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How to find all roots of an equation?

Jacques Ou

Posted 4 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 = 5.610^(-7); In[168]:= k = 2.5; In[169]:= v0 = 1; In[170]:= S = 1.410^(-8); In[171]:= cs0 = S/m Out[171]= 2.810^-7 In[172]:= t = 0.008 Out[172]= 0.008 In[173]:= R0 = r0/Sqrt[Dfb/m] Out[173]= 0.02 In[174]:= Rx = rx/Sqrt[Dfb/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 = r0k/(bDf) Out[177]= 1. In[178]:= Rv = r0v0/(bDf) Out[178]= 0.4 In[179]:= Jrmax = jrmax/(r0S) 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.

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

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Updating Name

Posted 4 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}}

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

Robert Nowak

Robert Nowak, IMS Nanofabrication

Posted 4 years ago

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

Daniel Lichtblau

Daniel Lichtblau, Wolfram Research

Posted 4 years ago

(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}} *)

(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

Robert Nowak

Robert Nowak, IMS Nanofabrication

Posted 4 years ago

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

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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