How to solve non linear equation for one variable?

Hi, I am trying to solve a non-linear equation for one variable but it's not working. Can somebody help me in this regards? your help would be highly appreciated. Thanks
if
temp=[24 78 139 194 297 397];
yieldstress=[45 36 37 33 30 28];
syms DD positive
solve((6.046.*sqrt(DD))./(1e6)+(9.79.*(1-((-4.7872e-5.*temp).*(log(6.667e-4./(286.*sqrt(DD))))).^(2/3)).^(3/2)).*(3.06) == yieldstress);
end

 Respuesta aceptada

njj1
njj1 el 24 de Abr. de 2018
Editada: njj1 el 24 de Abr. de 2018
If you want to solve for multiple values of temp and yieldstress, you will have to use a for loop.
for i=1:numel(temp)
for j=1:numel(yieldstress)
out{i,j} = solve((6.046.*sqrt(DD))./(1e6)+(9.79.*(1-((-4.7872e-5.*temp(i)).*(log(6.667e-4./(286.*sqrt(DD))))).^(2/3)).^(3/2)).*(3.06) == yieldstress(j));
end
end
The output variable "out" contains the values of DD that solve the equations as symbolic numbers (so long as the equation can be solved), and these can be evaluated using the command eval().

3 comentarios

Farhan Ashraf
Farhan Ashraf el 24 de Abr. de 2018
Now, Matlab is giving an error, specify a variable for which you solve. I introduce DD at the end of out{i,j} command. still same error is coming

If you type in the code exactly as I have then it should work. It will take a while but it should work.

for i=1:numel(temp)
  for j=1:numel(yieldstress)
    out{i,j} = solve((6.046.*sqrt(DD))./(1e6)+(9.79.*(1-((-4.7872e-5.*temp(i)).*(log(6.667e-4./(286.*sqrt(DD))))).^(2/3)).^(3/2)).*(3.06) == yieldstress(j));
    dd(i,j) = eval(out{i,j}); %DO NOT put DD here
  end
end

FYI, I've run this code for the first few values of temp and yieldstress that you posted above, and Matlab is not able to find an explicit solution.

Farhan Ashraf
Farhan Ashraf el 24 de Abr. de 2018
what this error mean: A null assignment can have only one non-colon index? can you please guide?

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Más respuestas (1)

Torsten
Torsten el 24 de Abr. de 2018

0 votos

The left-hand side of your equation (6.046.*sqrt(...)...) is a scalar, the right-hand side (yieldstress) is a vector. This is not compatible.
Best wishes
Torsten.

1 comentario

Farhan Ashraf
Farhan Ashraf el 24 de Abr. de 2018
Actually I want to calculate DD for each value of yieldstress and temp. Can you please suggest me something?

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