ABSTRACT
Multiplying the KdV equation (9) by a test function ϕ ∈ C1,3t,x (Q), after integration, we obtain
∫
u2ϕxdxdt =− ∫
u · Lϕdxdt+ ∫ Ω
uϕ ∣ ∣T 0 dx
+ T∫
B(u, ϕ) ∣ ∣L 0 dt.
(19)
Now we choose the test function ϕ as follows:
ϕ(x, t) = ϕ0(τ)ϕ1(ξ), with τ = t T and ξ =
x L . (20)
Here, ϕ0(τ) = (1− τ)2, with
c0 := 1∫
dτ = 4.
