Mth 538
50024849
Zhao Wu
Review for the lecture on A numerical Riemann-Hilbert approach for the Korteweg-de Vries
equation
First of all, let's introduce Korteweg-de Vires equation. The KdV equation is a mathematical
model of waves on shallow water surfaces. It is particularly notable as the prototypical example
of an exactly solvable model, that is , a non-linear partial differential equation whose solutions
can be exactly and precisely specified. The solutions in turn include prototypical examples of
solitons. Trogdon, the lecturer, said that the equation they are trying to solve in a numerical way
is:
𝜕𝑢
𝜕𝑢 𝜕 3 𝑢
−𝑢
+
=0
𝜕𝑡
𝜕𝑥 𝜕𝑥 3
𝑢(𝑥, 0) = 𝑢0 (𝑥)
Since evolving solutions of the KdV equation with standard time-stepping methods
the domain size needs to grow in time, and if using a pseudo-spectral method on the KdV
equation, the computational cost grows like 𝑂(𝑇 3 𝑙𝑜𝑔𝑇) with a very large constant, that’s why he
is using the inverse scattering transform to solve for KdV, with the computational cost O(1).
To solve the initial value problem for this equation where 𝑢(𝑥, 0) is a known function of x, one
associates to this equation the Schrödinger eigenvalue equation:
𝜕2𝜑
− 𝑢(𝑥, 𝑡)𝜑 = 𝛌𝜑
𝜕𝑥 2
where 𝜑 is an unknown function of t and x and u is the solution of the Korteweg–de Vries
equation that is unknown except at t=0 . The constant 𝛌 is an eigenvalue.
From the Schrödinger equation we obtain
𝑢=
1 𝜕 2𝜑
∗
−𝛌
𝜑 𝜕𝑥 2
Substituting this into the Korteweg–de Vries equation and integrating gives the equation
𝜕𝜑 𝜕 3 𝜑
𝜕𝜑
𝑑𝑥
+ 3 − 3 ∗ (𝑢 − 𝛌) ∗
= 𝐶𝜑 + 𝐷𝜑 ∫ 2
𝜕𝑡 𝜕𝑥
𝜕𝑥
𝜑
where C and D are constants.
Since evolving solutions of the KdV equation with standard time-stepping methods
the domain size needs to grow in time,
Mth 538
50024849
Zhao Wu
For solving linear equation like:
𝑞𝑡 + 𝑞𝑥𝑥𝑥 = 0, (𝑥, 𝑡) ∈ 𝑅 × (0, 𝑇]
𝑞(𝑥, 0) = 𝑞0 (𝑥)
We can just apply Fourier transform to solve, the solutions can be computed directly using standard
collocation techniques.
However, for nonlinear-equations like Riemann-Hilbert problems, we can’t simply follow the same way.
The detail is too difficult to understand, here I’ll just point out some key points that I remembered.
By enforcing Φ± − 𝐼 ∈ 𝐿2 (Γ), it follwos from the theory of Hardy spaces that
1
Φ(𝑘) = 𝐼 + 2∗𝑝𝑖∗𝑖 ∫Γ
𝑢(𝑠)
𝑑𝑠
𝑠−𝑘
= 𝐼 + 𝐶Γ 𝑢(𝑘) for some 𝑢 ∈ 𝐿2 (Γ)
Properties of the Cauchy integral operator imply
Φ+ (𝑠) = Φ− (𝑠)𝐺(𝑠) <==> 𝑢 − 𝐶Γ − 𝑢(𝐺 − 𝐼) = 𝐺 − 𝐼
Define 𝐶[𝐺; Γ]𝑢 = 𝑢 − 𝐶Γ − 𝑢(𝐺 − 𝐼)
Thus, the key point is to solve 𝐶[𝐺; Γ] = 𝐺 − 𝐼 numerically.
Define Pn = {cos(j*pi/n) : j = 0,…,n}, the Chebyshev points.Given function values at Mi(Pn), we
compute the mapped Chebyshev interpolant 𝑢𝑛 .We compute 𝐶[𝐺; Γ]* 𝑢𝑛 exactly at each point in
Mi(Pn). This gives the discretization 𝐶𝑛 [𝐺; Γ]* 𝑢𝑛 of 𝐶[𝐺; Γ]* 𝑢𝑛 .In general, the numerical
method returns a function 𝑢𝑛 of mapped Chebyshevpolynomials such that:
𝐶Γ ∗ 𝑢𝑛 is bounded, and
I+𝐶Γ ∗ 𝑢𝑛 satisfies the RHP exactly at Mi(Pn) for all i.
After finding the jump matrix G for KdV, we can recover the solution of the KdV equation from the RHP by using the
formula
𝑞(𝑥, 𝑡) = 2𝑖
lim
|𝑘|→𝑖𝑛𝑓𝑖𝑛𝑖𝑡𝑦
𝑘𝜕𝑥 (Φ11 + Φ21 )