common tangent of two curves with vpasolve.

The following code finds common tangents of fx and gx by symbolic expressions. I used first "solve" but sometimes it can not find all the roots of the denk1 and denk2 which are derived from common tangent geometry. vpasolve finds a pair of x,y but not both. I appreciate any help or suggestions.

best wishes,

clc, clear all, close all
disp(['        (t)      a        b']);
R=8.314 ;%j/mol K;
O1=-15000;
O2=0; %j/mol K;
for t=1300:50:1350
    syms x fx gx ek m(x) y1(x) a b y2(x) f(x) g(x) fu(x) gu(x) 
syms denk1 denk2 denk3 p yx
f(x)=R*t*((x*log(x))+((1-x)*log(1-x)))+O1*(x-x^2);
g(x)=x*(12000-10*t)+((1-x)*(8000-10*t))+R*t*((x*log(x))+((1-x)*log(1-x)));
fu(x)=diff(f(x),x);%1. differential
gu(x)=diff(g(x),x);%2.differential
denk1=(((g(b)-f(a))/(b-a))==(fu(a)));
denk2=(fu(a)==gu(b));
%[a b]=solve(denk1,denk2,a,b);
%[kola kolb] = vpasolve([denk1 denk2],[a b],[0.2;.99991])
[Sola Solb] = vpasolve([denk1 denk2],[a b],[0.1; 0.3]);
a=Sola;
b=Solb;
disp([t,a,b])
x=0:0.01:1;
plot(x ,f(x),'r-',x ,g(x),'-b')
hold on
p = linspace(a,b);
%y=mx+n graph of common tangent line between a and b
yx=((fu(a))*(p-a))+f(a);
plot(p,yx, 'm-')
pause(2)
hold off
figure
end

[darova]

Here the results i got from your code

Look good. What is your question again?

[muhsin ider]

there is another common tangent on the right side of the 1 st figure around 0.8-0.9 and also in the 2nd figure. Solve or vpasolve can not automatically find those points.

Answer 1

[Bjorn Gustavsson]

In your code you gove starting-points for the variable you search for. Perhaps vpasolve doesn't bother to look for all solutions of your equation (in case they are too complex one solution might be considered good enough). Try to look for a second solution starting from another point (~0.8 and 0.9). If that gives you a second solution you've found the second solution. I've not looked at your equations to know how many solutions you should have...

HTH

[muhsin ider]

I already know that vpasolve gives different answers with different starting points. If you look at the figures there are at least 2 common tangents. Actually my code gives the other solution where x is closer to 1 if I dont use vpasolve with specific starting points. I want to have all the solutions no matter what. my starting points are.... Is it possible non numerically. I know solve can not do it.

[Bjorn Gustavsson]

...and there you have a solution to your problem, right? Even tough perhaps not as neat and smooth as you'd hope for.

[muhsin ider]

If I have a solution to my problem Why would I write here? Do you really understand what I am asking help about?

[Bjorn Gustavsson]

Yes, I understand that you want two solutions to common tangents to two curves. If you can get two different solutions when starting vpasolve from different starting-points then that gives you your two different solutions. You don't explicitly need to get both solutions from one single call to vpasolve (or any other solver). You could do something like this:

a0b0 = [0.1;0.3];
[Sola Solb] = vpasolve([denk1 denk2],[a b],a0b0);
a=Sola;
b=Solb;
secondonefound = 0;
loopeduntillbored = 1;
while ~secondonefound && loopeduntillbored < 100
  [Sola2 Solb2] = vpasolve([denk1 denk2],[a b],a0b0+loopeduntilbored);
  a2 = Sola2;
  b2 = Solb2;
  if abs(a2-a)>eps || abs(b2-b)>eps
    break; % solution found where at least one of a2 and b2 differ from a and b respectively
  else
    [Sola2 Solb2] = vpasolve([denk1 denk2],[a b],a0b0-loopeduntilbored);
    a2 = Sola2;
    b2 = Solb2;
  end
  loopeduntillbored = loopeduntillbored + 1;
  if abs(a2-a)>eps || abs(b2-b)>eps
      break
  end
end

...and yes, I did put in typos in the above code, because reasons.

[muhsin ider]

Thanks for your help. I tried to correct undefined functions or typos but I could not. But I understand What you meant. My questions anwer is Matlab cannot give both answers for some cases at least symbolically. I know how to do with other methods such as newtons method or secant method etc. I wanted it to be very short and neat But I think matlab is not an unlimited software in terms of solving for 2 rooted non linear equations.

Thanks. best wishes,

[Bjorn Gustavsson]

That is most likely a design-choise by Mathworks. If, for example, vpasolve were to try to return all solutions of equations there might be problems when you try to solve equations with infinitely many solutions. Then there would need to be some arbitrary cut-off when the number of solutions will be considered "enough". Perhaps Mathworks then made the choise that one solution is enough.

[muhsin ider]

I am guessing and understanding now that vpasolve is also using a newton based or similar method or something like that because the same problem sometimes happen with that method. That method sometimes gives only one solution depending on the starting guess point.

Thanks for your help and comments.

best regards,