How to make graph that plot between Analytical Solution (Exact Solution) and Numerical Method Solution?

2022 3 5
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更新时间：2022 3 6
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Greetings!
Can someone help to guide me how to make graph that plot between Analytical Solution (Exact Solution) and Adams-Moulton Solution? Here I attached my script.
%% 2nd-order Adams-Moulton Solution
fun = @(t,y) ((1+4*t)*((y)^1/2));
y0 = 1;
tspan = [0,1];
N = 4;
%% Initial Values
h = (tspan(2) - tspan(1))/N;
exactY = @(t) (((3*t)/2)+3*t.^2).^(2/3);
t1 = tspan(1) + h; y1 = exactY(t1);
t2 = tspan(1) + 2*h; y2 = exactY(t2);
[t2,Y2] = AM2(fun,tspan,y0,y1,N);
%% Display Solution
Y = exactY(t2);
disp('-----------------------------');
disp('t_i y(t_i) AM2 Error')
disp('-----------------------------');
formatSpec = '%.2f %.5f %.5f %.5f\n';
fprintf(formatSpec,[t2';Y';Y2';abs(Y'-Y2')])
I really appreciate any help you can provide. Thank you.

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Torsten
Torsten 2022-3-5
编辑：Torsten 2022-3-5
The Adams-Moulton method is an implicit method.
You can see this by the fact that in your code, Fiplus1 depends on the unknown y(i+1).
As a consequence, you will have to iterate or use a nonlinear equations solver like "fsolve" or "fzero" to solve the equation
y(i+1) = y(i) + (h/12)*(5*fun(t(i+1),y(i+1)) + 8*Fi - Fi1)
for y(i+1).
In your code, you use y(i+1) = 0 on the right-hand side of the equation
y(i+1) = y(i) + (h/12)*(5*fun(t(i+1),y(i+1)) + 8*Fi - Fi1)
because you initialized the y-vector to zero. This is wrong.
Hi. Thank you for reviewing my script. Do you mind to guide me how to make the right script of Adams-Moulton? I get confused. Where do I supposed to put fsolve or fzero in my script?
Torsten
Torsten 2022-3-5
fun = @(t,y) (1+4*t)*sqrt(y);
y0 = 1;
tspan = [0,1];
N = 4;
%% Initial Values
h = (tspan(2) - tspan(1))/N;
exactY = @(t)(t/2 + t.^2 + 1).^2;
t1 = tspan(1) + h; y1 = exactY(t1);
[T,Ynum] = AM2(fun,tspan,y0,y1,N);
%% Display Solution
Yexact = exactY(T);
plot(T,[Ynum,Yexact])
function [t,y] = AM2(fun,tspan,y0,y1,N)
a = tspan(1);
b = tspan(2);
h = (b-a)/N;
t = zeros(N+1,1);
y = zeros(N+1,1);
t(1) = a; y(1) = y0;
t(2) = a+h; y(2) = y1;
for i = 2:N
t(i+1) = t(i) + h;
Fi = fun(t(i),y(i));
Fi1 = fun(t(i-1),y(i-1));
funh = @(yh) yh - y(i) - h/12*(5*fun(t(i+1),yh) + 8*Fi - Fi1);
y(i+1) = fsolve(funh,y(i));
end
end

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 采纳的回答

Alan Stevens
Alan Stevens 2022-3-5

0 个投票

After Y = exactY(t2) you probably need something like:
plot(t2,Y2,t2,Y)
legend('AM','Exact')
However, you haven't supplied the code for your AM function, so I can't check!

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Well, this is my AM2 function. I'll accept your answer if it succeed.
function [t,y] = AM2(fun,tspan,y0,y1,N)
a = tspan(1);
b = tspan(2);
h = (b-a)/N;
t = zeros(N+1,1);
y = zeros(N+1,1);
t(1) = a; y(1) = y0;
t(2) = a+h; y(2) = y1;
for i = 2:N
t(i+1) = t(i) + h;
Fi = fun(t(i),y(i));
Fi1 = fun(t(i-1),y(i-1));
Fiplus1 = fun(t(i+1),y(i+1));
y(i+1) = y(i) + (h/12)*(5*Fiplus1 + 8*Fi - Fi1)
end
Thank you.
Alan Stevens
Alan Stevens 2022-3-5
Ran in:
Ran in:
Like so:
%% 2nd-order Adams-Moulton Solution
fun = @(t,y) ((1+4*t)*((y)^1/2));
y0 = 1;
tspan = [0,1];
N = 4;
%% Initial Values
h = (tspan(2) - tspan(1))/N;
exactY = @(t) (((3*t)/2)+3*t.^2).^(2/3);
t1 = tspan(1) + h; y1 = exactY(t1);
t2 = tspan(1) + 2*h; y2 = exactY(t2);
[t2,Y2] = AM2(fun,tspan,y0,y1,N);
%% Display Solution
Y = exactY(t2);
plot(t2,Y2,t2,Y),grid
xlabel('t2'),ylabel('y2 and Y')
legend('AM','Exact')
disp('-----------------------------');
-----------------------------
disp('t_i y(t_i) AM2 Error')
t_i y(t_i) AM2 Error
disp('-----------------------------');
-----------------------------
formatSpec = '%.2f %.5f %.5f %.5f\n';
fprintf(formatSpec,[t2';Y';Y2';abs(Y'-Y2')])
0.00 0.00000 1.00000 1.00000 0.25 0.68142 0.68142 0.00000 0.50 1.31037 0.78457 0.52580 0.75 1.99248 0.96652 1.02596 1.00 2.72568 1.26418 1.46150
function [t,y] = AM2(fun,tspan,y0,y1,N)
a = tspan(1);
b = tspan(2);
h = (b-a)/N;
t = zeros(N+1,1);
y = zeros(N+1,1);
t(1) = a; y(1) = y0;
t(2) = a+h; y(2) = y1;
for i = 2:N
t(i+1) = t(i) + h;
Fi = fun(t(i),y(i));
Fi1 = fun(t(i-1),y(i-1));
Fiplus1 = fun(t(i+1),y(i+1));
y(i+1) = y(i) + (h/12)*(5*Fiplus1 + 8*Fi - Fi1);
end
end
Alan Stevens
Alan Stevens 2022-3-5
Looks like you have the exact solution wrong if your function is correct!
Assuming dy/dt = ((1+4*t)*((y)^1/2)) then y = (t/2 + t^2 + 1).^2 (with y(0) = 1)

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