bvodeS - simplified call of bvode
This interface program simplifies the call of bvode , a program for the numerical solution of multi-point boundary value problems for mixed order systems of ordinary differential equations. The Scilab program bvode is adapted from the fortran program colnew. See the paper by U. Asher, J. Christiansen, and R.D. Russel: Collocation software for boundary-value ODE's, ACM Trans. Math. Soft. 7:209-222, 1981. The following examples should demonstrate not only how such systems can be solved with the help of bvodeS, but also should emphazise some important problems which occur with boundary value problems for ordinary ode's.
// 1. Modified example from help bvode. // DE: y1''''(x)=(1-6*x*x*y1'''(x)-6*x*y1''(x))/(y2(x)^3) // y2'(x)=1 // z=[y1 ; y1' ; y1'' ; y1''' ; y2] // BV: y1(1)=0 y1''(1)=0 // BV: y1(2)=1 y1''(2)=0 y2(2)=2 function RhS=fsub(x,z) RhS=[(1-6*x*x*z(4)-6*x*z(3))/(z(5)^3);1] endfunction function g=gsub(i,z) g=[z(1) z(3) z(1)-1 z(3) z(5)-2] g=g(i) endfunction function [z,lhS]=ystart(x) z=zeros(5,1);z(5)=1; lhS=[0;1]; endfunction n=2; m=[4 1]; N=100; a=1; b=2; zeta=[a a b b b]; x=linspace(a,b,N); ltol=4; // We want to change the default error for y1'''. tol=1e-12; tic() z=bvodeS(x,m,n,a,b,fsub,gsub,zeta,ltol=ltol,tol=tol,ystart=ystart); // Try tol=1e-14 etc. toc() function z=yex(x) // True solution z=zeros(5,1); z(1)=0.25*(10*log(2)-3)*(1-x)+0.5*(1/x+(3+x)*log(x)-x)+(x-1) z(2)=-0.25*(10*log(2)-3)+0.5*(-1/x^2+(3+x)/x+log(x)-1)+1 z(3)=0.5*(2/x^3+1/x-3/x^2) z(4)=0.5*(-6/x^4-1/x/x+6/x^3) z(5)=x endfunction zex=[];for xx=x, zex=[zex yex(xx)]; end scf(0); clf(); plot2d(x,abs(z-zex)',style=[1 2 3 5 6]) xtitle('Absolute error','x',' ') legend(['z1(x)';'z2(x)';'z3(x)';'z4(x)';'z5(x)']) // example #2. An eigenvalue problem // y''(x)=-la*y(x) // BV: y(0)=y'(0); y(1)=0 // Eigenfunctions and eigenvalues are y(x,n)=sin(s(n)*(1-x)), la(n)=s(n)^2, // where s(n) are the zeros of f(s,n)=s+atan(s)-(n+1)*pi, n=0,1,2,... // To get a third boundary condition, we choose y(0)=1 // (With y(x) also c*y(x) is a solution for each constant c.) // We solve the following ode system: // y''=-la*y // la'=0 // BV: y(0)=y'(0), y(0)=1; y(1)=0 // z=[y(x) ; y'(x) ; la] function rhs=fsub(x,z) rhs=[-z(3)*z(1);0] endfunction function g=gsub(i,z) g=[z(1)-z(2) z(1)-1 z(1)] g=g(i) endfunction // The following start function is good for the first 8 eigenfunctions. function [z,lhs]=ystart(x,z,la0) z=[1;0;la0] lhs=[0;0] endfunction a=0;b=1; m=[2;1]; n=2; zeta=[a a b]; N=101; x=linspace(a,b,N)'; // We have s(n)-(n+1/2)*pi -> 0 for n to infinity. la0=input('n-th eigenvalue: n= ?');la0=(%pi/2+la0*%pi)^2; z=bvodeS(x,m,n,a,b,fsub,gsub,zeta,ystart=list(ystart,la0)); clf() plot2d(x,[z(1,:)' z(2,:)'],style=[5 1],axesflag=5) xtitle(['Startvalue = '+string(la0);'Eigenvalue = '+string(z(3,1))],'x',' ') legend(['y(x)';'y''(x)']) // example #3. A boundary value problem with more than one solution. // DE: y''(x)=-exp(y(x)) // BV: y(0)=0; y(1)=0 // This boundary value problem has more than one solution. // It is demonstrated how to find two of them with the help of // some preinformation of the solutions y(x) to build the function ystart. // z=[y(x);y'(x)] a=0;b=1;m=2;n=1; zeta=[a b]; N=101; tol=1e-8*[1 1]; x=linspace(a,b,N); function rhs=fsub(x,z),rhs=-exp(z(1));endfunction function g=gsub(i,z) g=[z(1) z(1)] g=g(i) endfunction function [z,lhs]=ystart(x,z,M) //z=[4*x*(1-x)*M ; 4*(1-2*x)*M] z=[M;0] //lhs=[-exp(4*x*(1-x)*M)] lhs=0 endfunction for M=[1 4] if M==1 z=bvodeS(x,m,n,a,b,fsub,gsub,zeta,ystart=list(ystart,M),tol=tol); else z1=bvodeS(x,m,n,a,b,fsub,gsub,zeta,ystart=list(ystart,M),tol=tol); end end // Integrating the ode yield e.g. the two solutions yex and yex1. function y=f(c),y=c.*(1-tanh(sqrt(c)/4).^2)-2;endfunction c=fsolve(2,f); function y=yex(x,c) y=log(c/2*(1-tanh(sqrt(c)*(1/4-x/2)).^2)) endfunction function y=f1(c1), y=2*c1^2+tanh(1/4/c1)^2-1;endfunction c1=fsolve(0.1,f1); function y=yex1(x,c1) y=log((1-tanh((2*x-1)/4/c1).^2)/2/c1/c1) endfunction disp(norm(z(1,:)-yex(x)),'norm(yex(x)-z(1,:))= ') disp(norm(z1(1,:)-yex1(x)),'norm(yex1(x)-z1(1,:))= ') clf(); subplot(2,1,1) plot2d(x,z(1,:),style=[5]) xtitle('Two different solutions','x',' ') subplot(2,1,2) plot2d(x,z1(1,:),style=[5]) xtitle(' ','x',' ') // example #4. A multi-point boundary value problem. // DE y'''(x)=1 // z=[y(x);y'(x);y''(x)] // BV: y(-1)=2 y(1)=2 // Side condition: y(0)=1 a=-1;b=1;c=0; // The side condition point c must be included in the array fixpnt. n=1; m=[3]; function rhs=fsub(x,z) rhs=1 endfunction function g=gsub(i,z) g=[z(1)-2 z(1)-1 z(1)-2] g=g(i) endfunction N=10; zeta=[a c b]; x=linspace(a,b,N); z=bvodeS(x,m,n,a,b,fsub,gsub,zeta,fixpnt=c); function y=yex(x) y=x.^3/6+x.^2-x./6+1 endfunction disp(norm(yex(x)-z(1,:)),'norm(yex(x)-z(1,:))= ')
Rainer von Seggern