# Problem in solving non linear set of equation?

I am trying to solve a set of non linear equation but using the function nsolve but mathematica is not giving me proper answer. The code is given below.

h = 0.8;
\[Epsilon]r = 2.94;
ae = 5;


The Equations are given below.

Equation Number 1....

1/we = (120*\[Pi])/(ae (w/h + 1.393 + 0.667*Log[w/h + 1.44]));


Equation Number 2....

1/we = (4.38/
ae)*(E^((-0.627*\[Epsilon]r)/(((\[Epsilon]r + 1)/
2) + (\[Epsilon]r - 1)/2 + 1/Sqrt[(1 + (12*h)/w)])));


Nsolve for solving both equations for "we" and "w" simultaneously.

NSolve[{1/we == (120*\[Pi])/(
ae (w/h + 1.393 + 0.667*Log[w/h + 1.44])),
1/we == 4.38/
ae*(E^((-0.627*\[Epsilon]r)/(((\[Epsilon]r + 1)/
2) + (\[Epsilon]r - 1)/2 + 1/Sqrt[(1 + (12*h)/w)])))}, {we,
w}, Reals]


I am not getting any solution for the above equations using NSOLve.

• NSolve deals primarily with linear and polynomial equations. In your case you might have more luck using e.g. FindRoot with an adequate estimate of the solution values. – MarcoB Jun 8 '15 at 5:02
• Using {{w, 100}, {we, 2}} for the starting values should do it with FindRoot as recommended by MarcoB. – JimB Jun 8 '15 at 5:17
• @JimBaldwin , I used the Find Root Method with the code given below .FindRoot[{1/we == (120*\[Pi])/( ae (w/h + 1.393 + 0.667*Log[w/h + 1.44])), 1/we == (4.38/ ae)*(E^((-0.627*\[Epsilon]r)/(((\[Epsilon]r + 1)/ 2) + (\[Epsilon]r - 1)/2 + 1/Sqrt[(1 + (12*h)/w)])))}, {{w, 100}, {we, 2}}] But that gives me some dimension related error . – AK K Khan Jun 8 '15 at 6:47

Using the starting value {w,100} and setting both sides equal (we can ignore the reciprocal 1/we) and then solve for w first. Apologies to Jim and MarcoB, this is a rip answer to clean things up.

h = 0.8;
\[Epsilon]r = 2.94;
ae = 5;

eqn = (120*\[Pi])/(ae (w/h + 1.393 + 0.667*Log[w/h + 1.44])) ==
(4.38/ae)*(E^((-0.627*\[Epsilon]r)/(((\[Epsilon]r + 1)/
2) + (\[Epsilon]r - 1)/2 + 1/Sqrt[(1 + (12*h)/w)])));

FindRoot[eqn, {w, 100}] (* Gives  w -> 106.761 *)
1/(eqn[] /. w -> 106.761) (* Implies we is 1.83182 *)

• @Histograms...Thanks for your answer..This was really helpful. – AK K Khan Jun 8 '15 at 8:26
• No need to apologize. Thank you for writing this up! – MarcoB Jun 8 '15 at 14:03