Solving 3 Simultaneous Exponential Equations

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Rory Thornton
Rory Thornton el 20 de Dic. de 2019
Comentada: Rory Thornton el 22 de Dic. de 2019
I am trying to solve these 3 simultaneous exponential equations for a,b and c (This is from Vogels Viscosity Equation):
159.2543 = a*exp(b/(291.15-c))
117.2699 = a*exp(b/(293.15-c))
63.8384 = a*exp(b/(299.15-c))
I would really appriciate it if someone could show me how to write the code to solve them please!
Thank you in advance!

Respuesta aceptada

Star Strider
Star Strider el 20 de Dic. de 2019
Try this:
x = [291.15; 293.15; 299.15];
y = [159.2543; 117.2699; 63.8384];
% % % MAPPING: a = b(1), b = b(2), c = b(3)
objfcn = @(b,x) b(1).*exp(b(2)./(x - b(3)));
B0 = [0.07; 450; 230];
[B,normresid] = fminsearch(@(b) norm(y - objfcn(b,x)), B0)
xv = linspace(min(x), max(x));
figure
plot(x, y, 'p')
hold on
plot(xv, objfcn(B,xv), '-r')
hold off
grid
Values of:
a = 0.970
b = 180
c = 255
give a reasonable fit to the data.
  2 comentarios
Rory Thornton
Rory Thornton el 20 de Dic. de 2019
Great, fits the rest of my data really well too.
Thank you so much!
Star Strider
Star Strider el 20 de Dic. de 2019
As always, my pleasure!

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Más respuestas (1)

David Goodmanson
David Goodmanson el 21 de Dic. de 2019
HI Rory,
just for completeness
y1 = 159.2543;
y2 = 117.2699;
y3 = 63.8384;
x1 = 291.15;
x2 = 293.15;
x3 = 299.15;
A1 = log(y1/y2)/log(y2/y3);
A2 = (x2-x1)/(x3-x2);
c = (x1*A1-x3*A2)/(A1-A2);
% back substitute
b = log(y1/y2)*(x1-c)*(x2-c)/(x2-x1);
a = y1/exp(b/(x1-c));
a
b
c
a =
10.6205
b =
42.5004
c =
275.4539
% these should be small
y1 - a*exp(b/(x1-c))
y2 - a*exp(b/(x2-c))
y3 - a*exp(b/(x3-c))
ans =
0
ans =
-4.2633e-14
ans =
-7.1054e-14
The c result is fairly close to Star Strider's, but for some reason a and b differ from that result by quite a bit. The checks here show agreement at all three y points, but the best fit isn't necessarily the one that goes through all three points exactly.
  1 comentario
Rory Thornton
Rory Thornton el 22 de Dic. de 2019
Thank you! It's just a line of best fit, there will be many viable solutions which may have very different a/b/c values. This fits the data well too.

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