※ 引述《qrewasd (shu)》之銘言：
: beta=-2 % 2,0,-2
: a=(omega-(beta^2))^0.5
: b=(omega+(beta^2))^0.5
: a*b*(1-cosh(b)*cos(a))+(beta^2)*sinh(b)*sin(a)=0
: 求解 a b omega
: 我自己試過
: 結果如下
: Warning: Explicit solution could not be found.
: > In solve at 81
: ans =
: [ empty sym ]
: 請問該如何改寫呢?
: 3Q
本題涉及非線性跟根號問題
我是用fsolve 解的
帶入的值影響解是有範圍的
------------------------------------------
function pttex144(t)
global beta
format long e
beta = t;
% beta=-2 % -2, 0, 2
omega = fsolve(@yourfun,1)
a=(omega-(beta.^2)).^0.5
b=(omega+(beta.^2)).^0.5
function f = yourfun(x)
global beta
a=(x-(beta.^2)).^0.5; % d1
b=(x+(beta.^2)).^0.5; % d2
f = a.*b.*(1-cosh(b).*cos(a))+(beta.^2).*sinh(b).*sin(a);
------------------------------------------------------------
執行結果:
>> pttex144(-2)
Equation solved.
fsolve completed because the vector of function values is near zero
as measured by the default value of the function tolerance, and
the problem appears regular as measured by the gradient.
<stopping criteria details>
omega =
3.724123341394964e-03 + 1.772674751964346e-13i
a =
4.451065079099797e-14 + 1.999068752359109e+00i
b =
2.000930814231565e+00 + 4.429625300775632e-14i
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