How to solve two non-linear equations simultaneously?

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...

[Bibhu Prasad Ganthia]

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.

Answer 1

[Matt J]

See FSOLVE if you have the Optimization Toolbox.

Answer 2

[Oonitejas Sahoo]

• %% 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)