How to find L from the given code?

Question

Akhtar Jan 2022-7-30

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1770990-how-to-find-l-from-the-given-code

编辑： Torsten 2022-7-30

采纳的回答： Torsten

clc

close all

lags = 0.6;

k1 = 3;

k_1 = 1;

k2 = 2.5;

k3 = 1;

k_3 = 1;

k4 = 2;

k5 = 1;

E1 = 1;

E2 = 2;

K1 = (k_1+k2)/k1;

K2 = (k_3+k4)/k3;

k3*E2*(L^2+L*(k1*K1+k4+k1*E1)+(E1+K1)*k1*k4+k1*k2*E1)+(L+k3*K2)*(k1*k2*E1+(L+k1*K1+k1*E1)*(L-k5*E1*e^(-L*lags))) = 0;

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Answer 1

Torsten 2022-7-30

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1770990-how-to-find-l-from-the-given-code#answer_1018065

编辑：Torsten 2022-7-30

在 MATLAB Online 中打开

lags = 0.6;

k1 = 3;

k_1 = 1;

k2 = 2.5;

k3 = 1;

k_3 = 1;

k4 = 2;

k5 = 1;

E1 = 1;

E2 = 2;

K1 = (k_1+k2)/k1;

K2 = (k_3+k4)/k3;

fun = @(L) k3*E2*(L.^2+L.*(k1*K1+k4+k1*E1)+(E1+K1)*k1*k4+k1*k2*E1)+(L+k3*K2).*(k1*k2*E1+(L+k1*K1+k1*E1).*(L-k5*E1*exp(-L*lags)));

L0 = -1.5;

L1 = fsolve(fun,L0)

Equation solved. fsolve completed because the vector of function values is near zero as measured by the value of the function tolerance, and the problem appears regular as measured by the gradient.

L1 = -1.6620

L0 = -2.5;

L2 = fsolve(fun,L0)

Equation solved. fsolve completed because the vector of function values is near zero as measured by the value of the function tolerance, and the problem appears regular as measured by the gradient.

L2 = -2.4088

plot((-2.5:0.1:-1),fun(-2.5:0.1:-1))

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Akhtar Jan 2022-7-30

Thanks alot sir

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Answer 2

Walter Roberson 2022-7-30

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1770990-how-to-find-l-from-the-given-code#answer_1018070

编辑：Torsten 2022-7-30

在 MATLAB Online 中打开

syms L

e = exp(sym(1))

e =

e

lags = 0.6;

k1 = 3;

k_1 = 1;

k2 = 2.5;

k3 = 1;

k_3 = 1;

k4 = 2;

k5 = 1;

E1 = 1;

E2 = 2;

K1 = (k_1+k2)/k1;

K2 = (k_3+k4)/k3;

eqn = k3*E2*(L^2+L*(k1*K1+k4+k1*E1)+(E1+K1)*k1*k4+k1*k2*E1)+(L+k3*K2)*(k1*k2*E1+(L+k1*K1+k1*E1)*(L-k5*E1*e^(-L*lags)))

eqn =

vpasolve(eqn)

ans =

There is a second solution near -2.3

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Akhtar Jan 2022-7-30

thank you sir

Akhtar Jan 2022-7-30

this one is also helpfull

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