Modified Korteweg–de Vries Equation
Posted 08-14-2008 at 12:18 PM by G-man
∂w
∂t
+
∂3w
∂x3 + 6ľw2 ∂w
∂x
= 0.
Modified Korteweg–de Vries equation.
1±. One solution on
solution for ľ = 1:
w(x, t) = a + k2
p4a2 + k2 cosh z + 2a
, z = kx - (6a2k + k3)t + b,
where a, b, and k are arbitrary constants.
2±.
solution for ľ = 1:
w(x, t) = 2 a1eµ1 + a2eµ2 + Aa2e2µ1+µ2 + Aa1eµ1+2µ2
1 + e2µ1 + e2µ2 + 2(1 - A)eµ1+µ2 + Ae2(µ1+µ2) ,
µ1 = a1x - a3
1t + b1, µ2 = a2x - a3
2 t + b2, A =
µ
a1 - a2
a1 + a2
¶2
,
where a1, a2, b1, and b2 are arbitrary constants.
3±. Rational solutions for ľ = 1:
w(x, t) = a -
4a
4a2z2 + 1
, z = x - 6a2t,
w(x, t) = a -
12a
ˇ
z4 + 3
2 a-2z2 - 3
16 a-4 - 24tz
˘
4a2
ˇ
z3 + 12t - 3
4 a-2z
˘2 + 3
ˇ
z2 + 1
4 a-2
˘2 ,
where a is an arbitrary constant.
4±. There is a selfsimilar
solution of the form w = t-1=3U(z), where z = t-1=3x.
5±. The modified Korteweg–de Vries equation is solved by the inverse scattering method.
∂t
+
∂3w
∂x3 + 6ľw2 ∂w
∂x
= 0.
Modified Korteweg–de Vries equation.
1±. One solution on
solution for ľ = 1:
w(x, t) = a + k2
p4a2 + k2 cosh z + 2a
, z = kx - (6a2k + k3)t + b,
where a, b, and k are arbitrary constants.
2±.
solution for ľ = 1:
w(x, t) = 2 a1eµ1 + a2eµ2 + Aa2e2µ1+µ2 + Aa1eµ1+2µ2
1 + e2µ1 + e2µ2 + 2(1 - A)eµ1+µ2 + Ae2(µ1+µ2) ,
µ1 = a1x - a3
1t + b1, µ2 = a2x - a3
2 t + b2, A =
µ
a1 - a2
a1 + a2
¶2
,
where a1, a2, b1, and b2 are arbitrary constants.
3±. Rational solutions for ľ = 1:
w(x, t) = a -
4a
4a2z2 + 1
, z = x - 6a2t,
w(x, t) = a -
12a
ˇ
z4 + 3
2 a-2z2 - 3
16 a-4 - 24tz
˘
4a2
ˇ
z3 + 12t - 3
4 a-2z
˘2 + 3
ˇ
z2 + 1
4 a-2
˘2 ,
where a is an arbitrary constant.
4±. There is a selfsimilar
solution of the form w = t-1=3U(z), where z = t-1=3x.
5±. The modified Korteweg–de Vries equation is solved by the inverse scattering method.
Total Comments 1
Comments
-
I agree 100%
Posted 08-15-2008 at 02:01 PM by abbey
