lecture 14, 19 oct 2010
TRANSCRIPT
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Part I
First and Second Order Partial
Differential Equations
1
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DRAFTCHAPTER 1Partial Differential Equations
(PDEs): Generalities
Consider a space and time, defined on a given domain and denoted by coor-
dinates x , y , z , . . . , t . For example, the domain can be a surface, a cube or acylinder as shown in Figure 1.1.
One of the coordinates, usually denoted by t is time.
Figure 1.1: domain
A vector function u = u(t , x , y , z , . . .) ; = 1, 2, ...d is defined on ad-dimensional space and, in general, describes some physical phenomena. For
example, u(x , y, z, t) can be the pressure gradient at a point, or the velocity ofa fluid, or the electric potential and so on. The function u(x , y, z, t) is usuallyspecified only implicitly by determining the rate of change of the quantity as one
moves around in space and time. For example, conservation of mass and mo-
mentum determines how the fluid velocity varies over space and time. Physical
considerations lead to a general equation, relating the rate of change with time
to rates of change in space. Define u = (u1, . . . , ud). Let
u
t= ut ,
u
x= ux ,
2u
tx= utx , etc. (1.1)
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4 1: Partial Differential Equations (PDEs): Generalities
A general PDE has the form
F[x, y, . . . , t, u, ut, ux, uxt, uxy, . . .] = 0 (1.2)
x , y , . . . , t : independent variables and u, ut, ux, . . . : dependent variablesA PDE depends on two or more variables, unlike an ordinary differential that
depends only on one variable. The highest partial derivative is the order of the
PDE.
The domain D of the PDE is defined by the range of the independent vari-
ables x , y , z , . . . , t for which u(x,y,..,t) is to be determined. See Figure 1.2.
u(x,y)
Figure 1.2: u(x, y)
All dependent functions u(x , y, z, t) are assumed to have finite partial deriva-tives, called a strong solution. Weak solutions are also allowed for which the
partial derivatives may become singular at certain regions, allowing for discon-
tinuities in the behavior of the dependent functions u(x , y, z, t).The PDE is linear if the function F[t, x, y, . . . , u, ut, ux, uxt, uxy, . . .] is a lin-
ear function ofu, ut, ux, uxt, . . .. A PDE is quasi-linear if it is linear in the high-est order derivative and the other coefficient functions of the terms u, ut, ux, uxt, . . .are nonlinear functions.
The PDE give in Eq. 1.2 yields a unique solution only if some conditions
are imposed on the solution, termed as boundary values. The solution of an
ordinary differential equation is unique if some constraints are specified, such
as the initial and final value of the sought for solution. In contrast, for a PDE,
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1.1: Diffusion Equation 5
an entire boundary value function has to be specified on a subdomain ofD
for the sought solution. For example, ifu(x, t) depends on time and space, theboundary (initial) value ofu(x, t) is specified at initial time t = 0 for all x, suchthat u(x, 0) = f(x). See Figure. 1.3.
u(t,x)
f(x)
Figure 1.3: u(t, x) and boundary
In this course, we focus only first order and second order partial differential
equations; to simplify the discussions to its bare essential we focus on study-ing scalar functions in two dimensions, with derivatives in higher dimensions
addressed only for special cases.
1.1 Diffusion Equation
To illustrate the content of a typical PDE, a derivation is given of the diffusion
equation, which is an important exemplar of parabolic PDEs. Consider the con-
centration of particles n(z, t) in a medium under the influence of gravity. Theparticles will tend to move downwards. The configurations depends only on
height z. The medium has a time distribution of the particles that we now deter-mine.
At each instant of time, the total number of particles is constant. The number
of particles in the shaded portion of Figure 1.4(b) is equal to n(t, z)Adz. Sincethere is conservation of mass, the rate of change of n(z, t)Adz must be equalto the difference between the fluxes of particles entering from surface at z andlearning through surface at z + dz, as shown in Figure 1.4. Hence
t[n(t, z)Adz] = AJ(z) AJ(z + dz) (1.3)
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6 1: Partial Differential Equations (PDEs): Generalities
n(t,z)Adz
AJ(z)
AJ(z + dz)
Figure 1.4: (a) Area A; (b) Particles moving in the upward direction.
where J(z) = rate per unit area and per unit time that particles cross surface atposition z. Simplifying yields,
n(z, t)
t= J
z(1.4)
J=-D n z
Figure 1.5: J(z)
Ifn =constant, we have J = 0; hence a good approximation is to set
J = D n(z, t)z
(1.5)
where D is the diffusion constant and is a characteristic of the medium. Hence
n(z, t)
t=
z Dn(z, t)
z (1.6)For constant diffusion constant D we have
n(z, t)
t= D
2n(z, t)
z2: Diffusion Equation (1.7)
1.2 Special cases of PDEs
Before starting the systematic study of PDEs, we mention a few special cases
whose solutions can be found by inspection.
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1.2: Special cases of PDEs 7
1. u(x, y)/y = 0 ; u(x, y) = (x)
2. 2u(x, y)/xy = 0 ; u(x, y) = (x) + v(y)
3.
uxy = f(x, y) (1.8)
u(x, y) = (x) + v(y) +
x0
d
y0
df(, ) (1.9)
4.
ux
uy
= 0 (1.10)
Clearly, this looks like case 1, i.e. u/z = 0, where x, y are related to z bya change of variables. Let
x =1
2( + ); y =
1
2( )
= x + y, = x yu(x, y)
x
=
x
u
+
x
u
=u
+u
u(x, y)
x=
y
u
+
y
u
=
u
u
ux uy = u = 0 u = f()u(x, y) = f(x + y) : wave moving to the left
and shown in Figure 1.6.
V(x)y = 0
V(x)y = 1
x= 1
Figure 1.6: (a) y = 0 (b) y = 1
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8 1: Partial Differential Equations (PDEs): Generalities
In general, for , constant, consider the PDE
ux + uy = 0 (1.11)
A change of variables
= x + y (1.12)
= x y (1.13)yields the solution
u = 0 ; u(x, y) = f(x y) (1.14)
5. Consider the wave equation
2u
x2
2u
y2= 0
or (
x+
y)(
x
y)u = 0
= x + y, = x y0 = 2xu 2yu = 4u u(x, y) = V1(x + y) + V2(x y)
which yields left-running and right-running waves.
Similarly
c2uxx uyy = 0 (1.15) u(x, y) = V1(x + cy) + V2(x cy) (1.16)
6. Consider the eikonal equation
u2x + u2y = 1 (1.17)
Assume, in analogy with the linear case, that
u(x, y) = (x) + (y) (1.18)and which yields (prime is differentiation)
((x))2 + ((y))2 = 1
((x))2 = 1 [(y)]2 = 2 : constant of separation (x) = ; (y) =
1 2
u(x, y) = x + y
1 2 + where is a constant of separation and is a constant of integration.
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DRAFTCHAPTER 2First order PDEs andCharacteristics
Consider a general first order partial differential equation given by
a(x, t)V
x+ b(x, t)
V
t= c(t, x)V + d(x, t) (2.1)
We want a solution subject to the solution equaling a given initial value along
some contour in the (t, x) plane. Let have coordinates x(), t(), as shownin Figure 2.1, and be defined parametrically, namely
: x(); t() : [a, b] (2.2)V() = f() (2.3)
Figure 2.1: : a curve in domain D.
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10 2: First order PDEs and Characteristics
We want to find the unique value of u(t, x) in the domain D so that the
solution agrees with the pre-fixed values on specified by f().Using the method of characteristics, we will study the following three special
types of first order PDEs.
(i) ut + cux = 0 : rigid translation with no damping
(ii) ut + cux = u : rigid translation with damping (no distortion)(iii) ut + uux = 0 : translation with distortion
2.1 Characteristic Curves
A wide class of linear and quasi-linear first order PDEs can be solved by finding
special curves in the xt-plane, called characteristic curves, or characteristics inshort. On these curves, the PDE simplifies to an ordinary differential equation.
The task of solving the PDE decomposes into the task of finding the characteris-
tics and then integrating the ODE along the characteristics from a parametrized
initial value.
Define the characteristic curve, parametrized by s, and defined as follow
dx(s)ds
= a(x, t) ; dt(s)ds
= b(x, t) (2.4)
Figure 2.2 shows a typical characteristic.
(x(s),t(s))
Figure 2.2: x(s) and t(s)
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2.1: Characteristic Curves 11
Then, from Eqs. 2.1 and 2.4
dV
ds=
dx(s)
ds
V
x+
dy(s)
ds
V
y
= aV
x+ b
V
y(2.5)
Hence, we have reduced solving the first order PDE to solving the following
system of ordinary differential equations
x(s)
ds= a ,
dt(s)
ds= b (2.6)
dVds
= cV + d (2.7)
The existence and uniqueness of ODEs, for smooth a,b,c, and d, guaran-tees that exactly one solution curve x(s), t(s), V(s) passes through a given pointx0, t0, V0. To obtain a unique solution, one needs to solve the initial value prob-lem. Recall the curve is characterized by parameter so that
x() ; t() ; V = f() (2.8)
(x(s), t(s))
Figure 2.3: (a) Labeling characteristic curves by intersection with . (b) Char-acteristic (x(s), t(s)) intersection at .
Typically, all the characteristics intersect with the curve , as shown in Figure2.1. For each value of, there is a unique characteristic curve intersecting , asshown in Figure. 2.3(a). 1 Let us label the different characteristic curves by the
different point at which they intersect , as shown for a given characteristic inFigure 2.3(b).
1The case when this is not true is discussed in the last Section of this Chapter.
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12 2: First order PDEs and Characteristics
The coordinates s, provide a coordinate system, equivalent to x, t, that
specify all points on the plane, as shown in Figure ??. For a non-singular coor-dinate transformation from x, t to s, , we need to have a non-zero Jacobian Jdefined by
dxdt = Jdsd
J =x
s
t
x
t
s(2.9)
(t(),x
())
= 1
= 2
= 3
= 4
s = 0
s =1
s = 2
s = 3
(x(, s),t(, s))
u(t,x)
(t, x
)
Figure 2.4: (a) The parameter labels the different characteristic curves by spec-ifying their intersection with . (b) Specifying u(x, t) = u(x(s, ), t(s, )).
2.2 Solution on Characteristics
As discussed in previous Section, every point of the xt-plane can be labeled by(s, ), as shown in Figure 2.4(a), which in turn yields the solution u(s, ), asshown in Figure 2.4(b). The solution of the PDE is uniquely specified on the
xt-plane by specifying V(s, ), as shown in Figure 2.5; the characteristic curvesare labeled by x(s, ), t(s, ) and the solution is V(s, ), such that
x(0, ) = x() (2.10)
t(0, ) = t() (2.11)
V(0, ) = f() (2.12)
On the characteristic curve we have
dV(s)
ds= d(x(s)t(s)) + c(t(s), x(s))V(s) (2.13)
d(s) + c(s)V(s) (2.14)
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2.2: Solution on Characteristics 13
V
t
0
V(0,0 ) = f(0 )
V(s,0)
Figure 2.5: The solution V(s, ) specified on the characteristics.
Using the integrating factor exp( s0
dscs(s)) yields
d
ds(e
s
0dsc(s)V(s)) = e
s
0dsc(s)d(s) (2.15)
Therefore, integrating from 0 to s yields
V(s, ) = V(0) + es0 ds
c(s
)s0
dses
0 ds
a(s
)d(s) (2.16)
Note the solution satisfies the boundary condition since
V(0) = V(x(, 0), t(, 0)) (2.17)
= f(t(), x()) (2.18)
= f() (2.19)
We invert x(, s), t(, s) to obtain = (t, x), s = s(t, x), and obtain the sought
after solution
V(x, t) = V(s(x, t), (x, t)) (2.20)
Consider the special case of c and d beings constants; we have
V(t, x) = f(t, x) +d
c(1 ecs) (2.21)
where s = s(t, x)
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14 2: First order PDEs and Characteristics
2.3 Rigid Translation
We discuss two cases of rigid translation of a wave form.
With damping.
Without damping.
Rigid translation with no damping
Consider the case of rigid translation, namely
cVx + Vt = 0 (2.22)
with initial vaue V(0, x) = f(x). The characteristic curve is given by
dx
ds= c ,
dt
ds= 1 (2.23)
and
dV
ds= 0 : constant on characteristics (2.24)
Let be the boundary value curve parameterized by such that
x = , t = 0 , V = f() (2.25)
Then, from Eqs. 2.23 and 2.25
x(s, ) = cs + ; t(s, ) = s (2.26)
and
V(s, ) = f() (2.27)
Solving for s, yields
s = t ; = x ct (2.28)
Hence the solution of the first order PDE is given by
V(s, ) = f((x, t)) = f(x ct) (2.29)
The initial value f() propagates unchanged on the characteristic curve.
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2.3: Rigid Translation 15
Rigid translation with damping
cux + uy = u ; u(x, 0) = f(x) (2.30)
Boundary curve and value is given by
= {(x, y) : y = 0, x = , } (2.31)u(x, 0) = u(, 0) = f() (2.32)
ey
Figure 2.6: Wave dissipating while propagating along the x-direction
The characteristic curves are given by
dx
ds= c ;
dy
ds= 1 (2.33)
x = cs + ; y = s (2.34)
s = y ; = x cy (2.35)
On the characteristic curve
duds
= u u(s, ) = u(0, )es = f()es
Converting from s, to x, t yields the solution
u(x, y) = f(x cy)ey
As shown in Figure 2.6, u(x, y) is a traveling wave that is exponentially damped.
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16 2: First order PDEs and Characteristics
2.4 Linear PDEs
Consider the linear PDE
ux + 3x2uy = 2xu (2.36)
The boundary curve is given by parametrized by , namely
= {x(), y()| [a, b]} (2.37)
The boundary value is given by
u(x, y) = g() (2.38)
The characteristic curve is fixed by
dx
ds= 1 ;
dy
ds= 3x2 (2.39)
The characteristic curves are given by
x(, s) = s + x() ; y(, s) = s3 + y() (2.40)
The characteristic curves intersect at s = 0 and hence
x(, 0) = x() (2.41)
y(, 0) = y() (2.42)
a
b
x
y
Figure 2.7: Boundary curve : x = 0; a < y < b.
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2.4: Linear PDEs 17
Define a specific by
= {(x, y) : x = 0, a < y < b} x = 0 , y() = ; a < < b
and hence the characteristics are given by
x(, s) = s (2.43)
y(, s) = s3 + ; a < < b (2.44)
Inverting Eq. 2.40 yields
s(x, y) = x (2.45)
(x, y) = y x3 (2.46)
On the characteristic curve
du
ds= 2su du
u= 2s (2.47)
Note the characteristic curves given in Eq. 2.40 are labeled by , which specifiesthe boundary curve . On characteristic curves
u = k()es2 = k()ex2 (2.48)
Boundary Condition on yields
u() = u(0, y) = g() = k (2.49)
or k = g(y x3) (2.50)
Hence,
u(x, y) = g(y x3)ex2 (2.51)
Verification
uy =u
y= ex
2
g(y x3) (2.52)
ux =u
x= [3g(y x3)x2 + 2xg(y x3)]ex2 (2.53)
ux + 3x2uy = 2xe
x2g(y x3) = 2xu (2.54)
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18 2: First order PDEs and Characteristics
2.5 Fan-like Characteristics
Consider the PDE
xux + yuy = 1 + y2 (2.55)
with boundary curve = {(x, y); y = 1, x = , } and boundaryvalue given by
u(x, 1) = f() = + 1 (2.56)
The characteristic curve equation is given by
dxds
= x , dyds
= y (2.57)
x(, s) = h1()es ; y(, s) = h2()e
s (2.58)
The boundary conditions are
: x(, 0) = = h1() (2.59)
y(, 0) = 1 = h2() (2.60)
x(, s) = es ; y(, s) = es (2.61)
and yields the characteristics
x(, s) = es ; y(, s) = es (2.62)
x = yThe characteristics are shown in Figure 2.8. Inverting Eq. 2.62 yields
s = ln y ; =x
y(2.63)
From Eq. 2.55
du
ds= 1 + y2 = 1 + e2s
u = c + s +1
2e2s
u(, 0) = c +1
2= + 1
c = + 12
=x
y+
1
2
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2.6: Expansion Wave 19
y = 1
x= y
Figure 2.8: Fan-like characteristic curves x = y.
Hence
u =x
y+
1
2+ ln y +
1
2y2 (2.64)
To verify the solution note
ux =1
y; uy =
1
y+ y x
y2(2.65)
Therefore
xux + yuy = 1 + y2 (2.66)
2.6 Expansion Wave
Consider the PDE
ut + uux = 0 (2.67)
with
u(x, 0) = h1, x > 0h2, x < 0.
where h2 < h1. This equation emerges in the study of a bursting dam (or mem-brane), with the water from the burst dam forming what is called an expansion
wave. The boundary curve is
=
(x, t) : t = 0, x = ,
(2.68)
Define x(s, ), t(s, ) by
dt
ds= 1,
dx
ds= u (2.69)
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20 2: First order PDEs and Characteristics
h2
h1
x
u(x,0)
Figure 2.9: Initial value u(x, 0).
and hence
du(s, )
ds= 0 u(s, ) = u(0, ) (2.70)
From Equation 2.69 we have
t = s, x(s, ) = su(s, ) + (2.71)
x
t x=h2t
x=h1t
u=h1
u=h2
Figure 2.10: Solution of the PDE with no characteristics in the wedge.
The initial condition u(x, 0) = f(x) yields
u(0, ) = f() =
h1; > 0h2; < 0.
Hence, the characteristic curves are
x(s, ) = su(0, ) + =
h1s + ; > 0h2s + ; < 0.
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2.6: Expansion Wave 21
The characteristic are hence straight lines and shown in Figure 2.10. The solution
is given by
u(x, t) =
h1, x > h1t ; ( > 0)h2, x < h2t ; ( < 0).
Note that domain bounded by x = h2t and x = h1t has no characteristics.The reason is the discontinuity at x = 0 in the initial condition. In particular, weneed characteristics for h2t < x < h1t.
x
t
x=h2t
x=h1t
t0
xx=h2t0 x=h1t0
u=h1
u=h2
u=x/t0
u
Figure 2.11: (a) Characteristic curves inside the fan. (b) The solution u(x, t0),including the portion inside the fan.
One assumes that all the values in interval [h1, h2] are present at for u(x, 0)at point x = 0 and thus yields the characteristics that fill up the blank domain.2
Namely
dx
ds= u x = us ; h1 < u < h2 (2.72)
Hence, along the characteristic
u =x
s=
x
t, h
2t < x < h
1t (2.73)
This yields a fan-like structure for characteristics given in Figure 2.11(a); the
solution is an expansion wave given by
u(x, t) =
h1 ; x > h1tx/t ; h2t < x < h1t
h2 ; x < h2t.2By considering the case of a continuous initial condition, it can be shown that our assump-
tion emerges by taking the limit of a discontinous initial condition at x = 0.
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22 2: First order PDEs and Characteristics
The values of u(x, t) for a value of t > 0 is shown in Figure 2.11. As can be
seen from Figure 2.11(a) many characteristics emerge from the origin x = 0 =t; Figure 2.11(b) shows how, for a typical time t0 > 0, the solution u(x, t0)smoothly increases from u = h2 for x < 0 to u = h1 for x > 0.
2.7 Traffic Flow
To exemplify the interpretation of characteristics, consider the simple model of
traffic flow, with density and velocity of of cars, denoted by , u respectively arerelated due to conservation of number of cars by the following first order
partial differential equation
t + (u)x = 0 ; t > 0 ; < x <
After a long red light at x = 0, the roadway is empty for x > 0, and is full ofstationary bumper-to-bumper cars for x < 0. At time t = 0 the light turns green.The initial data, as shown in Figure 2.12(a) is given by3
(x, 0) =
m x < 00 x > 0.
corresponds to a stop light at x = 0, which turns green at t = 0.
0 x
(x,0)
m
-1/n 1/n x
n(x,0)
m
Figure 2.12: (a) Initial traffic distribution. (b) Initial traffic distribution as a limit
of a continuous distribution.
As the signals effect propagates up road, the cars accelerate in turn. One canask: at what time will the car at x = a be able to start moving? and where andwhen will this car attain speed V /2? We analyze the characteristics to addressthese and similar questions.
The traffic flow equation is given by
t+
x(u) = 0, t > 0, < x < . (2.74)
3The initial discontinuous traffic distribution given in Figure 2.12(a) can be obtained as the
limit, ofn , of the initial distribution shown in Figure 2.12(b).
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2.7: Traffic Flow 23
We assume the linear velocity-density relation(empirical):
u = V(1 m
) (2.75)
Thus when = 0 (open road) the car travels at velocity V; when = m(traffic jam) the car does not move.
Substituting in and differentiating we have the standard form for the PDE
given by
t + c()x = 0 (2.76)
where
c() = V(1 2m
). (2.77)
The initial distribution ofc() is shown in Figure 2.13.
0 x
c/((x,0))=V[1-(2/m)(x,0)]
V
-V
Figure 2.13: The initial distribution ofc() = c((x, 0).
This quasi-linear first-order PDE has two simplifying features. The coeffi-
cient of unity in front of t allows us to forego parametrization, with t = s.The homogeneity of the equation means that along a characteristic (x, t) is aconstant.
Hence the equation of the characteristic in the (x, t)-plane yields
dt
ds= 1 ;
dx
dt= c();
d
dt= 0 ; (x, t) = (x, 0) (2.78)
dxdt
= c((x, 0)) = constant (2.79)
Hence, all the characteristics are straight lines, with slopes depending upon
the (constant) value of(x, 0).Graphs of(x, 0) and c
(x, 0)
versus x appear below, as well as a sketch of
the characteristics in the (x, t)-plane. From the sketch it is evident that (x, t) =
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24 2: First order PDEs and Characteristics
0 for x > V t and (x, t) = m for x x0
(2.125)
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32 2: First order PDEs and Characteristics
The boundary curve is given by
= {(x, 0)|t() = 0; x() = , +}and hence
f(x) = f() =
0, 0 < < x01, > x0
The characteristics, shown in Figure 2.19(b), are given by
dx(s, )
ds= et;
dt(s, )
ds= 1
x(s, ) = 1 + es
; t = s t = ln(x c) ; c = 1
On the characteristics, the PDE yields
du(s, )
ds= 0 ; u(s, ) = u(0, ) = f()
Expressing s, in terms ofx, t yields the solution
u(x, t) = 0, x < et + x0 11, x > et + x0
1
As shown in Figure 2.19(b), the discontinuity in f(x) at x0 creates a disconti-nuity in u(x, t) that propagates along the characteristic curve passing throughx0.
2.11 Intersection of Characteristic Curves
The intersection of characteristic curves creates discontinuities since on each
characteristic curve, the PDE propagates the value picked up from the boundary
value specified by , with a typical case shown in Figure 2.20.
Hence along the intersection, the value of u(x, y) has to become multi-valued. What this means is that the value ofu(x, y) has discontinuities acrossthe line of intersection.5
Consider the PDE
u
x+ u
u
y= 0 (2.126)
5Note for the case when the solution u(x, y) has discontinuities (shock waves) the equiva-lence of Eqs. 2.99 and 2.98, no longer holds. The intersection of characteristic curves give rise
to discontinuous solutions.
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2.11: Intersection of Characteristic Curves 33
Figure 2.20: Intersection of characteristic curves
with initial condition, shown in Figure 2.21(a), as follows (a > 1)
u(x, 0) = f(x) =
1 x2, < x < a1 a2, x > a (2.127)
f(x)
a
x=1
x= axs
= 1
x= 1
x< a x> a
= 0 = 1 = 2
ts
Figure 2.21: (a) Initial value f(x). (b) Intersecting characteristic curves.
From Eq. 2.105
u(x, y) = xy 1
1 + 4y(y x)2y2 (2.128)Recall, from Eq. 2.91, that the characteristic curves are
x(s, ) = su(s, ) + (2.129)
y(s, ) = s (2.130)
Since
u(s, ) = u(0, ) = f() (2.131)
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34 2: First order PDEs and Characteristics
we have
x(, s) = sf() + ; y(, s) = s (2.132)
Hence, for < a, the characteristic curves are given by
x(s, ) = s(1 2) + ; y(s, ) = s (2.133) x y(1 2) = (2.134)
and for > a the characteristic curves are given by
x y(1 a2) = , a > 1 (2.135)The intersecting characteristic curves are shown in Figure 2.21(b), with the first
two characteristic curves to intersect being the one for = a coming from theleft and the right.
2.12 Shock Waves
Consider the initial value problem
(x, t)
t+ 2
(x, t)
t= 0 (2.136)
with initial value
(x, 0) = f(x) =
4, x < 03, x > 0
(2.137)
3
4
Figure 2.22: Initial value of(x, 0).
The characteristics are given by
dt(, s)
ds= 1 ;
dx(, s)
ds= 2 (2.138)
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2.12: Shock Waves 35
Hence
d
ds= 0 (, s) = (, 0) = f() (2.139)
t(, s) = s + h1() ; x(, s) = 2f()s + h2() (2.140)
The initial curve is given by
= {(x, t) : x = , t = 0; } (2.141)Hence
t(, s) = s ; x(, s) = 2f()s + (2.142)
or, x(, s) = 2f()t(, s) + : characteristics (2.143)
The characteristics, shown in Figure 2.23(a), can be seen to intersect.
tS0
xS0
1
2
Figure 2.23: (a) Intersecting characteristic curves define the shock front. (b) The
time and position xs0 , ts0 at which the shock starts.
The explicit expression for the characteristics are given by
x = 2f()s + = 8t + , < 06t + , > 0
(2.144)
The shock starts at xs0, ts0 , as shown in Figure 2.23(b), where ts0 is the firstinstant when the characteristics intersect. To find ts0 note that, we have, in gen-eral
xs0 = g(1)ts0 + 1
xs0 = g(2)ts0 + 2
ts0 = 2 1
g(2) g(1)
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36 2: First order PDEs and Characteristics
The first characteristic curve, after 1, to intersect with 1 is clearly 1+. Hence,
ts0 = 1
g(1)
xs0 = 1 g(1)
g(1)= 1 + g(1)ts0
Note since ts0>0, for a shock wave to form we must have
g(1) < 0 (2.145)
Conversely, ifg()>0 for all , there is no shock front formation, and the waveis said to be an expansion wave.
For our case 1 < 0 and 2 > 0; hence, from Eq. 2.144
g(1) = 2f(1) = 8 ; 1 = 0 (2.146)g(2) = 2f(2) = 6 ; 1 = 0+ (2.147)
and hence
ts0 = 2 16 8 =
1
2(2 1) 0 (2.148)
and xs0 = 1 + 0 0 (2.149)In summary the shock starts at the origin, namely (xs = 0, ts = 0) and has aconstant velocity.
Shock front; shock velocity
We rewrite the PDE as follows
0 =
t+ c()
x=
t+
q
x(2.150)
where c =q
The PDE breaks down at the intersection of the characteristics giving mul-
tiple solutions at the intersection. However, it is assumed that the conserved
quantities continue to be conserved at the intersection. In effect, what this meansis that we assume the integrated version of the PDE continues to be valid at the
intersection.
Consider the PDE in the range of a < x < b. We expect that the integratedversion of Eq. 2.150 holds across the shock wave (discontinuity), namely
0 =
ba
dx
t+
q
x
q(a, t) q(b, t) =ba
dx
t=
d
dt
ba
dx(x, t)
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2.12: Shock Waves 37
xs
xs+
xs
Figure 2.24: Shock located at xs; value to the left xs and right xs+ ofxs.
xs(t) = 7t
Figure 2.25: Shock wave located at xs(t) = 7t.
Since there is a discontinuity in the density , we break up the integral andcompute the following
q(a, t) q(b, t) = ddt
xs(t)a
dx(x, t) +
bxs(t)
dx(x, t)
=dxs(t)
dt(xs, t) +
xs(t)a
dx
t dxs(t)
dt(xs+, t) +
bxs(t)
dx
t
=dxsdt
[(xs , t) (xs+, t)] +
q(a, t) q(xs, t) q(xs+, t) q(b, t)
where xs and xs+ are defined in Figure 2.24.Hence
dxsdt
= q(xs, t) q(xs+, t)(xs, t) (xs+, t)
(2.151)
Consider the following PDE
t+ 2
t= 0
We have q = 2 and hence
dxsdt
=2() 2(+)
() (+) =42 324 3 = 7
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38 2: First order PDEs and Characteristics
The shock front trajectory, shown in Figure 2.25, is a straight line given by
xs(t) =dxs(t)
dt(t ts0) + xs0
Since xs0 = 0 = ts0 and dxs/dt = 7, we obtain
xs(t) = 7t