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Applied Analysis and PDE7. Hyperbolic systems
Lehel Banjai (Heriot-Watt University)based on lecture notes by Jack Carr
December 4, 2018
1
Systems of conservation laws
In this chapter we consider systems of hyperbolic conservationslaws
Ut + f (U)x = 0 U(x , t) ∈ Rn, x ∈ R, f : Rn → Rn.
Definition: The system of n first order PDE Ut + A(U)Ux = 0 isstrictly hyperbolic if A(U) has n real, distinct eigenvalues
λ1(U) < λ2(U) < · · · < λn(U).
This implies the existence of a basis of left and right eigenvectors`i , ri .
2
Example: Let f (U) = AU for some matrix A ∈ Rn:
Ut + AUx = 0, U(0) = U0.
Suppose A has n distinct real eigenvalues λ1 < · · · < λn and rightand left eigenvectors ri , `i such that
ri · `j ={
0 i 6= j1 i = j
.
Denoting ui = `i · U, the system decouples to n scalar equations
(ui )t + λi (ui )x = 0
giving the solution
U =n∑
j=1
`j · U0(x − λj t)rj .
3
Example continued
Let now initial data be piecewise constant
U0 =
{U− if x < 0U+ if x > 0.
Note that the solution for λj < x/t < λj+1 is
u(x , t) =n∑
i=1
`i · U0(x − λi t)ri =j∑
i=1
`i · U+ri +n∑
i=j+1
`i · U−ri
=
j∑i=1
`i · (U+ − U−)ri +n∑
i=1
`i · U−ri
=
j∑i=1
`i · (U+ − U−)ri + U−.
4
Example: Shallow water wave equation
ht + uhx + hux = 0
ut + uux + ghx = 0
I h = h(x , t) – the depth of the water
I u = u(x , t) – the horizontal velocity of the water
I g – gravity
I Shallow water assumption: H � L.5
Shallow water wave equations are obtained either
I as the limiting case of general equations of fluid mechanicswhen the ratio of the water depth to the wavelength of atypical surface wave is small
I or the equations can be derived directly from the principles ofconservation of mass and momentum (see book by Logan).
6
Small-Amplitude Approximation
Assume that the surface does not deviate much from theundisturbed depth H and that the velocity is small, i.e.,
h(x , t) = H + η(x , t), u(x , t) = 0 + v(x , t)
with η and v small. Substituting into shallow water equations gives
ηt + vηx + (H + η)vx = 0 vt + vvx + gηx = 0
Leaving the quadratic terms out (assuming that, e.g., vηx is small)we have
ηt + Hvx = 0 vt + gηx = 0
Note both η and v satisfy the classical linear wave equation
ηtt = −Hvxt = Hgηxx vtt − Hgvxx = 0.
Note: speed of propagation is√gH.
7
Back to shallow water equations
Scaling the equations so that gravity is equal to 1:
ht + uhx + hux = 0 ut + uux + hx = 0.
The matrix form is(hu
)t
+
(u h1 u
)(hu
)x
=
(00
)For
A =
(u h1 u
)the eigenvalues are λ± = u ±
√h and the left eigenvectors are
`± =(1 ±
√h.)
8
Multiply from left by `± to get
`±
(hu
)t
+ λ±`±
(hu
)x
= 0
i.e.,
`±
(ht + λ±hxut + λ±ux
)= 0.
Hence along characteristics
dx
dt= λ±
we have
`±d
dt
(h(x(t), t)u(x(t), t)
)= 0.
9
In conclusion, along two curves of characteristics
dx
dt= u ±
√h
u and h satisfy an ODE:
dh
dt±√hdu
dt= 0.
10
General case
Back to the general case
Ut + A(U)Ux = 0
as before multiply by a left eigenvector
`iUt + `iAUx = 0
to obtain`i (Ut + λiUx) = 0
for eigenvalue λi = λi (U).Hence
`id
dtU(x(t), t) = 0 along
dx(t)
dt= λi .
11
Example: For the scalar equationut + f
′(u)ux = 0
the eigenvalue is f ′(u) so the equation is hyperbolic. On thecharacteristic given by
dx(t)
dt= f ′(u(x , t))
the PDE reduces to the ODE du/dt = 0.Example: Wave equation
ut − vx = 0 , vt − c2ux = 0
where c is a positive constant. Then U = (u, v)T and
A =
(0 −1−c2 0
).
12
Wave equation continued
Here λ = ±c and left eigenvectors(∓c 1
)so the system is
hyperbolic.
−c dudt
+dv
dt= 0 along
dx(t)
dt= c
and
cdu
dt+
dv
dt= 0 along
dx(t)
dt= −c
Integrating gives the Riemann invariants
±cu + v = constant along x = ∓ct + constant
Hence
cu + v = F (x + ct) , −cu + v = G (x − ct)
where F and G are arbitrary functions.
13
Example: For the shallow water waves equation,
A =
(u h1 u
)the eigenvalues are λ = u ±
√h and the left eigenvectors are(
1 ±√h), hence
h−1/2dh
dt± du
dt= 0 along
dx(t)
dt= u ±
√h
Integrating, we obtain
2√h ± u = constant on dx(t)
dt= u ±
√h
The expressions 2√h ± u are the Riemann invariants for this
problem.14
Example: Isentropic gas dynamics – The p systemLet p : R→ R be s.t. p′(w) < 0. The p-system is:
wt − vx = 0 , vt + p(w)x = 0Then U = (w , v)T and
A =
(0 −1
p′(w) 0
).
The eigenvalues are
λ1 = −√−p′(w) and λ2 =
√−p′(w)
so the system is hyperbolic. For λ = λ1 the left eigenvector is(√−p′(w) 1
)so√
−p′(w) dwdt
+dv
dt= 0 along
dx(t)
dt= λ1
Integrating, we obtain the Riemann invariant
R− = R−(w , v) = v +
∫ w0
√−p′(s) ds
which is constant on dx/dt = λ1.15
Conservation Laws and Shock Waves
With f : Rn → Rn we study the system of conservation laws:
Ut + f (U)x = 0
Note, in previous notation A = f ′(U) –Jacobian of f .
As in the scalar case the PDE is derived from the conservation law
d
dt
∫ ba
U(x , t) dx = f (U(a, t))− f (U(b, t))
Using the same calculations made for the scalar case the jumpcondition is
s[U] = [f (U)].
16
Example: Recall the linear example
Ut + AUx = 0, A ∈ Rn×n.
Recall that the solution jumped across
x/t = λj
from ωj−1 to ωj with
ωj =
j∑i=1
`i · (U+ − U−)ri + U−.
The jump condition requires
λj(ωj − ωj−1) = A(ωj − ωj−1).
Asωj − ωj−1 = `j · (U+ − U−)rj
this is indeed satisfied.17
Example: The conservation law for the shallow water equations is
ht + (hu)x = 0
(hu)t + (hu2 + h2/2)x = 0
so we have that (hhu
)t
+
(hu
hu2 + h2/2
)x
= 0
and the jump conditions are
s[h] = [hu] and s[hu] = [hu2 + h2/2]
Remark: the conservation of mass %h and momentum %hu wasused in the derivation of the conservation law (%–density).
18
Entropy conditionRecall for scalar conservation laws we had the characteristics
dx
dt(t) = f ′(u)
and the entropy condition
f ′(ul) > s > f′(ur ).
Now the characteristics aredx
dt(t) = λk(U)
where λk(U) are the eigenvalues of A and
λ1 < λ2 < · · · < λn.Lax’s entropy condition: For some k (so called k-shocks)
λk(U`) > s > λk(Ur )
andλj(U`) < s, λj(Ur ) < s; j < k ,
λj(U`) > s, λj(Ur ) > s; j > k .19
Example: (The p-system) Recall
wt − vx = 0 , vt + p(w)x = 0
with p′(w) < 0 . Assume, in addition, that p′′(w) > 0.The eigenvalues are
λ1(U) = −√−p′(w) and λ2(U) =
√−p′(w)
We fix a constant vector
U` =
(w`v`
)and ask which states
Ur =
(wv
)can be connected to U` by 1- and 2-shocks.
20
The jump conditions are
s(w − w`) = −(v − v`) s(v − v`) = p(w)− p(w`)
Eliminating s gives (v − v`)2 = (w − w`)(p(w`)− p(w)) so that
v − v` = ±√
(w − w`)(p(w`)− p(w))
We first consider 1-shocks.
−√−p′(w) < s < −
√−p′(w`)
so that p′(w`) > p′(w). Since p′′ > 0, this implies that w` > w .
21
Using s < 0 and w` > w in the jump condition
s(w − w`) = −(v − v`)
givesv < v`.
Hence the set of w and v which can be connected to U` by a1-shock must lie on the curve
v − v` = −√
(w − w`)(p(w`)− p(w)) , w` > w
A similar analysis shows that the set of w and v which can beconnected to U` by a 2-shock must lie on the curve
v − v` = −√
(w − w`)(p(w`)− p(w)) , w` < w
22
Rarefaction wavesWe generalize the scalar case to a system of n first order PDE.
Ut + A(U)Ux = 0
IfU(x , t) = Ū(x/t) = Ū(z) , z = x/t
is a solution then
(A− zI )Ūz = 0
We make the following conclusionsI
z = λk(Ū)
is an eigenvalue of the matrix AI Ūz is a multiple of the corresponding eigenvectorI The exact multiple of the eigenvector is determined by the
constraint z = λk(Ū) =⇒ 1 = Ūz · ∇λk .We call this the k-rarefaction wave.
23
Example: For the shallow water waves equation, U = (h, u)T and
A =
(u h1 u
).
The eigenvalues are λ = u ±√h. We find a rarefactive wave
h = h̄(z), u = ū(z) , z = x/t
corresponding to the eigenvalue
λ = u −√h
Since z = λ(Ū) we have that
ū(z) = λ+√h̄(z) = z +
√h̄(z)
To determine Ūz we find the (right) eigenvector
k
(−√h̄
1
)so that
h̄z = −k√h̄ , and ūz = k .
24
To find k we differentiate to get
ūz = 1 +h̄z
2√
h̄(z)
Combining this with above gives k = 2/3.Solving
ūz =2
3
gives ūz = 2z/3 + C where C is a constant and we find h̄
h̄ =(3C − x/t)2
9and ū =
2x
3t+ C .
25
The Dam breaking problem
Consider the Riemann problem
ht + uhx + hux = 0 , ut + uux + hx = 0
with initial data
u(x , 0) = 0, h(x , 0) =
{1 x < 00 x > 0
26
We anticipate that there will be three regions:
I a region x < s1t where h = 1 and u = 0 ,
I a rarefaction wave in the region s1t < x < s2t and
I a region x > s2t where h = u = 0.
We have to find s1 and s2.
The solution is continuous across x = s1t. On this line, h = 1 andu = 0, so
(3C − s1)2
9= 1 and
2s13
+ C = 0
Thus s21 = 1 and as height should decrease s1 = −1 andC = 2/3 .Hence in the region −t < x < s2t we have that
h(x , t) =(2− x/t)2
9and u(x , t) =
2(1 + x/t)
3
27
We use the jump conditions
s[h] = [hu] and s[hu] = [hu2 + h2/2]
to determine s2. For this case, hr = ur = 0 so that
s2h` = u`h` and s2u`h` = h`u2` + h
2`/2
It follows that h` = 0 and using the formula for h at x = s2t
(2− s2)2 = 0
so that s2 = 2.h(x,t)
2t
1
−tx
28
Note:We have connected the states (u, h) = (0, 1) and (u, h) = (0, 0)via the intermediate state (u∗, h∗) = (2, 0) and a 1-rarefactionwave and a 2-shock.
29
Hyperbolic systems