solve 4 equations with some unknowns parameters.
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I'm attempting to solve this problem which is attached. But I faced ERROR. Any suggestions?
12 Comments
Matt J
on 2 Nov 2018
This looks like only part of the message. The full error message should give a reason, as well as the error location.
madhan ravi
on 2 Nov 2018
Edited: madhan ravi
on 2 Nov 2018
when I ran your code it didn't have any errors instead it returned all zeros
Skill_s
on 2 Nov 2018
Skill_s
on 2 Nov 2018
madhan ravi
on 2 Nov 2018
do you have symbolic toolbox? type ver in command window to check whether you have that toolbox
Skill_s
on 2 Nov 2018
Star Strider
on 2 Nov 2018
The ver output disagrees with you:
MATLAB Version 7.11.0.584 (R2010b)
Operating System: Microsoft Windows 7 Version 6.2 (Build 9200)
...
Symbolic Math Toolbox Version 5.5 (R2010b)
...
Skill_s
on 2 Nov 2018
Accepted Answer
Bruno Luong
on 3 Nov 2018
Edited: Bruno Luong
on 3 Nov 2018
Your problem looks like you find the EM field with 3 layers and 2 interfaces. The coefficients A, B, C, D gives the strength of different field components, they are linearly related because EM is linear. In order to solve with without getting trivial solution
A=B=C=D=0
You must for example fix one of them arbitrary, e.g.
A = 1
That left you with 4 linear equations and 3 unknowns. Then in order such system to have solution, you must have the linear system to have dependent equation, meaning determinant of the matrix is 0.
Write it down this probably gives the equation you looks for.
I don't have symbolic tbx to do this kind of calculation to confirm, it would be possible to carry out numerically me think.
4 Comments
Skill_s
on 3 Nov 2018
Walter Roberson
on 3 Nov 2018
Edited: Walter Roberson
on 4 Nov 2018
If you solve the first three equations for B, C, D, then you are left with
-A*(-(eps1*k3+eps3*k1)*(eps1*k2+eps2*k1)*exp(a*(2*k1-k3))+exp(-a*(2*k1+k3))*(-eps1*k3+eps3*k1)*(-eps1*k2+eps2*k1))/(2*eps3*k1*eps2*eps1) == 0
The obvious solution is A == 0, which then causes B, C, D to all be 0. If, on the other hand, (-(eps1*k3+eps3*k1)*(eps1*k2+eps2*k1)*exp(a*(2*k1-k3))+exp(-a*(2*k1+k3))*(-eps1*k3+eps3*k1)*(-eps1*k2+eps2*k1))/(2*eps3*k1*eps2*eps1) can be 0, then the equations would hold for all finite A, in which case there would be a family of solutions
B = A*((eps1*k3+eps3*k1)*exp(4*a*k1)+eps3*k1-eps1*k3)*exp(-a*(2*k1-k2+k3))/(2*eps3*k1)
C = -A*(eps1*k3-eps3*k1)*exp(-a*(k1+k3))/(2*eps3*k1)
D = A*(eps1*k3+eps3*k1)*exp(a*(k1-k3))/(2*eps3*k1)
which are linear in A.
Skill_s
on 4 Nov 2018
Bruno Luong
on 4 Nov 2018
Actually solving linear system is not what Skill interested, what he is interested is the the condition under which the 4 linear equation is dependent thus solvable.
This boils down to single equation of determinant. I shows here the solution he gives makes DET(M) = 0.
% Fake k, epsilon 1,2,3
k = rand(1,3);
eps = rand(1,3);
p = k./eps;
% compute solution a
r = ((p(1)+p(2))*(p(1)+p(3)))/((p(1)-p(2))*(p(1)-p(3)));
a = log(r)/(-4*k(1))
% Forming linear matrix
ep = exp(k*a);
em = exp(-k*a);
M = [em(3) 0 -ep(1) -em(1);
p(3)*em(3) 0 p(1)*ep(1) -p(1)*em(1);
0 em(2) -em(1) -ep(1);
0 -p(2)*em(2) p(1)*em(1) -p(1)*ep(1)];
% verifies the 4 x 4 system is rank 3
rank(M)
det(M)
Now if Walter who has access to Symb Tbx can show the opposite, started from
det(M) = 0
shows the solution of it is
r = ((p(1)+p(2))*(p(1)+p(3)))/((p(1)-p(2))*(p(1)-p(3)));
a = log(r)/(-4*k(1))
Then the problem is completely solved.
More Answers (1)
Florian Augustin
on 2 Nov 2018
Hi,
I think you are using a modern syntax to call 'solve' that was not supported in R2010b. The equivalent call in R2010b would be
syms a A B C D eps1 eps2 eps3 k1 k2 k3;
eqn1 = (C*exp(k1*a))+(D*exp(-k1*a))-(A*exp(-k3*a));
eqn2 = ((-(C*k1)/eps1)*exp(k1*a))+(((D*k1)/eps1)*exp(-k1*a))-(((A*k3)/eps3)*exp(-k3*a));
eqn3 = (C*exp(-k1*a))+(D*exp(k1*a))-(B*exp(-k2*a));
eqn4 = ((-(C*k1)/eps1)*exp(-k1*a))+(((D*k1)/eps1)*exp(k1*a))+(((B*k2)/eps2)*exp(-k2*a));
sol = solve(eqn1, eqn2, eqn3, eqn4, A, B, C, D);
ASol = sol.A
BSol = sol.B
CSol = sol.C
DSol = sol.D
You can access the documentation for your release of MATLAB by typing 'doc solve' in the MATLAB interface.
Hope this helps,
-Florian
7 Comments
Skill_s
on 2 Nov 2018
Matt J
on 2 Nov 2018
@Skill93,
If Florian's answer was what you were looking for, then you should Accept-click it.
Walter Roberson
on 2 Nov 2018
If I recall correctly, in R2010b, the syntax used for solve() was available but not clearly documented.
The problem was elsewhere: before R2011b, using relational expressions with symbolic expressions caused the expressions to be evaluated as a logical immediately. So the line
eqn1 = (C*exp(k1*a))+(D*exp(-k1*a))-(A*exp(-k3*a))==0;
built the expression on the left side, and immediately compared it to 0, found it was not identical, and immediately returned false into eqn1.
The workaround in those days was to not do the ==0 part on the equations: convert
eqn = A == B
to
eqn = (A) - (B)
And in those days, if you had an inequality such as
eqn1 = (C*exp(k1*a))+(D*exp(-k1*a))-(A*exp(-k3*a)) > 3
then you would have to build it using character vectors,
eqn1 = '(C*exp(k1*a))+(D*exp(-k1*a))-(A*exp(-k3*a)) > 3'
This would pretty much force you to use separate entries in the solve() command, like the
sol = solve(eqn1, eqn2, eqn3, eqn4, A, B, C, D);
that Florian shows, since [eqn1, eqn2, eqn3, eqn4] with character vectors would result in a single long character vector instead of multiple equations.
So, Florian's suggested code should work for the purpose, just not for exactly the reason that Florian indicated.
Skill_s
on 2 Nov 2018
Stephan
on 2 Nov 2018
Executing your code in 2018b gives the same result. So you need a better approach.
Walter Roberson
on 2 Nov 2018
MATLAB gives all 0.
If you work through the equations one by one doing stepwise elimination, then all 0 is the only solution.
Skill_s
on 3 Nov 2018
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