Cole-Hopf transformation as numerical tool for the … transformation as numerical tool for the...

Cole-Hopf transformationas numerical tool for the Burgers Equation

A paper by Taku Ohwada

Alejandro Pozo

July 29th, 2011

Cole-Hopf transformation Numerical schemes Some results

Outline

1 Cole-Hopf transformation

2 Numerical schemes

3 Some results


Cole-Hopf transformation

The Burgers’ Equation∂u∂t + u ∂u

∂x = ν∂2u∂x2

using the Cole-Hopf transformation, given by

u(x , t) = −2ν∂∂x Θ(x , t)

Θ(x , t)

is transformed into the linear diffusion equation

∂Θ

∂t = ν∂2Θ

∂x2

that, for the initial value problem, has the solution

Θ(x , t) =1

2√πνt

∫ ∞−∞

Θ(y , 0)e−(y−x)2

4νt dy


Cole-Hopf transformation

Let us observe that

u(x , t) = −2ν∂∂x Θ(x , t)

Θ(x , t)=⇒ Θ(x , t) = e−

12ν

∫ x

αu(ξ,t)dξ

For convenience, we denote:

G(x , y) := e−(y−x)2

4νt

Then:

u(x , t) = −∫∞−∞(y − x)Θ(y , 0)G(x , y)dy

t∫∞−∞Θ(y , 0)G(x , y)dy


Outline


2 Numerical schemes

3 Some results


Numerical schemes

We have:

Θ(x , t) = e−1

2ν

∫ x

αu(ξ,t)dξ

u(x , t) = −∫∞−∞(y − x)Θ(y , 0)G(x , y)dy

t∫∞−∞Θ(y , 0)G(x , y)dy

The logical computation would be:1 Compute Θ(y , 0) using u(y , 0)

2 Compute u(x , t) using Θ(y , 0)

However, the magnitude of Θ(y , 0) may become huge or vanishingly small,usual programming languages cannot deal with it. That is why theCole-Hopf transformation may not seem really useful.


Numerical schemes

We can express the constant α in Θ(y , 0) depending on x . Taking α = x ,we can rewrite Θ(y , 0) as

Θ(y , 0; x) := Θ(y , 0) = e−1

2ν

∫ y

xu(ξ,0)dξ

It satisfies:

Θ(x , 0; x) = 1|u(y , 0)| ≤ C1 =⇒ Θ(y , 0; x) ≤ eC2|y−x |

On the other hand G(x , y) decays double-exponentially as |y − x | increases.

Therefore, ΘG and (y − x)ΘG are effectively zero for |y − x | � 1 and, ifνt � 1, we can compute u(x , t) using Θ(y , 0; x) in the neighborhood ofy = x .

Then, we will just use u(x , t) as the initial data for the following step.


Numerical schemes

Now, we just need approximate the initial data u(ξ, 0) with respect to space.

We will consider a uniform discretization xk = k∆x and a sufficiently smalltime-step ∆t that satisfies (∆x)2

4ν∆t � 1, so that only the data of Θ(y , 0; xk) in[xk−1, xk+1] is necessary for the computation of u(xk ,∆t).

Some additional notation:

η = ξ − xk

uk = u(xk , 0)

We shall consider three different approaches.

Using piecewise linear polynomials (u(xk + η, 0) = bη + a)Using cubic polynomials (u(xk + η, 0) = cν3 + bη + a)Using quartic polynomials (u(xk + η, 0) = eν4 + dν3 + cν2 + bν + a)


Numerical schemes

Scheme A:Piecewise linear polynomials

u(xk + η, 0) = ±uk±1 − uk∆x η + uk , 0 < ±η < ∆x

Therefore, the exponent of ΘG becomes a piecewise quadratic polynomialof ν and, if t satisfies that 1 + b±∆t > 1, the coefficient of η2 is negativeand the integration in the formula for u(x ,∆t) can be done analytically.Even the integration on (−∞,∞) can be done safely.


Numerical schemes

Scheme B:Piecewise cubic polynomials

u(xk + η, 0) =uk+1 − 2uk + uk−1

(∆x)2 η3 +uk+1 − uk−1

2∆x + uk , 0 < ±η < ∆x

For this case, no general analytical formula can be obtained, as

Θ(xk + ν, 0; xk)G(xk , xk + η) = e−cη36ν e− 1

4ν∆t [(1+b∆t)η2+2a∆tν]

For small ∆t, we can use the approximation e−cη36ν ≈ 1− cη3/6ν, so that

the integration for u(x ,∆t) can already be computed analytically.


Numerical schemes

So, we have:

u(xk ,∆t) = u(xk , 0) +s1∆t + s2∆t2 + s3∆t3 + s4∆t4

k0 + k1∆t + k2∆t2 + k3∆t3 + k4∆t4

with

k0 = 6ν, k1 = 24bν, k2 = (36b2 + 6ac)ν,

k3 = a3c + (24b3 + 12abc)ν, k4 = a3bc + 6(b4 + ab2c)ν,

s1 = −6abν + 12cν2, s2 = (−18ab2 + 6a2c)ν + 24bcν2,

s3 = −18ab3ν + 12b2cν2, −a4bc − 6(ab4 + a2b2c)ν


Numerical schemes

Scheme C:An extension of the scheme B, using piecewise quartic polynomialapproximation by a five point formula, that results on:

u(xk ,∆t) = u(xk , 0)+p0 + p1∆t + p2∆t2 + p3∆t3 + p4∆t4 + p5∆t5 + p6∆t6

q0 + q1∆t + q2∆t2 + q3∆t3 + q4∆t4 + q5∆t5 + q6∆t6


Outline


2 Numerical schemes

3 Some results


Some results

We will make two numerical tests:

P1: u(x , 0) =

{1, x ≤ 00, x > 0

P2: u(x , 0) = 1− sin(πx)


Some results

L1 error at t = 1 versus ∆x in P1 for ν = 0.1 and ∆t = ∆x2.

Scheme B and Scheme C are nearly second and four order approximations,whereas Scheme A error does not seem to converge.


Some results

Scheme Cν = 10−3, t = 10,

∆x = 0.1, ∆x = 0.01

Scheme Cν = 10−3, t = 10,

∆x = 10−3, ∆t = 10−4


Some results

Modification for Scheme B and Scheme C.

We increase ν to C1∆x locally in the computation of u(xk , t) if thecorresponding slopes b± = ± uk±1−uk

∆x satisfy b+b− < 0 or |b±| > C2∆x , forC1,C2 ∈ R+.

ν = 10−5, t = 10, ∆x = 0.1,∆t = 0.01, C1 = 1, C2 = 1.5


Some results

ν = 10−3, t = 1, ∆x = 0.02,∆t = 0.002, C1 = 1, C2 = 1.5


Bibliography

K. Sakai and I. Kimura, A numerical scheme based on a solutionof nonlinear advection-diffusion equations, Journal of Computationaland Applied Mathematics, vol. 173, 2005, pp. 39-55.


Thanks for your attention!

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