Scalar First Order Nonlinear PDEs

Initial Value Problem

ut + φ(u)ux = f(x, t) x ∈ R, t > 0, [PDE]

u(x, 0) = g(x) x ∈ R, [IC] (0.1)

where f(x, t) and g(x) are given functions. The function φ(u) represents the nonlinear term.If φ is a constant, the PDE reduces to a linear equation that we have already dealt with.

Solution Method

Step 1: Compute Characteristics. Solve the following ODE system:

dt

ds= 1, t(0) = 0 (0.2)

dx

ds= φ(U), x(0) = ξ (0.3)

dU

ds= f(x, t), U(0) = g(ξ) (0.4)

Notice that the first two equations alone do not form a closed system; U appears in the2nd equation. Three equations together, however, form a closed system of ODEs. Thesolution triplet t, x, U will depend on parameter s and the initial point ξ; so let us denotethem by t = t(s, ξ), x = x(s, ξ), U = U(s, ξ).

Step 2: Find the Inverse Function.

{

t = t(s, ξ),

x = x(s, ξ)⇒

{

s = s(x, t),

ξ = ξ(x, t)

Step 3: Solution Formula.u(x, t) = U(s(x, t), ξ(x, t)).

Example 1 (Burgers’ Equation)

Consider{

ut + uux = 0 x ∈ R, t > 0,

u(x, 0) = g(x) x ∈ R,

STEP 1: Consider

dt

ds= 1, t(0) = 0 (0.5)

dx

ds= U, x(0) = ξ (0.6)

dU

ds= 0, U(0) = g(ξ). (0.7)

From (0.5), t = s.From (0.7), U = g(ξ).Plug the last result in (0.6), dx/ds = g(ξ). Hence x = ξ + g(ξ)sIn summary,

t = s

x = ξ + g(ξ)s

U = g(ξ)

In particular, along each charateristic line, the solution U remains constant.STEP 2: Now we solve inverse function of t = s, x = ξ+g(ξ)s. We immediately know s = t.

Plug this into x-formula:x = ξ + g(ξ)t (0.8)

In general, we cannot express this solution explicitly. But if we can, we are done. In addition,even in the case we have no explicit formula for the inverse function, the above informationstill tells us a lot about the behavior of the solution.

Example 2

Let the initial data g(x) = x in Example 1. Then we can solve the inverse function. Eq.(0.8) becomes

x = ξ + ξt.

Hence, ξ = x/(1 + t). The inverse function is

s = t, ξ = x/(1 + t).

The solution is finally given by:

u(x, t) = U(s, ξ) = g(ξ) = ξ = x/(1 + t) x ∈ R, t ≥ 0.

Example 3

Let the initial data

g(x) =

0 x ≤ 0

x 0 ≤ x ≤ 1

1 x ≥ 1.

The characteristic lines through the initial point (x, t) = (ξ, 0) is parametrized by

{

t = s

x = ξ + g(ξ)s,

with parameter s. By considering various values of initial point ξ, we can sketch the picture ofthe characteristic lines.

x

76543210

t

-1

6

5

4

3

2

1

0

Characteristic ´ Lines

STEP 2: Now we solve inverse function of t = s, x = ξ + g(ξ)s. The whole upper half planeis divided into three parts. For x ≤ 0, we get

s = t, ξ = x.

For 0 < x ≤ t + 1,s = t, ξ = x/(1 + t).

For x > t + 1,s = t, ξ = x − t.

STEP 3: Now the solution is given by

u(x, t) =

0 x ≤ 0, t ≥ 0,

x/(1 + t) 0 < x ≤ t + 1, t ≥ 0,

1 x > t + 1, t ≥ 0..

Exercises

[1] Solve

ut + (1 − u)ux = 0 x ∈ R, t > 0,

u(x, 0) = g(x) =

1 x ≤ 0,

1 − x 0 < x < 1,

0 x ≥ 1.

[2] Solve

ut + uux = 0 x ∈ R, t > 0,

u(x, 0) = g(x) =

0 x ≤ 0,

x 0 < x < 1,

1 1 ≤ x ≤ 3,

(x − 1)/2 3 < x < 5,

2 x ≥ 5.

[3] Solve

ut + uux = x x ∈ R, t > 0,

u(x, 0) = 2x x ∈ R.

[4] Solve

ut + u2ux = 0 x ∈ R, t > 0,

u(x, 0) = g(x) =

0 x ≤ 0,

x 0 < x < 1,

1 x ≥ 1.

ANSWERS

[1] u(x, t) =

1 x ≤ 0, t ≥ 0,

1 − x/(1 + t) 0 < x < 1 + t, t ≥ 0,

0 x ≥ 1 + t, t ≥ 0.

[2] u(x, t) =

0 x ≤ 0, t ≥ 0,

x/(1 + t) 0 < x < 1 + t, t ≥ 0,

1 1 + t ≤ x ≤ 3 + t, t ≥ 0,

(x − 1)/(2 + t) 3 + t < x < 5 + 2t, t ≥ 0,

2 x ≥ 5 + 2t, t ≥ 0.

[3] u(x, t) = x(3et + e−t)/(3et − e−t) for x ∈ R, t ≥ 0.

[4] u(x, t) =

0 x ≤ 0, t ≥ 0,

2x/(1 +√

1 + 4tx) 0 < x < 1 + t, t ≥ 0,

1 x ≥ 1 + t, t ≥ 0.