# how to solve this non-linear equations?

6 visualizzazioni (ultimi 30 giorni)
M.Rameswari Sudha il 4 Gen 2020
I couldnot find the value of v and w. let me know what is the mistake in the following code.
function solveeqs()
guess=[200 110];
[result,fval,exit,output]= fsolve(@eqns,guess);
result
fval
eqns(guess)
output
end
function fcns=eqns(z)
v=z(1);
w=z(2);
p=30;d=1;lamda=0.8;c=10;r=35;h=0.05;chi=1000;k=0.02;b=7;a =0.01;T =107;
%w=k*(1-exp(-a*chi));
M=k*a;
fcns(1)=(1/T)*[p*d-((lamda*p*d*v)/T)-(d*exp(-(w-k)*v)-(d*v/T))*c-d*((v/T)-1)*r-(d/(w-k))*(1-exp(-(w-k)*v)*h)];
fcns(2)=(1/T)*[-(d*c*v*exp(-(w-k)*v)*M/(w-k))+(((d*c*(1-exp(-(w-k)*v))+h*d*v*(1+exp(-(w-k)*v)))*M)/(w-k)^2)+((2*h*d*(exp(-(w-k)*v)-1)*M)/(w-k)^3)-1];
end
##### 3 CommentiMostra 1 commento meno recenteNascondi 1 commento meno recente
John D'Errico il 4 Gen 2020
"how do u get ? I didnot get the answer. let me know what is the mistake made by me"
M.Rameswari Sudha il 18 Mar 2020
Thank you for giving the answer. sorry for the delay response.

Accedi per commentare.

### Risposta accettata

John D'Errico il 4 Gen 2020
Modificato: John D'Errico il 4 Gen 2020
It is pretty easy to find at least one solution. vpasolve will do so, and my guess is, fsolve will do the same.
syms v w
p=30;d=1;lamda=0.8;c=10;r=35;h=0.05;chi=1000;k=0.02;b=7;a =0.01;T =107;
M=k*a;
fcns(1)=(1/T)*[p*d-((lamda*p*d*v)/T)-(d*exp(-(w-k)*v)-(d*v/T))*c-d*((v/T)-1)*r-(d/(w-k))*(1-exp(-(w-k)*v)*h)];
fcns(2)=(1/T)*[-(d*c*v*exp(-(w-k)*v)*M/(w-k))+(((d*c*(1-exp(-(w-k)*v))+h*d*v*(1+exp(-(w-k)*v)))*M)/(w-k)^2)+((2*h*d*(exp(-(w-k)*v)-1)*M)/(w-k)^3)-1];
sol = vpasolve(fcns,v,w)
sol =
struct with fields:
v: [1×1 sym]
w: [1×1 sym]
sol.v
ans =
98.167804480477247473056536377199
sol.w
ans =
0.070053725915927256484131799766249
Nonlinear solvers find a solution based on the starting guess. Use different start guesses, and you will find other solutions. vpasolve uses a default starting guess at (0,0) as I recall.
So where do the solutions lie? This is most easily done using a contour plot.
f1 = @(v,w) (1/T)*[p*d-((lamda*p*d*v)/T)-(d*exp(-(w-k).*v)-(d*v/T))*c-d*((v/T)-1).*r-(d./(w-k)).*(1-exp(-(w-k).*v)*h)];
f2 = @(v,w) (1/T)*[-(d*c*v.*exp(-(w-k).*v).*M./(w-k))+(((d*c*(1-exp(-(w-k).*v))+h*d*v.*(1+exp(-(w-k).*v)))*M)./(w-k).^2)+((2*h*d*(exp(-(w-k).*v)-1)*M)./(w-k).^3)-1];
H1 = fcontour(f1,[-200,200,-.25 .25]);
H1.LineColor = 'b';
H1.LevelList = 0;
hold on
H2 = fcontour(f2,[-200,200,-.25 .25]);
H2.LineColor = 'r';
H2.LevelList = 0;
grid on
xlabel 'v'
ylabel 'w'
Intersections of the red and blue lines indicate all solutions. There appear to be 6 solutions. Start in the vicinity of each solution, and you will get one of them.
The shape of the curves suggests there may be more solutions for large negative values of v, but I doubt it, as that would force w to change sign as we follow along that curve. More likely, the red curve will approach the asymptote at w==0 from below.
So, what did you do incorrectly? If fsolve failed to find at least one solution, then you should show what you did, and what error was the result.
I would conjecture the problem lies in your starting guess, which is terrible.
guess=[200 110];
which has a ridiculous initial value for w. 110 is so far out in the weeds, that it makes no sense in context of your equations.
p=30;d=1;lamda=0.8;c=10;r=35;h=0.05;chi=1000;k=0.02;b=7;a =0.01;T =107;
f1 = @(v,w) (1/T)*[p*d-((lamda*p*d*v)/T)-(d*exp(-(w-k).*v)-(d*v/T))*c-d*((v/T)-1).*r-(d./(w-k)).*(1-exp(-(w-k).*v)*h)];
f2 = @(v,w) (1/T)*[-(d*c*v.*exp(-(w-k).*v).*M./(w-k))+(((d*c*(1-exp(-(w-k).*v))+h*d*v.*(1+exp(-(w-k).*v)))*M)./(w-k).^2)+((2*h*d*(exp(-(w-k).*v)-1)*M)./(w-k).^3)-1];
solve(@(vw) [f1(vw(1),vw(2));f2(vw(1),vw(2))],[200,110])
Equation solved.
fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.
<stopping criteria details>
ans =
141.92 88.652
So why do I suggest that w makes no sense? w always seems to be paired with k, which is 0.02 = 1/50.
pretty(fcns(1))
/ / 1 \ \
exp| -v | w - -- | |
\ \ 50 / / / / 1 \ \
-------------------- - 1 10 exp| -v | w - -- | |
20 \ \ 50 / / 49 v 65
------------------------ - ----------------------- - ----- + ---
/ 1 \ 107 11449 107
| w - -- | 107
\ 50 /
So w should be something on the order of k.
fsolve(@(vw) [f1(vw(1),vw(2));f2(vw(1),vw(2))],[100,.05])
Equation solved.
fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.
<stopping criteria details>
ans =
98.168 0.070054
This system of equations is a bit tricky, as fsolve can return garbage solutions if you are not careful with the start guess. That is often true for problems that contain exponentials.
##### 1 CommentoMostra -1 commenti meno recentiNascondi -1 commenti meno recenti
M.Rameswari Sudha il 18 Mar 2020
Thank you for giving the answer. sorry for the delay response. Both are give the same solution with different concept. Thanks a lot

Accedi per commentare.

### Più risposte (1)

M.Rameswari Sudha il 11 Dic 2020
Modificato: Walter Roberson il 11 Dic 2020
I am unable to find the answer. let me know what is the mistake in the following code.
clc
clear all
D=1000;
P=3200;
A=25;
S=200;
Cp=25;
Cv=20;
r=0.2;
k=2.33;
I=0.1;
g=15;
th0=0.0002;
R1=[0 5.6 22.4 57.4];
j=1;
l=1;
for i=1:length(R1)
M1=(S*(r*Cp-(1-2*D/P)*r*Cv))/((A+R1(i))*(r*Cv*(1-D/P)+g*D*th0));
M(j)=M1;
j=j+1;
end
M
for p=1:1:4
for m=2:1:6
if m*(m-1)<=M(p) && m*(m+1)>=M(p)
x1=m
else
x2=m;
end
end
end
x1;
m1=[3 3 2 2]
c=1;
z1=1;
b=1;
for z=2:20
for w=1:4
for y=1:1:20
Q1=sqrt((2*D*(A+(S/m1(w)))+R1(w))/(r*(Cv*(m1(w)*(1-D/P)-1+2*D/P)+Cp)+g*m1(w)*D*th2(y)));
q=400*log(0.0002/th2(y));
theta2=(2*I*q)/(g*m1(w)*D*Q1(y));
th2(b)=theta2;
Q(b)=Q1;
b=b+1;
end
b=b-1;
if ((Q(z)~=Q(z-1)) | (th2(z)~=th2(z-1)))
z1=z1+1
else
break
end
end
end
TC(m1)=(D/Q1(i))*(A+S/m1(i)+R)+(Q1(i)/2)*((m1(i)*(1-D/P)-1+2*D/P)*r*Cv+r*Cp+g*m1(i)*D*theta)+r*Cp*k*sigma*sqrt(L)+i*q*log(th0/theta);
##### 2 CommentiMostra NessunoNascondi Nessuno
Walter Roberson il 11 Dic 2020
You should start a new Question for this.
When you do, you should add more documentation as to what you are trying to solve.
for i=1:length(R1)
M1=(S*(r*Cp-(1-2*D/P)*r*Cv))/((A+R1(i))*(r*Cv*(1-D/P)+g*D*th0));
M(j)=M1;
j=j+1;
end
Why are you not just using
for i=1:length(R1)
M1=(S*(r*Cp-(1-2*D/P)*r*Cv))/((A+R1(i))*(r*Cv*(1-D/P)+g*D*th0));
M(i)=M1;
end
with no j ?
M.Rameswari Sudha il 18 Dic 2020
No. I couldn't get the answer.

Accedi per commentare.

### Categorie

Scopri di più su Solver Outputs and Iterative Display in Help Center e File Exchange

### Community Treasure Hunt

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

Start Hunting!

Translated by