ans = Empty sym: 0-by-1 error in solving equation

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Filippo
Filippo il 18 Gen 2023
Risposto: Nikhilesh il 18 Gen 2023
I am trying to solve a equation for a small project but experiencing
syms x
y= (0:1:100)
avd=5
s = 0.25
paco2=40
r= 0.8
hb=14
pb=740
eqn = ((hb*(((((pb-47)*y-paco2/r).^3+150*((pb-47)*y-paco2/r)).^(-1)*23400)+1).^(-1)*1.34)+((pb-47)*y-paco2/r)*0.0031)- (hb*(((x.^3+150*x).^(-1)*23400)+1).^(-1)*1.34+(x*0.0031))==(s/(1-s))*avd
solve(eqn, x)
and get this result
ans =
Empty sym: 0-by-1

Risposte (3)

John D'Errico
John D'Errico il 18 Gen 2023
Modificato: John D'Errico il 18 Gen 2023
So many times I see this mistake made. I can understand where it comes from. But it gets MATLAB (actually solve) confused.
When you define y in advance as a vector, then the symbolic toolbox thinks you have 101 equations, but only the 1 unknown x. Solve now fails. Instead, you need to leave y as an unknown parameter, and try to solve for x as a function of y.
syms x
% y= (0:1:100) DON'T DO THIS IN ADVANCE!!!!!!!
% instead, leave y as a symbolic parameter. You will subs in that vector at
% the end.
syms y
avd=5;
s = 0.25;
paco2=40;
r= 0.8;
hb=14;
pb=740;
eqn = ((hb*(((((pb-47)*y-paco2/r).^3+150*((pb-47)*y-paco2/r)).^(-1)*23400)+1).^(-1)*1.34)+((pb-47)*y-paco2/r)*0.0031)- (hb*(((x.^3+150*x).^(-1)*23400)+1).^(-1)*1.34+(x*0.0031))==(s/(1-s))*avd
eqn = 
So we have a rather messy expression for x and y together. Before we go any further, it is ALWAYS important to plot everything.
fimplicit(eqn,[-100 100 0 100])
So fimplicit finds some negative solutions for x, and some large positive solutions between 80 and 100, all depending on the value of y.
If y has a known value, we can clear the fractions, and find what is effectively a 4th degree polynomial in x.
xsol = solve(eqn,x,'returnconditions',true)
xsol = struct with fields:
x: [4×1 sym] parameters: [1×0 sym] conditions: [4×1 sym]
xsol.x
ans = 
Unfortunately, we cannot yet resolve the roots, until we substitute in for y. So it is only now that we provide the values for y. For example...
vpa(subs(xsol.x,y,2))
ans = 
So, as I said, a negative root, and a large positive root, and two complex roots.
plot(0:100,double(subs(xsol.x(1),y,0:100)))
xlabel y
ylabel x
Note that the abscissa here is the value of y, and the ordinate (y axis) is the computed value of x, for the first root.

Torsten
Torsten il 18 Gen 2023
syms x y
Y = 0:1:100;
avd=5;
s = 0.25;
paco2=40;
r= 0.8;
hb=14;
pb=740;
eqn = ((hb*(((((pb-47)*y-paco2/r).^3+150*((pb-47)*y-paco2/r)).^(-1)*23400)+1).^(-1)*1.34)+((pb-47)*y-paco2/r)*0.0031)- (hb*(((x.^3+150*x).^(-1)*23400)+1).^(-1)*1.34+(x*0.0031))-(s/(1-s))*avd==0;
[N,D] = numden(lhs(eqn));
for i=1:numel(Y)
sol{i} = vpa(root(subs(N,y,Y(i))));
end

Nikhilesh
Nikhilesh il 18 Gen 2023
It seems you are trying to iterate over the value of y and solve for each value of x.
It would be beneficial if you can use a for loop which iterate over values of y (1-100) and place the equation inside the for loop.
for y=0:100
eqn = ((hb*(((((pb-47)*y-paco2/r).^3+150*((pb-47)*y-paco2/r)).^(-1)*23400)+1).^(-1)*1.34)+((pb-47)*y-paco2/r)*0.0031)- (hb*(((x.^3+150*x).^(-1)*23400)+1).^(-1)*1.34+(x*0.0031))==(s/(1-s))*avd
S=solve(eqn, x)
end

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