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PARTIAL DIFFERENTIAL EQUATIONS Math 124A
创建日期：2024-04-15 15:59:26
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标签： Math 124A
PARTIAL DIFFERENTIAL EQUATIONS

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PARTIAL DIFFERENTIAL EQUATIONS
Math 124A

1 Introduction

Recall that an ordinary differential equation (ODE) contains an independent variable x and a dependent
variable u, which is the unknown in the equation. The defining property of an ODE is that derivatives
of the unknown function u′ = du
dx
enter the equation. Thus, an equation that relates the independent
variable x, the dependent variable u and derivatives of u is called an ordinary differential equation. Some
examples of ODEs are:
u′(x) = u
u′′ + 2xu = ex
u′′ + x(u′)2 + sinu = lnx
In general, and ODE can be written as F (x, u, u′, u′′, . . . ) = 0.
In contrast to ODEs, a partial differential equation (PDE) contains partial derivatives of the depen-
dent variable, which is an unknown function in more than one variable x, y, . . . . Denoting the partial
derivative of ?u
?x
= ux, and
?u
?y
= uy, we can write the general first order PDE for u(x, y) as
F (x, y, u(x, y), ux(x, y), uy(x, y)) = F (x, y, u, ux, uy) = 0. (1.1)
Although one can study PDEs with as many independent variables as one wishes, we will be primar-
ily concerned with PDEs in two independent variables. A solution to the PDE (1.1) is a function
u(x, y) which satisfies (1.1) for all values of the variables x and y. Some examples of PDEs (of physical
significance) are:
ux + uy = 0 transport equation (1.2)
ut + uux = 0 inviscid Burger’s equation (1.3)
uxx + uyy = 0 Laplace’s equation (1.4)
utt ? uxx = 0 wave equation (1.5)
ut ? uxx = 0 heat equation (1.6)
ut + uux + uxxx = 0 KdV equation (1.7)
iut ? uxx = 0 Shro¨dinger’s equation (1.8)
It is generally nontrivial to find the solution of a PDE, but once the solution is found, it is easy to
verify whether the function is indeed a solution. For example to see that u(t, x) = et?x solves the wave
equation (1.5), simply substitute this function into the equation:
(et?x)tt ? (et?x)xx = et?x ? et?x = 0.
1.1 Classification of PDEs
There are a number of properties by which PDEs can be separated into families of similar equations.
The two main properties are order and linearity.
Order. The order of a partial differential equation is the order of the highest derivative entering the
equation. In examples above (1.2), (1.3) are of first order; (1.4), (1.5), (1.6) and (1.8) are of second
order; (1.7) is of third order.
Linearity. Linearity means that all instances of the unknown and its derivatives enter the equation
linearly. To define this property, rewrite the equation as
Lu = 0, (1.9)
where L is an operator, which assigns u a new function Lu. For example L = ?2
?x2
+1, then Lu = uxx+u.
The operator L is called linear if
L(u+ v) = Lu+ Lv, and L(cu) = cLu (1.10)
1
for any functions u, v and constant c. The equation (1.9) is called linear, if L is a linear operator. In
our examples above (1.2), (1.4), (1.5), (1.6), (1.8) are linear, while (1.3) and (1.7) are nonlinear (i.e.
not linear). To see this, let us check, e.g. (1.6) for linearity:
L(u+ v) = (u+ v)t ? (u+ v)xx = ut + vt ? uxx ? vxx = (ut ? uxx) + (vt ? vxx) = Lu+ Lv,
and
L(cu) = (cu)t ? (cu)xx = cut ? cuxx = c(ut ? uxx) = cLu.
So, indeed, (1.6) is a linear equation, since it is given by a linear operator. To understand how linearity
can fail, let us see what goes wrong for equation (1.3):
L(u+v) = (u+v)t+(u+v)(u+v)x = ut+vt+(u+v)(ux+vx) = (ut+uux)+(vt+vvx)+uvx+vux 6= Lu+Lv.
You can check that the second condition of linearity fails as well. This happens precisely due to the
nonlinearity of the uux term, which is quadratic in “u and its derivatives”.
Notice that for a linear equation, if u is a solution, then so is cu, and if v is another solution, then
u+ v is also a solution. In general any linear combination of solutions
c1u1(x, y) + c2u2(x, y) + · · ·+ cnun(x, y) =
n∑
i=1
ciui(x, y)
will also solve the equation.
The linear equation (1.9) is called homogeneous linear PDE, while the equation
Lu = g(x, y) (1.11)
is called inhomogeneous linear equation. Notice that if uh is a solution to the homogeneous equation
(1.9), and up is a particular solution to the inhomogeneous equation (1.11), then uh+up is also a solution
to the inhomogeneous equation (1.11). Indeed
L(uh + up) = Luh + Lup = 0 + g = g.
Thus, in order to find the general solution of the inhomogeneous equation (1.11), it is enough to find
the general solution of the homogeneous equation (1.9), and add to this a particular solution of the
inhomogeneous equation (check that the difference of any two solutions of the inhomogeneous equation
is a solution of the homogeneous equation). In this sense, there is a similarity between ODEs and PDEs,
since this principle relies only on the linearity of the operator L.
1.2 Examples
Example 1.1. ux = 0
Remember that we are looking for a function u(x, y), and the equation says that the partial derivative
of u with respect to x is 0, so u does not depend on x. Hence u(x, y) = f(y), where f(y) is an arbitrary
function of y. Alternatively, we could simply integrate both sides of the equation with respect to x.
More on this in the following examples.
Example 1.2. uxx + u = 0
Similar to the previous example, we see that only the partial derivative with respect to one of the
variables enters the equation. In such cases we can treat the equation as an ODE in the variable in which
partial derivatives enter the equation, keeping in mind that the constants of integration may depend on
the other variables. Rewrite the equation as
uxx = ?u,
which, as an ODE, has the general solution
u = c1 cosx+ c2 sinx.
2
Since the constants may depend on the other variable y, the general solution of the PDE will be
u(x, y) = f(y) cosx+ g(y) sinx,
where f and g are arbitrary functions. To check that this is indeed a solution, simply substitute the
expression back into the equation.
Example 1.3. uxy = 0
We can think of this equation as an ODE for ux in the y variable, since (ux)y = 0. Then similar to
the first example, we can integrate in y to obtain
ux = f(x).
This is an ODE for u in the x variable, which one can solve by integrating with respect to x, arriving
at at the solution
u(x, y) = F (x) +G(y).
1.3 Conclusion
Notice that where the solution of an ODE contains arbitrary constants, the solution to a PDE contains
arbitrary functions. In the same spirit, while an ODE of order m has m linearly independent solutions, a
PDE has infinitely many (there are arbitrary functions in the solution!). These are consequences of the
fact that a function of two variables contains immensely more (a whole dimension worth) of information
than a function of only one variable.
3
Problem Set 1
1. (#1.1.2 in [Str]) Which of the following operators are linear?
(a) Lu = ux + xuy
(b) Lu = ux + uuy
(c) Lu = ux + u2y
(d) Lu = ux + uy + 1
(e) Lu = √1 + x2(cos y)ux + uyxy ? [arctan(x/y)]u
2. (#1.1.3 in [Str]) For each of the following equations, state the order and whether it is nonlinear,
linear inhomogeneous, or linear homogeneous; provide reasons.
(a) ut ? uxx + 1 = 0
(b) ut ? uxx + xu = 0
(c) ut ? uxxt + uux = 0
(d) utt ? uxx + x2 = 0
(e) iut ? uxx + u/x = 0
(f) ux(1 + u
2
x)
?1/2 + uy(1 + u2y)
?1/2 = 0
(g) ux + e
yuy = 0
(h) ut + uxxxx +
√
1 + u = 0
3. Show that cos(x? ct) is a solution of ut + cux = 0.
4. (#1.1.10 in [Str]) Show that the solutions of the differential equation u′′′ ? 3u′′ + 4u = 0 form a
vector space. Find a basis of it.
5. (#1.1.11 in [Str]) Verify that u(x, y) = f(x)g(y) is a solution of the PDE uuxy = uxuy for all pairs
of (differentiable) functions f and g of one variable.
6. (#1.1.12 in [Str]) Verify by direct substitution that
un(x, y) = sinnx sinhny
is a solution of uxx + uyy = 0 for every n > 0.
7. Find the general solution of
uxy + ux = 0.
(Hint: first treat it as an ODE for ux).
2 First-order linear equations
Last time we saw how some simple PDEs can be reduced to ODEs, and subsequently solved using ODE
methods. For example, the equation
ux = 0 (2.1)
has “constant in x” as its general solution, and hence u depends only on y, thus u(x, y) = f(y) is the
general solution, with f an arbitrary function of a single variable. Today we will see that any linear
first order PDE can be reduced to an ordinary differential equation, which will then allow as to tackle
it with already familiar methods from ODEs.
Let us start with a simple example. Consider the following constant coefficient PDE
aux + buy = 0. (2.2)
Here a and b are constants, such that a2 +b2 6= 0, i.e. at least one of the coefficients is nonzero (otherwise
this would not be a differential equation). Using the inner (scalar or dot) product in R2, we can rewrite
the left hand side of (2.2) as
(a, b) · (ux, uy) = 0, or (a, b) · ?u = 0.
Denoting the vector (a, b) = v, we see that the left hand side of the above equation is exactly Dvu(x, y),
the directional derivative of u in the direction of the vector v. Thus the solution to (2.2) must be
constant in the direction of the vector v = ai + bj.
v = Ha, bL
-5 5
x
-5
5
y
Figure 2.1: Characteristic lines bx? ay = c.
Ξ
Η
Ha, bL
H-b, aL
Hx, yL
Ξ = Hx, yL Ha, bL
Η = Hx, yL H-b, aL
?
-10 -5 5
x
-5
5
y
Figure 2.2: Change of coordinates.
The lines parallel to the vector v have the equation
bx? ay = c, (2.3)
since the vector (b,?a) is orthogonal to v, and as such is a normal vector to the lines parallel to v. In
equation (2.3) c is an arbitrary constant, which uniquely determines the particular line in this family of
parallel lines, called characteristic lines for the equation (2.2).
As we saw above, u(x, y) is constant in the direction of v, hence also along the lines (2.3). The line
containing the point (x, y) is determined by c = bx? ay, thus u will depend only on bx? ay, that is
u(x, y) = f(bx? ay), (2.4)
5
where f is an arbitrary function. One can then check that this is the correct solution by plugging it
into the equation. Indeed,
a?xf(bx? ay) + b?yf(bx? ay) = abf ′(bx? ay)? baf ′(bx? ay) = 0.
The geometric viewpoint that we used to arrive at the solution is akin to solving equation (2.1) simply
by recognizing that a function with a vanishing derivative must be constant. However one can approach
equation (2.2) from another perspective, by trying to reduce it to an ODE.
2.1 The method of characteristics
To have an ODE, we need to eliminate one of the partial derivatives in the equation. But we know that
the directional derivative vanishes in the direction of the vector (a, b). Let us then make a change of the
coordinate system to one that has its “x-axis” parallel to this vector, as in Figure 2. In this coordinate
system
(ξ, η) = ((x, y) · (a, b), (x, y) · (b,?a)) = (ax+ by, bx? ay).
So the change of coordinates is {
ξ = ax+ by,
η = bx? ay. (2.5)
To rewrite the equation (2.2) in this coordinates, notice that
ux = uξ
?ξ
?x
+ uη
?η
?x
= auξ + buη,
uy = uξ
?ξ
?y
+ uη
?η
?y
= buξ ? auη.
Thus,
0 = aux + buy = a(auξ + buη) + b(buξ ? auη) = (a2 + b2)uξ.
Now, since a2 + b2 6= 0, then, as we anticipated,
uξ = 0,
which is an ODE. We can solve this last equation just as we did in the case of equation (2.1), arriving
at the solution
u(ξ, η) = f(η).
Changing back to the original coordinates gives u(x, y) = f(bx? ay). This is the same solution that we
derived with the geometric deduction. This method of reducing the PDE to an ODE is called the method
of characteristics, and the coordinates (ξ, η) given by formulas (2.5) are called characteristic coordinates.
Example 2.1. Find the solution of the equation 3ux ? 5uy = 0 satisfying the condition u(0, y) = sin y.
From the above discussion we know that u will depend only on η = ?5x?3y, so u(x, y) = f(?5x?3y).
The solution also has to satisfy the additional condition (called initial condition), which we verify by
plugging in x = 0 into the general solution.
sin y = u(0, y) = f(?3y).
So f(z) = sin(? z
3
), and hence u(x, y) = sin
(
5x+3y
3
)
, which one can verify by substituting into the
equation and the initial condition.
6
2.2 General constant coefficient equations
We can easily adapt the method of characteristics to general constant coefficient linear first-order equa-
tions
aux + buy + cu = g(x, y). (2.6)
Recall that to find the general solution of this equation it is enough to find the general solution of the
homogeneous equation
aux + buy + cu = 0, (2.7)
and add to this a particular solution of the inhomogeneous equation (2.6). Notice that in the charac-
teristic coordinates (2.5), equation (2.7) will take the form
(a2 + b2)uξ + cu = 0, or uξ +
c
a2 + b2
u = 0,
which can be treated as an ODE in ξ. The solution to this ODE has the form
uh(ξ, η) = e
? c
a2+b2
ξ
f(η),
with f again being an arbitrary single-variable function. Changing the coordinates back to the original
(x, y), we will obtain the general solution to the homogeneous equation
uh(x, y) = e
? c(ax+by)
a2+b2 f(bx? ay).
You should verify that this indeed solves equation (2.7).
To find a particular solution of (2.6), we can use the characteristic coordinates to reduce it to the
inhomogeneous ODE
(a2 + b2)uξ + cu = g(ξ, η), or uξ +
c
a2 + b2
u =
g(ξ, η)
a2 + b2
.
Having found the solution to the homogeneous ODE, we can find the solution to this inhomogeneous
equation by e.g. variation of parameters. So the particular solution will be
up = e
? c
a2+b2
ξ
?
g(ξ, η)
a2 + b2
e
c
a2+b2
ξ
dξ.
The general solution of (2.6) is then
u(ξ, η) = uh + up = e
? c
a2+b2
ξ
(
f(η) +
?
g(ξ, η)
a2 + b2
e
c
a2+b2
ξ
dξ
)
.
To find the solution in terms of (x, y), one needs to first carry out the integration in ξ in the above
formula, then replace ξ and η by their expressions in terms of x and y.
Example 2.2. Find the general solution of ?2ux + 4uy + 5u = ex+3y.
The characteristic change of coordinates for this equation is given by{
ξ = ?2x+ 4y,
η = 4x+ 2y.
From these we can also find the expressions of x and y in terms of (ξ, η). In particular notice that
x+ 3y = ξ+η
2
. In the characteristic coordinates the equation reduces to
20uξ + 5u = e
ξ+η
2 .
7
The general solution of the homogeneous equation associated with the above equation is
uh = e
? 1
4
ξf(η),
and the particular solution will be
up = e
? 1
4
ξ
?
e
ξ+η
2
20
e
1
4
ξ dξ = e?
1
4
ξ · 1
15
e
η
2 e
3
4
ξ = e?
1
4
ξ · 1
15
e
1
4
(3ξ+2η).
Adding these two will give the general solution to the inhomogeneous equation
u(ξ, η) = e?
1
4
ξ
(
f(η) +
1
15
e
1
4
(3ξ+2η)
)
.
Finally, substituting the expressions for ξ and η in terms of (x, y), we will obtain the solution
u(x, y) = e?
1
4
(2x?4y)
(
f(4x+ 2y) +
1
15
e
1
4
(2x+16y)
)
.
You should check that this indeed solves the differential equation.
2.3 Variable coefficient equations
The method of characteristics can be generalized to variable coefficient first-order linear PDEs as well,
albeit the change of variables may no longer be orthogonal. The general variable coefficient linear
first-order equations is
a(x, y)ux + b(x, y)uy + c(x, y)u = d(x, y). (2.8)
Let us first consider the following simple particular case
ux + yuy = 0. (2.9)
Using our geometric intuition from the constant coefficient equations, we see that the directional deriva-
tive of u in the direction of the vector v = (1, y) is constant. Notice that the vector v itself is no longer
constant, and varies with y. The curves that have v as their tangent vector have slope y
1
, and thus satisfy
dy
dx
=
y
1
.
We can solve this equation as an ODE, and obtain the general solution
y = Cex, or e?xy = C. (2.10)
As in the case of the constant coefficients, the solution to the equation (2.9) will be constant along these
curves, called characteristic curves. This family of non-intersecting curves fills the entire coordinate
plane, and the curve containing the point (x, y) is uniquely determined by C = e?xy, which implies that
the general solution to (2.9) is
u(x, y) = f(C) = f(e?xy).
As always, we can check this by substitution.
ux + yuy = ?f ′(e?xy)e?xy + yf ′(e?xy)e?x = 0.
Let us now try to generalize the method of characteristics to the equation
a(x, y)ux + b(x, y)uy = 0. (2.11)
8
The idea is again to introduce new coordinates (ξ, η), which will reduce (2.11) to an ODE. Suppose{
ξ = ξ(x, y),
η = η(x, y)
(2.12)
gives such a change of variables. To rewrite the equation in this coordinates, we compute
ux = uξξx + uηηx,
uy = uξξy + uηηy,
and substitute these into equation (2.11) to get
(aξx + bξy)uξ + (aηx + bηy)uη = 0.
Since we would like this to give us an ODE, say in ξ, we require the coefficient of uη to be zero,
aηx + bηy = 0.
Without loss of generality, we may assume that a 6= 0 (locally). Notice that for curves y(x) that have
the slope dy
dx
= b
a
we have
d
dx
η(x, y(x)) = ηx + ηy
dy
dx
= ηx +
b
a
ηy = 0.
So the characteristic curves, just as before, are given by
dy
dx
=
b
a
. (2.13)
The general solution to this ODE will be η(x, y) = C, with ηy 6= 0 (otherwise ηx = 0 as well, and this
will not be a solution). This is how we find the new variable η, for which our PDE reduces to an ODE.
We choose ξ(x, y) = x as the other variable. For this change of coordinates the Jacobian determinant is
J =
?(ξ, η)
?(x, y)
=
ξx ξy
ηx ηy
= ηy 6= 0.
Thus, (2.12) constitutes a non-degenerate change of coordinates. In the new variables equation (2.11)
reduces to
a(ξ, η)uξ = 0, hence uξ = 0,
which has the solution
u = f(η).
The general variable coefficient equation (2.8) in these coordinates will reduce to
a(ξ, η)uξ + c(ξ, η)u = d(ξ, η),
which is called the canonical form of equation (2.8). This equation, as in previous cases, can be solved
by standard ODE methods.
Example 2.3. Find the general solution of the equation
xux ? yuy + y2u = y2, x, y 6= 0.
Condition (2.13) in this case is dy
dx
= ? y
x
. This is a separable ODE, which can be solved to obtain the
general solution xy = C. Thus, our change of coordinates will be{
ξ = x,
η = xy.
9
In these coordinates the equation takes the form
ξuξ +
η2
ξ2
u =
η2
ξ2
, or uξ +
η2
ξ3
u =
η2
ξ3
.
Using the integrating factor
e
′ η2
ξ3
dξ
= e
? η2
2ξ2 ,
the above equation can be written as (
e
? η2
2ξ2 u
)
ξ
= e
? η2
2ξ2
η2
ξ3
.
Integrating both sides in ξ, we arrive at
e
? η2
2ξ2 u =
?
e
? η2
2ξ2
η2
ξ3
dξ = e
? η2
2ξ2 + f(η).
Thus, the general solution will be given by
u(ξ, η) = e
η2
2ξ2
(
f(η) + e
? η2
2ξ2
)
= e
η2
2ξ2 f(η) + 1.
Finally, substituting the expressions of ξ and η in terms of (x, y) into the solution, we obtain
u(x, y) = f(xy)e
y2
2 + 1.
One should again check by substitution that this is indeed a solution to the PDE.
2.4 Conclusion
The method of characteristics is a powerful method that allows one to reduce any first-order linear PDE
to an ODE, which can be subsequently solved using ODE techniques. We will see in later lectures that
a subclass of second order PDEs – second order hyperbolic equations can be also treated with a similar
characteristic method.
10
3 Method of characteristics revisited
3.1 Transport equation
A particular example of a first order constant coefficient linear equation is the transport, or advection
equation ut + cux = 0, which describes motions with constant speed. One way to derive the transport
equation is to consider the dynamics of the concentration of a pollutant in a stream of water flowing
through a thin tube at a constant speed c.
Let u(t, x) denote the concentration of the pollutant in gr/cm (unit mass per unit length) at time t.
The amount of pollutant in the interval [a, b] at time t is then
? b
a
u(x, t) dx.
Due to conservation of mass, the above quantity must be equal to the amount of the pollutant after
some time h. After the time h, the pollutant would have flown to the interval [a+ ch, b+ ch], thus the
conservation of mass gives ? b
a
u(x, t) dx =
? b+ch
a+ch
u(x, t+ h) dx.
To derive the dynamics of the concentration u(x, t), differentiate the above identity with respect to b to
get
u(b, t) = u(b+ ch, t+ h).
Notice that this equation asserts that the concentration at the point b at time t is equal to the con-
centration at the point b + ch at time t + h, which is to be expected, due to the fact that the water
containing the pollutant particles flows with a constant speed. Since b is arbitrary in the last equation,
we replace it with x. Now differentiate both sides of the equation with respect to h, and set h equal to
zero to obtain the following differential equation for u(x, t).
0 = cux(x, t) + ut(x, t),
or
ut + cux = 0. (3.1)
Since equation (3.1) is a first order linear PDE with constant coefficients, we can solve it by the
method of characteristics. First, we rewrite the equation as
(1, c) · ?u = 0,
which implies that the slope of the characteristic lines is given by
dx
dt
=
c
1
.
Integrating this equation, one arrives at the equation for the characteristic lines
x = ct+ x(0), (3.2)
where x(0) is the coordinate of the point at which the characteristic line intersects the x-axis. The
solution to the PDE (3.1) can then be written as
u(t, x) = f(x? ct) (3.3)
for any arbitrary single-variable function f .
11
Let us now consider a particular initial condition for u(t, x)
u(0, x) =
{
x 0 < x < 1,
0 otherwise.
(3.4)
According to (3.3), u(0, x) = f(x), which determines the function f . Having found the function from the
initial condition, we can now evaluate the solution u(t, x) of the transport equation from (3.3). Indeed
u(t, x) = f(x? ct) =
{
x? ct 0 < x? ct < 1
0 otherwise
Noticing that the inequalities 0 < x ? ct < 1 imply that x is in-between ct and ct + 1, we can rewrite
the above solution as
u(t, x) =
{
x? ct ct < x < ct+ 1,
0 otherwise,
which is exactly the initial function u(0, x), given by (3.4), moved to the right along the x-axis by ct
units. Thus, the initial data u(0, x) travels from left to right with constant speed c.
We can alternatively understand the dynamics by looking at the characteristic lines in the xt coordi-
nate plane. From (3.2), we can rewrite the characteristics as
t =
1
c
(x? x(0)).
Along these characteristics the solution remains constant, and one can obtain the value of the solution
at any point (t, x) by tracing it back to the x-axis:
u(t, x) = u(t? t, x? ct) = u(0, x(0)).
t=0 t=3
x
u
Figure 3.1: u(t, x) at two different times t.
x
t
u
Figure 3.2: The graph of u(t, x) colored with re-
spect to time t.
Figure 1 gives the graphs of u(t, x) at two different times, while Figure 3.1 gives the three dimensional
graph of u(t, x) as a function of two variables.
3.2 Quasilinear equations
We next look at a simple nonlinear equation, for which the method of characteristics can be applied as
well. The general first order quasilinear equation has the following form
a(x, y, u)ux + b(x, y, u)uy = g(x, y, u).
12
We can see that the highest order derivatives, in this case the first order derivatives, enter the equation
linearly, although the coefficients depend on the unknown u. A very particular example of first order
quasilinear equations is the inviscid Burger’s equation
ut + uux = 0. (3.5)
As before, we can rewrite this equation in the form of a dot product, which is a vanishing condition for
a certain directional derivative
(1, u) · (ut, ux) = 0, or (1, u) · ?u = 0.
This shows that the tangent vector of the characteristic curves, v = (1, u), depends on the unknown
function u.
案例分类
标签云
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