The first four solutions are of traveling-wave type and the last one is a radial symmetric solution with center at the point –A, –B... This is a stationary heat equation with a nonlinear
Trang 11◦ Solutions:
w(x, y) = 1
β ln
2(A2
+ B2)
aβ(Ax + By + C)2
if aβ >0,
w(x, y) = 1
β ln
2(A2
+ B2)
aβsinh2(Ax + By + C)
if aβ >0,
w(x, y) = 1
β ln
–2(A2+ B2)
aβcosh2(Ax + By + C)
if aβ <0,
w(x, y) = 1
β ln
2(A2
+ B2)
aβcos2(Ax + By + C)
if aβ >0,
w(x, y) = 1
β ln
8C
aβ
– 2
β ln(x + A)2+ (y + B)2– C,
where A, B, and C are arbitrary constants The first four solutions are of traveling-wave type and the last one is a radial symmetric solution with center at the point (–A, –B).
2◦ Functional separable solutions:
w(x, y) = –2
β ln
C1e ky
2aβ 2k cos(kx + C2)
,
w(x, y) = 1
β ln 2k2(B2– A2)
aβ[A cosh(kx + C1) + B sin(ky + C2)]2,
w(x, y) = 1
β ln 2k2(A2+ B2)
aβ[A sinh(kx + C1) + B cos(ky + C2)]2,
where A, B, C1, C2, and k are arbitrary constants (x and y can be swapped to give another
three solutions)
3◦ General solution:
w(x, y) = –2
β ln 1–2aβΦ(z)Φ(z)
4|Φ
z (z)| ,
whereΦ = Φ(z) is an arbitrary analytic (holomorphic) function of the complex variable
z = x+iy with nonzero derivative, and the bar over a symbol denotes the complex conjugate.
4. ∂
2w
∂x2 + ∂
2w
∂y2 = ae βw + be2βw.
1◦ Traveling-wave solutions:
w(x, y) = –1
β ln
– aβ
C2
1+ C22
+ C3exp(C1x + C2y) + a
2β2– bβ(C2
1+ C22)
4C3(C12+ C22)2 exp(–C1x – C2y)
,
w(x, y) = –1
β ln
aβ
C2
1+ C22
+
a2β2+ bβ(C2
1+ C22)
C2
1+ C22
sin(C1x + C2y + C3)
,
where C1, C2, and C3are arbitrary constants
2◦ For other exact solutions of this equation, see equation T9.3.1.7 with f (w) = ae βw+
be2βw.
Trang 25. ∂
2w
∂x2 + ∂
2w
∂y2 = αw ln(βw).
1◦ Solutions:
w(x, y) = 1
β exp
* 1
4α(x + A)2+ 14α(y + B)2+1+,
w(x, y) = 1
β exp
*
A(x + B)2
Aα–4A2(x + B)(y + C) + (1
4α – A)(y + C)2+ 12
+
,
where A, B, and C are arbitrary constants.
2◦ There are exact solutions of the following forms:
w(x, y) = F (z), z = Ax + By, w(x, y) = G(r), r=
(x + C1)2+ (y + C2)2,
w(x, y) = f (x)g(y).
6. ∂
2w
∂x2 + ∂
2w
∂y2 = α sin(βw).
1◦ Functional separable solution for α = β =1:
w(x, y) =4arctan
cot A cosh F cosh G
, F = √ cos A
1+ B2(x – By), G=
sin A
√
1+ B2(y + Bx), where A and B are arbitrary constants.
2◦ Functional separable solution (generalizes the solution of Item1◦):
w(x, y) = 4
β arctan
f (x)g(y)
,
where the functions f = f (x) and g = g(y) are determined by the first-order autonomous
ordinary differential equations
(f x )2= Af4+ Bf2+ C, (g y)2= Cg4+ (αβ – B)g2+ A, and A, B, and C are arbitrary constants.
3◦ For other exact solutions of this equation, see equation T9.3.1.7 with f (w) = α sin(βw).
7. ∂
2w
∂x2 + ∂
2w
∂y2 = f (w).
This is a stationary heat equation with a nonlinear source.
1◦ Suppose w = w(x, y) is a solution of the equation in question Then the functions
w1= w( x + C1, y + C2),
w2= w(x cos β – y sin β, x sin β + y cos β),
where C1, C2, and β are arbitrary constants, are also solutions of the equation (the plus or minus signs in w1are chosen arbitrarily)
Trang 32◦ Traveling-wave solution in implicit form:
A2+ B2F (w)
–1 2
dw = Ax + By + D, F (w) =
f (w) dw, where A, B, C, and D are arbitrary constants.
3◦ Solution with central symmetry about the point (–C1, –C2):
w = w(ζ), ζ =
(x + C1)2+ (y + C2)2,
where C1and C2 are arbitrary constants and the function w = w(ζ) is determined by the ordinary differential equation w ζζ + ζ–1w
ζ = f (w).
T9.3.2 Equations of the Form
∂
∂x
*
f (x) ∂w
∂x
+
+ ∂
∂y
*
g(y) ∂w
∂y
+
= f (w)
1. ∂
∂x
ax n ∂w
∂x
+ ∂
∂y
by m ∂w
∂y
= f (w).
Functional separable solution for n≠ 2and m≠ 2:
w = w(r), r=
b(2 – m)2x2–n + a(2 – n)2y2–m1 2
Here, the function w(r) is determined by the ordinary differential equation
w
rr + Ar–1w r = Bf (w), where A = 4– nm
(2– n)(2 – m) , B =
4
ab(2 – n)2(2– m)2.
2. a ∂
2w
∂x2 + ∂
∂y
be μy ∂w
∂y
= f (w), ab> 0.
Functional separable solution for μ≠ 0:
w = w(ξ), ξ=
bμ2(x + C
1)2+4ae–μy1 2
,
where C1is an arbitrary constant and the function w(ξ) is defined implicitly by
C2+ 2
abμ2F (w)
–1 2
dw = C3 ξ, F (w) =
f (w) dw, with C2and C3being arbitrary constants
3. ∂
∂x
ae βx ∂w
∂x
+ ∂
∂y
be μy ∂w
∂y
= f (w), ab> 0.
Functional separable solution for βμ≠ 0:
w = w(ξ), ξ = bμ2e–βx + aβ2e–μy 1 2
,
where the function w(ξ) is determined by the ordinary differential equation
w
ξξ – ξ–1w ξ = Af (w), A=4/(abβ2μ2).
Trang 44. ∂
∂x
f (x) ∂w
∂x
+ ∂
∂y
g(y) ∂w
∂y
= kw ln w.
Multiplicative separable solution:
w(x, y) = ϕ(x)ψ(y), where the functions ϕ(x) and ψ(y) are determined by the ordinary differential equations
[f (x)ϕ x] x = kϕ ln ϕ + Cϕ, [g(y)ψ y ] y = kψ ln ψ – Cψ, and C is an arbitrary constant.
T9.3.3 Equations of the Form
∂
∂x
*
f (w) ∂w
∂x
+
+ ∂
∂y
*
g(w) ∂w
∂y
+
= h(w)
1. ∂
2w
∂x2 + ∂
∂y
(αw + β) ∂w
∂y
= 0.
Stationary Khokhlov–Zabolotskaya equation It arises in acoustics and nonlinear
mechan-ics
1◦ Solutions:
w(x, y) = Ay – 12A2αx2+ C
1x + C2,
w(x, y) = (Ax + B)y – α
12A2(Ax + B)4+ C1x + C2,
w(x, y) = –1
α
y + A
x + B
2
+ C1
x + B + C2(x + B)
2– β
α,
w(x, y) = –1
α
β + λ2
A(y + λx) + B
,
w(x, y) = (Ax + B)
C1y + C2– β
α,
where A, B, C1, C2, and λ are arbitrary constants.
2◦ Generalized separable solution quadratic in y (generalizes the third solution of Item1◦):
w(x, y) = – 1
α(x + A)2y
2+* B1
(x + A)2 + B2(x + A)
3+
y
+ C1
x + A + C2(x + A)
2– β
α – αB
2 1
4(x + A)2 –
1
2αB1B2(x + A)3–
1
54αB22(x + A)8,
where A, B1, B2, C1, and C2are arbitrary constants
3◦ See also equation T9.3.3.3 with f (w) =1and g(w) = αw + β.
Trang 52. ∂
2w
∂x2 + ∂
∂y
ae βw ∂w
∂y
= 0, a> 0.
1◦ Additive separable solutions:
w(x, y) = 1
β ln(Ay + B) + Cx + D, w(x, y) = 1
β ln(–aA2y2+ By + C) – 2
β ln(–aAx + D), w(x, y) = 1
β ln(Ay2+ By + C) + 1
β ln
p2
aAcosh2(px + q)
,
w(x, y) = 1
β ln(Ay2+ By + C) + 1
β ln
p2
–aA cos2(px + q)
,
w(x, y) = 1
β ln(Ay2+ By + C) + 1
β ln
p2
–aA sinh2(px + q)
,
where A, B, C, D, p, and q are arbitrary constants.
2◦ There are exact solutions of the following forms:
w(x, y) = F (r), r = k1x + k2y;
w(x, y) = G(z), z = y/x;
w(x, y) = H(ξ) –2(k +1)β–1ln|x|, ξ = y|x| k; w(x, y) = U (η) –2β–1ln|x|, η = y + k ln|x|;
w(x, y) = V (ζ) –2β–1x, ζ = ye x, where k, k1, and k2are arbitrary constants
3◦ For other solutions, see equation T9.3.3.3 with f (w) =1and g(w) = ae βw
3. ∂
∂x
f (w) ∂w
∂x
+ ∂
∂y
g(w) ∂w
∂y
= 0.
This is a stationary anisotropic heat (diffusion) equation.
1◦ Traveling-wave solution in implicit form:
A2f (w) + B2g(w)
dw = C1(Ax + By) + C2,
where A, B, C1, and C2are arbitrary constants
2◦ Self-similar solution:
w = w(ζ), ζ = x + A
y + B, where the function w(ζ) is determined by the ordinary differential equation
[f (w)w ζ ] ζ + [ζ2g(w)w ζ ] ζ =0 (1)
Integrating (1) and taking w to be the independent variable, one obtains the Riccati equation
Cζ
w = g(w)ζ2+ f (w), where C is an arbitrary constant.
Trang 63◦ The original equation can be represented as the system of the equations
f (w) ∂w
∂x = ∂v
∂y, –g(w) ∂w
∂y = ∂v
The hodograph transformation
x = x(w, v), y = y(w, v), where w, v are treated as the independent variables and x, y as the dependent ones, brings (2)
to the linear system
f (w) ∂y
∂v = ∂x
∂w, –g(w) ∂x
∂v = ∂y
Eliminating y yields the following linear equation for x = x(w, v):
∂
∂w
1
f (w)
∂x
∂w
+ g(w) ∂
2x
∂v2 =0
Likewise, we can obtain another linear equation for y = y(w, v) from system (3).
T9.4 Other Second-Order Equations
T9.4.1 Equations of Transonic Gas Flow
1. a ∂w
∂x
∂2w
∂x2 + ∂
2w
∂y2 = 0.
This is an equation of steady transonic gas flow.
1◦ Suppose w(x, y) is a solution of the equation in question Then the function
w1= C1–3C2
2w(C1x + C3, C2y + C4) + C5y + C6,
where C1, , C6are arbitrary constants, is also a solution of the equation
2◦ Solutions:
w(x, y) = C1xy + C2x + C3y + C4,
w(x, y) = – (x + C1)
3
3a(y + C2)2 + C3y + C4,
w(x, y) = a
2C3
1
39 (y + A)13+
2
3aC12(y + A)8(x + B) +3C1(y + A)3(x + B)2– (x + B)
3
3a(y + A)2,
w(x, y) = –aC1y2+ C
2y + C3 4
3C1(C1x + C4)3 2,
w(x, y) = –aA3y2– B2
aA2x + C1y + C2
4
3(Ax + By + C3)3 2,
w(x, y) = 1
3(Ay + B)(2C1x + C2)3 2–
aC3 1
12A2(Ay + B)4+ C3y + C4,
w(x, y) = – 9aA2
y + C1 +4A
x + C2
y + C1
3 2
– (x + C2)
3
3a(y + C1)2 + C3y + C4,
w(x, y) = –3
7aA2(y + C1)7+4A(x + C2)3 2(y + C1)5 2–
(x + C2)3 3a(y + C1)2 + C3y + C4,
where A, B, C1, , C4are arbitrary constants (the first solution is degenerate)
Trang 73◦ There are solutions of the following forms:
w(x, y) = y–3k–2U (z), z = xy k (self-similar solution, k is any number); w(x, y) = ϕ1(y) + ϕ2(y)x3 2+ ϕ3(y)x3 (generalized separable solution);
w(x, y) = ψ1(y) + ψ2(y)x + ψ3(y)x2+ ψ4(y)x3 (generalized separable solution);
w(x, y) = ψ1(y)ϕ(x) + ψ2(y) (generalized separable solution)
2. ∂
2w
∂y2 + a
y
∂w
∂y + b ∂w
∂x
∂2w
∂x2 = 0.
1◦ Suppose w(x, y) is a solution of the equation in question Then the function
w1= C1–3C2
2w(C1x + C3, C2y) + C4y1–a + C5,
where C1, , C5are arbitrary constants, is also a solution of the equation
2◦ Additive separable solution:
w(x, y) = – bC1
4(a +1)y2+ C2y1–a + C3
2 3C1(C1x + C4)3 2,
where C1, , C4are arbitrary constants
3◦ Generalized separable solutions:
w(x, y) = – 9A2b
16(n +1)(2n+1+ a) y
2n+2+ Ay n (x + C)3 2+ a–3
9b
(x + C)3
y2 ,
where A and C are arbitrary constants, and n = n1,2 are roots of the quadratic equation
n2+ (a –1)n + 5
4(a –3) =0
4◦ Generalized separable solution:
w(x, y) = (Ay1–a + B)(2C1x + C2)3 2+9bC3
1θ(y),
θ(y) = – B
2
2(a +1)y2–
AB
3– a y
3–a– A2
2(2– a)(3 – a) y
4–2a + C3y1–a + C4,
where A, B, C1, C2, C3, and C4are arbitrary constants
5◦ There are solutions of the following forms:
w(x, y) = y–3k–2U (z), z = xy k (self-similar solution, k is any number); w(x, y) = ϕ1(y) + ϕ2(y)x3 2+ ϕ3(y)x3 (generalized separable solution);
w(x, y) = ψ1(y) + ψ2(y)x + ψ3(y)x2+ ψ4(y)x3 (generalized separable solution);
w(x, y) = ψ1(y)ϕ(x) + ψ2(y) (generalized separable solution)
... Cg4+ (αβ – B)g2+ A, and A, B, and C are arbitrary constants.3◦ For other exact solutions of this equation, see equation T9.3.1.7 with...
+
,
where A, B, and C are arbitrary constants.
2◦ There are exact solutions of the following forms:
w(x, y) = F (z),... C1, C2, and C3are arbitrary constants
2◦ For other exact solutions of this equation, see equation T9.3.1.7 with