Usando Polinômios de Darboux na Obtenção de Integrais
Primeiras Liouvillianas de Equações Diferenciais Ordinárias
Racionais de Segunda Ordem
André Claudino
Como este trabalho está organizado?
Abordagem Darbouxiana
Dado uma 1EDO racional
dxdy=N(x,y)M(x,y)
\dfrac{dy}{dx} = \dfrac{M(x,y)}{N(x,y)}
tem fácil solução se...
∂y∂M=−∂x∂N
\dfrac{\partial M}{\partial y} = - \dfrac{\partial N}{\partial x}
Existe um fator integrante...
R=∏ipi(x,y)ni
R = \prod_i p_i(x,y)^{n_i}
ni∈Q
n_i \in \mathbb{Q}
Que satisfaz...
∂y∂(RM)=−∂x∂(RN)
\dfrac{\partial (RM)}{\partial y} = - \dfrac{\partial (RN)}{\partial x}
com...
D≡N(x,y)∂x+M(x,y)∂y
\mathbb{D} \equiv N(x,y) \partial_x + M(x,y) \partial_y
RD[R]=−(Nx+My)
\dfrac{\mathbb{D}[R]}{R} = -(N_x + M_y)
teremos...
D[pi]=piqi
\mathbb{D} [p_i] = p_i q_i
o problema se torna...
R=∏ipi(x,y)ni
R = \prod_i p_i(x,y)^{n_i}
onde
RD[R]=∑inipiD[pi]
\dfrac{\mathbb{D} [R]}{R} = \sum_i n_i \dfrac{\mathbb{D}[p_i]}{p_i}
substituindo...
qi≡piD[pi]
q_i \equiv \dfrac{\mathbb{D}[p_i]}{p_i}
para uma solução liouviliana...
D[Bi]=Bi(x,y)λi(x,y)
\mathbb{D}[B_i] = B_i(x,y) \lambda_i(x,y)
portanto...
R=eBA∏ipini
R = e^{\frac{A}{B}} \prod_i p_i^ {n_i}
BD[B]=∑imiBiD[Bi]
\dfrac{\mathbb{D} [B]}{B} = \sum_i m_i \dfrac{\mathbb{D}[B_i]}{B_i}
D[A]−A∑imiλi=−∏iBimi(∑iniqi+My+Nx)
\mathbb{D}[A]-A \sum_i m_i \lambda_i = -\prod_i B_i ^{m_i} \left( {\sum_i n_i q_i + M_y + N_x} \right)
D[pi]=piqi
\mathbb{D}[p_i] = p_i q_i
e o problema se torna...
D[Bi]=Biλi
\mathbb{D}[B_i] = B_i \lambda_i
D[pi]=piqi
\mathbb{D} [p_i] = p_i q_i
Observações
-
não sabemos os graus máximos dos polinômios de Darboux
-
O número de indeterminados cresce rápido com o grau dos polinômios de Darboux
-
resolver o sistema de equações é complicado, mesmo para graus baixos
Método da expansão
-
Gera um sistema de equações em formato de cadeira regular
- A complexidade do sistema cresce mais devagar
- Permite solucionar para graus maiores de polinômios de Darboux
- O algoritmo é mais comportado do ponto de vista de consumo de recursos
Ideia
dxdy=NM
\dfrac{dy}{dx} = \dfrac{M}{N}
M=∑iMi(y)xiN=∑iNi(y)xi
M = \sum_{i} M_{i}(y) x^i \\
N = \sum_{i} N_{i}(y) x^i
P=∑Pi(y)xiQ=∑Qi(y)xi
P = \sum P_{i}(y) x^i \\
Q = \sum Q_{i}(y) x^i
M0dydP0+N0P1=Q0P0
M_0 \dfrac{d P_0}{dy} + N_0 P_1 = Q_0 P_0
M0dydP1+M1dydP0+2N0P2+N1P1=Q0P1+Q1P0
M_0 \dfrac{d P_1}{dy} + M_1 \dfrac{d P_0}{dy} + 2 N_0 P_2 + N_1 P_1 = Q_0 P_1 + Q_1 P_0
⋯
\cdots
Exemplo
dxdy=−x43+y2x4
\dfrac{dy}{dx} = -\dfrac{3 + y^2 x^4}{x^4}
D=x4∂x−(3+y2x4)∂y
\mathbb{D} = x^4\partial_x - (3+y^2 x^4)\partial_y
P=P0(y)+P1(y)x+P2(y)x2+P3(y)x3+P4(y)x4
P = P_0(y) + P_1(y) x + P_2(y) x^2 + P_3(y) x^3 + P_4(y) x^4
Q=Q0(y)+Q1(y)x+Q2(y)x2+Q3(y)x3+Q4(y)x4
Q = Q_0(y) + Q_1(y) x + Q_2(y) x^2 + Q_3(y) x^3 + Q_4(y) x^4
g(P)≡6
g(P) \equiv 6
gx(Q)≡max(gx(M),gx(N)−1)=max(4,4−1)=max(4,3)=4
g_x(Q) \equiv max(g_x(M), g_x(N)-1) = max(4, 4 -1) = max(4, 3) = 4
gx(p)≡4
g_x(p) \equiv 4
g(Q)≡max(g(M),g(N))=5
g(Q) \equiv max(g(M), g(N)) = 5
−3dydP0−P0Q0=0−P0Q1−P1Q0−3dydP1=0−3dydP4−y2dydP0+P1−P0Q4−P1Q3−P2Q2−P3Q1−P4Q0=0−P0Q3−P1Q2−P2Q1−P3Q0−3dydP3=0−P0Q2−P1Q1−P2Q0−3dydP2=0
-3 \frac{d P_0}{dy} - P_0 Q_0 = 0 \\
-P_0 Q_1 - P_1 Q_0 -3 \frac{dP_1}{dy} = 0 \\
-3 \frac{dP_4}{dy} - y^2\frac{dP_0}{dy} + P_1 - P_0 Q_4 - P_1 Q_3 - P_2 Q_2 - P_3 Q_1 - P_4 Q_0 = 0 \\
-P_0 Q_3 - P_1 Q_2 - P_2 Q_1 -P_3Q_0 - 3 \frac{dP_3}{dy} = 0 \\
-P_0 Q_2 - P_1 Q_1 - P_2 Q_0 - 3 \frac{dP_2}{dy} = 0
2P2−P1Q4−P2Q3−P3Q2−P4Q1−y2dydP1=03P3−P2Q4−P3Q3−P4Q2−y2dydP2=04P4−P3Q4−P4Q3−y2dydP3=0−y2dydP4−P4Q4=0
2 P_2 - P_1Q_4 -P_2Q_3 -P_3 Q_2 - P_4 Q_1 - y^2 \frac{dP_1}{dy} = 0 \\
3 P_3 -P_2 Q_4 -P_3Q_3 - P_4Q_2 - y^2\frac{dP_2}{dy} = 0 \\
4 P_4 - P_3Q_4 - P_4Q_3 -y^2\frac{dP_3}{dy} = 0 \\
-y^2 \frac{dP_4}{dy} - P_4 Q_4 = 0
−y2dydP4−P4Q4=0
-y^2 \frac{dP_4}{dy} - P_4 Q_4 = 0
Resovendo
P4=Ce∫−y2Q4dy
P_4 = C e^{\int{-\frac{Q_4}{y^2}}dy}
temos
∫−y2Q4dy
\int{-\frac{Q_4}{y^2}}dy
deve ser logaritmo ou zero
max(g(P4))=max(gy(P4))=2
max(g(P_4)) = max(g_y(P_4)) = 2
g(Q4)=gy(Q4)=1⇒Q4=ky
g(Q_4) = g_y (Q_4) = 1 \Rightarrow Q_4 = k y
P4=Ce∫−y2kydy=Ce−k∫y2ydy=Ce−k∫ydy=Ce−kln(y)=Cy−k
P_4 = C e^{\int{-\frac{k y}{y^2}}dy} = C e^{- k \int{\frac{y}{y^2}}dy} = C e^{- k \int{\frac{dy}{y}}} = C e^{-kln(y)} = C y^{-k}
C≡1k∈{0,−1,−2}
C \equiv 1 \\
k \in \{0, -1, -2\}
P4=y2
P_4 = y^{2}
Q4=−2y
Q_4 = -2y
g(Q)=5
g(Q) = 5
g(P)≡6
g(P) \equiv 6
Resolvendo
Substituindo
4P4−P3Q4−P4Q3−y2dydP3=0
4 P_4 - P_3Q_4 - P_4Q_3 -y^2\frac{dP_3}{dy} = 0
4y2+2P3y−y2Q3−y2dydP3=0
4 y^2 + 2P_3y - y^2Q_3 - y^2\frac{dP_3}{dy} = 0
⇒P3=(∫y24−Q3dy+C1)y2
\Rightarrow P_3 = \left( \int{\frac{4-Q_3}{y^2}}dy + C_1 \right)y^2
Q3≡a0+a1y+a2y2
Q_3 \equiv a_0 + a_1y + a_2y^2
usando
P3=−a1y2ln(y)−a2y3+y2K1+(a0−4)y
P_3 = -a_1 y^2 ln(y) -a_2 y^3 + y^2 K_1 + (a_0 - 4)y
P3=−a2y3+y2K1+(a0−4)y
P_3 = -a_2 y^3 + y^2 K_1 + (a_0 - 4)y
P3
P_3
é polinômio
⇒a1≡0
\Rightarrow a_1 \equiv 0
Q3=a0+a2y2
Q_3 = a_0 + a_2 y^2
g(Q)=5
g(Q) = 5
g(P)≡6
g(P) \equiv 6
3P3−P2Q4−P3Q3−P4Q2−y2dydP2=0
3P_3 - P_2 Q_4 -P_3 Q_3 -P_4 Q_2 - y^2 \frac{d P_2}{dy} = 0
Resovendo
Temos
P2=y2∫y3a22y4−a2K1y3+a2y2+(3K1−Q2−a0k1)y−12+7a0−a02dy+K2y2
P_2 = y^2 \int {\dfrac{a_2^2 y^4 -a_2 K_1 y^3 + a_2 y^2 + (3K_1 -Q_2 -a_0 k_1)y -12 + 7a_0 -a_0^2}{y^3}} dy + K_2 y^2
Q2≡b0+b1y+b2y2+b3y3⇒
Q_2 \equiv b_0 + b_1 y + b_2 y^2 +b_3 y^3 \Rightarrow
P2=(a2−b1)y2ln(y)+(2a22−b3)y4−(b2+a2k1)y3+y2K2+(b0+a0K1−3K1)y+2a02−7a0+6
P_2 = (a_2 - b_1)y^2 ln(y) + \left( \frac{a_2^2 -b^3}{2} \right) y^4 - (b^2 + a_2 k_1)y^3 + y^2 K_2 + (b_0 + a_0 K_1 -3K_1)y
+ \frac{a_0^2-7a_0}{2} + 6
P2
P_2
é polinômio
P2=2a32−b3y4−(b2+a2K1)y3+K2y2+(K1(a0−3)+b0)y+2a02−7a0+6Q2=b0+a2y+b2y2+b3y3
P_2 = \frac{a_3^2-b^3}{2} y^4 - (b_2 + a_2 K_1)y^3 + K_2 y^2 + (K_1(a_0 - 3) + b_0 )y + \frac{a_0^2 - 7a_0}{2} + 6 \\
Q_2 = b_0 + a_2 y + b_2 y^2 + b_3 y^3
⇒b1=a2
\Rightarrow b_1 = a_2
g(Q)=5
g(Q) = 5
g(P)≡6
g(P) \equiv 6
2P2−P1Q4−P2Q3−P3Q2−P4Q1−y2dydP1=0
2 P_2 - P_1 Q_4 - P_2 Q_3 - P_3 Q_2 - P_4 Q_1 - y^2\frac{dP_1}{dy} = 0
Resovendo
Temos
P1=y2∫−2y4a2(a2−3b3)y6−2(a2(K1+2b2)−K1b3)y5+(a22(a0−4)+2a2K2+b3(a0−6)+2b2K1)y4−4b2y3dy+K3y2
P_1 = y^2 \int{\dfrac{a_2(a_2 -3 b_3)y^6 -2(a_2(K_1+2b_2) -K_1b_3)y^5 \\
+ (a_2^2(a_0-4) + 2 a_2K_2 + b_3(a_0 -6) + 2 b_2 K_1)y^4 \\
- 4b_2 y^3}{-2y^4}
}dy + K_3y^2
Q1≡c0+c1y+c2y2+c3y3+c4y4⇒
Q_1 \equiv c_0 + c_1 y + c_2 y^2 +c_3 y^3 + c_4 y^4 \Rightarrow
P1=−6a22−3a2b3+2c4y5+2K1a22+2a2b2−K1b3−c3y4
P_1 = - \frac{a_2^2 - 3 a_2 b_3 + 2c_4}{6}y^5
+ \frac{K_1 a_2^2 + 2a_2 b_2 - K_1 b_3 -c_3}{2}y^4
+y2∫−2y4(a2(a02−5a0+4)+2b0K1+2K2(a0−2)+2Q1)y2+2(2K1a02+a0(2b0−5K1)+6K16b0)ya03−9a02+26a0−24)dy+K3y2
+ y^2 \int{\dfrac{
(a_2(a_0^2 - 5 a_0 + 4) + 2 b_0K_1 + 2K_2(a_0 -2) + 2 Q_1)y^2 \\
+ 2(2 K_1 a_0^2 + a_0(2 b_0 - 5K_1) + 6K_1 6 b_0) y \\
a_0^3 - 9 a_0^2 + 26 a_0 -24)}{-2y^4}
}dy + K_3y^2
{2b2−c1=0⇒c1=2b2a03−9a02+26a0−24=0⇒a0∈{2,3,4}
\begin{cases}
2b_2 - c_1 = 0 \Rightarrow c_1 = 2b_2 \\
a_0^3 - 9 a_0^2 + 26 a_0 - 24 = 0 \Rightarrow a_0 \in \{ 2, 3, 4 \}
\end{cases}
+(a22−a2K2+2b3−2b2K1−c2)y3+K2y2+(b0K1−a2+c0)y−b0
+ (a_2^2 - a_2 K_2 + 2b_3 -2b_2 K_1 -c_2)y^3
+ K_2 y^2 + (b_0 K_1 -a_2 + c_0)y -b_0
Q1=c0+2b2y+c2y2+c3y3+c4y4
Q_1 = c_0 + 2 b_2 y + c_2 y^2 + c_3 y^3 + c_4y^4
−3dydP0−P0Q0=0
-3 \dfrac{d P_0}{dy} - P_0 Q_0 = 0
Resovendo
Temos
P0=K4e−∫3Q0dy
P_0 = K_4 e^{-\int\frac{Q_0}{3}dy}
−∫3Q0dy
{-\int\frac{Q_0}{3}dy}
Deve ser logaritmo ou zero
−∫3Q0dy=0⇒{Q0=0P0=K4
-\int\frac{Q_0}{3}dy = 0 \Rightarrow
\begin{cases}
Q_0 = 0 \\
P_0 = K_4
\end{cases}
continuando, teremos...
−6K3−2K4b2=0−3b3+3c2=0−1/2a2b3c4+1/3c42+1/6a23c4=0−6−b02+c0+2K4=06+b02−2K4−c0=03K1+b0c0−3b0−K4b0=0−K4c0+3a2−3c0−3b0K1=0−K4c4+5/2a23−15/2a2b3+5c4=01/6a23b3−1/2a22c4−1/2a2b32+5/6b3c4=04/3c4a2+1/6a24+1/2b32−a22b3=06c3+6b3K1−6K1a22−12a2b2−K4c3=0−K1c0−K3−b0K2+3b2+a2K1=0−2b2+2a2b0−6K1+K1c0−b02K1−2b0c0=0−K1c4+3/2b3K1a2−3/2b2a22−1/2K1a23+3/2c3a2+3/2b2b3=0−K4a2−6K2+2b0b2+a2c0−c02−b0K1c0=0−K4c2+9c2−18b3+9a2K2−9a22+9b2K1=01/2b3K1c4−1/2K1a22c4+1/6a23c3−a2b2c4−1/2a2b3c3+5/6c4c3=09b2−K4b2−K3c0+9a2K1−2b0K1b2+b0c2+2a2b2−2c0b2=01/6a23b2−1/2K1a22b3+4/3c4b2+a2K1c4−1/2a22c3−3/2a2b3b2+b3c3+1/2b32K1=0−K1c3+2b2K1a2+K2a22+2c2a2−2a2b3+7/3c4−b3K2−4/3a23+b22=0−K3a2+2a2b2+1/2K1a22−1/2b0b3−b2K2−K1c2+5/2c3−1/2a22b0+3/2b3K1=0a2K2b0+a2b2−b0b3−2b2K2−c3−a22b0−K3a2+K1c2+a2K1c0=09a2−K2c0−c2−b0b2−K4a2−a2c0−b0K1a2−K3b0+a22+2b2K1=0−6a22−a22c0−2b3c0+b2K1c0+b0c3+a2K2c0+6b3+c2a2−2K3b2−b0K1c2−K4b3=0−a22c4+b2K1c4−a2b2c3+a2K2c4−1/2a2b3c2+1/2c32−1/2K1a22c3++1/2b3K1c3+1/6a23c2+4/3c4c2−2b3c4=0−a22c3−a2b3b2+2/3c4b2+a2K2c3−2b3c3−a2b2c2+b2K1c3++1/2b3K1c2+3/2c3c2−1/2K1a22c2−K3c4+1/3a23b2=0−2b2a22−b0K1c3−4b2b3+2b22K1−a2b2c0−1/2c3c0+c3a2++b0c4+2a2K2b2−1/2K1a22c0+2c2b2−K3c2+1/2b3K1c0=03/2b3K1b2−2b32−a2b22−1/2a22c2+a2K1c3+3/2b3c2+1/6a24++1/3c4a2+3/2c3b2−1/2K1a22b2+a2K2b3−K2c4−3/2a22b3=0−2/3b0c4−K2c3+1/6a23b0−b2b3−1/2K1a23+1/2c3a2−1/2a2b3b0++b22K1+a2K2b2+1/2b3K1a2+2c2b2+K1c4−K3b3+a2K1c2−3b2a22=0−2b3c2+b2K1c2−b0K1c4−K1a22b2+a2K2c2+c3b2−1/2a2b3c0++1/6a23c0+b3K1b2−2/3c4c0−a22c2+c22−K3c3+c4a2−2a2b22=0c2a2−a2b3−1/2a22c0−c4−a23+2b22−K2c2−a2b2b0+K1c3+−1/2b0c3−K3b2−1/2K1a22b0+3b2K1a2−1/2b3c0−1/2b3K1b0+K2a22=0
-6 {K_3}-2 {K_4} {b_2}=0\\-3 {b_3}+3 {
c_2}=0\\-1/2 {a_2} {b_3} {c_4}+1/3 {{c_4}}^{2}+1/6 {{
a_2}}^{3}{c_4}=0\\
-6-{{b_0}}^{2}+{c_0}+2 {K_4}=0\\\,\,\,
6+{{b_0}}^{2}-2 {K_4}-{c_0}=0\\3 {K_1}+{b_0} {c_0}-3 {
b_0}-{K_4} {b_0}=0\\
-{K_4} {c_0}+3 {a_2}-3 {
c_0}-3 {b_0} {K_1}=0\\-{K_4} {c_4}+5/2 {{a_2}}^
{3}-15/2 {a_2} {b_3}+5 {c_4}=0\\
1/6 {{a_2}}^{3}{b_3}-1
/2 {{a_2}}^{2}{c_4}-1/2 {a_2} {{b_3}}^{2}+5/6 {b_3}
{c_4}=0\\4/3 {c_4} {a_2}+1/6 {{a_2}}^{4}+1/2 {{b_3}}
^{2}-{{a_2}}^{2}{b_3}=0\\
6 {c_3}+6 {b_3} {K_1}-6 {
K_1} {{a_2}}^{2}-12 {a_2} {b_2}-{K_4} {c_3}=0\\
-{K_1} {c_0}-{K_3}-{b_0} {K_2}+3 {b_2}+{
a_2} {K_1}=0\\
-2 {b_2}+2 {a_2} {b_0}-6 {K_1}+{
K_1} {c_0}-{{b_0}}^{2}{K_1}-2 {b_0} {c_0}=0\\ -{
K_1} {c_4}+3/2 {b_3} {K_1} {a_2}-3/2 {b_2}
{{a_2}}^{2}-1/2 {K_1} {{a_2}}^{3}+3/2 {c_3} {a_2}+3/2 {b_2} {b_3}=0\\ -{K_4} {a_2}-6 {K_2}+2 {
b_0} {b_2}+{a_2} {c_0}-{{c_0}}^{2}-{b_0} {K_1} {c_0}=0\\ -{K_4} {c_2}+9 {c_2}-18 {b_3}+9 {
a_2} {K_2}-9 {{a_2}}^{2}+9 {b_2} {K_1}=0\\ 1/2 {
b_3} {K_1} {c_4}-1/2 {K_1} {{a_2}}^{2}{c_4}+
1/6 {{a_2}}^{3}{c_3}-{a_2} {b_2} {c_4}-1/2 {a_2}
{b_3} {c_3}+5/6 {c_4} {c_3}=0\\ 9 {b_2}-{K_4} {
b_2}-{K_3} {c_0}+9 {a_2} {K_1}-2 {b_0} {
K_1} {b_2}+{b_0} {c_2}+2 {a_2} {b_2}-2 {c_0} {b_2}=0\\ 1/6 {{a_2}}^{3}{b_2}-1/2 {K_1} {{a_2}}^
{2}{b_3}+4/3 {c_4} {b_2}+{a_2} {K_1} {c_4}-1/2
{{a_2}}^{2}{c_3}-3/2 {a_2} {b_3} {b_2}+{b_3} {
c_3}+1/2 {{b_3}}^{2}{K_1}=0\\ -{K_1} {c_3}+2 {b_2
} {K_1} {a_2}+{K_2} {{a_2}}^{2}+2 {c_2} {a_2}-2 {a_2} {b_3}+7/3 {c_4}-{b_3} {K_2}-4/3 {{
a_2}}^{3}+{{b_2}}^{2}=0\\ -{K_3} {a_2}+2 {a_2} {b_2
}+1/2 {K_1} {{a_2}}^{2}-1/2 {b_0} {b_3}-{b_2} {
K_2}-{K_1} {c_2}+5/2 {c_3}-1/2 {{a_2}}^{2}{b_0}+3/2 {b_3} {K_1}=0\\ {a_2} {K_2} {b_0}+{a_2}
{b_2}-{b_0} {b_3}-2 {b_2} {K_2}-{c_3}-{{a_2}}^{2}{b_0}-{K_3} {a_2}+{K_1} {c_2}+{a_2} {
K_1} {c_0}=0\\ 9 {a_2}-{K_2} {c_0}-{c_2}-{b_0}
{b_2}-{K_4} {a_2}-{a_2} {c_0}-{b_0} {K_1
} {a_2}-{K_3} {b_0}+{{a_2}}^{2}+2 {b_2} {K_1
}=0\\ -6 {{a_2}}^{2}-{{a_2}}^{2}{c_0}-2 {b_3} {c_0}+{
b_2} {K_1} {c_0}+{b_0} {c_3}+{a_2} {K_2}
{c_0}+6 {b_3}+{c_2} {a_2}-2 {K_3} {b_2}-{
b_0} {K_1} {c_2}-{K_4} {b_3}=0\\ -{{a_2}}^{2}{
c_4}+{b_2} {K_1} {c_4}-{a_2} {b_2} {c_3}+{
a_2} {K_2} {c_4}-1/2 {a_2} {b_3} {c_2}+1/2 {
{c_3}}^{2}-1/2 {K_1} {{a_2}}^{2}{c_3}+ +1/2 {b_3} {
K_1} {c_3}+1/6 {{a_2}}^{3}{c_2}+4/3 {c_4} {c_2
}-2 {b_3} {c_4}=0\\ -{{a_2}}^{2}{c_3}-{a_2} {b_3} {
b_2}+2/3 {c_4} {b_2}+{a_2} {K_2} {c_3}-2 {
b_3} {c_3}-{a_2} {b_2} {c_2}+{b_2} {K_1} {
c_3}+ +1/2 {b_3} {K_1} {c_2}+3/2 {c_3} {c_2}-1/
2 {K_1} {{a_2}}^{2}{c_2}-{K_3} {c_4}+1/3 {{a_2}}^{3}{b_2}=0\\ -2 {b_2} {{a_2}}^{2}-{b_0} {K_1} {
c_3}-4 {b_2} {b_3}+2 {{b_2}}^{2}{K_1}-{a_2} {
b_2} {c_0}-1/2 {c_3} {c_0}+{c_3} {a_2}+ +{b_0}
{c_4}+2 {a_2} {K_2} {b_2}-1/2 {K_1} {{a_2
}}^{2}{c_0}+2 {c_2} {b_2}-{K_3} {c_2}+1/2 {b_3
} {K_1} {c_0}=0\\ 3/2 {b_3} {K_1} {b_2}-2 {{b_3}}^{2}-{a_2} {{b_2}}^{2}-1/2 {{a_2}}^{2}{c_2}+{a_2}
{K_1} {c_3}+3/2 {b_3} {c_2}+1/6 {{a_2}}^{4}+ +1/3 {c_4} {a_2}+3/2 {c_3} {b_2}-1/2 {K_1} {{a_2
}}^{2}{b_2}+{a_2} {K_2} {b_3}-{K_2} {c_4}-3/2
{{a_2}}^{2}{b_3}=0\\ -2/3 {b_0} {c_4}-{K_2} {c_3}
+1/6 {{a_2}}^{3}{b_0}-{b_2} {b_3}-1/2 {K_1} {{
a_2}}^{3}+1/2 {c_3} {a_2}-1/2 {a_2} {b_3} {b_0}
+ +{{b_2}}^{2}{K_1}+{a_2} {K_2} {b_2}+1/2 {b_3}
{K_1} {a_2}+2 {c_2} {b_2}+{K_1} {c_4}-{
K_3} {b_3}+{a_2} {K_1} {c_2}-3 {b_2} {{a_2}}^{2}=0\\ -2 {b_3} {c_2}+{b_2} {K_1} {c_2}-{b_0
} {K_1} {c_4}-{K_1} {{a_2}}^{2}{b_2}+{a_2} {
K_2} {c_2}+{c_3} {b_2}-1/2 {a_2} {b_3} {c_0}+ +1/6 {{a_2}}^{3}{c_0}+{b_3} {K_1} {b_2}-2/3 {
c_4} {c_0}-{{a_2}}^{2}{c_2}+{{c_2}}^{2}-{K_3} {
c_3}+{c_4} {a_2}-2 {a_2} {{b_2}}^{2}=0\\ {c_2} {a_2}-{a_2} {b_3}-1/2 {{a_2}}^{2}{c_0}-{c_4}-{{a_2}}
^{3}+2 {{b_2}}^{2}-{K_2} {c_2}-{a_2} {b_2} {b_0}+{K_1} {c_3}+ -1/2 {b_0} {c_3}-{K_3} {b_2}-
1/2 {K_1} {{a_2}}^{2}{b_0}+3 {b_2} {K_1} {a_2}-1/2 {b_3} {c_0}-1/2 {b_3} {K_1} {b_0}+{K_2} {{a_2}}^{2}=0
{K1,K2,K3,K4,a2,b0,b2,b3,c0,c2,c3,c4}
\{K_1, K_2, K_3, K_4, a_2, b_0, \\b_2, b_3, c_0, c_2, c_3, c_4\}
K4=3K1=K2=K3=0a2=0b0=b2=b3=0c0=c2=c3=c4=0
K_4=3 \\
K_1=K_2=K_3=0\\
a_2=0\\
b_0=b_2=b_3=0\\
c_0=c_2=c_3=c_4=0
P=3+x2−2x3y+x4y2Q=2x3−2yx4
P = 3 + x^2 - 2 x^3 y + x^4 y^2 \\
Q = 2 x^3 - 2 y x^4
⇒
\Rightarrow
Método da expansão
Simultâneo
dxdy=NM
\dfrac{dy}{dx} = \dfrac{M}{N}
M=∑iM(y)ixiN=∑iN(y)ixi
M = \sum_{i} M(y)_{i} x^i \\
N = \sum_{i} N(y)_{i} x^i
p=∑p(y)ixiq=∑q(y)ixi
p = \sum p(y)_{i} x^i \\
q = \sum q(y)_{i} x^i
M=∑iM(x)jyjN=∑iN(x)jyj
M = \sum_{i} M(x)_{j} y^j \\
N = \sum_{i} N(x)_{j} y^j
p=∑p(x)jyjq=∑q(x)jyj
p = \sum p(x)_{j} y^j \\
q = \sum q(x)_{j} y^j
Expande em
x
x
Expande em
y
y
Procedimento
- Expande em ambas as variáveis
- Resolve o sistema de menor complexidade (sistema A)
- Resolva o sistema restante (sistema B) usando os resultados encontrados no sistema A
- Caso haja indeterminados, resolva no sistema remanescente
Exemplo
dxdy=−x43+y2x4
\dfrac{dy}{dx} = -\dfrac{3 + y^2 x^4}{x^4}
D=x4∂x−(3+y2x4)∂y
\mathbb{D} = x^4\partial_x - (3+y^2 x^4)\partial_y
P=p0(x)+p1(x)y+p2(x)y2
P = p_0(x) + p_1(x) y + p_2(x) y^2
Q=q0(x)+q1(x)y
Q = q_0(x) + q_1(x) y
g(P)≡6
g(P) \equiv 6
gy(Q)≡max(gy(M)−1,gy(N))=max(2−1,0)=max(1,0)=1
g_y(Q) \equiv max(g_y(M) - 1, g_y(N)) = max(2-1, 0) = max(1, 0) = 1
gy(p)≡2
g_y(p) \equiv 2
g(Q)≡max(g(M),g(N))=5
g(Q) \equiv max(g(M), g(N)) = 5
−3p1(x)−p0(x)q0(x)+x4dxdp0(x)=0−6p2(x)+x4dxdp1(x)−p0(x)q1(x)−p1(x)q0(x)=0−x4p1(x)+x4dxdp2(x)−p1(x)q1(x)−p2(x)q0(x)=0−2x4p2(x)−p2(x)q1(x)=0
-3 p_1(x) -p_0(x) q_0(x) +x^4\frac{d}{dx}p_0(x) = 0\\
-6 p_2(x) + x^4\frac{d}{dx}p_1(x) - p_0(x) q_1(x) - p_1(x) q_0(x) = 0 \\
-x^4 p_1(x) + x^4 \frac{d}{dx}p_2(x) - p_1(x) q_1(x) -p_2(x) q_0(x) = 0 \\
-2 x^4 p_2(x) -p_2(x) q_1(x) = 0
Resulta no sistema B
−2x4p2(x)−p2(x)q1(x)=0
-2 x^4 p_2(x) -p_2(x) q_1(x) = 0
Resolvendo
q1(x)=−2x4
q_1(x) = -2 x^4
temos...
Com a informação do sistema A
q1(x)=−2x4
q_1(x) = -2 x^4
P4=y−kQ4=ky1
P_4 = y^{-k} \\
Q_4 = k y^1
E do sistema B
k=−2=>
k = -2 =>
=y2
= y^2
Q=q0(x)−2x4y
Q = q_0(x) -2 x^4 y
{P4=y2p2(x)≡a0+a1x+a2x2+a3x3+a4x4⇒a4=1
\begin{cases}
P_4 = y^2 \\
p_2(x) \equiv a_0 + a_1 x + a_2 x^2 + a_3 x^3 + a_4 x^4
\end{cases}
\Rightarrow a_4 = 1
p2(x)=a0+a1x+a2x2+a3x3+x4
p_2(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + x^4
P3=(∫y24−Q3dy+C1)y2
P_3 = \left( \int{\frac{4-Q_3}{y^2}}dy + C_1 \right)y^2
g(Q3)=1⇒Q3≡A0+A1y
g(Q_3) = 1 \Rightarrow Q_3 \equiv A_0 + A_1y
P3=(K1−A1ln(y))y2+(A0−4)y
P_3 = (K_1-A_1 ln(y))y^2 + (A_0 - 4)y
P3=K1y2+(A0−4)y
P_3 = K_1 y^2 + (A_0 -4)y
P3
P_3
é polinômio
⇒A1≡0
\Rightarrow A_1 \equiv 0
Q3=A0
Q_3 = A_0
P2=(B1ln(y)+K2)y2+(B0+A0K1−3K1)+2A02−7A0+6
P_2 = (B_1 ln(y) + K_2) y^2 + (B_0 + A_0 K_1 -3K_1) + \frac{A_0^2 - 7A_0}{2} + 6
B1=0
B_1 = 0
P2=K2y2+(B0+A0K1−3K1)+2A02−7A0+6
P_2 = K_2 y^2 + (B_0 + A_0 K_1 -3K_1) + \frac{A_0^2 - 7A_0}{2} + 6
Q2=B0
Q_2 = B_0
Q2=B0+B1y⇒
Q_2 = B_0 + B_1 y \Rightarrow
P1=(−C1ln(y)+K3)y2+(−2K2+A0K2+B0K1+C0)y+3K1−5/2A0K1−3B0+1/2A02K1+A0B0+y−4−3/2A02+1/6A03+13/3A0
P_1 = ( -C_1\,\ln(y)+K_3) {y}^{2}+ ( -2\,K_2+A_0\,K_2+B_0\,K_1+C_0) y+3\,K_1-5/2\,A_0\,K_1 \\
-3\,B_0+1/2\,{A_0}^{2} K_1 + A_0\,B_0+{\frac{-4-3/2\,{A_0}^{2}+1/6\,{A_0}^{3}+13/3\,A_0}{y}}
P1=K3y2+(−2K2+A0K2+B0K1+C0)y+3K1−5/2A0K1−3B0+1/2A02K1+A0B0
P_1 = K_3 {y} ^ {2} + (-2 \, K_2 + A_0 \, K_2 + B_0 \, K_1 + C_0) y + 3 \, K_1-5 / 2 \, A_0 \, K_1-3 \, B_0 + 1/2 \, {A_0} ^ {2} K_1 + A_0 \, B_0
Q1=C0
Q_1 = C_0
Q1=C0+C1y⇒
Q_1 = C_0 + C_1 y \Rightarrow
{C1=0y−4−3/2A02+1/6A03+13/3A0=0
\begin{cases}
C_1 = 0 \\
{\frac {-4-3 / 2 \, {A_0} ^ {2} +1/6 \, {A_0} ^ {3} + 13 / 3 \, A_0} {y}} = 0
\end{cases}
⇒{C1=0A0∈{2,3,4}
\Rightarrow \begin{cases}
C_1 = 0 \\
A_0 \in \{ 2, 3, 4\}
\end{cases}
P0=K4e∫−3Q0dy
P_0 = K_4 e^{\int-\frac{Q_0}{3} dy}
P=x4y2+(−4+A0)yx3+K1y2x3+26−7A0+A02x2+(−3K1+A0K1+B0)yx2+K2y2x2+26K1−5A0K1−6B0+A02K1+2A0B0x+(−2K2+A0K2+B0K1+C0)yx+K3y2x+D0,
P = {x}^{4}{y}^{2}+ (-4+A_0)y{x}^{3}+K_1{y}^{2}{x}^{3}+ \frac{6-7A_0+{A_0}^{2}}{2} {x}^{2}+(-3K_1+\\
A_0K_1+B_0)y{x}^{2}+K_2{y}^{2}{x}^{2}+\frac{6K_1-5A_0K_1-6B_0+{A_0}^{2}K_1+2A_0B_0}{2}\,x+ \\ (-2\,K_2+A_0\,K_2+B_0\,K_1+C_0)yx+K_3\,{y}^{2}x+D_0,
Q=C0x+B0x2+A0x3−2x4y
Q = C_0\,x+B_0\,{x}^{2}+A_0\,{x}^{3}-2\,{x}^{4}y
⇒{Q0=0P0=K4
\Rightarrow \begin{cases}
Q_0 = 0 \\
P_0 = K_4
\end{cases}
Substituindo
−x4p1(x)+x4dxdp2(x)−p1(x)q1(x)−p2(x)q0(x)=0
-x^4 p_1(x) + x^4 \frac{d}{dx}p_2(x) - p_1(x) q_1(x) -p_2(x) q_0(x) = 0
x4p1(x)+x4(a1+2a2x+3a3x2+4x3)−(a1x+a2x2+a3x3+x4)q0(x)=0
{x}^{4}p_1(x) +{x}^{4} (a_1+2\,a_2\,x+3\,a_3\,{x}^{2}+4\,{x}^{3}) - (a_1\,x+a_2\,{x}^{2}+a_3\,{x}^{3}+{x}^{4})q_0(x) = 0
Obtemos
p2(x)=a0+a1x+a2x2+a3x3+x4
p_2(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + x^4
em
Q=C0x+B0x2+A0x3−2x4y⇒
Q = C_0\,x+B_0\,{x}^{2}+A_0\,{x}^{3}-2\,{x}^{4}y \Rightarrow
q0(x)≡b0+b1x+b2x2+b3x3
q_0(x) \equiv b_0+b_1\,x+b_2\,{x}^{2}+b_3\,{x}^{3}
q0(x)
q_0(x)
q0(x)
q_0(x)
teria grau 5, porém:
p1(x)=(b3−4)x3+(b2+a3b3−3a3)x2+(b1−2a2+a2b3a3b2)x+b0+a2b2+a1b3+a3+a3b1−a1+xa1b2+a2b1+a3b0+x2a1b1+a2a2b0+x3a1b0
p_1(x) = (b_3 -4)x^3 + (b_2 + a_3 b_3 -3 a_3)x^2 + (b_1 -2 a_2 + a_2 b_3 a_3 b_2)x + b_0 + a_2b_2 + a_1 b_3 + a_3 + a_3 b_1 -a_1 + \\
\frac{a_1 b_2 + a_2 b_1 + a_3 b_0}{x} + \frac{a_1b_1 + a_2a_2 b_0}{x^2} + \frac{a_1 b_0}{x^3}
continuando, teremos...
⎩⎨⎧29A02−A03+12−13A0=0215A0B0−3A02B0−9B0=0−23A0B02−6A0+3B02+12=02B03+6B0=0
\begin{cases}
\frac{9{A_0}^{2}-{A_0}^{3}}{2} +12 -13 A_0 = 0 \\
\frac{15 A_0 B_0 - 3 A_0^2B_0}{2} -9 B_0 = 0 \\
-\frac{3 A_0 B_0^2}{2} -6 A_0 + 3 B_0^2 + 12 = 0 \\
\frac{B_0^3}{2}+6B_0 = 0
\end{cases}
P=x4y2−2yx3+3+x2
P = x^4 y^2 - 2yx^3 + 3 + x^2
Q=2x3−2x4y
Q = 2 x^3 - 2 x^4 y
Método da expansão para segunda ordem
y′′=z′=N(x,y,z)M(x,y,z)
y'' = z' = \dfrac{M(x,y,z)}{N(x,y,z)}
(z≡y′)
(z \equiv y')
D≡N∂x+zN∂y+M∂z
\mathbb{D} \equiv N\partial_x + zN \partial_y + M\partial_z
D=(N0+N1z+N2z2)∂x+z((N0+N1z+N2z2)∂y+(M0+M1z+M2z2)∂z
\mathbb{D} = (N_0 + N_1 z + N_2 z^2) \partial_x + z((N_0 + N_1 z + N_2 z^2)\partial_y + (M_0 + M_1 z + M_2z^2)\partial_z
P=P0(x,y)+P1(x,y)z+P2(x,y)z2
P = P_0(x,y) + P_1(x,y)z + P_2(x,y)z^2
N≡N0(x,y)+N1(x,y)z+N2(x,y)z2
N \equiv N_0(x,y) + N_1(x,y) z + N_2(x,y) z^2
M≡M0(x,y)+M1(x,y)z+M2(x,y)z2
M \equiv M_0(x,y) + M_1(x,y) z + M_2(x,y) z^2
Q=Q0(x,y)+Q1(x,y)z+Q2(x,y)z2+Q3(x,y)z3
Q = Q_0(x,y) + Q_1(x,y)z + Q_2(x,y)z^2 + Q_3(x,y) z^3
gz(Q)=max(gz(M)−1,gz(N)+1)
g_z(Q) = max(g_z(M)-1, g_z(N)+1)
=max(1,3)=3
= max(1, 3) = 3
Teremos o sistema...
M0P1+N0∂x∂P0−P0Q0=02M0P2+M1P1+N0∂x∂P1+N0∂y∂P0+N1∂x∂P0−P1Q0−P0Q1=02M1P2+M2P1+N0∂x∂P2+N0∂y∂P1+N1∂x∂P1+N1∂y∂P0+N2∂x∂P0−P1Q1−P2Q0−P0Q2=02M2P2+N0∂y∂P2+N1∂x∂P2+N1∂y∂P1+N2∂x∂P1+N2∂y∂P0−P1Q2−P2Q1−P0Q3=0N1∂y∂P2+N2∂x∂P2+N2∂y∂P1−P1Q3−P2Q2=0N2∂y∂P2−P2Q3=0.
M_{{0}} P_{{1}} +N_{{0}}
{\frac {\partial }{\partial x}}P_{{0}}
-P_{{0}} Q_{{0}}
= 0 \\
2M_{{0}} P_{{2}} +M_{{1}}
P_{{1}} +N_{{0}}
{\frac {\partial }{\partial x}}P_{{1}} +
N_{{0}} {\frac {\partial }{\partial y}}P_{{0}}
+N_{{1}} {\frac {\partial }{
\partial x}}P_{{0}} -P_{{1}} Q_{
{0}} -
P_{{0}} Q_{{1}}
= 0 \\
2M_{{1}} P_{{2}} +M_{{2}}
P_{{1}} +N_{{0}}
{\frac {\partial }{\partial x}}P_{{2}} +
N_{{0}} {\frac {\partial }{\partial y}}P_{{1}}
+N_{{1}} {\frac {\partial }{
\partial x}}P_{{1}} +N_{{1}} {\frac {\partial }{\partial y}}P_{{0}} +
N_{{2}} {\frac {\partial }{\partial x}}P_{{0}}
-P_{{1}} Q_{{1}} -P_{{2
}} Q_{{0}} -
P_{{0}} Q_{{2}} = 0 \\
2M_{{2}} P_{{2}} +N_{{0}}
{\frac {\partial }{\partial y}}P_{{2}}
+N_{{1}} {\frac {\partial }{\partial x}}P_{{2}} +
N_{{1}} {\frac {\partial }{\partial y}}P_{{1}} +N_{{2}} {
\frac {\partial }{\partial x}}P_{{1}} +N_{{2}}
{\frac {\partial }{\partial y}}P_{{0}} -
P_{{1}} Q_{{2}} -P_{{2
}} Q_{{1}} -P_{{0
}} Q_{{3}} = 0 \\
N_{{1}} {\frac {\partial }{\partial y}}P_{{2}}
+N_{{2}} {\frac {\partial}{
\partial x}}P_{{2}} +N_{{2}} {
\frac {\partial }{\partial y}}P_{{1}} -
P_{{1}} Q_{{3}} - P_{{2}} Q_{{2}} = 0 \\
N_{{2}} {\frac {\partial }{\partial y}}P_{{2}}
- P_{{2}} Q_{{3}} =0.
z′=−2(xy+z2−x)zx2y2+x2yz+2xyz2+xz3+z4−x2z+xy2+yz2−xy
z'=-{\frac {{x}^{2}{y}^{2}+{x}^{2}yz+2\,xy{z}^{2}+x{z}^{3}+{z}^{4}-{
x}^{2}z+x{y}^{2}+y{z}^{2}-xy}{ 2\,\left( xy+{z}^{2}-x \right) z}}
Exemplo
D=(2(xy+z2−x)z)∂x+z(2(xy+z2−x)z)∂y+(−x2y2−x2yz−2xyz2−xz3−z4+x2z−xy2−yz2+xy)∂z
D =(2\, \left( xy+{z}^{2}-x \right) z)\,\partial_x + z\,(2\, \left( xy+{z}^{2}-x \right) z)\,\partial_y +
(-{x}^{2}{y}^{2}-{x}^{2}yz-2\,xy{z}^{2}-x{z}^{3}-{z}^{4}+{x}^{2}z-x{y}^{2}-y{z}^{2}+xy)\,\partial_z
P=P0(x,y)+P1(x,y)z+P2(x,y)z2
P = P_{{0}} \left( x,y \right) +P_{{1}} \left( x,y \right) z+P_{{2}} \left( x,y \right) {z}^{2}
Q=Q0(x,y)+Q1(x,y)z+Q2(x,y)z2+Q3(x,y)z3+Q4(x,y)z4
Q = Q_{{0}} \left( x,y \right) +Q_{{1}} \left( x,y \right) z+Q_{{2}} \left( x,y \right) {z}^{2}+Q_{{3}} \left( x,y \right) {z}^{3}+Q_{{4}}
\left( x,y \right) {z}^{4}
g(P)=2g(Q)≡max(g(M),g(N))=4
g(P) = 2 \\
g(Q) \equiv max(g(M), g(N)) = 4
Exemplo
(−y2x2−y2x+xy)P1(x,y)−P0(x,y)Q0(x,y)=02(xy−x)∂x∂P0(x,y)−P1(x,y)Q0(x,y)−P0(x,y)Q1(x,y)+2(−y2x2−y2x+xy)P2(x,y)+(−yx2+x2)P1(x,y)=0
\left( -{y}^{2}{x}^{2}-{y}^{2}x+xy \right) P_{{1}} \left( x,y
\right) -P_{{0}} \left( x,y \right) Q_{{0}} \left( x,y \right) = 0 \\
2\, \left( xy-x \right) {\frac {\partial }{\partial x}}P_{{0}} \left(x,y \right) -P_{{1}} \left( x,y \right) Q_{{0}} \left( x,y \right) -P_
{{0}} \left( x,y \right) Q_{{1}} \left( x,y \right) + 2\, \left( -{y}^{2}{x}^{2}-{y}^{2}x+xy \right) P_{{2}} \left( x,y \right) + \left( -y{x
}^{2}+{x}^{2} \right) P_{{1}} \left( x,y \right) = 0 \\
\\
2(−2xy−y)P2(x,y)−xP1(x,y)+2(xy−x)∂x∂P2(x,y)+2∂x∂P0(x,y)−P1(x,y)Q2(x,y)−P2(x,y)Q1(x,y)−P0(x,y)Q3(x,y)+2(xy−x)∂y∂P1(x,y)=0
2\, \left( -2\,xy-y \right) P_{{2}} \left( x,y \right) -xP_{{1}}
\left( x,y \right) +2\, \left( xy-x \right) {\frac {\partial }{
\partial x}}P_{{2}} \left( x,y \right) + 2\,{\frac {\partial }{\partial x}}P_{{0}} \left( x,y \right) \\
-P_{{1}} \left( x,y \right) Q_{
{2}} \left( x,y \right) -P_{{2}} \left( x,y \right) Q_{{1}} \left( x,y
\right) -P_{{0}} \left( x,y \right) Q_{{3}} \left( x,y \right) + 2\, \left( xy-x \right) {\frac {\partial }{\partial y}}P_{{1}} \left( x,y
\right) = 0 \\
−P1(x,y)Q4(x,y)−P2(x,y)Q3(x,y)+2∂x∂P2(x,y)+2∂y∂P1(x,y)−2P2(x,y)=0−P2(x,y)Q4(x,y)+2∂y∂P2(x,y)=0
-P_{{1}} \left( x,y \right) Q_{{4}} \left( x,y \right) -P_{{2}}
\left( x,y \right) Q_{{3}} \left( x,y \right) +2\,{\frac {\partial }{
\partial x}}P_{{2}} \left( x,y \right) + 2\,{\frac {\partial }{\partial y}}P_{{1}} \left( x,y \right) -2\,P_{{2}} \left( x,y \right)
=0 \\
-P_{{2}} \left( x,y \right) Q_{{4}} \left( x,y \right) +2\,{\frac {
\partial }{\partial y}}P_{{2}} \left( x,y \right)
= 0
2∂x∂P1(x,y)+2(xy−x)∂y∂P2(x,y)+2∂y∂P0(x,y)−P1(x,y)Q3(x,y)−P2(x,y)Q2(x,y)−P0(x,y)Q4(x,y)−2xP2(x,y)−P1(x,y)=0
2\,{\frac {\partial }{\partial x}}P_{{1}} \left( x,y \right) +2\,
\left( xy-x \right) {\frac {\partial }{\partial y}}P_{{2}} \left( x,y
\right) +2\,{\frac {\partial }{\partial y}}P_{{0}} \left( x,y
\right) - P_{{1}} \left( x,y \right) Q_{{3}} \left( x,y \right) \\
-P_{{2}} \left( x,y \right) Q_{{2}} \left( x,y \right) -P_{{0}} \left( x,y
\right) Q_{{4}} \left( x,y \right) - 2\,xP_{{2}} \left( x,y \right) -P_{{1}} \left( x,y \right)
= 0
2(xy−x)∂x∂P1(x,y)+2(xy−x)∂y∂P0(x,y)−P1(x,y)Q1(x,y)−P2(x,y)Q0(x,y)−P0(x,y)Q2(x,y)+2(−yx2+x2)P2(x,y)+(−2xy−y)P1(x,y)=0
2\, \left( xy-x \right) {\frac {\partial }{\partial x}}P_{{1}} \left(
x,y \right) +2\, \left( xy-x \right) {\frac {\partial }{\partial y}}P_
{{0}} \left( x,y \right) -P_{{1}} \left( x,y \right) Q_{{1}} \left( x,
y \right) \\
- P_{{2}} \left( x,y \right) Q_{{0}} \left( x,y \right) -P_{{0
}} \left( x,y \right) Q_{{2}} \left( x,y \right) +2\, \left( -y{x}^{2}
+{x}^{2} \right) P_{{2}} \left( x,y \right) + \left( -2\,xy-y \right) P_{{1}} \left( x,y \right)
= 0
g(P)=2g(Q)=4
g(P) = 2 \\
g(Q) = 4
Resolvendo
−P2(x,y)Q4(x,y)+2∂y∂P2(x,y)=0
-P_{{2}} \left( x,y \right) Q_{{4}} \left( x,y \right) +2\,{\frac {
\partial }{\partial y}}P_{{2}} \left( x,y \right) = 0
P2=F0(x)e∫2Q4dy
P_2 = F_0(x) e^{\int{\frac{Q_4}{2}}dy}
∫2Q4dy
\int{\frac{Q_4}{2}}dy
deve ser logaritmo ou zero
P2=F0(x)e0=F0(x)
P_2 = F_0(x) e^0 = F_0(x)
Q4=0
Q_4 = 0
P tem grau 2 e temos um grau de liberdade
⇒F0(x)≡1∴
\Rightarrow F_0(x) \equiv 1 \therefore
P2=1
P_2 = 1
g(P)=2g(Q)=4
g(P) = 2 \\
g(Q) = 4
Resolvendo
−P1(x,y)Q4(x,y)−P2(x,y)Q3(x,y)+2∂x∂P2(x,y)+2∂y∂P1(x,y)−2P2(x,y)=0
-P_{{1}} \left( x,y \right) Q_{{4}} \left( x,y \right) -P_{{2}}
\left( x,y \right) Q_{{3}} \left( x,y \right) +2\,{\frac {\partial }{
\partial x}}P_{{2}} \left( x,y \right) + 2\,{\frac {\partial }{\partial y}}P_{{1}} \left( x,y \right) -2\,P_{{2}} \left( x,y \right)
=0 \\
P1=21∫Q3(x,y)dy+y+F1(x)
P_1 = \frac{1}{2}\int{Q_3(x,y)}dy + y + F_1(x)
{Q3=a0F1=f0+f1x⇒P1=21a0y+y+f0+f1x
\begin{cases}
Q_3 = a_0 \\
F_1 = f_0 + f_1 x
\end{cases}
\Rightarrow P_1 = \dfrac{1}{2}a_0 y + y + f_0 + f_1 x
g(P)=2g(Q)=4
g(P) = 2 \\
g(Q) = 4
Resolvendo
P0=∫(4(a02+3a0+2)y+2Q2)dy+((a0+1)(f0+f1x)−f1)y+F2(x)
P_0 = \int{(\frac{(a_0^2 + 3a_0 + 2)y + 2Q_2}{4})}dy + ((a_0 + 1)(f_0 + f_1 x) - f_1)y + F_2(x)
{Q2=b0+b1x+b2yF2=g0+g1x+g2x2⇒
\begin{cases}
Q_2 = b_0 + b_1x + b_2y \\
F_2 = g_0 + g_1 x + g_2 x^2
\end{cases} \Rightarrow
2∂x∂P1(x,y)+2(xy−x)∂y∂P2(x,y)+2∂y∂P0(x,y)−P1(x,y)Q3(x,y)−P2(x,y)Q2(x,y)−P0(x,y)Q4(x,y)−2xP2(x,y)−P1(x,y)=0
2\,{\frac {\partial }{\partial x}}P_{{1}} \left( x,y \right) +2\,
\left( xy-x \right) {\frac {\partial }{\partial y}}P_{{2}} \left( x,y
\right) +2\,{\frac {\partial }{\partial y}}P_{{0}} \left( x,y
\right) - P_{{1}} \left( x,y \right) Q_{{3}} \left( x,y \right) \\
-P_{{2}} \left( x,y \right) Q_{{2}} \left( x,y \right) -P_{{0}} \left( x,y
\right) Q_{{4}} \left( x,y \right) - 2\,xP_{{2}} \left( x,y \right) -P_{{1}} \left( x,y \right)
= 0
P1=g0+g1x+g2x2+2b0+f0+a0f0−2f1y+22+b1+a0f1+f1xy+82+2b2+a02+3a0y2
P_1 =
g_0 + g_1 x + g_2 x^2
+ \frac{b_0 + f_0 + a_0 f_0 - 2 f_1}{2}y
+ \frac{2 + b_1 + a_0 f_1 + f_1}{2} xy
+ \frac{2 + 2 b_2 + a_0^2 + 3 a_0}{8} y^2
g(P)=2g(Q)=4
g(P) = 2 \\
g(Q) = 4
Resolvendo
2(−2xy−y)P2(x,y)−xP1(x,y)+2(xy−x)∂x∂P2(x,y)+2∂x∂P0(x,y)−P1(x,y)Q2(x,y)−P2(x,y)Q1(x,y)−P0(x,y)Q3(x,y)+2(xy−x)∂y∂P1(x,y)=0
2\, \left( -2\,xy-y \right) P_{{2}} \left( x,y \right) -xP_{{1}}
\left( x,y \right) +2\, \left( xy-x \right) {\frac {\partial }{
\partial x}}P_{{2}} \left( x,y \right) + 2\,{\frac {\partial }{\partial x}}P_{{0}} \left( x,y \right) \\
-P_{{1}} \left( x,y \right) Q_{
{2}} \left( x,y \right) -P_{{2}} \left( x,y \right) Q_{{1}} \left( x,y
\right) -P_{{0}} \left( x,y \right) Q_{{3}} \left( x,y \right) + 2\, \left( xy-x \right) {\frac {\partial }{\partial y}}P_{{1}} \left( x,y
\right) = 0 \\
Apenas ainda é desconhecido
Q1(x,y)
Q_1(x,y)
P=xy+z2
P = xy + z^2
Q=−2xyz−2z3
Q = -2 xyz - 2z^3
g(P)=2g(Q)=4
g(P) = 2 \\
g(Q) = 4
Função S
Na busca de polinômios de Darboux
O que é
z′=dxdz=ϕ=N(x,y,z)M(x,y,z)
z' = \dfrac{dz}{dx} = \phi = \dfrac{M(x,y,z)}{N(x,y,z)}
z≡y′
z \equiv y'
dydz=−S(x,y,z)
\dfrac{dz}{dy} = -S(x,y,z)
S(x,y,z)≡IzIy
S(x,y,z) \equiv \dfrac{I_y}{I_z}
S≡N(x,y,z)P(x,y,z)⇒{dydz=−NPRDA[R]=Pz−Ny
S \equiv \dfrac{P(x,y,z)}{N(x,y,z)} \Rightarrow
\begin{cases}
\frac{dz}{dy} = -\frac{P}{N} \\
\frac{D_A[R]}{R} = P_z - N_y
\end{cases}
DA≡N∂y−P∂z
D_A \equiv N\partial_y - P\partial_z
Nos casos
Dado que...
dxdz=N(x,y,z)M(x,y,z)
\dfrac{dz}{dx} = \dfrac{M(x,y,z)}{N(x,y,z)}
dydz=−S(x,y,z)
\dfrac{dz}{dy} = -S(x,y,z)
}
\}
Tem
R
R
como fator integrante
∑inipiDA[pi]=Pz−Ny
\sum_i n_i \dfrac{D_A[p_i]}{p_i} = P_z - N_y
⇓
\Downarrow
Cenário 1
- Solução Liouviliana não elementar
- Só existe um polinômio no fator integrante
R=ep11p1n1
R = e^{\frac{1}{p_1}}p_1^{n_1}
RDA[R]=n1p1DA[p1]−p12DA[p1]=Pz−Ny
\dfrac{D_A [R]}{R} = n_1 \dfrac{D_A[p_1]}{p_1} - \dfrac{D_A[p_1]}{p_1^2} = P_z - N_y
n1DA[p1]−p1DA[p1]=(Pz−Ny)p1
n_1 D_A[p_1] - \dfrac{D_A[p_1]}{p_1} = (P_z - N_y) p_1
⇓
\Downarrow
qA1≡p1DA[p1]⇒
q_{A1} \equiv \dfrac{D_A[p_1]}{p_1} \Rightarrow
n1DA[p1]−qA1=(Pz−Ny)p1
n_1 D_A[p_1] - q_{A1} = (P_z - N_y) p_1
⇓
\Downarrow
Cenário 2
- Solução elementar
- Existem vários polinômios no fator integrante, mas só um tem grau elevado
R=∏ipini
R = \prod_i p_i ^{n_i}
RDA[R]=n1p1DA[p1]+∑i=2knipiDA[pi]=Pz−Ny
\dfrac{D_A [R]}{R} = n_1 \dfrac{D_A[p_1]}{p_1} + \sum^{k}_{i=2} n_i \dfrac{D_A[p_i]}{p_i} = P_z - N_y
n1qA1+∑i=2kniqAi=(Pz−Ny)
n_1 q_{A1} + \sum_{i=2}^k n_i q_{Ai} = (P_z - N_y)
⇓
\Downarrow
DA[p1]=qA1p1
D_A[p_1] = q_{A1} p_1
⇓
\Downarrow
com o cofator determinado
Problema dos graus máximos
- Não existe limite superior teórico para os graus dos polinômios de Darboux
- Todos os métodos são semi-algorítmicos
Problema
Objetivo
Estimar os graus em que existam polinômios de Darboux
Modelo semântico
Semântica
- Representação vetorial de objetos com métrica que faça sentido para comparações no problema, baseada nas relações entre os objetos
Frases e Palavras
- m1: Rock and Roll Music in the 1960’s
- m2: Recipe for a Drum Roll, a Demonstration of Techniques
- m3: Drum and Bass Composition
- m4: A Perspective of Rock Music in the 90’s
- m5: Music and Composition of Popular Bands
- b1: How to make Bread and Rolls, a Demonstration
- b2: Ingredients for Crescent Roll Dough
- b3: A Recipe for Sourdough Bread
- b4: A Quick Recipe for Pizza Dough with Natural Ingredients
L=0.01.00.00.00.00.00.01.01.01.00.00.01.00.00.00.00.00.00.01.01.00.00.00.00.00.00.00.00.00.00.00.01.00.01.00.01.00.00.01.01.00.00.00.00.00.00.00.00.00.00.01.01.01.00.01.00.00.00.01.00.00.00.00.00.00.00.00.00.01.00.01.00.00.01.00.01.00.00.00.00.0
L = \left[\begin{matrix}0.0 & 1.0 & 0.0 & 0.0 & 1.0 & 0.0 & 0.0 & 0.0 & 0.0\\1.0 & 0.0 & 1.0 & 0.0 & 0.0 & 0.0 & 1.0 & 0.0 & 0.0\\0.0 & 0.0 & 1.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 1.0\\0.0 & 1.0 & 0.0 & 0.0 & 1.0 & 0.0 & 0.0 & 0.0 & 0.0\\0.0 & 0.0 & 0.0 & 0.0 & 1.0 & 0.0 & 0.0 & 0.0 & 1.0\\0.0 & 0.0 & 0.0 & 1.0 & 0.0 & 0.0 & 1.0 & 0.0 & 0.0\\0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 1.0 & 0.0 & 1.0 & 0.0\\1.0 & 0.0 & 0.0 & 1.0 & 0.0 & 1.0 & 0.0 & 0.0 & 0.0\\1.0 & 0.0 & 0.0 & 0.0 & 0.0 & 1.0 & 0.0 & 1.0 & 0.0\end{matrix}\right]
- m1: Rock and Roll Music in the 1960’s
- m2: Recipe for a Drum Roll, a Demonstration of Techniques
- m3: Drum and Bass Composition
- m4: A Perspective of Rock Music in the 90’s
- m5: Music and Composition of Popular Bands
- b1: How to make Bread and Rolls, a Demonstration
- b2: Ingredients for Crescent Roll Dough
- b3: A Recipe for Sourdough Bread
- b4: A Quick Recipe for Pizza Dough with Natural Ingredients
L=UnWnVnT
L = U_nW_nV_n^T
Análise de componentes principais...
U8=
U_8 =
V8=
V_8 =
W=2.60.00.00.00.00.00.00.00.00.02.190.00.00.00.00.00.00.00.00.01.910.00.00.00.00.00.00.00.00.01.590.00.00.00.00.00.00.00.00.01.260.00.00.00.00.00.00.00.00.00.990.00.00.00.00.00.00.00.00.00.720.00.00.00.00.00.00.00.00.00.240.00.00.00.00.00.00.00.00.00.0
W = \displaystyle \left[\begin{matrix}2.6 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0\\0.0 & 2.19 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0\\0.0 & 0.0 & 1.91 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0\\0.0 & 0.0 & 0.0 & 1.59 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0\\0.0 & 0.0 & 0.0 & 0.0 & 1.26 & 0.0 & 0.0 & 0.0 & 0.0\\0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.99 & 0.0 & 0.0 & 0.0\\0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.72 & 0.0 & 0.0\\0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.24 & 0.0\\0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0\end{matrix}\right]
Vetores semânticos
U4=0.0−0.38−0.080.0−0.02−0.19−0.36−0.56−0.59−0.59−0.07−0.19−0.59−0.49−0.010.040.030.040.16−0.63−0.40.16−0.05−0.350.390.030.31−0.210.00.59−0.210.31−0.520.23−0.280.2
U_4 = \displaystyle \left[\begin{matrix}0.0 & -0.59 & 0.16 & -0.21\\-0.38 & -0.07 & -0.63 & 0.0\\-0.08 & -0.19 & -0.4 & 0.59\\0.0 & -0.59 & 0.16 & -0.21\\-0.02 & -0.49 & -0.05 & 0.31\\-0.19 & -0.01 & -0.35 & -0.52\\-0.36 & 0.04 & 0.39 & 0.23\\-0.56 & 0.03 & 0.03 & -0.28\\-0.59 & 0.04 & 0.31 & 0.2\end{matrix}\right]
Semântica de polinômios
p1=a1x2y1+a2x2y2
p_1 = a_1 x^2 y^1 + a_2 x^2 y^2
p2=b1xy1+b2x2y2
p_2 = b_1 x y^1 + b_2 x^2 y^2
⇒
\Rightarrow
0xy1+a1x2y1+a2x2y2
0 x y^1 + a_1 x^2 y^1 + a_2 x^2 y^2
b1xy1+0x2y1+b2x2y2
b_1 x y^1 + 0 x^2 y^1 + b_2 x^2 y^2
(0b1a10a2a2)
\begin{pmatrix}
0 & a_1 & a_2 \\
b_1 & 0 & a_2
\end{pmatrix}
xy1
x y^1
x2y1
x^2 y^1
x2y2
x^2 y^2
p1
p_1
p2
p_2
Preparação dos dados
- Separar numeradores e denominadores das equações
- Cria semânticas separadas para numeradores e denominadores
- Normaliza os vetores
Taxa de acerto em torno de 80%
Problemas
- É preciso treinar um modelo para cada grau
- Adicionar um novo grau implica num modelo treinado do zero
- Treinamento muito demorado
- Necessita de muitas amostras
- Semântica não evolui para novos graus, deve ser reconstruída.
Modelo de transferência
- Classes se baseiam no maior polinômio conhecido que compõe o fator integrante
- Redução de dimensionalidade se dá com base na distância entre classes
- Aprende a diferença entre as equações antes de aprender o padrão comum
Rede triplice siamesa
- Ancora e Positivo têm o mesmo grau máximo no PD, negativo têm grau diferente
- f(E) é a equação que determina o modelo
- S é o a função objetivo a ser minimizada
∣∣f(Ea)−f(Ep)∣∣<∣∣f(Ea)−f(En)∣∣
|| f(E_a) - f(E_p)|| < ||f(E_a) - f(E_n)||
S(Ea,Ep,En)≡max(∣∣f(Ea)−f(Ep)∣∣−∣∣f(Ea)−f(En)∣∣+α,0)
S(E_a, E_p, E_n) \equiv max(|| f(E_a) - f(E_p)|| - ||f(E_a) - f(E_n)|| + \alpha, 0)
S(Ea,Ep,En)≡∣∣f(Ea)−f(Ep)∣∣−∣∣f(Ea)−f(En)∣∣+α
S(E_a, E_p, E_n) \equiv || f(E_a) - f(E_p)|| - ||f(E_a) - f(E_n)|| + \alpha
ou
⇓
\Downarrow
Modelo de classificação
Resultados gerais
Caso excepcional
Considerações finais
Usando Polinômios de Darboux na Obtenção de Integrais Primeiras Liouvillianas de Equações Diferenciais Ordinárias Racionais de Segunda Ordem André Claudino