

\begin{eqnarray*}
\int_{0}^{\infty}e^{-st}t^{n\alpha+\beta-1}E_{\alpha,\beta}^{(n)}(t^{\alpha})dt
&=&\int_{0}^{\infty}e^{-st}t^{n\alpha+\beta-1}\sum_{k=0}^{\infty}\frac{k(k-1)(k-2)
\cdots (k-n+1)t^{(k-n)\alpha}}{\Gamma(k\alpha+\beta)}dt\\
&=&\sum_{k=0}^{\infty}k(k-1)(k-2)
\cdots (k-n+1)\int_{0}^{\infty}e^{-st}\frac{t^{k\alpha+\beta-1}}{\Gamma(k\alpha+\beta)}dt\\
&=&\sum_{k=0}^{\infty}\frac{k(k-1)(k-2)\cdots (k-n+1)}{s^{k\alpha+\beta}}\\
&=&\frac{1}{s^{\beta}}\sum_{k=0}^{\infty}\frac{k(k-1)(k-2)\cdots(k-n+1)}{(s^{a})^{k}}\\
&=&\frac{1}{s^{\beta}}\sum_{k=0}^{\infty}k(k-1)(k-2)\cdots(k-n+1)\lambda ^{k}\ \ (\lambda=s^{-\alpha})\\
&=&\frac{\lambda^{n}}{s^{\beta}}\sum_{k=0}^{\infty}k(k-1)(k-2)\cdots(k-n+1)\lambda ^{k-n}\\
&=&\frac{\lambda^{n}}{s^{\beta}}\frac{d^{n}}{d\lambda^{n}}(\frac{1}{1-\lambda})\\
&=&\frac{\lambda^{n}}{s^{\beta}}\frac{n!}{(1-\lambda)^{n+1}}\\
&=&\frac{n!s^{\alpha-\beta}}{(s^{\alpha}-1)^{n+1}}\\
\end{eqnarray*}

$$\mathscr{L}\{t^{n\alpha+\beta-1}E_{\alpha,\beta}^{(n)}(\gamma t^{\alpha})\}=\frac{n!s^{\alpha-\beta}}{(s^{\alpha}-\gamma)^{n+1}}$$

$$\mathscr{L}\{t^{(n+1)\alpha-1}E_{\alpha,\alpha}^{(n)}(\gamma t^{\alpha})\}=\frac{n!}{(s^{\alpha}-\gamma)^{n+1}}$$

$$\mathscr{L}^{-1}\{\frac{k!}{(s^{\alpha}-\gamma)^{k+1}}\}=t^{(k+1)\alpha-1}E_{\alpha,\alpha}^{(k)}(\gamma t^{\alpha})$$
$$\mathscr{L}^{-1}\{\frac{k!s^{\alpha-\beta}}{(s^{\alpha}-\gamma)^{k+1}}\}=t^{k\alpha+\beta-1}E_{\alpha,\beta}^{(k)}(\gamma t^{\alpha})$$

$$_{0}D_{t}^{\alpha}f(x)=\lambda f(x)$$

$$s^{\alpha}F(s)-\sum_{k=0}^{n-1}s^{k}D^{\alpha-k-1}f(0)=\lambda F(s)$$

$$F(s)=\sum_{k=0}^{n-1}\frac{s^{k}}{s^{\alpha}-\lambda}b_{k}$$

$$f(t)=\sum_{k=0}^{n-1}b_{k}\mathscr{L}^{-1}\{\frac{s^{k}}{s^{\alpha}-\lambda}\}=\sum_{k=0}^{n-1}b_{k}t^{a-k-1}E_{\alpha,\alpha-k}(\lambda t^{\alpha})$$

$$_{0}D_{t}^{\alpha}f(x)-\lambda f(x)=h(x)$$

$$s^{\alpha}F(s)-\sum_{k=0}^{n-1}s^{k}D^{\alpha-k-1}f(0)-\lambda F(s)=H(s)$$

$$F(s)=\frac{H(s)}{s^{\alpha}-\lambda}+\sum_{k=0}^{n-1}\frac{s^{k}}{s^{\alpha}-\lambda}b_{k}$$

$$f(x)=h(x)\ast (t^{\alpha-1}E_{\alpha,\alpha}(\lambda t^{\lambda}))+\sum_{k=0}^{n-1}b_{k}t^{\alpha-k-1}E_{\alpha,\alpha-k}(\lambda t^{\alpha})$$

$$_{0}D_{t}^{p}f(x)+_{0}D_{t}^{q}f(x)=h(x)$$

$$[s^{p}F(s)-s^{0}D^{p-1}f(0)]+[s^{q}F(s)-s^{0}D^{q-1}f(0)]=H(s)$$

$$F(s)=\frac{C+H(s)}{s^{p}+s^{q}}=(C+H(s))\frac{s^{-p}}{s^{q-p}+1}$$

$$f(t)=C t^{q-1}E_{q-p,q}(-t^{q-p})+ t^{q-1}E_{q-p,q}(-t^{q-p})\ast h(t)$$

$$_{0}^{RL}D_{t}^{\alpha}u(x,t)=\lambda^{2}\frac{\partial ^{2}u(x,t)}{\partial x^{2}}\ \ \ \ \ \ \ (0<\alpha<1,t>0,-\infty<x<+\infty)$$
$$\lim_{x \to \infty} u(x,t)=0,_{0}D_{t}^{\alpha-1}u(x,t)|_{t=0}=\varphi(x)$$

$$_{0}D_{t}^{\alpha}u(\omega,t)+\lambda^{2}\omega^{2}u(\omega,t)=0$$
$$_{0}D_{t}^{\alpha-1}u(\omega,t)|_{t=0}=\varphi(\omega)$$

$$s^{\alpha}u(\omega,s)-\varphi(w)+\lambda^{2}\omega^{2}u(\omega,s)=0$$

$$u(\omega,s)=\frac{\varphi(\omega)}{s^{\alpha}+\lambda^{2}\omega^2}$$

$$u(\omega,t)=\varphi(w)t^{\alpha-1}E_{\alpha,\alpha}(-\lambda^{2}\omega^{2}t^{\alpha})$$

$$u(x,t)=\varphi(x) \ast \mathscr{F}^{-1}\{t^{\alpha-1}E_{\alpha,\alpha}(-\lambda^{2}\omega^{2}t^{\alpha})\}$$

$$_{0}^{C}D_{t}^{\alpha}u(x,t)=\lambda^{2}\frac{\partial ^{2}u(x,t)}{\partial x^{2}}\ \ \ \ \ \ \ (0<\alpha<1,t>0,-\infty<x<+\infty)$$
$$\lim_{x \to \infty} u(x,t)=0,u(x,t)|_{t=0}=\varphi(x)$$

$$_{0}D_{t}^{\alpha}u(\omega,t)+\lambda^{2}\omega^{2}u(\omega,t)=0$$
$$u(\omega,t)|_{t=0}=\varphi(\omega)$$

$$s^{\alpha}u(\omega,s)-s^{\alpha-1}\varphi(w)+\lambda^{2}\omega^{2}u(\omega,s)=0$$

$$u(\omega,s)=\frac{s^{\alpha-1}}{s^{\alpha}+\lambda^{2}\omega^2}\varphi(\omega)$$

$$u(\omega,t)=\varphi(w)E_{\alpha,1}(-\lambda^{2}\omega^{2}t^{\alpha})$$

$$u(x,t)=\varphi(x) \ast \mathscr{F}^{-1}\{E_{\alpha,1}(-\lambda^{2}\omega^{2}t^{\alpha})\}$$

$$Ay''(t)+B _{0}^{RL}D_{t}^{3/2}y(t)+C y(t)=f(t)$$

$$A [s^{2} Y(s)-sy(0)-y'(0)]+B [s^{3/2}Y(s)-s^{0}D^{1/2}y(0)-sD^{-1/2}y(0)]+CY(s)=F(s)$$

$$Y(s)=\frac{F(s)}{A\ s^{2}+B\ s^{3/2}+C}$$
同樣可用 Mellin變換解分數階初值問題，用變換解方程的思想都是將微分方程轉換為代數方程，代數方程較好求解，然后利用逆變換得到原方程的解.可能用到的特殊函數還有Wright 函數和 Fox函數.

Wright函數:
$$W_{\alpha,\beta}(z)=\sum_{k=0}^{\infty}\frac{z^{k}}{k!\Gamma(\alpha k+\beta)}$$
Fox函數:
$$H_{p,q}^{m,n}[z|_{(a_{1},A_{1}),\cdots ,(a_{p},A_{p})}^{(b_{1},B_1),\cdots ,(b_p,B_p)}]=\frac{1}{2\pi i}\int_{L}\mathscr{H}_{p,q}^{m,n}(s)z^{s}ds$$

$$\mathscr{H}_{p,q}^{m,n}(s)=\frac{\prod_{k=1}^{m}\Gamma(b_{k}-B_{k}s)\prod_{j=1}^{n}\Gamma(1-a_{j}+A_{j}s)}{\prod_{k=m+1}^{q}\Gamma(1-b_{k}+B_{k}s) \prod_{j=n+1}^{p}\Gamma(b_{k}+B_{k}s)}$$

Green函數法

Sturm-Liouville問題

Fourier級數波解

(Banach)   完備的度量空間的壓縮映射有唯一的不動點。
(Banach)  度量空間 $R$完備，$B$是$R$到$R$映射，若有$B^{n}$是$R$上的壓縮映射,則$B$在$R$中有唯一的不動點。

(schauder)  設$R$是賦范線性空間，$A$是$R$中凸緊集，$f$是$A$到$A$的連續映射，那么$f(x)$在$A$中有不動點。
(schauder)  設$X$是Banach空間,$S$是空間$S$的凸閉集，$f$是$S$到$S$的連續映射且$f(S)$致密, 那么$f(x)$在$S$中有不動點。

(Lax-Milgram) 設$a(x,y)$是Hilbert空間H上的共軛雙線性函數,滿足
(1)相容性，$\exists M>0$,使得$|a(x,y)|\leq M \|x\|\|y\|$
(2)正定性，$\exists \delta>0$,使得$|a(x,y)|\geq \delta \|x\|^{2}$

$$a(x,y)=(x,Ay),\mbox{且}||A^{-1}||\leq \frac{1}{\delta}$$