How to solve two non-linear equations simultaneously?

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Dr. Siva Malla
Dr. Siva Malla 2014-5-4
编辑： Walter Roberson 2024-3-10
I have two non-linear equations, which are having two unknowns. It is not possible to make it one equation with one variable. now I want the solution of these two equations. please help me to solve this by iteration methods, I want that how to code for iterations. I write for solve command with syms, but I got empty solution. i think it can be solved by iteration methods. Hence please tell me how to write a code...
  2 个评论
Bibhu Prasad Ganthia
Bibhu Prasad Ganthia 2022-7-3
This siva ganesh malla and her wife priyanka malla and rajesh Koilada are fraudsters. Cheating people by taking money for research help. After payment they are not responding to call and messages. Be aware.
Digvijay Kanase
Digvijay Kanase 2024-3-10
Rajesh koilada and his team takes money from people in name of research paper publication and after that they never respond at all. They cheat people. Be aware all of you

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采纳的回答

Matt J
Matt J 2014-5-5
See FSOLVE if you have the Optimization Toolbox.

更多回答（1 个）

Oonitejas Sahoo
Oonitejas Sahoo 2018-4-10
编辑：Walter Roberson 2024-3-10
%% its the mathematical approach by which i sloved the thermal problems
clear all
close all
%Specify grid size
Nx = 10;
Ny = 10;
%Specify boundary conditions
Tbottom = 50
Ttop = 150
Tleft = 50
Tright = 50
% initialize coefficient matrix and constant vector with zeros
A = zeros(Nx*Ny);
C = zeros(Nx*Ny,1);
% initial 'guess' for temperature distribution
T(1:Nx*Ny,1) = 100;
% Build coefficient matrix and constant vector
% inner nodes
for n = 2:(Ny-1)
for m = 2:(Nx-1)
i = (n-1)*Nx + m;
A(i,i+Nx) = 1;
A(i,i-Nx) = 1;
A(i,i+1) = 1;
A(i,i-1) = 1;
A(i,i) = -4;
end
end
% Edge nodes
% bottom
for m = 2:(Nx-1)
%n = 1
i = m;
A(i,i+Nx) = 1;
A(i,i+1) = 1;
A(i,i-1) = 1;
A(i,i) = -4;
C(i) = -Tbottom;
end
%top:
for m = 2:(Nx-1)
% n = Ny
i = (Ny-1)*Nx + m;
A(i,i-Nx) = 1;
A(i,i+1) = 1;
A(i,i-1) = 1;
A(i,i) = -4;
C(i) = -Ttop;
end
%left:
for n=2:(Ny-1)
%m = 1
i = (n-1)*Nx + 1;
A(i,i+Nx) = 1;
A(i,i+1) = 1;
A(i,i-Nx) = 1;
A(i,i) = -4;
end
%right:
for n=2:(Ny-1)
%m = Nx
i = (n-1)*Nx + Nx;
A(i,i+Nx) = 1;
A(i,i-1) = 1;
A(i,i-Nx) = 1;
A(i,i) = -4;
C(i) = -Tright;
end
% Corners
%bottom left (i=1):
i=1;
A(i,Nx+i) = 1;
A(i,2) = 1;
A(i,1) = -4;
C(i) = -(Tbottom + Tleft);
%bottom right:
i = Nx;
A(i,i+Nx) = 1;
A(i,i-1) = 1;
A(i,i) = -4;
C(i) = -(Tbottom + Tright);
%top left:
i = (Ny-1)*Nx + 1;
A(i,i+1) = 1;
A(i,i) = -4;
A(i,i-Nx) = 1;
C(i) = -(Ttop + Tleft);
%top right:
i = Nx*Ny;
A(i,i-1) = 1;
A(i,i) = -4;
A(i,i-Nx) = 1;
C(i) = -(Tright + Ttop);
%Solve using Gauss-Seidel
residual = 100;
iterations = 0;
while (residual > 0.0001) % The residual criterion is 0.0001 in this example
% You can test different values
iterations = iterations+1;
%Transfer the previously computed temperatures to an array Told
Told = T;
%Update estimate of the temperature distribution
for n=1:Ny
for m=1:Nx
i = (n-1)*Nx + m;
Told(i) = T(i);
end
end
% iterate through all of the equations
for n=1:Ny
for m=1:Nx
i = (n-1)*Nx + m;
%sum the terms based on updated temperatures
sum1 = 0;
for j=1:i-1
sum1 = sum1 + A(i,j)*T(j);
end
%sum the terms based on temperatures not yet updated
sum2 = 0;
for j=i+1:Nx*Ny
sum2 = sum2 + A(i,j)*Told(j);
end
% update the temperature for the current node
T(i) = (1/A(i,i)) * (C(i) - sum1 - sum2);
end
end
residual = max(T(i) - Told(i));
end
%compute residual
deltaT = abs(T - Told);
residual = max(deltaT);
iterations; % report the number of iterations that were executed
%Now transform T into 2-D network so it can be plotted.
delta_x = 0.03/(Nx+1);
delta_y = 0.03/(Ny+1);
for n=1:Ny
for m=1:Nx
i = (n-1)*Nx + m;
T2d(m,n) = T(i);
x(m) = m*delta_x;
y(n) = n*delta_y;
end
end
T2d;
surf(x,y,T2d)
figure
contour(x,y,T2d)

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