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Shikhnazar Ismailov
Shikhnazar Ismailov el 10 de Feb. de 2022
Comentada: Shikhnazar Ismailov el 13 de Feb. de 2022
n=41; h=1/(n-1); nt=360; tau=0.001;
for i=1:n
x(i)=(i-1)*h;
y(i)=(i-1)*h;
end
for j=1:n
for i=1:n
p0(i,j)=1;
end
end
for jt=1:nt
t(jt)=jt*tau;
for j=1:n % Anlitik yechimni hisoblash
for i=1:n
pt(i,j)=exp(-2*x(i)*y(j)*t(jt));
end
end
end
tic
for jt=1:nt/2 % Vaqt bo`yicha tsikl}
[po,pp]=fxy(n,tau,h,p0,x,y,t,jt);
[po,pp]=fyx(n,tau,h,p0,x,y,t,jt);
end % jt
toc
for j=1:n
px(j)=pp(21,j);
pxt(j)=pt(21,j);
end
x=0:h:1;
y=0:h:1;
figure ('Position',[600 160 700 600]); %Natijalarni figuraga chiqarish oynasi: 750-eni, 650-boyi, 700-chapdan masofa, 75-yuqoridan masofa
subplot(3,2,1);
meshc(x,y,pt) %Uch o'lchovli 3D grafika
title('АНИК ЕЧИМ 3D ГРАФИГИ')
subplot(3,2,2);
meshc(x,y,pp) %Uch o'lchovli 3D grafika
title('CОНЛИ ЕЧИМ 3D ГРАФИГИ')
subplot(3,2,3);
contour(x,y,pt,'ShowText','on','LineWidth',2); %kontur chizish
title('АНИК ЕЧИМ КОНТУР ГРАФИГИ')
subplot(3,2,4);
contour(x,y,pp,'ShowText','on','LineWidth',2); %kontur chizish
title('CОНЛИ ЕЧИМ КОНТУР ГРАФИГИ')
subplot(3,2,5);
plot(x,px,x,pxt); %Ikki o'lchovli grafika
title('КЕСИМДА АНИК ВА CОНЛИ ЕЧИМ ЎЗГАРИШИ X БЎЙИЧА')
function [p0,p] = fxy(n,tau,h,p0,x,y,t,jt)
% k+0.5 vaqt qatlamida hisoblash
for j=2:n-1 % Tenglamalar koeffitsientini hisoblash
for i=2:n-1
%chekli ayirma tenglama koeffitsentlarini hisoblash
a(i)=1;
c(i)=1;
b(i)=2+2*h*h/tau;
ff=h*h*(-2*(2*t(jt)*t(jt)*(x(i)*x(i)+y(j)*y(j))+x(i)*y(j))*exp(-2*x(i)*y(j)*t(jt)));
d(i)=2*h*h/tau*p0(i,j)+(p0(i,j-1)-2*p0(i,j)+p0(i,j+1))+ff;
end
%Pragonka koeffitsentlarini hisoblash
aa1(1)=0; bb1(1)=1;
for i=2:n-1
aa1(i)=c(i)/(b(i)-aa1(i-1)*a(i));
bb1(i)=(d(i)+bb1(i-1)*a(i))/(b(i)-aa1(i-1)*a(i));
end
% bosim funktsiyasini hisoblash
p(n,j)=exp(-2*y(j)*t(jt));
for i=n-1:-1:1
p(i,j)=aa1(i)*p(i+1,j)+bb1(i);
end
end %j
for i=1:n % Chegaraviy qiymatlarni hisoblash
p(i,n)=(4.0*p(i,n-1)-p(i,n-2))/3;
p(i,1)=(4.0*p(i,2)-p(i,3))/3;
end
for j=1:n % k+1 qatlam uchun boshlangich qiymatni hisoblash
for i=1:n
p0(i,j)=p(i,j);
end
end
% k+1 vaqt qatlamida hisoblash
for i=2:n-1 % Tenglamalar koeffitsientini hisoblash
for j=2:n-1
%chekli ayirma tenglama koeffitsentlarini hisoblash
a(j)=1;
c(j)=1;
b(j)=2+2*h*h/tau;
ff=h*h*(-2*(2*t(jt)*t(jt)*(x(i)*x(i)+y(j)*y(j))+x(i)*y(j))*exp(-2*x(i)*y(j)*t(jt)));
d(j)=2*h*h/tau*p0(i,j)+(p0(i-1,j)-2*p0(i,j)+p0(i+1,j))+ff;
end
%Pragonka koeffitsentlarini hisoblash
aa1(1)=0; bb1(1)=1;
for j=2:n-1
aa1(j)=c(j)/(b(j)-aa1(j-1)*a(j));
bb1(j)=(d(j)+bb1(j-1)*a(j))/(b(j)-aa1(j-1)*a(j));
end
% bosim funktsiyasini hisoblash
p(i,n)=exp(-2*x(i)*t(jt));
for j=n-1:-1:1
p(i,j)=aa1(j)*p(i,j+1)+bb1(j);
end
end %i
for j=1:n % Chegaraviy qiymatlarni hisoblash
p(n,j)=(4.0*p(n-1,j)-p(n-2,j))/3;
p(1,j)=(4.0*p(2,j)-p(3,j))/3;
end
for i=1:n % keyingi qatlam uchun boshlangich qiymatni hisoblash
for j=1:n
p0(i,j)=p(i,j);
end
end
end
function [p0,p] = fyx(n,tau,h,p0,x,y,t,jt)
% k+1 vaqt qatlamida hisoblash
for i=2:n-1 % Tenglamalar koeffitsientini hisoblash
for j=2:n-1
%chekli ayirma tenglama koeffitsentlarini hisoblash
a(j)=1;
c(j)=1;
b(j)=2+2*h*h/tau;
ff=h*h*(-2*(2*t(jt)*t(jt)*(x(i)*x(i)+y(j)*y(j))+x(i)*y(j))*exp(-2*x(i)*y(j)*t(jt)));
d(j)=2*h*h/tau*p0(i,j)+(p0(i-1,j)-2*p0(i,j)+p0(i+1,j))+ff;
end
%Pragonka koeffitsentlarini hisoblash
aa1(1)=0; bb1(1)=1;
for j=2:n-1
aa1(j)=c(j)/(b(j)-aa1(j-1)*a(j));
bb1(j)=(d(j)+bb1(j-1)*a(j))/(b(j)-aa1(j-1)*a(j));
end
% bosim funktsiyasini hisoblash
p(i,n)=exp(-2*x(i)*t(jt));
for j=n-1:-1:1
p(i,j)=aa1(j)*p(i,j+1)+bb1(j);
end
end %i
for j=1:n % Chegaraviy qiymatlarni hisoblash
p(n,j)=(4.0*p(n-1,j)-p(n-2,j))/3;
p(1,j)=(4.0*p(2,j)-p(3,j))/3;
end
for i=1:n % keyingi qatlam uchun boshlangich qiymatni hisoblash
for j=1:n
p0(i,j)=p(i,j);
end
end
% k+0.5 vaqt qatlamida hisoblash
for j=2:n-1 % Tenglamalar koeffitsientini hisoblash
for i=2:n-1
%chekli ayirma tenglama koeffitsentlarini hisoblash
a(i)=1;
c(i)=1;
b(i)=2+2*h*h/tau;
ff=h*h*(-2*(2*t(jt)*t(jt)*(x(i)*x(i)+y(j)*y(j))+x(i)*y(j))*exp(-2*x(i)*y(j)*t(jt)));
d(i)=2*h*h/tau*p0(i,j)+(p0(i,j-1)-2*p0(i,j)+p0(i,j+1))+ff;
end
%Pragonka koeffitsentlarini hisoblash
aa1(1)=0; bb1(1)=1;
for i=2:n-1
aa1(i)=c(i)/(b(i)-aa1(i-1)*a(i));
bb1(i)=(d(i)+bb1(i-1)*a(i))/(b(i)-aa1(i-1)*a(i));
end
% bosim funktsiyasini hisoblash
p(n,j)=exp(-2*y(j)*t(jt));
for i=n-1:-1:1
p(i,j)=aa1(i)*p(i+1,j)+bb1(i);
end
end %j
for i=1:n % Chegaraviy qiymatlarni hisoblash
p(i,n)=(4.0*p(i,n-1)-p(i,n-2))/3;
p(i,1)=(4.0*p(i,2)-p(i,3))/3;
end
for j=1:n % k+1 qatlam uchun boshlangich qiymatni hisoblash
for i=1:n
p0(i,j)=p(i,j);
end
end
end
The graph looks like this.
The graphic should actually be like that!

Respuestas (1)

Benjamin Thompson
Benjamin Thompson el 10 de Feb. de 2022
Here you are drawing two lines:
subplot(3,2,5);
plot(x,px,x,pxt); %Ikki o'lchovli grafika
title('КЕСИМДА АНИК ВА CОНЛИ ЕЧИМ ЎЗГАРИШИ X БЎЙИЧА')
If you only intend to draw one line, remove the second dataset:
plot(x, px);
  2 comentarios
Shikhnazar Ismailov
Shikhnazar Ismailov el 11 de Feb. de 2022
Both graphics need to be released!
Shikhnazar Ismailov
Shikhnazar Ismailov el 13 de Feb. de 2022
The second function takes the values of p and p0 derived from the first function. The first function takes the values of p and p0 derived from the second function. Only in the beginning does the first function work with the above values of p and p0, otherwise it must calculate the mutual exchange. How can I do this?

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R2017b

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