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How to find all roots of an equation?

Posted 2 years ago

In[162]:= b = 1;

In[163]:= Df = 0.05;

In[164]:= r0 = 0.02;

In[165]:= rx = 2;

In[166]:= m = 0.05;

In[167]:= jrmax = 5.6*10^(-7);

In[168]:= k = 2.5;

In[169]:= v0 =  1;

In[170]:= S = 1.4*10^(-8);

In[171]:= cs0 = S/m

Out[171]= 2.8*10^-7

In[172]:= t = 0.008

Out[172]= 0.008

In[173]:= R0 = r0/Sqrt[Df*b/m]

Out[173]= 0.02

In[174]:= Rx = rx/Sqrt[Df*b/m]

Out[174]= 2.

In[175]:= Cs0 = cs0/cs0

Out[175]= 1.

In[176]:= T = t/(r0^2/Df)

Out[176]= 1.

In[177]:= Rr =  r0*k/(b*Df)

Out[177]= 1.

In[178]:= Rv =  r0*v0/(b*Df)

Out[178]= 0.4

In[179]:= Jrmax = jrmax/(r0*S)

Out[179]= 2000.

In[180]:= h1 =  Rv/R0

Out[180]= 20.

In[181]:= h2 = Rv/Rx

Out[181]= 0.2

In[182]:= (*characteristic root*)

In[188]:= eq1 = ((h1 + (1 - Rv/2)/R0)*
      BesselY[Rv/2, \[Lambda] R0] + \[Lambda]*
      BesselY[Rv/2 - 1, \[Lambda] R0])*((h2 + (1 - Rv/2)/Rx)*
      BesselJ[Rv/2, \[Lambda]*Rx] +
     \[Lambda]*
      BesselJ[Rv/2 - 1, \[Lambda]*Rx]) - ((h1 + (1 - Rv/2)/R0)* 
      BesselJ[Rv/2, \[Lambda]*R0] +
     \[Lambda]*BesselJ[Rv/2 - 1, \[Lambda] R0])*((h2 + (1 - Rv/2)/Rx)*
      BesselY[Rv/2, \[Lambda]*Rx] +
     \[Lambda]*BesselY[Rv/2 - 1, \[Lambda] Rx])

Out[188]= (\[Lambda] BesselJ[-0.8, 2. \[Lambda]] + 
    0.6 BesselJ[0.2, 2. \[Lambda]]) (\[Lambda] BesselY[-0.8, 
      0.02 \[Lambda]] + 
    60. BesselY[0.2, 0.02 \[Lambda]]) - (\[Lambda] BesselJ[-0.8, 
      0.02 \[Lambda]] + 
    60. BesselJ[0.2, 0.02 \[Lambda]]) (\[Lambda] BesselY[-0.8, 
      2. \[Lambda]] + 0.6 BesselY[0.2, 2. \[Lambda]])

Find all roots of eq1.

POSTED BY: Jacques Ou
4 Replies
Posted 2 years ago
In[214]:= eq3 = NSolve[{eq1 == 0, 0 < \[Lambda] < 100}, \[Lambda]]

Out[214]= {{\[Lambda] -> 0.901317}, {\[Lambda] -> 
   2.38046}, {\[Lambda] -> 3.96149}, {\[Lambda] -> 
   7.16348}, {\[Lambda] -> 8.76828}, {\[Lambda] -> 
   10.3734}, {\[Lambda] -> 11.9785}, {\[Lambda] -> 
   13.5832}, {\[Lambda] -> 15.1875}, {\[Lambda] -> 
   16.7915}, {\[Lambda] -> 18.3949}, {\[Lambda] -> 
   19.9979}, {\[Lambda] -> 21.6003}, {\[Lambda] -> 
   23.2024}, {\[Lambda] -> 24.8039}, {\[Lambda] -> 
   26.405}, {\[Lambda] -> 28.0056}, {\[Lambda] -> 
   29.6057}, {\[Lambda] -> 31.2055}, {\[Lambda] -> 
   32.8047}, {\[Lambda] -> 34.4036}, {\[Lambda] -> 
   36.002}, {\[Lambda] -> 37.6001}, {\[Lambda] -> 
   39.1977}, {\[Lambda] -> 40.795}, {\[Lambda] -> 
   42.392}, {\[Lambda] -> 43.9885}, {\[Lambda] -> 
   45.5848}, {\[Lambda] -> 47.1807}, {\[Lambda] -> 
   48.7762}, {\[Lambda] -> 50.3715}, {\[Lambda] -> 
   51.9665}, {\[Lambda] -> 53.5612}, {\[Lambda] -> 
   55.1557}, {\[Lambda] -> 56.7499}, {\[Lambda] -> 
   58.3438}, {\[Lambda] -> 59.9375}, {\[Lambda] -> 
   61.531}, {\[Lambda] -> 63.1242}, {\[Lambda] -> 
   64.7172}, {\[Lambda] -> 66.3101}, {\[Lambda] -> 
   67.9027}, {\[Lambda] -> 69.4951}, {\[Lambda] -> 
   71.0874}, {\[Lambda] -> 72.6795}, {\[Lambda] -> 
   74.2714}, {\[Lambda] -> 75.8632}, {\[Lambda] -> 
   77.4548}, {\[Lambda] -> 79.0463}, {\[Lambda] -> 
   80.6376}, {\[Lambda] -> 82.2288}, {\[Lambda] -> 
   83.8198}, {\[Lambda] -> 85.4108}, {\[Lambda] -> 
   87.0016}, {\[Lambda] -> 88.5923}, {\[Lambda] -> 
   90.1828}, {\[Lambda] -> 91.7733}, {\[Lambda] -> 
   93.3637}, {\[Lambda] -> 94.954}, {\[Lambda] -> 
   96.5441}, {\[Lambda] -> 98.1342}, {\[Lambda] -> 99.7242}}
POSTED BY: Updating Name

Hi Daniel,

NSolve[eq1 == 0,  \\[Lambda]]

Did not work, now I am aware that it is possible to specify additional conditions. Thank you.

POSTED BY: Robert Nowak

(1) Copying and pasting line by line is tedious. Supplying only the input, sans In[...]=, is preferred.

(2) It can be solved if the variable is restricted to a finite range.

NSolve[{eq1 == 0, -10 <= \[Lambda] <= 10}, \[Lambda]]

(* Out[26]= {{\[Lambda] -> -8.76828}, {\[Lambda] -> -7.16348}, \
{\[Lambda] -> -3.96149}, {\[Lambda] -> -2.38046}, {\[Lambda] -> \
-0.901317}, {\[Lambda] -> 0.901317}, {\[Lambda] -> 
   2.38046}, {\[Lambda] -> 3.96149}, {\[Lambda] -> 
   7.16348}, {\[Lambda] -> 8.76828}} *)
POSTED BY: Daniel Lichtblau

Hi Jaques

You could do this by iteration using the fact that three successive roots have closely the same distance.

Below the code to find the first 20 positive roots:

r1 = \\[Lambda] /. FindRoot[eq1 == 0, {\\[Lambda], 1}];

r2 = \\[Lambda] /. FindRoot[eq1 == 0, {\\[Lambda], 2}];


roots = Join[{r1, r2}, 
   Table[r = 2 r2 - r1; r1 = r2; 
    r2 = \\[Lambda] /. FindRoot[eq1 == 0, {\\[Lambda], r}], {20}]];

Plot[eq1, {\\[Lambda], 0, 35},  Epilog -> {Red, Point[{#, 0}] & /@ roots}]

Best Regards Robert

POSTED BY: Robert Nowak
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