我有两个非线性方程，有三个变量，这些变量是在论文中得出的 . 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; theta1=w/c*l1 和 theta2=w/c*l2 与 c=3*10^8 和 l1,l2 是长度 . 这两个方程处于工作频率 . 本文的作者已经将 z 给定值的这两个方程从一个特定值（如1.2）解析为另一个值（如3），并绘制了另外两个变量 theta1 和 theta2 ，相对于 z . w 被选为 500*10^6 . 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 theta1 和 theta2 （ theta1 和 theta2 的电子长度，在度数范围内）获得的值在 z=1.2 to 3 范围内，约为5度 . 到30度对于 theta2 （曲线如 sqrt(x) ）和约50度 . 到10度for theta1 （曲线如 exp(x) ） .
为了解决这些方程，我开始使用 z 的固定值，如 z=1.5 和相同的工作频率 w=500*10^6 . 具有这些考虑因素的两个结果方程是：
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)
根据该论文的结果，对于 z ， l1 和 l2 的值为1.5，大约为0.02和0.06 . 我在MATLAB中使用了 solve 函数，但是它提供的解决方案是-96和220.我使用 ezplot 函数来查看两个方程之间的交叉点，并看到交叉点对于 l1 和 l2 的实际值很多，而根据纸张溶液，只有一种溶液，其值为0.02和0.06 .
Plz指导我该怎么办 .