Repeated integration by parts introduces a minus sign for each derivative. Therefore, a functionu ∈L1loc(Ω)admits a weak derivativeDαu∈ L1loc(Ω)if
Ω(Dαu)ψdnx =(−1)|α|
Ωu Dαψdnx (10.7) for allψ∈Ccpt∞(Ω).
It might seem that we should distinguish between classical and weak derivatives in the notation. This is made unnecessary by the following:
Theorem 10.4 (Consistency of weak derivatives)If u ∈ Cm(Ω)then u is weakly differentiable to order k and the weak derivatives equal the classical derivatives.
Conversely, if u ∈ L1loc(Ω)admits weak derivatives Dαu for|α| ≤m, and each Dαu can be represented by a continuous function, then u is equivalent to a function in Cm(Ω)whose classical derivatives match the weak derivatives.
In one direction the argument is straightfoward. Classical derivatives satisfy the criterion (10.7) by integration by parts, so they automatically qualify as weak deriv- atives. Lemma10.1shows that weak derivatives are uniquely defined.
The argument for the converse statement, that continuity of the weak derivatives Dαu implies classical differentiability, is much more technical and we will not be able to give the details. The basic idea is to show that one can approximateu by a sequenceψk∈Ccpt∞(Ω)such thatDαψk→ Dαuuniformly on every compact subset of Ω for |α| ≤ m. The fact that the weak derivatives Dαu are continuous makes this possible. Uniform convergence then allows the classical derivatives ofu to be computed as limits of the functions Dαψk.
On the other hand, the spatial integration parts has no boundary term, ∞
−∞
∂q
∂xψd x= − ∞
−∞q∂ψ
∂x d x.
When the left-hand side of (10.8) is paired withψand integrated over botht andx, the result is thus
∞
0
∞
−∞
∂u
∂t +∂q
∂x
ψd x dt = − ∞
0
∞
−∞
u∂ψ
∂t +q∂ψ
∂x
ψd x dt
− ∞
−∞uψ|t=0d x.
Ifuis a classical solution of (10.8), then ∞
0
∞
−∞
u∂ψ
∂t +q∂ψ
∂x
d x dt+ ∞
−∞gψ|t=0d x =0 (10.9) for allψ∈Ccpt∞([0,∞)×R).
Thet =0 integral in (10.9) makes sense forg∈L1loc(R). Under this assumption, we defineu ∈ L1loc((0,∞)×R;R)to be a weak solution of (10.8) providedq ∈ L1loc((0,∞)×R;R)and (10.9) holds for all test functions.
Example 10.5 Consider the linear conservation equation with constant velocity, which means q = cu in (10.8). By the method of characteristics (Theorem3.2), the solution is
u(t,x)=g(x−ct).
Let us check that this defines a weak solution forg∈L1loc(R).
Forψ∈Ccpt∞(R)the first term in (10.9) is ∞
0
∞
−∞g(x−ct)∂ψ
∂t (t,x)−c∂ψ
∂x(t,x)
d x dt. (10.10) To evaluate the integral, introduce the variables
τ=t, y=x−ct, and define
ψ˜(τ,y):=ψ(τ,y+cτ).
By the chain rule,
∂ψ˜
∂τ = ∂ψ
∂t −c∂ψ
∂x. (10.11)
The Jacobian determinant of the transformation(τ,y)→(t,x)is 1, so that ∞
0
∞
−∞u ∂ψ
∂t −c∂ψ
∂x
d x dt= ∞
0
∞
−∞
g(y)∂ψ˜
∂τ dydτ. Theτintegration can now be done directly,
∞
0
∂ψ˜
∂τ(τ,y)dτ= − ˜ψ(0,y).
This gives ∞
0
∞
−∞u ∂ψ
∂t −c∂ψ
∂x
d x dt = − ∞
−∞g(y)ψ˜(0,y)dy
= − ∞
−∞
g(y)ψ(0,y)dy,
which verifies (10.9). ♦
We saw in Example10.3that in one dimension a jump discontinuity precludes the existence of a weak derivative. Example10.5shows that this is not the case in higher dimension. Forg∈L1loc(R), the solutionu(t,x)=g(x−ct)could be highly discontinuous. The direction of the derivative is crucial here; regularity is required only along the characteristics.
As an application of the weak formulation (10.9), let us return to an issue that arose in the traffic model in Sect.3.4. For certain initial conditions the characteristic lines crossed each other, ruling out a classical solution of the PDE. We will see that weak solutions can still exist in this case.
Consider a one-dimensional quasilinear equation of the form
∂u
∂t − ∂
∂xq(u)=0 (10.12)
with the fluxq(u)a smooth function ofuwhich is independent oftandx. As we saw in Sect.3.4, the characteristics are straight lines whose slope depends on the initial conditions. Let us study the situation pictured in Fig.3.11, where a shock forms as characteristic lines cross at some point. For simplicity we assume that the initial crossing occurs at the origin.
One possible way to resolve the issue of crossing characteristics is to subdivide the (t,x)plane into two regions by drawing ashock curve C, as illustrated in Fig.10.3.
Suppose that classical solutionsu±are derived by the method of characteristics above and below this curve. We will show that this combination yields a weak solution provided a certain jump condition is satisfied along C. The jump condition was discovered in the 19th century by engineers William Rankine and Pierre Hugoniot, who developed the first theories of shock waves in the context of gas dynamics.
Fig. 10.3 Shock curve with solutionsu±on either side
t x
u− u+
C
Theorem 10.6 (Rankine-Hugoniot condition) Let C be a curve parametrized as x=σ(t)withσ∈C1[0,∞). Suppose that u is a weak solution of (10.12)given by
u(t,x)= u−(t,x), x<σ(t), u+(t,x), x>σ(t), where u±are classical solutions. Then, at each point of C,
q(u+)−q(u−)=(u+−u−)σ. (10.13) Proof Since we are not concerned with the boundary conditions, we consider a test functionψ∈Ccpt∞((0,∞)×R), for which (10.9) specializes to
∞
0
∞
−∞
u∂ψ
∂t +q(u)∂ψ
∂x
d x dt =0. (10.14)
Since the solutionsu±are classical andσisC1, we can separate the integral (10.14) at the shock curve and integrate by parts on either side.
Consider first theu−side. For the term involving thexderivative the integration by parts is straightforward:
∞
0
σ(t)
−∞ q(u−)∂ψ
∂x d x dt= − ∞
0
σ(t)
−∞
ψ ∂
∂xq(u−)d x dt +
∞
0
ψq(u−)
x=σ(t)dt.
For thet-derivative term we start by using the fundamental theorem of calculus to derive
d dt
σ(t)
−∞
ψu−d x= σ(t)
−∞
∂ψ
∂t u−+ψ∂u−
∂t
d x+σ(t)ψu−
x=σ(t). By the compact support of ψ, the integral over t of the left-hand side vanishes,
yielding ∞
0
σ(t)
−∞
∂ψ
∂t u−d x dt = − ∞
0
σ(t)
−∞ ψ∂u−
∂t d x dt
− ∞
0
σ(t)ψu−
x=σ(t)dt.
Combining these integration by parts formulas gives ∞
0
σ(t)
−∞
u−∂ψ
∂t +q(u−)∂ψ
∂x
d x dt
= − ∞
0
σ(t)
−∞ ψ∂u−
∂t − ∂
∂xq(u−) d x dt
− ∞
0
σ(t)u−−q(u−) ψ
x=σ(t)dt.
The first term on the right vanishes by the assumption thatu−is a classical solution, leaving
∞
0
σ(t)
−∞
u−∂ψ
∂t +q(u−)∂ψ
∂x
d x dt= − ∞
0
σu−−q(u−) ψ
x=σ(t)dt.
The corresponding calculation on theu+side yields ∞
0
∞
σ(t)
u+∂ψ
∂t +q(u+)∂ψ
∂x
d x dt = ∞
0
σu+−q(u+) ψ
x=σ(t)dt.
By (10.14) the sum of theu−andu+integrals is zero, which implies ∞
0
(u+−u−)σ−(q(u+)−q(u−)) ψ
x=σ(t)dt.
Since this holds for allψ∈C∞((0,∞)×R), we conclude that (u+−u−)σ−(q(u+)−q(u−))
x=σ(t)=0
for allt >0.
Example 10.7 Consider the traffic equation introduced in Sect.3.4,
∂u
∂t +(1−2u)∂u
∂x =0,
for whichq(u)=u−u2. For the initial condition take a step function, g(x):= a, x<0,
b, x>0. (10.15)
From3.28the characteristic lines are given by
x(t)= x0+(1−2a)t, x0 <0, x0+(1−2b)t, x0 >0.
These intersect to form a shock provideda<b.
The solutions above and below the shock line are given by constants, u−(t,x)=a, u+(t,x)=b.
The Rankine-Hugoniot condition (10.13) thus reduces to q(b)−q(a)=(b−a)σ. Substituting withq(u)=u−u2reduces this condition to
σ=1−b−a.
Since the discontinuity starts at the origin, the shock curve is thus given by σ(t)=(1−b−a)t.
Hence the weak solution is
u(t,x)= a, x< (1−b−a)t, b, x> (1−b−a)t.
Some cases are illustrated in Fig.10.4. In the plot on the left, the shock wave
propagates backwards. ♦
For certain initial conditions, the definition (10.9) of a weak solution is not suf- ficient to determine the solution uniquely. For example, if we had takena > bin (10.15), then instead of overlapping the characteristics originating fromt=0 would separate, leaving a triangular region with no characteristic lines. An additional physi- cal condition is required to specify the solution uniquely in this case. We will discuss this further in the exercises.
t x
u−= 0.4 u+= 0.7
t x
u−= 0.3 u+= 0.5
Fig. 10.4 Characteristic lines meeting at the shock wave