0 个投票

syms c
delta = 0.0158;
alp = 6;
a = alp*delta;
Re = 1000;
y = 0.5;
y1 = y*delta;
U = (-357554.879*y1^2 + 11298.734*y1)/89.26;
ddU = -2;
l3 = -((-((a*Re*c*1i)-(2*a^2)-(a*Re*U*1i))+((2*a^2*Re^2*U*c)-(a^2*Re^2*(U^2+c^2))-(4*a*Re*ddU*1i))^(1/2))/2)^(1/2);
l4 = -((-((a*Re*c*1i)-(2*a^2)-(a*Re*U*1i))-((2*a^2*Re^2*U*c)-(a^2*Re^2*(U^2+c^2))-(4*a*Re*ddU*1i))^(1/2))/2)^(1/2);
eqn = subs((U-c)*((l3^2*exp(l3*y) - l4^2*exp(l4*y)) - (a^2*(exp(l3*y)-exp(l4*y))))-(ddU*(exp(l3*y)-exp(l4*y)))-((1/(a*Re*1i))*(((l3^4*exp(l3*y))-(l4^4*exp(l4*y)))-(2*a^2*(l3^2*exp(l3*y) - l4^2*exp(l4*y))) + a^4*(exp(l3*y)-exp(l4*y)))));
answer = vpasolve(eqn,c)
In this particular code, for alp = 6,8,12,13,, the output is an error, "empty sym 0-by-1". But for alp values like 1,2,3,4,5,7,..I am getting values of c. I actualy want all the values of c when alp varies from 1 to 20
Can somebody help me with the solution?
Thank you!!

4 个评论
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Walter Roberson
Walter Roberson 2020-10-7
Are you expecting your solutions to be real-valued or complex valued?
There might be a solution close to -1.23106389616001 + 0.287678549584008i -- it can be difficult to tell whether errors on the order of 6E-16 are places the function evaluates to slightly different than 0, or just a result of round-off and precision limitations.
Anyhow: if you simplify(eqn) then vpasolve() can find an answer fairly quickly.
Maya Venugopalan
Maya Venugopalan 2020-10-7
编辑：Maya Venugopalan 2020-10-7
c will be complex values only.
I tried to use
simplify(eqn);
vpasolve(eqn,c)
Still the output is empty sym: 0-by-1
Walter Roberson
Walter Roberson 2020-10-7
Ran in:
Ran in:
syms c
delta = 0.0158;
alp = 6;
a = alp*delta;
Re = 1000;
y = 0.5;
y1 = y*delta;
U = (-357554.879*y1^2 + 11298.734*y1)/89.26;
ddU = -2;
l3 = -((-((a*Re*c*1i)-(2*a^2)-(a*Re*U*1i))+((2*a^2*Re^2*U*c)-(a^2*Re^2*(U^2+c^2))-(4*a*Re*ddU*1i))^(1/2))/2)^(1/2);
l4 = -((-((a*Re*c*1i)-(2*a^2)-(a*Re*U*1i))-((2*a^2*Re^2*U*c)-(a^2*Re^2*(U^2+c^2))-(4*a*Re*ddU*1i))^(1/2))/2)^(1/2);
eqn = subs((U-c)*((l3^2*exp(l3*y) - l4^2*exp(l4*y)) - (a^2*(exp(l3*y)-exp(l4*y))))-(ddU*(exp(l3*y)-exp(l4*y)))-((1/(a*Re*1i))*(((l3^4*exp(l3*y))-(l4^4*exp(l4*y)))-(2*a^2*(l3^2*exp(l3*y) - l4^2*exp(l4*y))) + a^4*(exp(l3*y)-exp(l4*y)))));
answer = vpasolve(simplify(eqn),c)
answer = 
−0.80865908850082764227953517414956−0.012297044380461873020713794371614 i
Which release are you using? It works in R2020a and R2020b
Maya Venugopalan
Maya Venugopalan 2020-10-7
Yes!! This worked!!!
I'm using R2019a.
Thank you soooo much!!!!

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 采纳的回答

Walter Roberson
Walter Roberson 2020-10-7
编辑：Walter Roberson 2020-10-7

0 个投票

syms c
delta = 0.0158;
alp = 6;
a = alp*delta;
Re = 1000;
y = 0.5;
y1 = y*delta;
U = (-357554.879*y1^2 + 11298.734*y1)/89.26;
ddU = -2;
l3 = -((-((a*Re*c*1i)-(2*a^2)-(a*Re*U*1i))+((2*a^2*Re^2*U*c)-(a^2*Re^2*(U^2+c^2))-(4*a*Re*ddU*1i))^(1/2))/2)^(1/2);
l4 = -((-((a*Re*c*1i)-(2*a^2)-(a*Re*U*1i))-((2*a^2*Re^2*U*c)-(a^2*Re^2*(U^2+c^2))-(4*a*Re*ddU*1i))^(1/2))/2)^(1/2);
eqn = subs((U-c)*((l3^2*exp(l3*y) - l4^2*exp(l4*y)) - (a^2*(exp(l3*y)-exp(l4*y))))-(ddU*(exp(l3*y)-exp(l4*y)))-((1/(a*Re*1i))*(((l3^4*exp(l3*y))-(l4^4*exp(l4*y)))-(2*a^2*(l3^2*exp(l3*y) - l4^2*exp(l4*y))) + a^4*(exp(l3*y)-exp(l4*y)))));
answer = vpasolve(simplify(eqn),c)
Note: my research suggested that there might be up to three solutions, with the real and imaginary parts all within +/- 2 . It was difficult to tell whether some of the locations reached zero or just came close to zero.

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