필터 지우기
필터 지우기

yy3 is generating a straight line while it should generate a curve, i don't understand what is wrong!!!

조회 수: 6 (최근 30일)
function ABC
% Initialization of paramters
Pr=6.2;
M=4;
Bi=0.3;
beta=0.2;
ks1=0.5;
ks2=0.1;
rhos1=0.2;
rhos2=0.3;
n=3;
kbf=0.598;
phi1=0.1;
phi2=0.04;
rhocps1=0.3;
rhocps2=0.1;
rhof=997.1;
kf=0.613;
rhocpf=4179;
A1=((1-phi1)^(2.5)).*((1-phi2)^(2.5));
B1=(1-phi2).*((1-phi1).*rhof + rhos1.*phi1) + rhos2.*phi2;
A2=B1*(1/rhof); % rho_Hnf/rho_f
B2=ks2+(n-1)*kbf-(n-1)*(kbf-ks2).*phi2;
B3=ks2+(n-1)*kbf+(kbf-ks2).*phi2;
C5=B2/B3; % k_Hnf/k_bf Nanofluid Constant
B4=ks1+(n-1)*kf-(n-1)*(kf-ks1).*phi1;
B5=ks1+(n-1)*kf+(kf-ks1).*phi1;
C6=B4/B5; % k_bf/k_f
A4=C5*C6; % k_Hnf/k_f
B6=(1-phi2)*((1-phi1)*rhocpf+phi1*rhocps1)+phi2*rhocps2; % rhocp_Hnf
B7=1/rhocpf;
A3=B6*B7; % rhocp_Hnf/rhocp_f
% Initial Condition Input
sol = bvpinit(linspace(0,8,100), [1 0 0 0 0 0 0 0]);
% solution in structure form
sol1 = bvp4c(@bvpexam2, @bcexam2, sol);
x1 = sol1.x;
y1 = sol1.y;
%%% Plotting of the temperature
plot(x1, y1(7, :))
hold on
function res=bcexam2(y0, yinf)
res=[y0(2)-1;y0(5)-beta;y0(1)+y0(4);y0(8)+(Bi/A4)*(1-y0(7)); yinf(2);yinf(5);yinf(7); yinf(4)];
end
function dydx = bvpexam2(~,y)
yy1=A1*(A2*(y(2)^2-(y(1)+y(4))*y(3))+M*y(2));
yy2 = A1*(A2*(y(5)^2-(y(1)+y(4))*y(6))+M*y(5));
yy3 =-(A3/A4)*Pr*(y(1)+y(4))*y(8);
dydx= [y(2);y(3);yy1;y(5);y(6);yy2;y(7);yy3];
end
end
  댓글 수: 4

댓글을 달려면 로그인하십시오.

채택된 답변

Torsten
Torsten 2023년 8월 11일
편집: Torsten 2023년 8월 11일
As initial condition for y(7), you set y(7) = 0, and you define the differential equation for y(7) as dy(7)/dx = y(7). The solution is y(7) = 0 for all x - and that's what is plotted.
The line
dydx= [y(2);y(3);yy1;y(5);y(6);yy2;y(7);yy3];
must read
dydx= [y(2);y(3);yy1;y(5);y(6);yy2;y(8);yy3];
  댓글 수: 3
Fareeha
Fareeha 2023년 8월 11일
@Torsten i have another question
in function
function res=bcexam2(y0, yinf)
res=[y0(2)-1;y0(5)-beta;y0(1)+y0(4);y0(8)+(Bi/A4)*(1-y0(7)); yinf(2);yinf(5);yinf(7)];
end
I removed one BC, now i have 8 vectors in dydx and 7 BC, in such condition how could code work?
Torsten
Torsten 2023년 8월 11일
You need as many boundary conditions as there are first-order differential equations. Otherwise, you get a one-dimensional solution manifold.
Consider
y '' = c with y(0) = 1
The solution is
y(x) = 1 + a1*x + c/2*x^2
with arbitrary parameter a1.

댓글을 달려면 로그인하십시오.

추가 답변 (0개)

카테고리

Help CenterFile Exchange에서 Visual Exploration에 대해 자세히 알아보기

태그

제품


릴리스

R2018a

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by