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1、《粘性流体力学》程序平板边界层问题求解编程思路平面边界层问题可以归结为在已知边界层条件下解一个高阶微分方程，即解厂+0.5〃、0。Matlab提供了解微分方程的方法，运用换元法将高阶微分方程降阶，然后运用“ode45〃函数进行求解。函数其难点在于如何将边界条件中“tco,广t1运用好，由四阶龙格■库塔方法知其核心是换元试算匹配，故在运用函数时通过二分法实现〃T8,广T1是可行的。1.2m函数functiondy=rigid(x,y)dy=zeros(3,1);dy(1)=y(2);dy(2)=y(3);dy(
2、3)=-0.5*y(l)*y(3);%main程序[X,Y]=ode45(〒igid',[05],[000]);plot(X,Y(:,1)HX,Y(:,2),'*',X,Y(:,3),'+')%二分法试算f〃的初始值以满足F趋向无穷时的边界条件，图像上可以清晰看出F无穷时的结果»[X,Y]=ode45('rigid',[05],[001]);plot(X,Y(：M=X,Y(:,2),*：X,Y(:,3),屮)»[X,Y]=ode45('ngid'z[05],[000.5]);plot(X,Y(:J),=X,Y
3、(:,2)广,X,Y(:,3),屮)»[X,Y]=ode45('rigid',[05],[000.25]);plot(X,Y(:,—X,Y(:,2),TX,Y(:,3)”)»[X,Y]=ode45('ngid'z[05],[000.375]);plot(X,YWX,Y(:,2)严,X,Y(:,3)”)»gridon»[X,Y]=ode45('rigid,,[05],[000.3125]);plot(X,Y(:,—X,Y(:,2),TX,Y(:,3)”)»gridon»[X,Y]=ode45(,rigid'/[
4、05],[000.34375]);plot(X,Y(:Y(:,2),匕X,Y(⑶，屮)gridon»[X,Y]=ode45('rigid：[05],[000.328125]);plot(X,Y(:,l)yx,Y(:,2),匕X,Y(:,3),屮)gridon»[X,Y]=ode45('rigid',[010],[000.328125]);%当f〃为0.328125时，逼近结果已经很好,在0到5的变化范围内已经非常接近精确解plot(X,WJX,Y(:,2)广,X,Y(:,3),屮)gridon»[X,Y]=o
5、de45('rigid',[05],[000.335975]);plot(X,Y(J)P,X,Y(:,2),匕X,Y(:,3),屮)gridon1.3数据展示选取r=0.335975时的数据展示，X代表〃的变化，Y的笫一二三列代表仁的值。数据如下表：ffr0000.33600.00010.00000.00010.33600.00030.00000.00010.33600.00040.00000.00020.33600.00060.00000.00020.33600.00130.00000.00050.3360
6、0.00210.00000.00070.33600.00280.00000.00100.33600.00360.00000.00120.33600.00730.00000.00250.33600.01110.00000.00370.33600.01480.00000.00500.33600.01850.00010.00620.33600.03720.00020.01250.33600.05590.00050.01880.33600.07460.00090.02510.33600.09330.00150.031
7、30.33600.18680.00590.06270.33590.28020.01320.09410.33580.37370.02350.12550.33550.46710.03660.15680.33500.59210.05890.19870.33400.71710.08630.24030.33250.84210.11890.28180.33040.96710.15670.32290.32761.09210.19960.36360.32401.21710.24760.40380.31951.34210.30
8、060.44340.31411.46710.35840.48230.30771.59210.42110.52030.30031.71710.48850.55730.29191.84210.56040.59320.28251.96710.63670.62790.27212.09210.71730.66120.26082.21710.80200.69310.24872.34210.89050.72340
