Caputo 分数阶一维问题基于 L2-1 σ _\sigma σ 逼近的空间二阶方法
Caputo 分数阶一维问题基于 L2-1 σ _\sigma σ 逼近的快速差分方法(附Matlab程序)
文章目录
考虑如下时间分数阶慢扩散方程初边值问题
{ 0 C D t α u ( x , t ) = u x x ( x , t ) + f ( x , t ) , x ∈ ( 0 , L ) , t ∈ ( 0 , T ] u ( x , 0 ) = φ ( x ) , x ∈ ( 0 , L ) u ( 0 , t ) = μ ( t ) , u ( L , t ) = ν ( t ) , t ∈ [ 0 , T ] \left\{\begin{array}{l} { }_{0}^{C} D_{t}^{\alpha} u(x, t)=u_{x x}(x, t)+f(x, t), \quad x \in(0, L), t \in(0, T] \\ u(x, 0)=\varphi(x), \quad x \in(0, L) \\ u(0, t)=\mu(t), \quad u(L, t)=\nu(t), \quad t \in[0, T] \end{array}\right. ⎩⎨⎧0CDtαu(x,t)=uxx(x,t)+f(x,t),x∈(0,L),t∈(0,T]u(x,0)=φ(x),x∈(0,L)u(0,t)=μ(t),u(L,t)=ν(t),t∈[0,T]
其中 α ∈ ( 0 , 1 ) , f , φ , μ , ν \alpha \in(0,1), f, \varphi, \mu, \nu α∈(0,1),f,φ,μ,ν 为已知函数, 且 φ ( 0 ) = μ ( 0 ) , φ ( L ) = ν ( 0 ) \varphi(0)=\mu(0), \varphi(L)=\nu(0) φ(0)=μ(0),φ(L)=ν(0).
设解函数 u ∈ C ( 4 , 2 ) ( [ 0 , L ] × [ 0 , T ] ) u \in C^{(4,2)}([0, L] \times[0, T]) u∈C(4,2)([0,L]×[0,T]).
差分格式的建立
记
σ = 1 − α 2 , t n − 1 + σ = ( n − 1 + σ ) τ , s = τ α Γ ( 2 − α ) , f i n − 1 + σ = f ( x i , t n − 1 + σ ) . \sigma=1-\frac{\alpha}{2}, \quad t_{n-1+\sigma}=(n-1+\sigma) \tau, \quad s=\tau^{\alpha} \Gamma(2-\alpha), \quad f_{i}^{n-1+\sigma}=f\left(x_{i}, t_{n-1+\sigma}\right) \text {. } σ=1−2α,tn−1+σ=(n−1+σ)τ,s=ταΓ(2−α),fin−1+σ=f(xi,tn−1+σ).
在点 ( x i , t n − 1 + σ ) \left(x_{i}, t_{n-1+\sigma}\right) (xi,tn−1+σ) 处考虑微分方程, 得到
0 C D t α u ( x i , t n − 1 + σ ) = u x x ( x i , t n − 1 + σ ) + f i n − 1 + σ , 1 ⩽ i ⩽ M − 1 , 1 ⩽ n ⩽ N . \begin{array}{r} { }_{0}^{C} D_{t}^{\alpha} u\left(x_{i}, t_{n-1+\sigma}\right)=u_{x x}\left(x_{i}, t_{n-1+\sigma}\right)+f_{i}^{n-1+\sigma}, \\ 1 \leqslant i \leqslant M-1,1 \leqslant n \leqslant N . \end{array} 0CDtαu(xi,tn−1+σ)=uxx(xi,tn−1+σ)+fin−1+σ,1⩽i⩽M−1,1⩽n⩽N.
对上式中时间分数阶导数应用 L2- 1 σ _{\sigma} σ 公式离散, 可得
0 C D t α u ( x i , t n − 1 + σ ) = τ − α Γ ( 2 − α ) ∑ k = 0 n − 1 c k ( n , α ) ( U i n − k − U i n − k − 1 ) + O ( τ 3 − α ) { }_{0}^{C} D_{t}^{\alpha} u\left(x_{i}, t_{n-1+\sigma}\right)=\frac{\tau^{-\alpha}}{\Gamma(2-\alpha)} \sum_{k=0}^{n-1} c_{k}^{(n, \alpha)}\left(U_{i}^{n-k}-U_{i}^{n-k-1}\right)+O\left(\tau^{3-\alpha}\right) 0CDtαu(xi,tn−1+σ)=Γ(2−α)τ−αk=0∑n−1ck(n,α)(Uin−k−Uin−k−1)+O(τ3−α)
对空间二阶导数应用二阶中心差商离散, 可得
u x x ( x i , t n − 1 + σ ) = σ u x x ( x i , t n ) + ( 1 − σ ) u x x ( x i , t n − 1 ) + O ( τ 2 ) = σ δ x 2 U i n + ( 1 − σ ) δ x 2 U i n − 1 + O ( h 2 ) + O ( τ 2 ) \begin{aligned} u_{x x}\left(x_{i}, t_{n-1+\sigma}\right) &=\sigma u_{x x}\left(x_{i}, t_{n}\right)+(1-\sigma) u_{x x}\left(x_{i}, t_{n-1}\right)+O\left(\tau^{2}\right) \\ &=\sigma \delta_{x}^{2} U_{i}^{n}+(1-\sigma) \delta_{x}^{2} U_{i}^{n-1}+O\left(h^{2}\right)+O\left(\tau^{2}\right) \end{aligned} uxx(xi,tn−1+σ)=σuxx(xi,tn)+(1−σ)uxx(xi,tn−1)+O(τ2)=σδx2Uin+(1−σ)δx2Uin−1+O(h2)+O(τ2)
于是得到
1 s ∑ k = 0 n − 1 c k ( n , α ) ( U i n − k − U i n − k − 1 ) = σ δ x 2 U i n + ( 1 − σ ) δ x 2 U i n − 1 + f i n − 1 + σ + ( r 6 ) i n 1 ⩽ i ⩽ M − 1 , 1 ⩽ n ⩽ N , (2.6.4) \begin{array}{r} \frac{1}{s} \sum\limits_{k=0}^{n-1} c_{k}^{(n, \alpha)}\left(U_{i}^{n-k}-U_{i}^{n-k-1}\right)=\sigma \delta_{x}^{2} U_{i}^{n}+(1-\sigma) \delta_{x}^{2} U_{i}^{n-1}+f_{i}^{n-1+\sigma}+\left(r_{6}\right)_{i}^{n} \\ 1 \leqslant i \leqslant M-1,1 \leqslant n \leqslant N, \text { (2.6.4) } \end{array} s1k=0∑n−1ck(n,α)(Uin−k−Uin−k−1)=σδx2Uin+(1−σ)δx2Uin−1+fin−1+σ+(r6)in1⩽i⩽M−1,1⩽n⩽N, (2.6.4)
且存在正常数 c 6 c_{6} c6 使得
∣ ( r 6 ) i n ∣ ⩽ c 6 ( τ 2 + h 2 ) , 1 ⩽ i ⩽ M − 1 , 1 ⩽ n ⩽ N . \left|\left(r_{6}\right)_{i}^{n}\right| \leqslant c_{6}\left(\tau^{2}+h^{2}\right), \quad 1 \leqslant i \leqslant M-1,1 \leqslant n \leqslant N . ∣(r6)in∣⩽c6(τ2+h2),1⩽i⩽M−1,1⩽n⩽N.
注意到初边值条件, 有
{ U i 0 = φ ( x i ) , 1 ⩽ i ⩽ M − 1 , U 0 n = μ ( t n ) , U M n = ν ( t n ) , 0 ⩽ n ⩽ N \begin{cases}U_{i}^{0}=\varphi\left(x_{i}\right), & 1 \leqslant i \leqslant M-1, \\ U_{0}^{n}=\mu\left(t_{n}\right), & U_{M}^{n}=\nu\left(t_{n}\right), \quad 0 \leqslant n \leqslant N\end{cases} {
Ui0=φ(xi),U0n=μ(tn),1⩽i⩽M−1,UMn=ν(tn),0⩽n⩽N
在上式中略去小量项 ( r 6 ) i n \left(r_{6}\right)_{i}^{n} (r6)in, 并用数值解 u i n u_{i}^{n} uin 代替精确解 U i n U_{i}^{n} Uin, 可得求解问题的如下差分格式:
{ 1 s ∑ k = 0 n − 1 c k ( n , α ) ( u i n − k − u i n − k − 1 ) = σ δ x 2 u i n + ( 1 − σ ) δ x 2 u i n − 1 + f i n − 1 + σ , 1 ⩽ i ⩽ M − 1 , 1 ⩽ n ⩽ N , u i 0 = φ ( x i ) , 1 ⩽ i ⩽ M − 1 , u 0 n = μ ( t n ) , u M n = ν ( t n ) , 0 ⩽ n ⩽ N . \left\{\begin{array}{l} \frac{1}{s} \sum\limits_{k=0}^{n-1} c_{k}^{(n, \alpha)}\left(u_{i}^{n-k}-u_{i}^{n-k-1}\right)=\sigma \delta_{x}^{2} u_{i}^{n}+(1-\sigma) \delta_{x}^{2} u_{i}^{n-1}+f_{i}^{n-1+\sigma}, \\ 1 \leqslant i \leqslant M-1,1 \leqslant n \leqslant N, \\ u_{i}^{0}=\varphi\left(x_{i}\right), \quad 1 \leqslant i \leqslant M-1, \\ u_{0}^{n}=\mu\left(t_{n}\right), \quad u_{M}^{n}=\nu\left(t_{n}\right), \quad 0 \leqslant n \leqslant N . \end{array}\right. ⎩⎪⎪⎪⎪⎨⎪⎪⎪⎪⎧s1k=0∑n−1ck(n,α)(uin−k−uin−k−1)=σδx2uin+(1−σ)δx2uin−1+fin−1+σ,1⩽i⩽M−1,1⩽n⩽N,ui0=φ(xi),1⩽i⩽M−1,u0n=μ(tn),uMn=ν(tn),0⩽n⩽N.
分数阶微分方程数值算例
数值算例
考虑问题:
0 C D t α u ( x , t ) = ∂ 2 u ∂ x 2 ( x , t ) + f ( x , t ) , ( x , t ) ∈ ( 0 , 1 ) × ( 0 , 1 ) , { }_{0}^{C} D_{t}^{\alpha} u(x, t)=\frac{\partial^{2} u}{\partial x^{2}}(x, t)+f(x, t),(x, t) \in(0,1) \times(0,1), 0CDtαu(x,t)=∂x2∂2u(x,t)+f(x,t),(x,t)∈(0,1)×(0,1),
其中 0 < α < 1 0<\alpha<1 0<α<1, 右端源项 f ( x , t ) f(x, t) f(x,t) 和相应的初边值条件由真解 u ( x , t ) = sin ( x ) t 2 u(x, t)=\sin (x) t^{2} u(x,t)=sin(x)t2 确定.
源项和初边值条件
由真解 u ( x , t ) = sin ( x ) t 2 . u(x, t)=\sin (x) t^{2}. u(x,t)=sin(x)t2.
则有
{ u x = t 2 cos x u x x = − t 2 sin x { u t = 2 t sin x u t t = 2 sin x . \left\{\begin{array}{l}u_{x}=t^{2} \cos x \\ u_{x x}=-t^{2} \sin x\end{array} \quad\left\{\begin{array}{l}u_{t}=2 t \sin x \\ u_{t t}=2 \sin x .\end{array}\right.\right. {
ux=t2cosxuxx=−t2sinx{
ut=2tsinxutt=2sinx.
将真解对应部分带入微分方程从而反解右端源项 f ( x , t ) f(x, t) f(x,t)
0 c D t α u ( x , t ) = 1 Γ ( 1 − α ) ∫ 0 t 2 s sin x ( t − s ) α d s = − 2 sin x Γ ( 1 − α ) ∫ 0 t ( t − s − t ) ( t − s ) − α d s = − 2 sin x Γ ( 1 − α ) ∫ 0 t [ ( t − s ) 1 − α − t ( t − s ) − α ] d s = − 2 sin x Γ ( 1 − α ) [ ∫ 0 t − ( t − s ) 1 − α d ( t − s ) + ∫ 0 1 t ( t − s ) − α d ( t − s ) ] = − 2 sin x Γ ( 1 − α ) [ 1 2 − α ( t − s ) 2 − α ∣ t 0 + t 1 − α ( t − s ) 1 − α ∣ 0 t ] = − 2 sin x Γ ( 1 − α ) [ t 2 − α 2 − α + t 1 − α ⋅ ( − t 1 − α ) ] = − 2 sin x Γ ( 1 − α ) ⋅ − t 2 − α ( 2 − α ) ( 1 − α ) = 2 sin x Γ ( 3 − α ) t 2 − α \begin{aligned} { }_{0}^{c} D_{t}^{\alpha} u(x, t)&=\frac{1}{\Gamma(1-\alpha)} \int_{0}^{t} \frac{2s \sin x}{(t-s)^{\alpha}} d s\\ &=\frac{-2 \sin x}{\Gamma(1-\alpha)} \int_{0}^{t}(t-s-t)(t-s)^{-\alpha} d s\\ &=\frac{-2 \sin x}{\Gamma(1-\alpha)} \int_{0}^{t}\left[(t-s)^{1-\alpha}-t(t-s)^{-\alpha}\right] d s\\ &=\frac{-2 \sin x}{\Gamma(1-\alpha)}\left[\int_{0}^{t}-(t-s)^{1-\alpha} d(t-s)+\int_{0}^{1} t(t-s)^{-\alpha} d(t-s)\right]\\ &=\frac{-2 \sin x}{\Gamma(1-\alpha)}\left[\left.\frac{1}{2-\alpha}(t-s)^{2-\alpha}\right|_{t} ^{0}+\left.\frac{t}{1-\alpha}(t-s)^{1-\alpha}\right|_{0} ^{t}\right]\\ &=\frac{-2 \sin x}{\Gamma(1-\alpha)}\left[\frac{t^{2-\alpha}}{2-\alpha}+\frac{t}{1-\alpha} \cdot\left(-t^{1-\alpha}\right)\right]\\ &=\frac{-2 \sin x}{\Gamma(1-\alpha)} \cdot \frac{-t^{2-\alpha}}{(2-\alpha)(1-\alpha)}\\ &=\frac{2 \sin x}{\Gamma(3-\alpha)} t^{2-\alpha} \end{aligned} 0cDtαu(x,t)=Γ(1−α)1∫0t(t−s)α2ssinxds=Γ(1−α)−2sinx∫0t(t−s−t)(t−s)−αds=Γ(1−α)−2sinx∫0t[(t−s)1−α−t(t−s)−α]ds=Γ(1−α)−2sinx[∫0t−(t−s)1−αd(t−s)+∫01t(t−s)−αd(t−s)]=Γ(1−α)−2sinx[2−α1(t−s)2−α∣∣∣∣t0+1−αt(t−s)1−α∣∣∣∣0t]=Γ(1−α)−2sinx[2−αt2−α+1−αt⋅(−t1−α)]=Γ(1−α)−2sinx⋅(2−α)(1−α)−t2−α=Γ(3−α)2sinxt2−α
由于 0 C D t α u ( x , t ) = ∂ 2 u ∂ x 2 ( x , t ) + f ( x , t ) { }_{0}^{C} D_{t}^{\alpha} u(x, t)=\frac{\partial^{2} u}{\partial x^{2}}(x, t)+f(x, t) 0CDtαu(x,t)=∂x2∂2u(x,t)+f(x,t)
于是 f ( x , t ) = 0 C D t α u ( x , t ) − ∂ 2 u ∂ x 2 ( x , t ) = 2 sin x Γ ( 3 − α ) t 2 − α + t 2 sin x . \begin{aligned} f(x, t)&={ }_{0}^{C} D_{t}^{\alpha} u(x, t)-\frac{\partial^{2} u}{\partial x^{2}}(x, t)\\ &=\frac{2 \sin x}{\Gamma(3-\alpha)} t^{2-\alpha}+t^{2} \sin x. \end{aligned} f(x,t)=0CDtαu(x,t)−∂x2∂2u(x,t)=Γ(3−α)2sinxt2−α+t2sinx.
综上所述, 完整的数值算例为:
{ 0 C D t α u ( x , t ) = u x x ( x , t ) + 2 sin x Γ ( 3 − α ) t 2 − α + t 2 sin x , x ∈ ( 0 , L ) , t ∈ ( 0 , T ] u ( x , 0 ) = 0 , x ∈ ( 0 , L ) u ( 0 , t ) = 0 , u ( L , t ) = t 2 sin L , t ∈ [ 0 , T ] \left\{\begin{array}{l} { }_{0}^{C} D_{t}^{\alpha} u(x, t)=u_{x x}(x, t)+\frac{2 \sin x}{\Gamma(3-\alpha)} t^{2-\alpha}+t^{2} \sin x, \quad x \in(0, L), t \in(0, T] \\ u(x, 0)=0, \quad x \in(0, L) \\ u(0, t)=0, \quad u(L, t)=t^2\sin L, \quad t \in[0, T] \end{array}\right. ⎩⎨⎧0CDtαu(x,t)=uxx(x,t)+Γ(3−α)2sinxt2−α+t2sinx,x∈(0,L),t∈(0,T]u(x,0)=0,x∈(0,L)u(0,t)=0,u(L,t)=t2sinL,t∈[0,T]
代数系统
由差分格式:
1 s ∑ k = 0 n − 1 c k ( n , α ) ( u i n − k − u i n − k − 1 ) = σ δ x 2 u i n + ( 1 − σ ) δ x 2 u i n − 1 + f i n − 1 + σ \frac{1}{s} \sum\limits_{k=0}^{n-1} c_{k}^{(n, \alpha)}\left(u_{i}^{n-k}-u_{i}^{n-k-1}\right)=\sigma \delta_{x}^{2} u_{i}^{n}+(1-\sigma) \delta_{x}^{2} u_{i}^{n-1}+f_{i}^{n-1+\sigma} s1k=0∑n−1ck(n,α)(uin−k−uin−k−1)=σδx2uin+(1−σ)δx2uin−1+fin−1+σ
得到如下代数系统:
A ( u 1 n u 2 n ⋮ u M − 2 n u M − 1 n ) = b \mathbf{A}\left(\begin{array}{c}u_{1}^{n} \\ u_{2}^{n} \\ \vdots \\ u_{M-2}^{n} \\ u_{M-1}^{n}\end{array}\right)=\mathbf{b} A⎝⎜⎜⎜⎜⎜⎛u1nu2n⋮uM−2nuM−1n⎠⎟⎟⎟⎟⎟⎞=b
其中
A = ( 2 σ h 2 + τ − α Γ ( 2 − α ) c 0 ( n , α ) − σ h 2 − σ h 2 2 σ h 2 + τ − α Γ ( 2 − α ) c 0 ( n , α ) − σ h 2 ⋱ ⋱ 2 σ h 2 + τ − α Γ ( 2 − α ) c 0 ( n , α ) − σ h 2 ) \mathbf{A}=\left(\begin{array}{ccc}\frac{2\sigma}{h^{2}}+\frac{\tau^{-\alpha}}{\Gamma(2-\alpha)} c_{0}^{(n, \alpha)} & -\frac{\sigma}{h^{2}} &\\ -\frac{\sigma}{h^{2}} & \frac{2\sigma}{h^{2}}+\frac{\tau^{-\alpha}}{\Gamma(2-\alpha)} c_{0}^{(n, \alpha)} & -\frac{\sigma}{h^{2}} \\ &\ddots & \ddots &\\ & & \frac{2\sigma}{h^{2}}+\frac{\tau^{-\alpha}}{\Gamma(2-\alpha)} c_{0}^{(n, \alpha)}& -\frac{\sigma}{h^{2}}\end{array}\right) A=⎝⎜⎜⎜⎛h22σ+Γ(2−α)τ−αc0(n,α)−h2σ−h2σh22σ+Γ(2−α)τ−αc0(n,α)⋱−h2σ⋱h22σ+Γ(2−α)τ−αc0(n,α)−h2σ⎠⎟⎟⎟⎞
b = ( − 2 ( 1 − σ ) h 2 + τ − α Γ ( 2 − α ) c 0 ( n , α ) 1 − σ h 2 1 − σ h 2 − 2 ( 1 − σ ) h 2 + τ − α Γ ( 2 − α ) c 0 ( n , α ) 1 − σ h 2 ⋱ ⋱ − 2 ( 1 − σ ) h 2 + τ − α Γ ( 2 − α ) c 0 ( n , α ) 1 − σ h 2 ) ( u 1 n − 1 u 2 n − 1 ⋮ u M − 2 n − 1 u M − 1 n − 1 ) + σ h 2 ⋅ ( u 0 n 0 ⋮ 0 u M n ) + 1 − σ h 2 ⋅ ( u 0 n − 1 0 ⋮ 0 u M n − 1 ) − τ − α Γ ( 2 − α ) ∑ k = 1 n − 1 c k ( n , α ) ( u i n − k − u i n − k − 1 ) + f i n − 1 + σ \begin{aligned} \mathbf{b}&=\left(\begin{array}{ccc}-\frac{2(1-\sigma)}{h^{2}}+\frac{\tau^{-\alpha}}{\Gamma(2-\alpha)} c_{0}^{(n, \alpha)} & \frac{1-\sigma}{h^{2}} &\\ \frac{1-\sigma}{h^{2}} & -\frac{2(1-\sigma)}{h^{2}}+\frac{\tau^{-\alpha}}{\Gamma(2-\alpha)} c_{0}^{(n, \alpha)} & \frac{1-\sigma}{h^{2}} \\ &\ddots & \ddots &\\ & & -\frac{2(1-\sigma)}{h^{2}}+\frac{\tau^{-\alpha}}{\Gamma(2-\alpha)} c_{0}^{(n, \alpha)} & \frac{1-\sigma}{h^{2}} \end{array}\right)\left(\begin{array}{c}u_{1}^{n-1} \\ u_{2}^{n-1} \\ \vdots \\ u_{M-2}^{n-1} \\ u_{M-1}^{n-1}\end{array}\right)\\ &+\frac{\sigma}{h^{2}} \cdot\left(\begin{array}{c}u_{0}^{n} \\ 0 \\ \vdots \\ 0 \\ u_{M}^{n}\end{array}\right)+\frac{1-\sigma}{h^{2}} \cdot\left(\begin{array}{c}u_{0}^{n-1} \\ 0 \\ \vdots \\ 0 \\ u_{M}^{n-1}\end{array}\right)-\frac{\tau^{-\alpha}}{\Gamma(2-\alpha)} \sum_{k=1}^{n-1} c_{k}^{(n, \alpha)}\left(u_{i}^{n-k}-u_{i}^{n-k-1}\right)+f_{i}^{n-1+\sigma} \end{aligned} b=⎝⎜⎜⎜⎛−h22(1−σ)+Γ(2−α)τ−αc0(n,α)h21−σh21−σ−h22(1−σ)+Γ(2−α)τ−αc0(n,α)⋱h21−σ⋱−h22(1−σ)+Γ(2−α)τ−αc0(n,α)h21−σ⎠⎟⎟⎟⎞⎝⎜⎜⎜⎜⎜⎛u1n−1u2n−1⋮uM−2n−1uM−1n−1⎠⎟⎟⎟⎟⎟⎞+h2σ⋅⎝⎜⎜⎜⎜⎜⎛u0n0⋮0uMn⎠⎟⎟⎟⎟⎟⎞+h21−σ⋅⎝⎜⎜⎜⎜⎜⎛u0n−10⋮0uMn−1⎠⎟⎟⎟⎟⎟⎞−Γ(2−α)τ−αk=1∑n−1ck(n,α)(uin−k−uin−k−1)+fin−1+σ
数值结果
时间方向参数选取:
时间方向数值结果:
空间方向参数选取:
空间方向数值结果:
程序运行时间:
时间步长为 0.000100 空间步长为 0.012500 时 L1 插值逼近历时 166.610357 秒.
误差Error =6.9756e-07
而相同的网格剖分及精度下, 基于 L2-1 σ _\sigma σ 逼近的快速差分方法仅需历时 4.232766 秒.
误差Error =6.6569e-06
源代码
主程序:
clc,clear
tic
%% 初始化定解问题
alpha=0.2;
sigma=1-alpha/2;
tau=1/10000; T_a=0; T_b=1;
h=1/80; L_a=0; L_b=1;
M=(L_b-L_a)/h;
N=(T_b-T_a)/tau;
x=L_a:h:L_b; t=T_a:tau:T_b;
u=zeros(M+1,N+1); % @u 初始化数值解
u(:,1)=f_ic(x,T_a,0,0); % 定义初值条件
u(1,:)=f_bc(L_a,t,0,0); % 定义左边界条件
u(M+1,:)=f_bc(L_b,t,0,0); % 定义右边界条件
s=tau^alpha*gamma(2-alpha);
%% 两层格式（启动层）
n=2;
m=n-1;
% 创建代数系统
% 系数矩阵 Coefficient_Matrix_A
Coefficient_Matrix_A=toeplitz([1/s*c(0,m,alpha,sigma)+2*sigma/h^2,-sigma/h^2,zeros(M-3,1)']);
% 右端 Right_Term_B
Matrix_B=toeplitz([1/s*c(0,m,alpha,sigma)-2*(1-sigma)/h^2,(1-sigma)/h^2,zeros(M-3,1)']);
Right_Term_B=Matrix_B*u(2:M,n-1)...
+f_source(x(2:M),t(n-1)+sigma*tau,alpha)'...
+(1-sigma)/h^2*[u(1,n-1);zeros(M-3,1);u(M+1,n-1)]...
+sigma/h^2*[u(1,n);zeros(M-3,1);u(M+1,n)];
% 求解代数系统（由 k-2 层及第 k-1 层求第 k 层的值）
u(2:M,n)=Coefficient_Matrix_A\Right_Term_B;
fprintf('进程:\t%d/%d\n',n,N+1)
%% 三层格式
for n=3:N+1
% 创建代数系统
m=n-1;
% 生成求和项
S=zeros(M-1,1);
for k=1:m-1
S=S+c(k,m,alpha,sigma).*(u(2:M,m-k+1)-u(2:M,m-k-1+1));
end
% 系数矩阵 Coefficient_Matrix_A
Coefficient_Matrix_A=toeplitz([1/s*c(0,m,alpha,sigma)+2*sigma/h^2,-sigma/h^2,zeros(M-3,1)']);
Matrix_B=toeplitz([1/s*c(0,m,alpha,sigma)-2*(1-sigma)/h^2,(1-sigma)/h^2,zeros(M-3,1)']);
% 右端 Right_Term_B
Right_Term_B1=Matrix_B*u(2:M,n-1)...
+f_source(x(2:M),t(n-1)+sigma*tau,alpha)'...
+(1-sigma)/h^2*[u(1,n-1);zeros(M-3,1);u(M+1,n-1)]...
+sigma/h^2*[u(1,n);zeros(M-3,1);u(M+1,n)]...
-1/s*S;
% 求解代数系统（由 k-2 层及第 k-1 层求第 k 层的值）
u(2:M,n)=Coefficient_Matrix_A\Right_Term_B1;
fprintf('进程:\t%d/%d\n',n,N+1)
end
% clc
fprintf('时间步长为 %f 空间步长为 %f 时 L2_1_sigma 插值逼近 %d\n',tau,h)
toc
%% 误差分析
U=zeros(size(u));
for m=1:M+1
for n=1:N+1
U(m,n)=u_exact(x(m),t(n),0,0);
end
end
Error=max(max(abs(u-U)))
%% 绘图
figure
plot(t,u(7,:))
hold on
plot(t,U(7,:))
legend('数值解','精确解')
此处由于字数限制, 仅展示了主程序代码部分, 若需要绘图及误差分析等完整代码, 请在评论区留下邮箱.
参考文献
孙志忠,高广花.分数阶微分方程的有限差分方法(第二版).北京：科学出版社,2021.
本人水平有限, 若有不妥之处, 恳请批评指正.
作者:图灵的猫
作者邮箱: [email protected]