在MATLAB中求解两个困难的非线性方程

Question

我有两个非线性方程，其中三个变量是在一篇论文中得出的。在本文中，第三个变量从一定值变化到另一个值，另外两个变量相对于它绘制。这两个方程是（1） abs(C(z,theta1,theta2))=0.1和（2） diff(abs(C(z,theta1,theta2)),w)=0其中C(z,theta1,theta2)=A/B与

A=j*(2*(z-1/z)*tan(theta1)+(1/z^2-z^2)*tan(theta1)^2*tan(theta2));

B=2-2*(z+1/z)*tan(theta1)*tan(theta2)-2*tan(theta1)^2+j*(2*(z+1/z)*tan(theta1)+2*tan(theta2)-(1/z^2+z^2)*tan(theta1)^2*tan(theta2));

在上面的两个等式中，j=sqrt(-1)和w是角频率;具有theta1=w/c*l1和theta2=w/c*l2的{​​{1}}和c=3*10^8是长度。这两个方程处于工作频率。本文的作者已经将l1,l2的给定值从某个值（如1.2）解析为另一个值（如3），并绘制了另外两个变量z和{{1} }，关于theta1。 theta2被选为z。因为这是我的第一篇文章，我无法发布图片以显示论文中给出的结果，但是获得的值w和500*10^6（theta1和{{1在theta2范围内的电气长度（度），约为5度。到30度对于theta1（像theta2这样的曲线）和大约50度。到10度对于z=1.2 to 3（类似theta2的曲线）。

为了解决这些方程，我开始使用固定值sqrt(x)，如theta1和相同的工作频率exp(x)。具有这些考虑因素的两个结果方程是：

z

根据论文的结果，z=1.5和w=500*10^6的解equation1=(11.390625*(-3.333333334*tan(10/3*pi*l1)+3.611111112*tan(10/3*pi*l1)^2*tan(10/3*pi*l2))^2*(3.25*tan(10/3*pi*l1)*tan(10/3*pi*l2)+1.5*tan(10/3*pi*l1)^2-1.5)^2/(36.75390625*tan(10/3*pi*l1)^4*tan(10/3*pi*l2)^2-30.4687500*tan(10/3*pi*l1)^3*tan(10/3*pi*l2)+20.2500*tan(10/3*pi*l1)^4+40.5000*tan(10/3*pi*l1)^2*tan(10/3*pi*l2)^2+54.562500*tan(10/3*pi*l1)^2+20.2500*tan(10/3*pi*l2)^2+20.2500)^2+5.0625*(-1.666666667*tan(10/3*pi*l1)+1.805555556*tan(10/3*pi*l1)^2*tan(10/3*pi*l2))^2*(6.0625*tan(10/3*pi*l1)^2*tan(10/3*pi*l2)-9.750*tan(10/3*pi*l1)-4.50*tan(10/3*pi*l2))^2/(36.75390625*tan(10/3*pi*l1)^4*tan(10/3*pi*l2)^2-30.4687500*tan(10/3*pi*l1)^3*tan(10/3*pi*l2)+20.2500*tan(10/3*pi*l1)^4+40.5000*tan(10/3*pi*l1)^2*tan(10/3*pi*l2)^2+54.562500*tan(10/3*pi*l1)^2+20.2500*tan(10/3*pi*l2)^2+20.2500)^2)^(1/2)-.1 equation2=1/2*(22.781250*(-3.333333334*tan(10/3*pi*l1)+3.611111112*tan(10/3*pi*l1)^2*tan(10/3*pi*l2))*(3.25*tan(10/3*pi*l1)*tan(10/3*pi*l2)+1.5*tan(10/3*pi*l1)^2-1.5)^2*(-.1111111111e-7*(1+tan(10/3*pi*l1)^2)*l1+.2407407408e-7*tan(10/3*pi*l1)*tan(10/3*pi*l2)*(1+tan(10/3*pi*l1)^2)*l1+.1203703704e-7*tan(10/3*pi*l1)^2*(1+tan(10/3*pi*l2)^2)*l2)/(36.75390625*tan(10/3*pi*l1)^4*tan(10/3*pi*l2)^2-30.4687500*tan(10/3*pi*l1)^3*tan(10/3*pi*l2)+20.2500*tan(10/3*pi*l1)^4+40.5000*tan(10/3*pi*l1)^2*tan(10/3*pi*l2)^2+54.562500*tan(10/3*pi*l1)^2+20.2500*tan(10/3*pi*l2)^2+20.2500)^2+22.781250*(-3.333333334*tan(10/3*pi*l1)+3.611111112*tan(10/3*pi*l1)^2*tan(10/3*pi*l2))^2*(3.25*tan(10/3*pi*l1)*tan(10/3*pi*l2)+1.5*tan(10/3*pi*l1)^2-1.5)*(.1083333333e-7*(1+tan(10/3*pi*l1)^2)*l1*tan(10/3*pi*l2)+.1083333333e-7*tan(10/3*pi*l1)*(1+tan(10/3*pi*l2)^2)*l2+.1000000000e-7*tan(10/3*pi*l1)*(1+tan(10/3*pi*l1)^2)*l1)/(36.75390625*tan(10/3*pi*l1)^4*tan(10/3*pi*l2)^2-30.4687500*tan(10/3*pi*l1)^3*tan(10/3*pi*l2)+20.2500*tan(10/3*pi*l1)^4+40.5000*tan(10/3*pi*l1)^2*tan(10/3*pi*l2)^2+54.562500*tan(10/3*pi*l1)^2+20.2500*tan(10/3*pi*l2)^2+20.2500)^2-22.781250*(-3.333333334*tan(10/3*pi*l1)+3.611111112*tan(10/3*pi*l1)^2*tan(10/3*pi*l2))^2*(3.25*tan(10/3*pi*l1)*tan(10/3*pi*l2)+1.5*tan(10/3*pi*l1)^2-1.5)^2*(.4900520833e-6*tan(10/3*pi*l1)^3*tan(10/3*pi*l2)^2*(1+tan(10/3*pi*l1)^2)*l1+.2450260417e-6*tan(10/3*pi*l1)^4*tan(10/3*pi*l2)*(1+tan(10/3*pi*l2)^2)*l2-.3046875000e-6*tan(10/3*pi*l1)^2*tan(10/3*pi*l2)*(1+tan(10/3*pi*l1)^2)*l1-.1015625000e-6*tan(10/3*pi*l1)^3*(1+tan(10/3*pi*l2)^2)*l2+.2700000000e-6*tan(10/3*pi*l1)^3*(1+tan(10/3*pi*l1)^2)*l1+.2700000000e-6*tan(10/3*pi*l1)*tan(10/3*pi*l2)^2*(1+tan(10/3*pi*l1)^2)*l1+.2700000000e-6*tan(10/3*pi*l1)^2*tan(10/3*pi*l2)*(1+tan(10/3*pi*l2)^2)*l2+.3637500000e-6*tan(10/3*pi*l1)*(1+tan(10/3*pi*l1)^2)*l1+.1350000000e-6*tan(10/3*pi*l2)*(1+tan(10/3*pi*l2)^2)*l2)/(36.75390625*tan(10/3*pi*l1)^4*tan(10/3*pi*l2)^2-30.4687500*tan(10/3*pi*l1)^3*tan(10/3*pi*l2)+20.2500*tan(10/3*pi*l1)^4+40.5000*tan(10/3*pi*l1)^2*tan(10/3*pi*l2)^2+54.562500*tan(10/3*pi*l1)^2+20.2500*tan(10/3*pi*l2)^2+20.2500)^3+10.1250*(-1.666666667*tan(10/3*pi*l1)+1.805555556*tan(10/3*pi*l1)^2*tan(10/3*pi*l2))*(6.0625*tan(10/3*pi*l1)^2*tan(10/3*pi*l2)-9.750*tan(10/3*pi*l1)-4.50*tan(10/3*pi*l2))^2*(-.5555555557e-8*(1+tan(10/3*pi*l1)^2)*l1+.1203703704e-7*tan(10/3*pi*l1)*tan(10/3*pi*l2)*(1+tan(10/3*pi*l1)^2)*l1+.6018518520e-8*tan(10/3*pi*l1)^2*(1+tan(10/3*pi*l2)^2)*l2)/(36.75390625*tan(10/3*pi*l1)^4*tan(10/3*pi*l2)^2-30.4687500*tan(10/3*pi*l1)^3*tan(10/3*pi*l2)+20.2500*tan(10/3*pi*l1)^4+40.5000*tan(10/3*pi*l1)^2*tan(10/3*pi*l2)^2+54.562500*tan(10/3*pi*l1)^2+20.2500*tan(10/3*pi*l2)^2+20.2500)^2+10.1250*(-1.666666667*tan(10/3*pi*l1)+1.805555556*tan(10/3*pi*l1)^2*tan(10/3*pi*l2))^2*(6.0625*tan(10/3*pi*l1)^2*tan(10/3*pi*l2)-9.750*tan(10/3*pi*l1)-4.50*tan(10/3*pi*l2))*(.4041666667e-7*tan(10/3*pi*l1)*tan(10/3*pi*l2)*(1+tan(10/3*pi*l1)^2)*l1+.2020833333e-7*tan(10/3*pi*l1)^2*(1+tan(10/3*pi*l2)^2)*l2-.3250000000e-7*(1+tan(10/3*pi*l1)^2)*l1-.1500000000e-7*(1+tan(10/3*pi*l2)^2)*l2)/(36.75390625*tan(10/3*pi*l1)^4*tan(10/3*pi*l2)^2-30.4687500*tan(10/3*pi*l1)^3*tan(10/3*pi*l2)+20.2500*tan(10/3*pi*l1)^4+40.5000*tan(10/3*pi*l1)^2*tan(10/3*pi*l2)^2+54.562500*tan(10/3*pi*l1)^2+20.2500*tan(10/3*pi*l2)^2+20.2500)^2-10.1250*(-1.666666667*tan(10/3*pi*l1)+1.805555556*tan(10/3*pi*l1)^2*tan(10/3*pi*l2))^2*(6.0625*tan(10/3*pi*l1)^2*tan(10/3*pi*l2)-9.750*tan(10/3*pi*l1)-4.50*tan(10/3*pi*l2))^2*(.4900520833e-6*tan(10/3*pi*l1)^3*tan(10/3*pi*l2)^2*(1+tan(10/3*pi*l1)^2)*l1+.2450260417e-6*tan(10/3*pi*l1)^4*tan(10/3*pi*l2)*(1+tan(10/3*pi*l2)^2)*l2-.3046875000e-6*tan(10/3*pi*l1)^2*tan(10/3*pi*l2)*(1+tan(10/3*pi*l1)^2)*l1-.1015625000e-6*tan(10/3*pi*l1)^3*(1+tan(10/3*pi*l2)^2)*l2+.2700000000e-6*tan(10/3*pi*l1)^3*(1+tan(10/3*pi*l1)^2)*l1+.2700000000e-6*tan(10/3*pi*l1)*tan(10/3*pi*l2)^2*(1+tan(10/3*pi*l1)^2)*l1+.2700000000e-6*tan(10/3*pi*l1)^2*tan(10/3*pi*l2)*(1+tan(10/3*pi*l2)^2)*l2+.3637500000e-6*tan(10/3*pi*l1)*(1+tan(10/3*pi*l1)^2)*l1+.1350000000e-6*tan(10/3*pi*l2)*(1+tan(10/3*pi*l2)^2)*l2)/(36.75390625*tan(10/3*pi*l1)^4*tan(10/3*pi*l2)^2-30.4687500*tan(10/3*pi*l1)^3*tan(10/3*pi*l2)+20.2500*tan(10/3*pi*l1)^4+40.5000*tan(10/3*pi*l1)^2*tan(10/3*pi*l2)^2+54.562500*tan(10/3*pi*l1)^2+20.2500*tan(10/3*pi*l2)^2+20.2500)^3)/(11.390625*(-3.333333334*tan(10/3*pi*l1)+3.611111112*tan(10/3*pi*l1)^2*tan(10/3*pi*l2))^2*(3.25*tan(10/3*pi*l1)*tan(10/3*pi*l2)+1.5*tan(10/3*pi*l1)^2-1.5)^2/(36.75390625*tan(10/3*pi*l1)^4*tan(10/3*pi*l2)^2-30.4687500*tan(10/3*pi*l1)^3*tan(10/3*pi*l2)+20.2500*tan(10/3*pi*l1)^4+40.5000*tan(10/3*pi*l1)^2*tan(10/3*pi*l2)^2+54.562500*tan(10/3*pi*l1)^2+20.2500*tan(10/3*pi*l2)^2+20.2500)^2+5.0625*(-1.666666667*tan(10/3*pi*l1)+1.805555556*tan(10/3*pi*l1)^2*tan(10/3*pi*l2))^2*(6.0625*tan(10/3*pi*l1)^2*tan(10/3*pi*l2)-9.750*tan(10/3*pi*l1)-4.50*tan(10/3*pi*l2))^2/(36.75390625*tan(10/3*pi*l1)^4*tan(10/3*pi*l2)^2-30.4687500*tan(10/3*pi*l1)^3*tan(10/3*pi*l2)+20.2500*tan(10/3*pi*l1)^4+40.5000*tan(10/3*pi*l1)^2*tan(10/3*pi*l2)^2+54.562500*tan(10/3*pi*l1)^2+20.2500*tan(10/3*pi*l2)^2+20.2500)^2)^(1/2) 的值为1.5，约为0.02和0.06。我在MATLAB中使用了l1函数，但是它提供的解决方案是-96和220.我使用l2函数来查看两个方程之间的交叉点，并看到交叉点对于实数值来说很多z和solve，而根据论文的解决方案，只有一个解决方案的值为0.02和0.06。

Plz指导我该怎么做。