Fundamental Concepts: Linearity and Homogeneity
This is probably the most abstract section of the course, and also the most
important, since the procedures followed in solving PDEs will be simply a bewil-
dering welter of magic tricks to you unless you learn the general principles behind
them. We have already seen the tricks in use in a few examples; it is time to
extract and formulate the principles. (These ideas will already be familiar if you
have had a good linear algebra course.)
Linear equations and linear operators
I think that you already know how to recognize linear and nonlinear equa-
tions, so let’s look at some examples before I give the official definition of “linear”
and discuss its usefulness.
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Algebraic equations:
Linear Nonlinear
( )
x + 2y = 0, x5 = 2x
x − 3y = 1
Ordinary differential equations:
Linear Nonlinear
dy dy
+ t3 y = cos 3t = t2 + ey
dt dt
Partial differential equations:
2
Linear Nonlinear
2
∂2u
∂u ∂u ∂u
= =
∂t ∂x2 ∂t ∂x
What distinguishes the linear equations from the nonlinear ones? The most
visible feature of the linear equations is that they involve the unknown quantity
(the dependent variable, in the differential cases) only to the first power. The
unknown does not appear inside transcendental functions (such as sin and ln), or
in a denominator, or squared, cubed, etc. This is how a linear equation is usually
recognized by eye. Notice that there may be terms (like cos 3t in one example)
which don’t involve the unknown at all. Also, as the same example term shows,
there’s no rule against nonlinear functions of the independent variable.
The formal definition of “linear” stresses not what a linear equation looks like,
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but the properties that make it easy to describe all its solutions. For concreteness
let’s assume that the unknown in our problem is a (real-valued) function of one
or more (real) variables, u(x) or u(x, y). The fundamental concept is not “linear
equation” but “linear operator”:
Definition: An operation, L, on functions is linear if it satisfies
L(u + v) = L(u) + L(v) and L(λu) = λL(u) (∗)
for all functions u and v and all numbers λ.
Examples of linear operations are
du
• differentiation of u: L(u) ≡ ,
dx
• multiplication of u by a given function of x: L(u) ≡ x2 u(x),
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• evaluation of u at a particular value of x: L(u) ≡ u(2),
R1
• integration of u L(u) ≡ 0
u(x) dx.
In each example it’s easy to check that (∗) is satisfied, and we also see the char-
acteristic first-power structure of the formulas (without u-independent terms this
time). In each case L is a function on functions, a mapping which takes a function
as input and gives as output either another function (as in the first two exam-
ples) or a number (as in the last two). Such a superfunction, considered as a
mathematical object in its own right, is called an operator.
Now we can return to equations:
Definition: A linear equation is an equation of the form
L(u) = g,
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where L is a linear operator, g is a “given” or “known” function (or number, as
the case may be), and u is the unknown to be solved for.
So the possible u-independent terms enter the picture in the role of g. This
leads to an absolutely crucial distinction:
Homogeneous vs. nonhomogeneous equations
Definition: A linear equation, L(u) = g, is homogeneous if g = 0 (i.e., all
terms in the equation are exactly of the first degree in u); it is nonhomogeneous
if g 6= 0 (i.e., “constant” terms also appear).
In the second parenthetical clause, “constant” means independent of u. The
“constant” term g may be a nontrivial function of the independent variable(s) of
the problem.
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Among our original examples, the linear ODE example was nonhomogeneous
(because of the cos 3t) and the PDE example was homogeneous. The algebraic
example is nonhomogeneous because of the 1. Here we are thinking of the sys-
tem of simultaneous equations as a single linear equation in which the unknown
quantity is a two-component vector,
x
~u ≡ .
y
The linear operator L maps ~u onto another vector,
0
~g = .
1
As you probably know, the system of equations can be rewritten in matrix nota-
tion as
1 2 x 0
= .
1 −3 y 1
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The linear operator is described by the square matrix
1 2
M= .
1 −3
In solving a differential equation we usually need to deal with initial or
boundary conditions in addition to the equation itself. The main reason is that
initial or boundary data need to be specified to give the problem a unique an-
swer. Usually these conditions are themselves linear equations — for example, a
standard initial condition for the heat equation:
u(0, x) = f (x).
Often the differential equation will be homogeneous but at least one of the bound-
ary conditions will be nonhomogeneous. (The reverse situation also occurs.)
Therefore, I think it’s helpful to introduce one more bit of jargon:
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Definitions: A linear problem consists of one or more linear conditions
(equations) to be satisfied by the unknown, u. A linear problem is homogeneous
if all of its conditions are homogeneous, nonhomogeneous if one or more of the
conditions are nonhomogeneous.
Example A: The ODE problem
u′′ + 4u = 0, u(0) = 1, u′ (0) = 0
is a nonhomogeneous linear problem. The ODE by itself is homogeneous, however.
Example B: The PDE problem
∂u ∂2u
= + j(x), u(x, 0) = 0, u(0, t) = 0, u(t, 1) = 0
∂t ∂x2
is a nonhomogeneous linear problem. The boundary conditions and the initial
condition are homogeneous, but the heat equation itself is nonhomogeneous in
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this case; the function j represents generation of heat inside the bar (perhaps by
combustion or radioactivity), a possibility not considered in the discussion of the
heat-conduction problem in Appendix A.
Remark: It is easy to see that every homogeneous linear equation has u = 0
as a solution. (One proof: L(0) = L(u − u) (for any u) = L(u) − L(u) = 0,
QED.) Therefore, any homogeneous linear problem has 0 as a solution. Therefore,
if a linear problem has a unique solution and that solution is nontrivial (not
just the 0 function), then that linear problem must be nonhomogeneous. That
is, an interesting, well-posed problem always has at least one nonhomogeneous
condition.
Solving linear problems
The importance of linear problems is that solving them is made easy by the
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superposition principles (which don’t apply to nonlinear problems):
Principles of Superposition:
1. A linear combination of solutions of a homogeneous problem is a new solution
of that problem. That is, if L(u1 ) = 0 and L(u2 ) = 0, then L(c1 u1 +c2 u2 ) = 0
for any numbers c1 and c2 (and similarly for more than two solutions, and
for more than one homogeneous linear equation defining the problem).
Example: Let Problem 1 be the homogeneous ODE u′′ + 4u = 0. Two
solutions of this problem are
u1 ≡ cos 2x, u2 ≡ sin 2x.
Then u = u1 + 3u2 , for example, is also a solution. (In fact, we know
that the most general solution is c1 u1 + c2 u2 where the c’s are arbi-
trary constants. But for this we need a deeper existence-and-uniqueness
theorem for second-order ODEs; it doesn’t just follow from linearity.)
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2. The sum of a solution of a nonhomogeneous problem and a solution of the
corresponding homogeneous problem is a new solution of the original non-
homogeneous problem. (“Corresponding homogeneous problem” means the
one with the same L’s, but with all g’s replaced by 0.)
Example: Let Problem 2 be the nonhomogeneous equation u′′ + 4u =
ex . One solution is up ≡ 15 ex . (This has to be found by the method
of undetermined coefficients, or by luck. Again, general principles of
linearity by themselves can’t solve the whole problem.) Now if we add
a solution of Problem 1 we get a new solution of Problem 2: u3 ≡
1 x
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e + cos 2x.
3. The difference of two solutions of a nonhomogeneous problem is a solution
of the corresponding homogeneous problem. Therefore, every solution of a
nonhomogeneous problem can be obtained from one particular solution of
that problem by adding some solution of the homogeneous problem.
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Example: The general solution of Problem 2 is
u = 15 ex + c1 cos 2x + c2 sin 2x.
4. The sum of solutions to two nonhomogeneous problems with the same L’s
is a solution of a new nonhomogeneous problem, for which the g’s are the
sums of the corresponding g’s of the two original problems. (Similarly for
more than two nonhomogeneous problems.)
Example 1: The sum of two solutions of Problem 2, up and u3 , is
z ≡ 52 ex + cos 2x, which is a solution of z ′′ + 4z = 2ex . The important
lesson to be learned from this example is that the right-hand side of
this new equation is not ex , the nonhomogeneous term of the two old
equations. Do not superpose solutions of a nonhomogeneous problem
in the hope of getting a solution of that same problem.
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Example 2: Note that up is the unique solution of Problem 3:
u′′ + 4u = ex , u(0) = 15 , u′ (0) = 1
5 .
Suppose that we really want to solve Problem 4:
u′′ + 4u = ex , u(0) = 0, u′ (0) = 0.
Recalling Principles 2 and 3 as applied to the differential equation alone
(not the initial conditions), we see that u = up + y, where y is some
solution of y ′′ + 4y = 0. A moment’s further thought shows that the
correct y is the solution of Problem 5:
y ′′ + 4y = 0, y(0) = − 15 , y ′ (0) = − 15 .
A standard calculation shows that y = − 15 cos 2x − 10
1
sin 2x, and from
this and up we can get the solution of Problem 4. (Of course, in solving
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such problems we usually don’t write out Problem 5 as an intermediate
step; the standard procedure is to impose the initial data of Problem 4 on
the general solution found earlier. That is just a different way of organiz-
ing the same algebra. However, consciously splitting a nonhomogeneous
problem into two nonhomogeneous problems, as I’ve demonstrated here
for an ODE, is a common technique for solving PDEs.)
In summary, these principles provide the basic strategies for solving linear
problems. If the problem is nonhomogeneous and complicated, you split it into
simpler nonhomogeneous problems and add the solutions. If the solution is not
unique, the nonuniqueness resides precisely in the possibility of adding a solution
of the corresponding homogeneous problem. (In particular, if the original problem
is homogeneous, then you seek the general solution as a linear combination of
some list of basic solutions.) If the problem statement contains enough initial
and boundary conditions, the solution will be unique; in that case, the only
solution of the homogeneous problem is the zero function.
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An important example application of this strategy is the solution of the
heat-conduction problem in a bar with fixed end temperatures:*
∂u ∂2u
PDE: = ,
∂t ∂x2
IC: u(x, 0) = f (x),
BC: u(0, t) = T1 , u(1, t) = T2 .
Here we have a homogeneous PDE, a nonhomogeneous initial condition, and two
nonhomogeneous boundary conditions. The trick is to treat the two types of
nonhomogeneity separately. One writes u = v + w, where
* See Appendix A, or Haberman’s book.
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(1) v is to be a solution of the problem consisting of the PDE and the nonhomo-
geneous BC, with no particular IC assumed. It is possible to find a solution
of this problem which is independent of t: v(x, t) = V (x).
(2) w is to be a solution of the problem consisting of the PDE, the homogeneous
Dirichlet boundary conditions
w(0, t) = 0, w(1, t) = 0,
and the initial condition needed to make u satisfy the original IC. Namely,
w(x, 0) = f (x) − V (x).
It is very important that the only nonhomogeneity in this second problem is
the IC. This makes it possible to solve for w by the method of separation of
variables and then add the solutions without falling into the trap I warned
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you against earlier (Example 1). The solution is completed by finding the
Fourier series of the function f − V .
The details of steps (1) and (2) are carried out in Appendix A.
For the processes of separating variables and calculating Fourier coefficients
to work here, it was absolutely crucial to make the boundary conditions homoge-
neous first. In the calculation of normal modes, no nonhomogeneous conditions
at all are imposed. The appropriate nonhomogeneous IC is imposed on a super-
position (w) of normal modes. Then still another term, v, is added to satisfy the
nonhomogeneous BC.
One more time:
Impose only HOMOGENEOUS conditions on normal modes
(separated solutions).
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Impose nonhomogeneous conditions only on a SUPERPOSI-
TION (sum or integral) of normal modes.
A related principle is
Handle only one nonhomogeneity at a time!
This principle is handled in practice by different strategies in different prob-
lems. Let’s consider a doubly nonhomogeneous problem with the structure
L1 (u) = f1 , L2 (u) = f2 .
The two principal strategies are these:
1. “Zero out” the other condition. Solve
L1 (u1 ) = f1 , L2 (u1 ) = 0,
L1 (u2 ) = 0, L2 (u2 ) = f2 .
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Then u = u1 + u2 .
Examples where this strategy is used include
∂u
(a) treatment of the initial data u and ∂t in the wave equation;
(b) Laplace’s equation in a rectangle with boundary values given on two
perpendicular sides.
2. Temporarily ignore the other condition. Solve L1 (u1 ) = f1 and let
L2 (u1 ) be whatever it turns out to be, say L2 (u1 ) ≡ h. Next solve
L1 (u2 ) = 0, L2 (u2 ) = f2 − h.
Then u = u1 + u2 .
Examples where this strategy is used include
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(a) the method of undetermined coefficients for an ordinary differential
equation with initial conditions;
(b) finding a steady-state solution for the wave or heat equation with
nonzero, but time-independent, boundary conditions.
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