AppendTo@$Path, ToFileName@8$HomeDirectory, "Dropbox", "mathematica"<DD;
<< TrigInt`
Upper bounds
T2@a_, b_D = 88- 1, 0<, 81, 0<, 8a, b<<; H* general triangle *L
f = Equilateral@Neumann, SymmetricD@0, 1D
f = Transplant@f, 880, Sqrt@3D<, 81, 0<, 8- 1, 0<<, Equilateral@DD
H* scale and rotate to put nodal line near vertex H-1,0L *L
- HD@ef1, x, xD + D@ef1, y, yDL ê ef1 êê FullSimplify
And üü HÒ@@2DD § Ò@@1DD § Ò@@3DD & êü 8Limits@T2@0, Sqrt@3DDD<L
RegionPlot@8f > 0, %<, 8x, - 2, 2<, 8y, - 2, 2<D
tf = Transplant@f, T2@a, bD, T2@0, Sqrt@3DDD êê FullSimplify
TrigInt@tf, Limits@T2@a, bDDD êê FullSimplify@Ò, b > 0D &
H* average is zero, good test function *L
U1 = Rayleigh@tf, T2@a, bDD êê FullSimplify@Ò, b > 0D &
H* U1 is the upper bound for near equilateral *L
U2 = BesselJZero@0, 1D ^ 2
H* U2 is the upper bound for near degenerate HCheng's boundL
We could also use Rayleigh@x-aê3,T2@a,bDD. *L
lin = x;H*+bê2 y;*L
c = TrigInt@lin, Limits@T2@a, bDDD ê b êê FullSimplify@Ò, b > 0D &
TrigInt@lin - c, Limits@T2@a, bDDD êê FullSimplify@Ò, b > 0D &
U2 = Rayleigh@lin - c, T2@a, bDD êê FullSimplify@Ò, b > 0D &
-cos
1
3
-cos
1
3
p J 3 - 2 yN
p H2 x + 2L cos
+ cos
3
p 5-4
x
3 y
-
4
+
4
p K 3 - 2 K-
3 x
4
y
4
+
3
4
OO
3
0
y
3
3 x
4
-
y
4
3
+
4
O
cos
3
cos
3 Ì
2y
-1§x§1-
y
3
1
3
- cos
3
+ cos p 1 -
cos
0§y§
3
p H2 x - 1L cos p 1 -
2 K-
3
4
-
1
p 2
x
4
-
3 y
4
+
3
4
+2
1
3
1
3
p 2
p H5 - 4 xL
x
4
-
3 y
4
+
3
4
-1
-
2
simple.nb
sin
p H2 Ha + 3L y - 2 b x + bL
6b
- 2 sin
p H-a y + b x + yL
0
64 p2 Ha2 + b2 + 3L + 243 HHa - 6L a + b2 - 3L
288 b2
H j0,1 L2
a
3
0
18
2
a +3
2b
cos
p Hb Hx + 1L - Ha + 3L yL
6b
simple.nb
Lower bounds
H* H0,Sqrt@3DL as reference *L
ineq1 = Ha ^ 2 + 3L H1 - gL + 2 a b d + b ^ 2 g < 4 p ^ 2 ê 3 ê u;
H* H0,1L as reference *L
ineq2 = Ha ^ 2 + 1L H1 - gL + 2 a b d + b ^ 2 g < 3 p ^ 2 ê 4 ê u;
H* H1ê2,Sqrt@3Dê2L as reference Hhalf of equilateralL *L
ineq3 = HHa - 1 ê 2L ^ 2 + 3 ê 4L H1 - gL + 2 Ha - 1 ê 2L b d + b ^ 2 g < 2 p ^ 2 ê 3 ê u;
H* H1,2êSqrt@3DL as reference Hhalf of equilateral, longest side on x-axis L*L
ineq4 = HHa - 1L ^ 2 + 4 ê 3L H1 - gL + 2 Ha - 1L b d + b ^ 2 g < 8 p ^ 2 ê 9 ê u;
H* Near equilateral we need 1 2 3. Near degenerate 3 4. *L
neareq = ineq1 »» ineq2 »» ineq3 ê. u Ø U1
nearde = ineq3 »» ineq4 ê. u Ø U2
middle = ineq3 »» ineq4 ê. u Ø U1
g1@c_ ? NumberQ, d_ ? NumberQD :=
FullSimplify@neareq ê. a Ø c ê. b Ø d, 0 § g § 1 && - 1 ê 2 § d § 1 ê 2D
g2@c_ ? NumberQ, d_ ? NumberQD :=
FullSimplify@nearde ê. a Ø c ê. b Ø d, 0 § g § 1 && - 1 ê 2 § d § 1 ê 2D
Ia2 + 3M H1 - gL + 2 a b d + b2 g <
384 p2 b2
64 p2 Ha2 + b2 + 3L + 243 HHa - 6L a + b2 - 3L
Ia2 + 1M H1 - gL + 2 a b d + b2 g <
2 a-
2 a-
2 a-
1
2
1
2
1
2
bd+
a-
bd+
a-
bd+
a-
1
1
2
+
2
4
2
+
2
1
3
3
4
2
+
2
2 Ha - 1L b d + Ha - 1L2 +
3
4
4
3
Î
216 p2 b2
64 p2 Ha2 + b2 + 3L + 243 HHa - 6L a + b2 - 3L
H1 - gL + b2 g <
H1 - gL + b2 g <
H1 - gL + b2 g <
H1 - gL + b2 g <
Î
192 p2 b2
64 p2 Ha2 + b2 + 3L + 243 HHa - 6L a + b2 - 3L
1
27
p2 Ia2 + 3M Î 2 Ha - 1L b d + Ha - 1L2 +
4
3
192 p2 b2
64 p2 Ha2 + b2 + 3L + 243 HHa - 6L a + b2 - 3L
256 p2 b2
64 p2 Ha2 + b2 + 3L + 243 HHa - 6L a + b2 - 3L
H1 - gL + b2 g <
Î
4
81
p2 Ia2 + 3M
3
4
simple.nb
tri = Ha + 1L ^ 2 + b ^ 2 § 4 && a > 0 && 2 a ^ 2 + 2 b ^ 2 > 2;
RegionPlot@8g1@a, bD, g2@a, bD, tri, a > 1 ê 2<,
8a, 0, 1<, 8b, 0, Sqrt@3D<, PlotPoints Ø 10, MaxRecursion Ø 2D
Ha + 1L2 + b2 § 4 Ï a > 0 Ï 2 a2 + 2 b2 > 2
Clearly neareq and nearde cover every acute triangle.
Proof for nearly degenerate triangles (with 1/2 £ a £ 1 and 0 < b £ 1.08).
RegionPlot@8tri, 1 ê 2 § a § 1 && 0 < b § 1.08<, 8a, 0, 1<, 8b, 0, Sqrt@3D<D
nearde H* both inequalities *L
2 a-
1
2
bd+
a-
1
2
2
+
3
4
H1 - gL + b2 g <
1
27
p2 Ia2 + 3M Î 2 Ha - 1L b d + Ha - 1L2 +
4
3
H1 - gL + b2 g <
At least one inequality will be true if a positive linear combination of inequalities is true.
4
81
p2 Ia2 + 3M
simple.nb
5
nearde ê. Less Ø Subtract ê. Or Ø List
left = 81 - a, a - 1 ê 2<.% êê Collect@Ò, g, FullSimplifyD &
H* linear combination, d eliminated, linear in g *L
:-
1
27
-
1
81
p2 Ia2 + 3M + 2 a -
p2 Ha + 1L Ia2 + 3M +
1
bd+
2
1
6
a-
1
2
+
2
3
H1 - gL + b2 g, -
4
g Ia H3 a - 8L + 3 b2 + 1M +
1
6
4
81
p2 Ia2 + 3M + 2 Ha - 1L b d + Ha - 1L2 +
4
3
H1 - gL + b2 g>
HH8 - 3 aL a - 1L
Linear in g, hence we need to check endpoints g = 0 and g = 1.
both = 8left < 0 ê. g Ø 1, left < 0 ê. g Ø 0< êê Collect@Ò, 8a, b<, FullSimplifyD &
:-
1
81
p2 a3 -
p2 a2
-
p2 a
81
+
27
b2
-
2
p2
27
< 0, -
1
81
p2 a3 + -
1
-
2
p2
81
a2 -
1
27
Hp - 6L H6 + pL a +
1
54
I-9 - 2 p2 M < 0>
The first inequality has negative signs for a, so we can put a = 1/2. The second inequality is a cubic in a.
both@@1DD ê. a Ø 1 ê 2 êê Reduce êê N
both@@2DD êê Reduce êê N
-1.08996 < b < 1.08996
a > -6.44176
Hence we are done.
Algorithm for polynomial inequalities
PolyNeg[P,{x,y},{dx,dy}]: show that polynomial P with variables x, y is nonpositive on rectangle (0, 0) (dx, dy). True
means it is, False means algorithm failed. See http://www.ams.org/mathscinet-getitem?mr=MR2779073 for detail.
We can use this on large polynomials in all other cases.
CumFun@f_, l_D := Rest@FoldList@f, 0, lDD;
PolyNeg@P_, 8x_, y_<, 8dx_, dy_<D :=
HHFold@CumFun@Min@Ò1, 0D ê dy + Ò2 &, Map@Max@Ò1, 0D &, Ò1D dx + Ò2D &,
0, Reverse@CoefficientList@P, 8x, y<DDD êê MaxL § 0L;
Proof for triangles with 1/2 £ a £ 1 and b ³≥ 1.08.
neareq
Ia2 + 3M H1 - gL + 2 a b d + b2 g <
384 p2 b2
64 p2 Ha2 + b2 + 3L + 243 HHa - 6L a + b2 - 3L
Ia2 + 1M H1 - gL + 2 a b d + b2 g <
2 a-
1
2
bd+
a-
1
2
2
+
3
4
Î
216 p2 b2
64 p2 Ha2 + b2 + 3L + 243 HHa - 6L a + b2 - 3L
H1 - gL + b2 g <
Î
192 p2 b2
64 p2 Ha2 + b2 + 3L + 243 HHa - 6L a + b2 - 3L
Coefficient for d is positive, hece we can put d = 1/2. Now we need to show that at least one inequolity is true for any
g. Start with inequality 3.
6
simple.nb
neareq ê. Or Ø List ê. Less Ø Subtract;
lin = %@@3DD ê. d Ø 1 ê 2 êê Collect@Ò, g, FullSimplifyD &
RegionPlot@8D@lin, gD > 0, tri && b > 1.08 && a > 1 ê 2<, 8a, 0, 1<, 8b, 0, Sqrt@3D<D
H* red are is where we need a proof, blue has positive coefficient for g *L
D@lin, gD
g I-a2 + a + b2 - 1M -
192 p2 b2
2
2
2
2
64 p Ha + b + 3L + 243 HHa - 6L a + b - 3L
+ a2 + a Hb - 1L -
b
2
+1
-a2 + a + b2 - 1
Derivative is clearly minimal for b = 1.08, and it has no zeros in the red area. Hence we need to check only large
values of g. Unfortunately g=1 does not work.
simple.nb
lin ê. g Ø 2 ê 3
RegionPlot@Evaluate@8% < 0, tri && b > 1.08 && a > 1 ê 2<D, 8a, 0, 1<, 8b, 0, Sqrt@3D<D
2
3
I-a2 + a + b2 - 1M -
192 p2 b2
2
2
2
2
64 p Ha + b + 3L + 243 HHa - 6L a + b - 3L
+ a2 + a Hb - 1L -
b
2
+1
It seems that inequality is true. We will use PolyNeg to prove this. First we move the origin to the vertex with a=1/2.
Together@lin ê. g Ø 2 ê 3 ê. b Ø Sqrt@7D ê 2 - b ê. a Ø a + 1 ê 2D
poly = poly2 = Numerator@%D Denominator@%D;
PolyNeg@poly, 8b, a<, 81 ê 4, 99 ê 1000<D
poly = poly ê. a Ø a + 99 ê 1000 ê. b Ø b + 1 ê 9;
PolyNeg@poly, 8b, a<, 815 ê 100, 1 ê 10<D
K256 p2 a4 + 972 a4 - 2916 a3 b - 768 p2 a3 b + 384
2916
7 a2 b + 14 580 a2 b - 768
2916 a b3 - 768 p2 a b3 + 4374
512
3888
2
2
7 p2 a2 b - 768 p2 a2 b + 384
7 a b2 - 9720 a b2 + 1152
7 p a b - 6528 p a b + 1920
7 b3 - 1024
7 p2 a3 + 256 p2 a3 + 1458
2
7 a3 - 4860 a3 + 2916 a2 b2 + 768 p2 a2 b2 -
7 p2 a2 + 2368 p2 a2 - 7290
7 p2 a b2 + 512 p2 a b2 + 9720
2
7 p a + 1088 p a - 5832
7 p2 b3 + 9963 b2 + 4928 p2 b2 + 3645
K12 K64 p2 a2 + 243 a2 + 64 p2 a - 1215 a + 243 b2 + 64 p2 b2 - 243
7 a2 + 243 a2 -
7 a b + 1458 a b -
4
7 a - 20 655 a + 1944 b + 512 p2 b4 -
7 b - 1344
7 b - 64
7 p2 b + 1408 p2 - 16 524O ì
7 p2 b + 320 p2 - 972OO
True
True
Hence inequality is true on the whole parameter space. Below is the plot of the rectangles used for the algorithm.
7
8
simple.nb
params = tri && b > 1.08 && a > 1 ê 2 ê. b Ø Sqrt@7D ê 2 - b ê. a Ø a + 1 ê 2;
H* transformed parameter space *L
RegionPlot@8poly2 < 0, params, 0 < a < 99 ê 1000 && 0 < b < 1 ê 4,
99 ê 1000 < a < 199 ê 1000 && 1 ê 9 < b < 1 ê 9 + 15 ê 100<, 8a, - 0.1, 0.25<, 8b, - 0.05, 0.3<D
Now we need to handle 1 >= g > 2/3. This time we will take a linear combination of inequalities 1 and 3.
simple.nb
neareq ê. Or Ø List ê. Less Ø Subtract
lin = %@@3DD + Hb - aL %@@1DD ê. d Ø 1 ê 2 êê Collect@Ò, g, FullSimplifyD &
RegionPlot@8D@%, gD > 0, tri && b > 1.08 && a > 1 ê 2<, 8a, 0, 1<, 8b, 0, Sqrt@3D<D
D@lin, gD
:-
384 p2 b2
64 p2 Ha2 + b2 + 3L + 243 HHa - 6L a + b2 - 3L
-
-
+ Ia2 + 3M H1 - gL + 2 a b d + b2 g,
216 p2 b2
64 p2 Ha2 + b2 + 3L + 243 HHa - 6L a + b2 - 3L
192 p2 b2
64 p2 Ha2 + b2 + 3L + 243 HHa - 6L a + b2 - 3L
+ Ia2 + 1M H1 - gL + 2 a b d + b2 g,
+2 a-
1
2
bd+
a-
1
2
2
+
3
4
H1 - gL + b2 g>
I-64 p2 I-a Ia2 + 3M b + a HHa - 1L a + 4L Ia2 + 3M - a b4 - Ha - 6L b3 - Ha Ha + 5L - 3L b2 M 243 a IHa - 6L a + b2 - 3M HHa - 1L a - b Hb + 1L + 4LM ë I64 p2 Ia2 + b2 + 3M + 243 IHa - 6L a + b2 - 3MM +
g Ia3 - a2 Hb + 1L - a Ib2 - 4M + b Ib2 + b - 3M - 1M +
5b
2
+1
a3 - a2 Hb + 1L - a Ib2 - 4M + b Ib2 + b - 3M - 1
Here it is not so clear that coefficient for g is negative, but we can use the algorithm to prove this. Then we can put
g=2/3.
9
10
simple.nb
lin ê. g Ø 2 ê 3
RegionPlot@8% < 0, tri && b > 1.08 && a > 1 ê 2<, 8a, 0, 1<, 8b, 0, Sqrt@3D<D
I-64 p2 I-a Ia2 + 3M b + a HHa - 1L a + 4L Ia2 + 3M - a b4 - Ha - 6L b3 - Ha Ha + 5L - 3L b2 M 243 a IHa - 6L a + b2 - 3M HHa - 1L a - b Hb + 1L + 4LM ë I64 p2 Ia2 + b2 + 3M + 243 IHa - 6L a + b2 - 3MM +
2
3
Ia3 - a2 Hb + 1L - a Ib2 - 4M + b Ib2 + b - 3M - 1M +
5b
2
+1
Hence it appears that inequality is true.
poly = D@lin, gD ê. b Ø Sqrt@7D ê 2 - b ê. a Ø a + 1 ê 2
PolyNeg@poly, 8b, a<, 81 ê 4, 1 ê 5<D
- a+
1
2
2
-b +
7
2
+1 - a+
True
Hence derivative is negative.
1
7
2
2
2
-b
-4 + a+
1
2
3
+
7
2
-b
7
2
2
-b
-b+
7
2
-3 -1
simple.nb
params = tri && b > 1.08 && a > 1 ê 2 ê. b Ø Sqrt@7D ê 2 - b ê. a Ø a + 1 ê 2;
H* transformed parameter space *L
RegionPlot@8poly < 0, params, 0 < a < 1 ê 5 && 0 < b < 1 ê 4<, 8a, - 0.1, 0.25<, 8b, - 0.05, 0.3<D
Together@lin ê. g Ø 2 ê 3 ê. b Ø Sqrt@7D ê 2 - b ê. a Ø a + 1 ê 2D
poly = poly2 = Numerator@%D Denominator@%D;
PolyNeg@poly, 8b, a<, 81 ê 4, 5 ê 100<D
poly = poly ê. a Ø a + 5 ê 100 ê. b Ø b + 5 ê 100;
PolyNeg@poly, 8b, a<, 81 ê 4 - 5 ê 100, 9 ê 100<D
poly = poly ê. a Ø a + 9 ê 100 ê. b -> b + 12 ê 100;
PolyNeg@poly, 8b, a<, 81 ê 4 - 17 ê 100, 6 ê 100<D
K-128 p2 a5 - 486 a5 + 972 a4 b + 256 p2 a4 b - 128
7 p2 a4 - 192 p2 a4 - 486
128 p2 a3 b - 64
7 p2 a3 - 960 p2 a3 + 2673
4374 a2 b - 384
7 p2 a2 b + 384 p2 a2 b - 192
128 p2 a b4 - 972
576
7 p2 a b2 + 3904 p2 a b2 + 8991
3200 p2 a - 8262
2430
7 a b3 + 4374 a b3 - 256
4288 p2 b2 + 3159
7 a3 + 2187 a3 - 1458 a2 b2 + 384 p2 a2 b2 + 1458
7 p2 a2 - 128 p2 a2 + 2187
7 p2 a b3 - 384 p2 a b3 - 6561
7 a b + 31 833 a b - 3008
7 a - 4617 a - 972 b5 - 256 p2 b5 + 2430
7 b3 - 12 636 b3 - 640
7 p2 b3 - 3328 p2 b3 + 1944
7 b + 15 066 b - 2048
True
True
7 a2 b -
7 a2 + 7533 a2 + 486 a b4 +
7 a b2 - 5589 a b2 +
7 p2 a b - 2752 p2 a b + 704
7 p2 a +
7 b4 + 1215 b4 + 640
7 p2 b4 + 320 p2 b4 -
7 b2 + 5346 b2 + 512
7 p2 b2 +
7 p2 b + 1088 p2 b - 96
K6 K64 p2 a2 + 243 a2 + 64 p2 a - 1215 a + 243 b2 + 64 p2 b2 - 243
True
7 a4 + 2187 a4 - 5346 a3 b +
7 p2 + 2240 p2 - 5832
7 b - 64
7 - 6804O ì
7 p2 b + 320 p2 - 972OO
11
12
simple.nb
params = tri && b > 1.08 && a > 1 ê 2 ê. b Ø Sqrt@7D ê 2 - b ê. a Ø a + 1 ê 2;
H* transformed parameter space *L
RegionPlot@
8poly2 < 0, params, 0 < a < 1 ê 20 && 0 < b < 1 ê 4, 5 ê 100 < a < 14 ê 100 && 5 ê 100 < b < 1 ê 4,
14 ê 100 < a < 20 ê 100 && 17 ê 100 < b < 1 ê 4<, 8a, - 0.1, 0.25<, 8b, - 0.05, 0.3<D
Hence a mixed inequality is true, so must be one of the inequalities.
simple.nb
13
Proof for nearly equilateral triangles (a <= 1/2).
RegionPlot@8tri, 0 < a < 1 ê 2 && Sqrt@3D ê 2 < b < Sqrt@3D<, 8a, 0, 1<, 8b, 0, Sqrt@3D<D
neareq ê. Or Ø List ê. Less Ø Subtract
:-
384 p2 b2
64 p2 Ha2 + b2 + 3L + 243 HHa - 6L a + b2 - 3L
-
-
+ Ia2 + 3M H1 - gL + 2 a b d + b2 g,
216 p2 b2
64 p2 Ha2 + b2 + 3L + 243 HHa - 6L a + b2 - 3L
192 p2 b2
64 p2 Ha2 + b2 + 3L + 243 HHa - 6L a + b2 - 3L
+ Ia2 + 1M H1 - gL + 2 a b d + b2 g,
+2 a-
1
2
bd+
a-
1
2
2
+
3
4
H1 - gL + b2 g>
Here we have 3 reference triangles, and no easy way of handling all cases together. In order to show that at least one
inequality is true, we must show that the same is true on the boundary of 0 <= g <= 1 and -1/2 <= d <= 1/2. Since three
lines (in g and d) giving the three inequalities may for a trinagle, we also need to check a point inside this triangle. The
easiest way to get a point inside is to find vertices and average them.
ü Case g = 1 :
Take the first inequality and put d = 1/2 (worst case, positive coefficient).
14
simple.nb
neareq ê. Or Ø List ê. Less Ø Subtract;
%@@1DD ê. g Ø 1
rational = % ê. d Ø 1 ê 2
RegionPlot@8rational < 0, tri && a § 1 ê 2 && b ¥ Sqrt@3D ê 2<, 8a, 0, 1<, 8b, 0, Sqrt@3D<D
-
-
384 p2 b2
64 p2 Ha2 + b2 + 3L + 243 HHa - 6L a + b2 - 3L
384 p2 b2
64 p2 Ha2 + b2 + 3L + 243 HHa - 6L a + b2 - 3L
+ 2 a b d + b2
+ a b + b2
Note that denominator is just the upper bound, so it is positive. Hence we can take a numerator of the common fraction. We can also add any expression that is positive on the parameter space (red area).
simple.nb
poly = Numerator@Together@rationalDD + 2000 H4 - Ha + 1L ^ 2 - b ^ 2L Hb - Sqrt@3D ê 2L
RegionPlot@8poly < 0, tri && a § 1 ê 2 && b ¥ Sqrt@3D ê 2<, 8a, 0, 1<, 8b, 0, 2<D
243 a3 b + 64 p2 a3 b + 243 a2 b2 + 64 p2 a2 b2 - 1458 a2 b + 243 a b3 + 64 p2 a b3 - 1458 a b2 +
2000 b -
3
2
I-Ha + 1L2 - b2 + 4M - 729 a b + 192 p2 a b + 64 p2 b4 + 243 b4 - 192 p2 b2 - 729 b2
15
16
simple.nb
pol = pol2 = poly ê. b Ø Sqrt@3D - b; H* equilateral gives 0 for this polynomial,
so it is a good starting pooint for the algorithm *L
PolyNeg@pol, 8b, a<, 82 ê 3, 1 ê 2<D
pol = pol ê. b Ø b + 2 ê 3;
PolyNeg@pol, 8b, a<, 81 ê 3, 1 ê 2<D
8tri && a < 1 ê 2 && b > Sqrt@3D ê 2< ê. b Ø Sqrt@3D - b H* transformed parameters *L
RegionPlot@8pol < 0, %, 0 < a < 1 ê 2 && 0 < b < 2 ê 3, 0 < a < 1 ê 2 && 2 ê 3 < b < 1<,
8a, - 0.1, 1 ê 2 + 0.1<, 8b, - 0.1, Sqrt@3D ê 2 + 0.1<D
True
True
:Ha + 1L2 + K
2
3 - bO § 4 Ì a > 0 Ì 2 a2 + 2 K
2
3 - bO > 2 Ì a <
1
Ì
2
3 -b>
3
2
ü Case d = 1/2 :
neareq ê. d Ø 1 ê 2 ê. Or Ø List ê. Less Ø Subtract
lin = %@@3DD êê Collect@Ò, g, FullSimplifyD &
D@lin, gD
:-
384 p2 b2
64 p2 Ha2 + b2 + 3L + 243 HHa - 6L a + b2 - 3L
-
-
+ Ia2 + 3M H1 - gL + a b + b2 g,
216 p2 b2
64 p2 Ha2 + b2 + 3L + 243 HHa - 6L a + b2 - 3L
192 p2 b2
64 p2 Ha2 + b2 + 3L + 243 HHa - 6L a + b2 - 3L
g I-a2 + a + b2 - 1M -
+ Ia2 + 1M H1 - gL + a b + b2 g,
+ a-
1
2
b+
a-
192 p2 b2
2
2
2
2
64 p Ha + b + 3L + 243 HHa - 6L a + b - 3L
-a2 + a + b2 - 1
Hence coefficient for g is positive. Put g =2/3.
1
2
2
+
3
4
H1 - gL + b2 g>
+ a2 + a Hb - 1L -
b
2
+1
>
simple.nb
17
lin ê. g Ø 2 ê 3
RegionPlot@8% < 0, tri && a < 1 ê 2<, 8a, 0, 1<, 8b, 0, Sqrt@3D<D
2
3
I-a2 + a + b2 - 1M -
192 p2 b2
64 p2 Ha2 + b2 + 3L + 243 HHa - 6L a + b2 - 3L
+ a2 + a Hb - 1L -
b
2
+1
Inequality appears to hold for g <= 2/3. Note that denominator is positive (upper bound for eigenvalue) hance we can
take just the numerator of common fraction. We can also add anything that is positive on our parameter space and
negative elsewhere.
18
simple.nb
Together@lin ê. g Ø 2 ê 3D
p = Numerator@%D + 10 ^ 4 H4 - Ha + 1L ^ 2 - b ^ 2L Hb - Sqrt@3D ê 2L
RegionPlot@8p < 0, tri && a < 1 ê 2<, 8a, - 0.1, 1 ê 2 + 0.1<, 8b, Sqrt@3D ê 2 - 0.1, Sqrt@3D + 0.1<D
I128 p2 a4 + 486 a4 + 1458 a3 b + 384 p2 a3 b - 128 p2 a3 - 3402 a3 + 1458 a2 b2 + 384 p2 a2 b2 - 9477 a2 b 192 p2 a2 b + 512 p2 a2 + 1944 a2 + 1458 a b3 + 384 p2 a b3 - 6318 a b2 - 128 p2 a b2 + 1152 p2 a b - 384 p2 a 1458 a + 972 b4 + 256 p2 b4 - 729 b3 - 192 p2 b3 - 2430 b2 - 256 p2 b2 + 2187 b - 576 p2 b + 384 p2 - 1458M ë
I6 I64 p2 a2 + 243 a2 - 1458 a + 243 b2 + 64 p2 b2 + 192 p2 - 729MM
128 p2 a4 + 486 a4 + 1458 a3 b + 384 p2 a3 b - 128 p2 a3 - 3402 a3 + 1458 a2 b2 + 384 p2 a2 b2 - 9477 a2 b - 192 p2 a2 b +
512 p2 a2 + 1944 a2 + 1458 a b3 + 384 p2 a b3 - 6318 a b2 - 128 p2 a b2 + 10 000 b -
3
2
I-Ha + 1L2 - b2 + 4M + 1152 p2 a b -
384 p2 a - 1458 a + 972 b4 + 256 p2 b4 - 729 b3 - 192 p2 b3 - 2430 b2 - 256 p2 b2 + 2187 b - 576 p2 b + 384 p2 - 1458
Again inequality seems to be true. Note that for eqilateral traingle we have equality here.
simple.nb
19
poly = poly2 = p ê. b Ø Sqrt@3D - b;
PolyNeg@poly, 8b, a<, 814 ê 100, 25 ê 100<D
poly = poly ê. b Ø b + 14 ê 100;
PolyNeg@poly, 8b, a<, 831 ê 100, 1 ê 2<D
poly = poly ê. b Ø b + 31 ê 100;
PolyNeg@poly, 8b, a<, 845 ê 100, 1 ê 2<D
8tri && a < 1 ê 2< ê. b Ø Sqrt@3D - b;
RegionPlot@8poly2 < 0, %, 0 < b < 14 ê 100 && 0 < a < 25 ê 100, 14 ê 100 < b < 45 ê 100 && 0 < a < 1 ê 2,
45 ê 100 < b < 90 ê 100 && 0 < a < 1 ê 2<, 8a, - 0.1, 1 ê 2 + 0.1<, 8b, - 0.1, Sqrt@3D ê 2 + 0.1<D
True
True
True
For the case g > 2/3 we can take the first inequality.
neareq ê. d Ø 1 ê 2 ê. Or Ø List ê. Less Ø Subtract
lin = %@@1DD êê Collect@Ò, g, FullSimplifyD &
D@lin, gD
:-
384 p2 b2
64 p2 Ha2 + b2 + 3L + 243 HHa - 6L a + b2 - 3L
-
-
+ Ia2 + 3M H1 - gL + a b + b2 g,
216 p2 b2
64 p2 Ha2 + b2 + 3L + 243 HHa - 6L a + b2 - 3L
192 p2 b2
2
2
2
2
64 p Ha + b + 3L + 243 HHa - 6L a + b - 3L
g I-a2 + b2 - 3M -
+ Ia2 + 1M H1 - gL + a b + b2 g,
+ a-
1
2
b+
384 p2 b2
64 p2 Ha2 + b2 + 3L + 243 HHa - 6L a + b2 - 3L
-a2 + b2 - 3
Hence derivative is clearly negative. Put g = 2/3.
a-
1
2
2
+
3
4
+ a2 + a b + 3
H1 - gL + b2 g>
20
simple.nb
lin ê. g Ø 2 ê 3
RegionPlot@8% < 0, tri && a < 1 ê 2<, 8a, 0, 1<, 8b, 0, Sqrt@3D<D
2
3
I-a2 + b2 - 3M -
384 p2 b2
64 p2 Ha2 + b2 + 3L + 243 HHa - 6L a + b2 - 3L
+ a2 + a b + 3
Inequality appears to hold for g >= 2/3. Denominator is positive (upper bound for eigenvalue) hance we can take just
the numerator of common fraction. We can also add anything that is positive on our parameter space and negative
elsewhere.
simple.nb
Together@lin ê. g Ø 2 ê 3D
p = Numerator@%D + 7 * 10 ^ 3 H4 - Ha + 1L ^ 2 - b ^ 2L Hb - 1L ^ 2
RegionPlot@8p < 0, tri && a < 1 ê 2<, 8a, - 0.1, 1 ê 2 + 0.1<, 8b, Sqrt@3D ê 2 - 0.1, Sqrt@3D + 0.1<D
I64 p2 a4 + 243 a4 + 729 a3 b + 192 p2 a3 b - 1458 a3 + 729 a2 b2 + 192 p2 a2 b2 - 4374 a2 b + 384 p2 a2 + 729 a b3 + 192 p2 a b3 2916 a b2 - 2187 a b + 576 p2 a b - 4374 a + 486 b4 + 128 p2 b4 - 729 b2 - 576 p2 b2 + 576 p2 - 2187M ë
I3 I64 p2 a2 + 243 a2 - 1458 a + 243 b2 + 64 p2 b2 + 192 p2 - 729MM
64 p2 a4 + 243 a4 + 729 a3 b + 192 p2 a3 b - 1458 a3 + 729 a2 b2 + 192 p2 a2 b2 4374 a2 b + 384 p2 a2 + 729 a b3 + 192 p2 a b3 - 2916 a b2 + 7000 Hb - 1L2 I-Ha + 1L2 - b2 + 4M 2187 a b + 576 p2 a b - 4374 a + 486 b4 + 128 p2 b4 - 729 b2 - 576 p2 b2 + 576 p2 - 2187
21
22
simple.nb
poly = poly2 = p ê. b Ø Sqrt@3D - b;
PolyNeg@poly, 8b, a<, 823 ê 100, 33 ê 100<D
poly = poly ê. b Ø b + 23 ê 100;
PolyNeg@poly, 8b, a<, 843 ê 100, 1 ê 2<D
poly = poly ê. b Ø b + 43 ê 100;
PolyNeg@poly, 8b, a<, 815 ê 100, 1 ê 2<D
poly = poly ê. b Ø b + 15 ê 100 ê. a Ø a + 1 ê 3;
PolyNeg@poly, 8b, a<, 88 ê 100, 1 ê 6<D
8tri && a < 1 ê 2< ê. b Ø Sqrt@3D - b;
RegionPlot@8poly2 < 0, %, 0 < b < 23 ê 100 && 0 < a < 33 ê 100, 23 ê 100 < b < 66 ê 100 && 0 < a < 1 ê 2,
66 ê 100 < b < 81 ê 100 && 0 < a < 1 ê 2, 81 ê 100 < b < 89 ê 100 && 1 ê 3 < a < 1 ê 2<,
8a, - 0.1, 1 ê 2 + 0.1<, 8b, - 0.1, Sqrt@3D ê 2 + 0.1<D
True
True
True
True
ü Case d = -1/2 :
Take a positive linear combination of the first 2 inequalities. Note that b+a-1 is positive.
simple.nb
23
neareq ê. d Ø - 1 ê 2 ê. Or Ø List ê. Less Ø Subtract;
l1 = %@@2DD
l2 = %%@@1DD
lin = Hb + a - 1L l2 + 8 ê 7 HSqrt@3D - bL l1 êê Collect@Ò, g, FullSimplifyD &
-
-
216 p2 b2
64 p2 Ha2 + b2 + 3L + 243 HHa - 6L a + b2 - 3L
384 p2 b2
64 p2 Ha2 + b2 + 3L + 243 HHa - 6L a + b2 - 3L
8
g - Ia2 + 1M K
7
1
7
+ Ia2 + 1M H1 - gL - a b + b2 g
+ Ia2 + 3M H1 - gL - a b + b2 g
3 - bO - Ia2 + 3M Ha + b - 1L + b2 Ha + b - 1L +
7 a3 + a b -
2688 p2 b
64 p2 Ha2 + b2 + 3L + 243 HHa - 6L a + b2 - 3L
192 p2 b2 J5 b + 9
8
3 - bO b2 +
K
7
+b-8
3 + 7 + 21 -
3 - 14N
64 p2 Ha2 + b2 + 3L + 243 HHa - 6L a + b2 - 3L
+ a2 K-8 b + 8
3 - 7O + 13 b + 8
3 - 21
D@lin, gD êê Collect@Ò, a, FullSimplifyD &
% < 0 ê. a Ø 0 êê Reduce êê FullSimplify
%% < 0 ê. a Ø 1 - b êê Reduce êê FullSimplify
-a3 +
1
7
1<b<
a2 Kb - 8
3 + 7O + a Ib2 - 3M -
3 Îb+8>7
b<1Îb>
1
7
Hb - 1L Kb Kb - 8
3 + 8O - 8
3 + 21O
3
3
Note that all coefficients for powers of a are negative (knowing that b<=Sqrt[3]). We want to show that the coefficient
of g is less than 0, so plug a=0. This gives 1<b<Sqrt[3]. For a=1-b we get b<1.
Hence coefficient of g is negative. Put g=0.
24
simple.nb
Together@lin ê. g Ø 0D
p = Numerator@%D Denominator@%D
Flatten@8% < 0, tri && a < 1 ê 2<D;
RegionPlot@%, 8a, 0, 1<, 8b, 0, Sqrt@3D<D
K448 p2 a5 + 1701 a5 - 1944 a4 b - 512 p2 a4 b + 512
1944
3 a3 b + 13 365 a3 b - 512
1944 a2 b3 - 512 p2 a2 b3 + 1944
1215 a2 b - 704 p2 a2 b + 2048
3 p2 a4 - 448 p2 a4 + 1944
3 p2 a3 b + 448 p2 a3 b + 2688 p2 a3 - 11 664
3 a2 b2 - 3159 a2 b2 + 512
3 p2 a2 - 2688 p2 a2 - 3888
1944
3 a b3 + 1701 a b3 - 512
1536
3 p2 a b + 1344 p2 a b + 4032 p2 a - 11 664
5103 b2 - 1216
3 a4 - 11 907 a4 + 1944 a3 b2 + 512 p2 a3 b2 3 a3 + 10 206 a3 -
3 p2 a2 b2 - 448 p2 a2 b2 + 11 664
3 a2 b -
3 a2 - 30 618 a2 + 243 a b4 + 64 p2 a b4 -
3 p2 a b3 + 448 p2 a b3 + 4374 a b2 - 1152 p2 a b2 + 5832
3 a b - 24 057 a b -
3 a + 15 309 a + 3159 b3 - 128 p2 b3 + 1944
3 p2 b2 + 1344 p2 b2 - 9477 b + 2496 p2 b + 1536
3 p2 - 4032 p2 - 5832
3 b2 -
3 + 15 309O ì
I7 I64 p2 a2 + 243 a2 - 1458 a + 243 b2 + 64 p2 b2 + 192 p2 - 729MM
7 I64 p2 a2 + 243 a2 - 1458 a + 243 b2 + 64 p2 b2 + 192 p2 - 729M
K448 p2 a5 + 1701 a5 - 1944 a4 b - 512 p2 a4 b + 512
512 p2 a3 b2 - 1944
3 p2 a4 - 448 p2 a4 + 1944
3 a3 b + 13 365 a3 b - 512
1944 a2 b3 - 512 p2 a2 b3 + 1944
1215 a2 b - 704 p2 a2 b + 2048
3 p2 a2 - 2688 p2 a2 - 3888
3 a b3 + 1701 a b3 - 512
1536
3 p2 a b + 1344 p2 a b + 4032 p2 a - 11 664
5103 b2 - 1216
3 p2 a3 b + 448 p2 a3 b + 2688 p2 a3 - 11 664
3 a2 b2 - 3159 a2 b2 + 512
1944
3 a4 - 11 907 a4 + 1944 a3 b2 +
3 p2 a2 b2 - 448 p2 a2 b2 + 11 664
3 a3 + 10 206 a3 3 a2 b -
3 a2 - 30 618 a2 + 243 a b4 + 64 p2 a b4 -
3 p2 a b3 + 448 p2 a b3 + 4374 a b2 - 1152 p2 a b2 + 5832
3 a b - 24 057 a b -
3 a + 15 309 a + 3159 b3 - 128 p2 b3 + 1944
3 p2 b2 + 1344 p2 b2 - 9477 b + 2496 p2 b + 1536
3 p2 - 4032 p2 - 5832
3 b2 -
3 + 15 309O
simple.nb
Hence inequality should hold.
poly = p ê. b Ø Sqrt@3D - b;
PolyNeg@poly, 8b, a<, 81 ê 5, 12 ê 100<D
poly = poly ê. b Ø b + 1 ê 5;
PolyNeg@poly, 8b, a<, 835 ê 100, 12 ê 100<D
poly = p ê. b Ø Sqrt@3D ê 2 + b ê. a Ø 1 ê 2 - a;
PolyNeg@poly, 8b, a<, 8Sqrt@3D ê 2, 38 ê 100<D
poly = p ê. b Ø Sqrt@3D ê 2 + b ê. a Ø 12 ê 100 - a;
PolyNeg@poly, 8b, a<, 832 ê 100, 12 ê 100<D
Flatten@8p < 0, tri, Sqrt@3D - 20 ê 100 < b < Sqrt@3D && 0 < a < 12 ê 100,
Sqrt@3D - 55 ê 100 < b < Sqrt@3D - 20 ê 100 && 0 < a < 12 ê 100,
Sqrt@3D ê 2 < b < Sqrt@3D ê 2 + 32 ê 100 && 0 < a < 12 ê 100,
Sqrt@3D ê 2 < b < Sqrt@3D && 12 ê 100 < a < 1 ê 2<D;
RegionPlot@%, 8a, 0, 1<, 8b, 0, Sqrt@3D<D
True
True
True
True
ü Case g = 0 :
Here we can take a linear combination of inequalities 2 and, in such a way that d gets cancelled.
25
26
simple.nb
neareq ê. g -> 0 ê. r Ø List ê. Less Ø Subtract;
l1 = %@@2DD
l2 = %%@@3DD
lin = a l2 + H1 ê 2 - aL l1 êê Collect@Ò, g, FullSimplifyD &
RegionPlot@8% < 0, tri && a < 1 ê 2<, 8a, - 0.1, 1 ê 2 + 0.1<, 8b, Sqrt@3D ê 2 - 0.1, Sqrt@3D + 0.1<D
-
-
216 p2 b2
64 p2 Ha2 + b2 + 3L + 243 HHa - 6L a + b2 - 3L
192 p2 b2
64 p2 Ha2 + b2 + 3L + 243 HHa - 6L a + b2 - 3L
+ a2 + 2 a b d + 1
+2 a-
1
2
bd+ a-
1
2
2
+
3
4
-243 Ha2 - 1L HHa - 6L a + b2 - 3L - 8 p2 HH8 a2 - 6 a + 19L b2 + 8 Ha4 + 2 a2 - 3LL
128 p2 Ha2 + b2 + 3L + 486 HHa - 6L a + b2 - 3L
Inequality seems true. Again we take just the numerator.
simple.nb
27
p = Numerator@linD;
poly = p ê. b Ø 97 ê 100 + b;
PolyNeg@poly, 8b, a<, 88 ê 10, 1 ê 2<D
poly = p ê. b Ø Sqrt@3D ê 2 + b ê. a Ø a + 1 ê 5;
PolyNeg@poly, 8b, a<, 81 ê 5, 3 ê 10<D
RegionPlot@8p < 0, tri && a < 1 ê 2, 97 ê 100 < b < 177 ê 100 && 0 < a < 1 ê 2,
Sqrt@3D ê 2 < b < Sqrt@3D ê 2 + 1 ê 5 && 1 ê 5 < a < 1 ê 2<,
8a, - 0.1, 1 ê 2 + 0.1<, 8b, - 0.1 + Sqrt@3D ê 2, Sqrt@3D + 0.1<D
True
True
ü Interior point :
Here we take all pairs of inequlities as equations of lines in (g, d) and set them equal 0. We get vertices of the triangle
formed by the lines. Average of the vertices gives a center of mass which is inside of the triangle. One of the inequalities must be true at this point if the point is in the parameter space.
lines = neareq ê. Less Ø Subtract ê. Or Ø List
8g, d< ê. Hlines@@1DD ã lines@@2DD ã 0 êê Solve@Ò, 8g, d<D@@1DD & êê FullSimplifyL
8g, d< ê. Hlines@@2DD ã lines@@3DD ã 0 êê Solve@Ò, 8g, d<D@@1DD & êê FullSimplifyL
8g, d< ê. Hlines@@1DD ã lines@@3DD ã 0 êê Solve@Ò, 8g, d<D@@1DD & êê FullSimplifyL
H% + %% + %%%L ê 3 êê FullSimplify
neareq ê. g Ø %@@1DD ê. d Ø %@@2DD ê. Or Ø List êê FullSimplify
p = %@@3, 1DD
RegionPlot@8p > 0, tri<, 8a, 0, 1<, 8b, 0, Sqrt@3D<D
:-
384 p2 b2
64 p2 Ha2 + b2 + 3L + 243 HHa - 6L a + b2 - 3L
-
-
+ Ia2 + 3M H1 - gL + 2 a b d + b2 g,
216 p2 b2
64 p2 Ha2 + b2 + 3L + 243 HHa - 6L a + b2 - 3L
192 p2 b2
2
2
2
2
64 p Ha + b + 3L + 243 HHa - 6L a + b - 3L
+ Ia2 + 1M H1 - gL + 2 a b d + b2 g,
+2 a-
1
2
bd+
a-
1
2
2
+
3
4
H1 - gL + b2 g>
28
simple.nb
:1 -
:
84 p2 b2
64 p2 Ha2 + b2 + 3L + 243 HHa - 6L a + b2 - 3L
,
4 p2 b H5 Hb2 - 3L - 37 a2 L - 243 b HHa - 6L a + b2 - 3L
128 p2 a Ha2 + b2 + 3L + 486 a HHa - 6L a + b2 - 3L
243 Ha2 - 1L HHa - 6L a + b2 - 3L + 8 p2 HH8 a2 - 6 a + 19L b2 + 8 Ha4 + 2 a2 - 3LL
Ha2 + b2 - 1L H64 p2 Ha2 + b2 + 3L + 243 HHa - 6L a + b2 - 3LL
-243 a b HHa - 6L a + b2 - 3L - 8 p2 b Ha Ha H8 a + 3L + 8 b2 - 3L - 3 b2 + 3L
Ha2 + b2 - 1L H64 p2 Ha2 + b2 + 3L + 243 HHa - 6L a + b2 - 3LL
:
,
>
64 p2 Ha4 + 4 a3 + HHa - 2L a + 3L b2 + 12 a - 9L + 243 Ha Ha + 4L - 3L HHa - 6L a + b2 - 3L
Ha Ha + 4L + b2 - 3L H64 p2 Ha2 + b2 + 3L + 243 HHa - 6L a + b2 - 3LL
-243 Ha + 2L b HHa - 6L a + b2 - 3L - 64 p2 b Ha Ha Ha + 5L + b2 - 3L - b2 + 3L
Ha Ha + 4L + b2 - 3L H64 p2 Ha2 + b2 + 3L + 243 HHa - 6L a + b2 - 3LL
>
,
>
9I4 p2 I48 Ha - 1L Ha + 1L Ia2 + 3M Ha Ha + 4L - 3L + 2 Ha H19 a - 32L + 77L b4 + Ha Ha Ha H91 a + 128L + 58L + 624L - 561L b2 - 5 b6 M +
243 IHa - 6L a + b2 - 3M I4 Ha Ha + 2L - 2L b2 + 3 Ha - 1L Ha + 1L Ha Ha + 4L - 3L + b4 MM ë
I3 Ia2 + b2 - 1M Ia Ha + 4L + b2 - 3M I64 p2 Ia2 + b2 + 3M + 243 IHa - 6L a + b2 - 3MMM,
-I243 b IHa - 6L a + b2 - 3M I2 a H3 a + 4L b2 + a Ha Ha H5 a + 16L - 12L - 8L + b4 - 4 b2 + 3M +
4 p2 b IHa H91 a - 64L + 35L b4 + a Ha Ha H197 a + 384L - 382L + 256L b2 +
a Ha Ha Ha Ha H101 a + 448L - 321L - 224L + 231L - 192L - 5 b6 - 75 b2 + 45MM ë
2
I6 a Ia + b2 - 1M Ia Ha + 4L + b2 - 3M I64 p2 Ia2 + b2 + 3M + 243 IHa - 6L a + b2 - 3MMM=
9b2 Ia2 + b2 - 1M I64 p2 Ia2 + b2 + 3M + 243 IHa - 6L a + b2 - 3MM I4 p2 Ia H5 a - 12L + 5 Ib2 - 3MM - 243 IHa - 6L a + b2 - 3MM > 0,
b2 Ia Ha + 4L + b2 - 3M I64 p2 Ia2 + b2 + 3M + 243 IHa - 6L a + b2 - 3MM I4 p2 Ia H5 a - 12L + 5 Ib2 - 3MM - 243 IHa - 6L a + b2 - 3MM < 0,
a b2 I64 p2 Ia2 + b2 + 3M + 243 IHa - 6L a + b2 - 3MM I4 p2 Ia H5 a - 12L + 5 Ib2 - 3MM - 243 IHa - 6L a + b2 - 3MM > 0=
a b2 I64 p2 Ia2 + b2 + 3M + 243 IHa - 6L a + b2 - 3MM I4 p2 Ia H5 a - 12L + 5 Ib2 - 3MM - 243 IHa - 6L a + b2 - 3MM
Hence the third inequyality should be true at the center of mass (even if it is not inside of the parameter space). Note
that this polynomial is a product of ab^2, upper bound for eigenvalue and the last factor which should be positive.
simple.nb
29
poly = - p@@4DD ê. b Ø Sqrt@3D - b êê Collect@Ò, 8b, a<, FullSimplifyD &
PolyNeg@poly, 8b, a<, 8Sqrt@3D ê 2, 1 ê 2<D
I243 - 20 p2 M a2 + 6 I8 p2 - 243M a + I243 - 20 p2 M b2 + 2
3 I20 p2 - 243M b
True
Spectral gap for an acute triangle
In[1]:=
AppendTo@$Path, ToFileName@8$HomeDirectory, "Dropbox", "mathematica"<DD;
<< TrigInt`
Here we can numerically optimize a test function, rationalize coefficient and find symbolic upper bound. We can also
find optimal lower bound based on at least 3 reference triangles.
ü Tables of eigenfunctions
Sorted eigenfunctions for known cases.
? Equilateral
? Square
Eigenfunctions of Neumann and Dirichlet Laplacian
on the equilateral triangle with vertices H0,0L, H1,0L, H1ê2,Sqrt@3Dê2L:
Equilateral@Dirichlet,SymmetricD @m,nD - 1<=m<=n
Equilateral@Dirichlet,AntisymmetricD @m,nD - 1<=m<n
Equilateral@Neumann,SymmetricD @m,nD - 0<=m<=n
Equilateral@Neumann,AntisymmetricD @m,nD - 0<=m<n
Equilateral@EigenvalueD @m,nD
Eigenfunctions of the right triangle with vertices H0,0L, H1,0L, H0,Sqrt@3DL:
Equilateral@Dirichlet,HalfD @m,nD - 1<=m<n
Equilateral@Neumann,HalfD @m,nD - 0<=m<=n
Equilateral@Eigenvalue,HalfD @m,nD
Vertices:
Equilateral@D
Equilateral@HalfD
Eigenfunctions of Neumann and Dirichlet Laplacian on the square with vertices H0,0L, H1,0L, H1,1L, H0,1L:
Square@DirichletD @m,nD - m>=1,n>=1
Square@NeumannD @m,nD - m>=0,n>=0
Square@EigenvalueD @m,nD
Eigenfunctions of the right isosceles triangle with vertices H0,0L, H1,0L, H0,1L:
Square@Dirichlet,HalfD @m,nD - 1<=m<n
Square@Neumann,HalfD @m,nD - 0<=m<=n
Vertices:
Square@HalfD à
ü Equilateral
In[3]:=
Out[9]=
num = 13;
Table@8Equilateral@EigenvalueD@a, bD, a, b<, 8a, 0, num<, 8b, 0, num<D;
max = %@@1, - 1, 1DD;
%% êê Flatten@Ò, 1D & êê Sort@Ò, Ò1@@1DD <= Ò2@@1DD &D & êê Select@Ò, Ò@@1DD § max &D &;
Rest êü % êê Rest;
eqeigfun = If@Ò1 § Ò2, Equilateral@Neumann, SymmetricD@Ò1, Ò2D,
Equilateral@Neumann, AntisymmetricD@Ò1, Ò2DD & üü Ò & êü %;
Length@
eqeigfunD
115
30
simple.nb
ü 30 - 60 - 90
In[10]:=
Out[16]=
num = 17;
Table@8Equilateral@Eigenvalue, HalfD@a, bD, a, b<, 8a, 0, num<, 8b, 0, num<D;
max = %@@1, - 1, 1DD;
%% êê Flatten@Ò, 1D & êê Sort@Ò, Ò1@@1DD <= Ò2@@1DD &D & êê
Select@Ò, Ò@@2DD § Ò@@3DD && Ò@@1DD § max &D &;
Rest êü % êê Rest;
heigfun = Equilateral@Neumann, HalfD üü Ò & êü %;
Length@heigfunD
100
ü Right isosceles
In[17]:=
num = 15;
Table@8Square@EigenvalueD@a, bD, a, b<, 8a, 0, num<, 8b, 0, num<D;
max = %@@1, - 1, 1DD
%% êê Flatten@Ò, 1D & êê Sort@Ò, Ò1@@1DD <= Ò2@@1DD &D & êê
Select@Ò, Ò@@2DD § Ò@@3DD && Ò@@1DD § max &D &;
Rest êü % êê Rest;
rieigfun = Square@Neumann, HalfD üü Ò & êü %;
Length@rieigfunD
Out[19]=
225 p2
Out[23]=
101
ü Optimal upper bound.
Take the first ? eigenfunctions for the 3 known cases and transplant them. I should also implement rotated transplantations.
In[673]:=
T2@a_, b_D = 88- 1, 0<, 81, 0<, 8a, b<<;
T2@a_, b_, cond_D = 88- 1, 0<, 81, 0<, 8a, b<, cond<;
ri = h = eq = 5;
Take@eqeigfun, eqD;
Transplant@Ò, RotateLeft@T2@a, bDD, Equilateral@DD & êü %;
eqc = Table@ceq@iD, 8i, eq<D;
eqtest = eqc.%%;
Take@heigfun, hD;
Transplant@Ò, Reverse@T2@a, bDD, Equilateral@HalfDD & êü %;
hc = Table@ch@iD, 8i, h<D;
htest = hc.%%;
Take@rieigfun, riD;
Transplant@Ò, Reverse@T2@a, bDD, Square@HalfDD & êü %;
ric = Table@cri@iD, 8i, ri<D;
ritest = ric.%%;
Clear@cD
test = eqtest + htest + ritest;
TrigInt@test, Limits@T2@a, b, b > 0DDD H* this integrates to 0 *L
H*Integrate@test,Limits@T2@a,b,b>0DD,AssumptionsØb>0D H* this is very slow *L*L
Timing@bound = Rayleigh@test, T2@a, b, b > 0DDD@@1DD
Out[690]=
0
Out[691]=
36.7235
simple.nb
In[378]:=
Out[379]=
In[622]:=
Clear@trifunD
triangle = HÒ@@2DD § Ò@@1DD § Ò@@3DD & êü 8Limits@T2@a, b, b > 0DD<L ê. List Ø And
trifun@c_, d_D := Function@8xx, yy<, Evaluate@triangle ê. a Ø c ê. b Ø d ê. x Ø xx ê. y Ø yyDD
0§y§bÌ
Ha + 1L y
-1§x§
Ha - 1L y
+1
b
b
optimal@c_ ? NumericQ, d_ ? NumericQD :=
NMinimize@8bound, Max@Abs@ceq@1DD, Abs@ch@1DD, Abs@cri@1DDD ã 1 && ceq@1D + ceq@2D ã 1< ê.
a Ø c ê. b Ø d, Flatten@8eqc, hc, ric<DD@@2DD êê Rationalize@Ò, 0.000001D &
optupper@c_, d_D := 8test, bound< ê. optimal@c, dD ê. a Ø c ê. b Ø d;
optupper@2 ê 1000, 17 ê 10D;
ContourPlot@%@@1DD, 8x, - 1, 1<, 8y, 0, 2<,
RegionFunction Ø trifun@aa, bbD, Contours Ø 40, AspectRatio Ø AutomaticD
4 %%@@2DD êê N
Out[625]=
Out[626]=
17.6212
ü Optimal lower bound.
In[64]:=
ineq1
ineq2
ineq3
ineq4
=
=
=
=
Ha ^ 2 + 3L H1 - gL + 2 a b d + b ^ 2 g < 4 p ^ 2 ê 3 ê u;
Ha ^ 2 + 1L H1 - gL + 2 a b d + b ^ 2 g < 3 p ^ 2 ê 4 ê u;
HHa - 1 ê 2L ^ 2 + 3 ê 4L H1 - gL + 2 Ha - 1 ê 2L b d + b ^ 2 g < 2 p ^ 2 ê 3 ê u;
HHa - 1L ^ 2 + 4 ê 3L H1 - gL + 2 Ha - 1L b d + b ^ 2 g < 8 p ^ 2 ê 9 ê u;
31
32
simple.nb
In[275]:=
Out[275]=
ineqs = ineq1 »» ineq2 »» ineq3 »» ineq4 ê. u Ø p ^ 2 ê u ê. Less Ø Equal êê
Collect@Ò, 8g, d<, FullSimplifyD &
lessu = u ê. H% êê Solve@Ò, uD &L
optimal2@c_ ? NumericQ, d_ ? NumericQD :=
Maximize@Evaluate@8Min@lessuD, 0 § g § 1 && - 1 ê 2 § d § 1 ê 2< ê. a Ø c ê. b Ø dD, 8g, d<D@@1DD
optlower@c_, d_D := lower = 2 u ê. u Ø p ^ 2 ê optimal2@c, dD
optlower@2 ê 1000, 17 ê 10D êê N
g I-a2 + b2 - 3M + a2 + 2 a b d + 3 ‡
4u
3
Î g I-a2 + b2 - 1M + a2 + 2 a b d + 1 ‡
g I-a2 + a + b2 - 1M + H2 a - 1L b d + Ha - 1L a + 1 ‡
Out[276]=
Out[279]=
2u
3
Î g -Ha - 2L a + b2 -
4
7
3
Î
+ 2 Ha - 1L b d + Ha - 1L2 +
4
3
‡
8u
9
3
4
:- Ia2 g - a2 - 2 a b d - b2 g + 3 g - 3M, - Ia2 g - a2 - 2 a b d - b2 g + g - 1M,
4
3
3
3
- Ia2 g - a2 - 2 a b d - a g + a - b2 g + b d + g - 1M, - I3 a2 g - 3 a2 - 6 a b d - 6 a g + 6 a - 3 b2 g + 6 b d + 7 g - 7M>
2
8
8.8855
ü Lower bound for the gap for triangle (-1,0), (1,0) and (a,b)
Gives exact value for equilateral, right isosceles and 30 - 60 - 90.
In[692]:=
3u
gap@c_, d_D := optlower@c, dD - 2 optupper@c, dD@@2DD
4 gap@0, Sqrt@3DD êê N
4 gap@0, Sqrt@3D * 9999 ê 10 000D êê N
4 gap@2 ê 1000, 17 ê 10D êê N
4 gap@1 ê 4, Sqrt@2DD êê N
optupper@1 ê 4, Sqrt@2DD@@2DD 4 êê N
Out[693]=
0.
Out[694]=
0.000257758
Out[695]=
0.299767
Out[696]=
5.29651
Out[697]=
17.4199