Applied Analysis and PDE7. Hyperbolic systems

Lehel Banjai (Heriot-Watt University)based on lecture notes by Jack Carr

December 4, 2018

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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 .

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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 .

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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−.

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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).

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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.

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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.)

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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.

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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 .

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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

).

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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.

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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)].

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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).

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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.

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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 .

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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

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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.

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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 .

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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 .

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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

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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

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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

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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.

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Hyperbolic systems