1、數學物理建模與計算機仿真考試試題(20112012第二學期1、 計算機仿真計算積分: (n 為自然數 ;積分方向為正方向(10分。解:利用留數定理計算閉合路徑積分, 設g=1+zn ,f=z(2*n/(1+zn求出所有n 個奇點并放在矩陣A 中,A=solve(g再求出每個奇點k A 對應的留數,Res(k=limit(f*(z-A(k,z,A(k故所求的積分結果為:INT=2i Res(k對應程序為:n=input('n='syms zg=1+zn; f=z(2*n/(1+zn;A=solve(g;sum=0;for k=1:nRes=limit(f*(z-A(k,z,A(k

2、;sum=sum+Res;endint=2*pi*i*sumINT=simplify(int2、 在同一幅圖中繪制函數 和平面 的圖像(10分。解:參照教材第二頁,用cplxgrid 指令繪圖,去掉投影,程序如下:function cplxmap2(z,w,Bblue = 0.2; x = real(z;221n n z z dz z =+ 0x =253(f z z =-(1d 2i Res ,nk C k f z z f z z =y = imag(z;u = real(w;v = imag(w;if nargin > 2k = find(abs(w > B | isnan(a

3、bs(w;if length(k > 0u(k = B*sign(u(k;v(k = zeros(size(k;v = v/max(max(abs(v;v(k = NaN*ones(size(k;endendM = max(max(u;m = min(min(u;axis(-1 1 -1 1 m M;caxis(-1 1;s = ones(size(z;surf(x,y,u,v;colormap(hsv(64再參照第8頁繪出2/(5-3z的圖像及x=0的圖像,程序如下:z=cplxgrid(30;cplxmap2(z, 2./(5-3.*z+eps*(z=5/3;colorbar(

4、9;vert'title('2/(5-3z'hold ony,z=meshgrid(-1:0.01:1,-1:0.01:1;a=0; x2=a*ones(size(y;mesh(x2,y,z;hold off輸出圖像為: 3、 繪制貝塞爾函數 的圖形并標注出零點(10分。解:參考書上29頁繪圖,程序如下:y=besselj(0:3,(0:0.2:10'figure(1;plot(0:0.2:10',y;legend('J0','J1','J2','J3'hold onx=0:0.05:10;

5、LD=;for n=0:3y=besselj(n,x;for k=1:length(y-1if y(k*y(k+1<0h = interp1(y(k:k+1,x(k:k+1,0;LD=LD,h;endendendplot(LD,0,'o'hold off輸出圖像為:0123,J J J J 4、求任意兩個圓相交的區域,并求出相交區域的面積。(注:如果兩個圓不相交,輸出“兩個圓不相交”(20分。解:先寫一個繪制任意圓的函數,程序如下:function circle(x0,y0,Rfai=0:2*pi/100:2*pi;y=y0+R*sin(fai;x=x0+R*cos(fa

6、i;plot(x,y,'b','LineWidth',5;%fill(x,y,'c'hold on;axis equalend根據輸入繪制兩個圓并計算面積:function twocirclex1=input('X1='y1=input('Y1='r1=input('R1='x2=input('X2='y2=input('Y2='r2=input('R2='circle(x1,y1,r1;circle(x2,y2,r2;Num=0;for x=x1-r

7、1:0.1:x1+r1;for y=y1-r1:0.1:y1+r1;p1=sqrt(x-x12+(y-y12;p2=sqrt(x-x22+(y-y22;if (p1<=r1&&(p2<=r2plot(x,y,'c.'hold onNum=Num+1;endendendhold offArea=Num/100;if Area=0disp('兩個圓不相交'elseArea=Areaend5、 兩端固定的均勻弦的自由振動的定解問題是它的解是 其中系數是設初始位移為零,初速度為=-(,(,(,(,(,(x f t x u x t x u t

8、l x u t x u u a u t xx tt 0000002l x n l at n B l at n A t x u n n n =+=1sin sin (cos ,(,sin (d l n l A l n =02d ln f a n B l n =02sin (=(其余053521(l x l x f(1根據給出的解的表達式進行仿真,畫出動畫圖形(10分。 解:初位移為零初速不為零,故參照教材167頁繪圖:function strN=50;t=0:0.001:2.0;x=0:0.001:1;ww=string(N,0;h=plot(x,ww,'linewidth',3

9、;axis(0,1,-0.12,0.12s=;for n=2:length(tww=string(N,t(n;set(h,'ydata',ww;drawnow;s=s,sum(ww;endendfunction wtx=string(N,tx=0:0.001:1; a=1;wtx=0;for k=1:Nbk=2/(k*k*pi*pi*(cos(2*k*pi/5-cos(3*k*pi/5;wtx=wtx+bk*sin(k*pi*t*sin(k*pi*x;endend 輸出圖像為: (2用差分方程求出數值結果,畫出動畫圖形。(10分。 解:令x=ix,t=jt,將微分方程寫成差分方

10、程,即有:u i,j+1=c(u i+1,j+ u i-1,j+2(1-c u i,j- u i,j-1 其中初始條件可以表示成u i,1=u i,2=所用的程序是:clearN=4025;dx=0.0024;dt=0.0005;c=dt*dt/dx/dx;u(1:420,1=0;x=linspace(0,1,420'u(1:420,1=0;u(180:240,2=dt*0.5;h=plot(x,u(:,1,'linewidth',5;axis(0,1,-0.05,0.05;set(h,'EraseMode','xor','Mar

11、kerSize',18for k=2:Nset(h,'XData',x,'YData',u(:,2;drawnow;%pauseu(2:419,3=2*u(2:419,2-u(2:419,1+c*(u(3:420,2.-2*u(2:419,2+u(1:418,2;u(2:419,1=u(2:419,2;u(2:419,2=u(2:419,3;end 輸出圖像為:(3用PDE工具箱進行仿真,寫出仿真步驟。(10分。步驟:1、在Option/Axes Limits下選擇x軸的范圍為-0.11,y軸的范圍為-0.10.1。2、以原點為頂點畫一個長為1寬為0.0

12、8的矩形,矩形頂點為(0,-0.04,(1,-0.04,(1,0.04,(0,0.04。3、選擇Boundary/Boundary Mode選項4、矩形的左右邊界是齊次的狄里克利邊界條件,可取h=1,r=0;上下邊界則取齊次的諾依曼邊界條件,即g=0,q=0。5、選擇PDE/PDE Specification選項,選擇Hyperbolic選項,參數是c=1,a=0,f=0,d=1;6、選擇Mesh/Initialize Mesh選項,再連續兩次選擇Mesh/Refine Mesh選項,將初始化的網格作兩次細分。7、選擇Solve/Parameters選項,將Time改為0:0.01:2,u(t0設為0,u(t0設為tanh(10000*(x-0.4-tanh(10000*(x-0.6,然后按OK鍵。8、選擇plot/Parameters選項,勾選Height(3-D plot和Animation選項,按OK鍵。9、輸出圖像: 6、本題20分 已知一個波包從左邊入射,它在這5個區間的波函數分別為數學物理建模與