Chapter 6 Partial Differential Equations

Chapter 6 

Partial Differential Equations 

6.1 Introduction 

Let R denote the real numbers, C the complex numbers, and n-dimensional Euclidean space 

is denoted by R n with points denoted by x (or sometimes by ⃗x) as tuples x =(x 1 , ··· ,x n ). 

Often in the case of R 2 or R 3 we will use (x, y)or(x, y, z) to denote points instead of subscript 

notation. Ocassionally when we are dealing with dynamical problems where time is one of 

the variables we will also use notations like (x, t), (⃗x, t), (x, y, t), or (x, y, z, t). The usual 

Euclidean inner product is given by ⃗x · ⃗z = ∑ n 

j=1 x jy j and the norm generated by this inner 

product we denote by single bars just like the absolute value, i.e., |⃗x| = √ ⃗x · ⃗x. 

An n-tuple α =(α 1 , ··· ,α n ) of nonnegative integers is called a multi-index. We define 

|α| = 

n∑ 

k=1 

α k ,α!=α 1 ! ···α n !, ∀ x ∈ R n , x α = x α 1 

1 ···x αn 

n . 

We will use several notations for partial derivatives of a function u of x ∈ R n : 

u j = ∂ j u = ∂u 

∂x j 

, 

for fist order partials and for higher order partials we have 

u α = ∂ α u =(∂ 1 ) α1 ···(∂ n ) αn u = 

∂ |α| u 

∂x α . 

1 

1 ···∂x αn 

n 

In particular when α = ⃗0 then ∂ α is the identity operator. We will agree to order the set of 

multi-indices α =(α 1 , ··· ,α n ), by requiring that α comes before β if |α| ≤|β| or |α| = |β| 

and α i

2 CHAPTER 6. PARTIAL DIFFERENTIAL EQUATIONS 

A partial differential equation of order k is an equation of the form 

F (x 1 ,x 2 , ··· ,x n ,u,∂ 1 u,...,∂ n u, ∂ 2 1u, ··· ,∂ k nu) = 0 (6.1.1) 

relating a function u of x =(x 1 , ··· ,x n ) ∈ R n and its partial derivatives of order ≤ k. 

Given numbers a α with |α| ≤k, we denote by (a α ) |α|≤k . the element in R N(k) given by 

ordering the α’s in any fashion, where N(k) is the cardinality of {α : |α| ≤k}. Similarly, if 

S ⊂{α : |α| ≤k} we can consider the ordered (card S)-tuple (a α ) α∈S . 

NowletΩbeanopensetinR n , and let F be a function of the variables x ∈ Ω and 

(u α ) |α|≤k . Then we can write the partial differential equatio of order k as 

F (x, (u α ) |α|≤k )=0. (6.1.2) 

A function u is called a classical solution of this equation if ∂ α u exists for each α in F , and 

F (x, (u α (x)) |α|≤k )=0, for every x ∈ Ω. 

We denote by C(Ω) the space of continuous functions on Ω. If Ω is open and k is a 

positive integer, C k (Ω) will denote the space of functions possessing continuous derivatives 

up to order k on Ω, and C k (Ω) will denote the space of all u ∈ C k (Ω) such that ∂ α u extends 

continuously to the closure of Ω denoted by Ω for all 0 ≤|α| ≤k. We also define 

C ∞ (Ω) = ∩ ∞ k=1C k (Ω), 

C ∞ (Ω) = ∩ ∞ k=1C k (Ω). 

If Ω ⊂ R n is open, a function u ∈ C ∞ (Ω) is said to analytic in Ω if it can be expanded 

in a convergent power series about every point of Ω. That is, u is analytic in Ω if for each 

x ∈ Ω there exist an r>0 so that for all y ∈ B r (x) ={y : |y − x|

6.1. INTRODUCTION 3 

In this case we define the differential operator L = ∑ |α|≤k a α(x)∂ α and write Lu = f. More 

generally, we have the quasi-linear equations which have the form 

∑ 

a α (x, (∂ β u) |β|≤(k−1) )∂ α u = b(x, (∂ β u) |β|≤(k−1) ). (6.1.4) 

|α|≤k 

Thus the partial differential equation is linear if u and all of its derivatives appear in a 

linear fashion. For example, the general form of a linear 2 nd order partial differential equation 

is 

Au xx + Bu xy + Cu yy + Du x + Eu y + Fu = G 

For instance, 

yu xx + u yy + u x =0 

is linear and 

uu xx + u yy + u x =0 

is nonlinear but quasi-linear. If G = 0 the equation is homogeneous. 

Some general concerns in the study of partial differential equation’s are: 

1. existence of solutions, 

2. uniqueness of solutions, 

3. dependence of solutions on the ‘data’, 

4. smoothness, or regularity of the solution, and 

5. representations for solutions and behavior of solutions. 

The first three of these issues are related to the notion of a well-posed problem. In 

particular, a problem is well posed in the sense of Hadamard if there exists a unique solution 

that depends continuously on the data. 

Consider the following examples in R 2 , 

1. The general solution of ∂ 1 u =0isu(x 1 ,x 2 )=f(x 2 ) where f is arbitrary. Thus the 

equations gives complete information about the behavior of the solution with respect 

to x 1 (it is a constant) but it gives no information with respect to the x 2 variable. 

2. The general solution of ∂ 1 ∂ 2 u =0isu(x 1 ,x 2 )=f(x 1 )+g(x 2 ) where f and g are 

arbitrary. Thus, other that the fact that the dependences on x 1 and x 2 are “uncoupled” 

we learn nothing about the dependence on the variables. 

3. Any complex valued solution u of the “Cauchy-Riemann” equation ∂ 1 u + i∂ 2 u = 0 is a 

holomorphic function of z = x 1 + ix 2 . Thus they are in particular C ∞ . The equation 

imposes very strong conditions on all derivatives of the solutions.


Linear partial differential equation’s are often classified as being of elliptic, hyperbolic, or 

parabolic type. What follows is a brief and heuristic discussion of some of the features that 

characterize these classifications of partial differential equations. Most of what is presented 

in this introduction will be made precise in subsequent chapters. 

Elliptic Equations 

An example of an elliptic partial differential equation is Laplace’s equation, 

div grad u =0 

If u = u(x 1 , ··· ,x n ) and (x 1 , ··· ,x n ) are rectangular Cartesian coordinates, and w(x) = 

(w 1 , ··· ,w n ), 

grad u = ∇u =(u x1 , ··· ,u xn ) 

div ⃗w = ∂w + ···+ ∂w 

∂x 1 ∂x n 

and so 

div grad u = ∇·∇u = ∇ 2 u =∆u = u x1 x 1 

+ ···+ u xnx n 

=0. 

The Laplacian of u, ∆u, provides a comparison of values of u at a point with values at 

neighboring points. To illustrate this idea consider the simplest case that u = u(x) and 

assume u xx (x) > 0. Let ũ denote the tangent line approximation to u. For h>0 and 

sufficiently small, 

u(x + h) > ũ(x + h) =u(x)+u ′ (x)h 

u(x − h) > ũ(x − h) =u(x) − u ′ (x)h 

and so 

u(x + h)+u(x − h) 

>u(x). 

2 

Roughly stated, u(x) is smaller than its average value at nearby points if u xx (x) > 0. In 

higher dimensions, say n = 2, the analogous statement is that if ∆u(x, y) > 0, then the 

average value of u at neighboring points, say on a circle about (x, y), is greater than u(x, y). 

If ∆u(x, y) < 0, then u(x, y) is greater than the average value of u on a circle about (x, y). 

If ∆u(x, y) = 0 then u(x, y) is equal to its average value on a circle about (x, y). (We 

will subsequently prove a theorem that makes these ideas rigorous.) More generally, if 

∆u(x) =0,x∈ Ω ⊂ R n , then u is equal to its average value at neighboring points everywhere 

in Ω. In a certain sense, this says that if u satisfies Laplace’s equation then u represents a 

state of equilibrium.


Solutions to Laplace’s equation ∆u(x) = 0 are said to be harmonic. Suppose f(z) = 

f(x + iy) =u(x, y)+iv(x, y) is analytic. The Cauchy-Riemann equations state that 

∂u 

∂x = ∂v 

∂y , 

∂u 

∂y = −∂v ∂x 

and so 

u xx + u yy =0, v xx + v yy =0. 

Thus the real and imaginary parts of an analytic function are harmonic. There is a converse 

statement of this result known as Weyl’s Theorem that depends on the notion of a weak 

solution. Roughly stated, ∆u(x) = 0 in a weak sense if for all smooth functions φ with 

compact support in Ω, ∫ 

u(x)(∆φ)(x) dx =0. 

We will not prove this next result. 

Ω 

Weyl’s Theorem If u is a weak solution of ∆u(x) = 0 on Ω, then u ∈ C ∞ (Ω) and u 

satisfies Laplace’s equation in a classical sense. 

The so-called Cauchy problem for Laplace’s equation, 

u xx + u yy =0, |x| < ∞,y >0 

u(x, 0) = f(x), 

u y (x, 0) = g(x), 

is not well-posed. Indeed, consider the specific problem 

u xx + u yy =0, |x| < ∞,y >0 

cos nx 

u(x, 0) = 

n 

u y (x, 0)=0 

with the solution 

u n (x, y) = 1 cosh(ny) cos(nx). 

n 

For n sufficiently large, the data of this problem can be made uniformly small. However, 

lim u n(x, y) =∞. 

n→∞ 

In other words, small changes in the data do not correspond to small changes in a solution.


Another general feature of elliptic problems is that solutions are smoother that data. For 

instance, we will develop the following solution of Poisson’s equation ∆u(x) =f(x), x∈ R 3 , 

u(x) = −1 

4π 

∫ 

R 3 

f(y) 

|x − y| dy. 

(Here |x| = √ ∑ x 

2 

i .) From this representation of the solution, we see that the effect of 

integrating the data f against the kernel, 1/|x − y|, mollifies, or smoothes the data and so 

we expect that a lack of regularity in f(x) would be eliminated. This is indeed the case and 

in the spirit of Weyl’s Theorem, we will show that the solution u is in fact C ∞ . 

Hyperbolic Equations 

One of the simplest examples of a hyperbolic equation is the wave equation, 

u xx = u tt , −∞


t 

u=0 

u=1/2 

u=1/2 

u=0 u=1 

x 

u=0 

The propagation of a disturbance in dimension 1. 

In general, for hyperbolic equations 

1. solutions are no smoother than data, 

2. there is a finite speed of propagation, 

3. solutions exhibit a strong dependence on spatial dimension, 

4. many quantities are preserved, and 

5. the Cauchy problem is well-posed. 

Parabolic Equations 

The heat equation, 

u t (x, t) =∆u(x, t), x ∈ R n ,t>0 

is an example of a parabolic equation. If we think of u(x, t) as being the temperature at a 

point x at time t, this equation describes the flow or diffusion of heat. In view of our earlier 

discussion of the interpretation of the Laplacian, we see that if say, ∆u(x, t) < 0, then the 

temperature at position x is greater than that at surrounding points. From Fourier’s Law of 

Cooling, heat would ‘flow’ away from the position x, and from the differential equation we 

see that u t < 0, corresponding to the decrease in temperature at that point. 

The Cauchy problem for the heat equation on R n is 

u t =∆u(x, t), |x| < ∞, t>0. 

u(x, 0) = f(x).


The solution may be expressed as 

u(x, t) = 

∫ 

1 

(4πt) n/2 

e −|x−y|2 

4t 

R n 

f(y) dy. 

From this formula, it is reasonable to expect the solution u(x, t) to be infinitely differentiable. 

In addtion, we see that if even if the data f is compactly supported, the temperature u(x, t) 

will be positive for all x when t > 0. These observations suggest the following general 

features of parabolic equations: 

1. solutions are smooth, and 

2. there is an infinite speed of conduction. 

6.2 Linear and Quasilinear equations of First Order 

|α|≤k 

In the study of linear partial differential equations a measure of the “strength” of a differential 

operator in a certain direction is provided by the notopn of charafcteristics. If 

L = ∑ 

a α (x)∂ α is a linear differential operator of order k onΩinR n , then its characteristic 

form (or principal symbol) atx ∈ Ω is the homogeneous polynomial of degree k on R n 

defined by 

χ L (x, ξ) = ∑ 

a α (x)ξ α . 

A vector ξ is characteristic for L at x if 

|α|=k 

χ L (x, ξ) =0. 

The characteristic variety is the set of all characteristic vectors ξ, i.e., 

Char x (L) ={ξ ≠0:χ L (x, ξ) =0}. 

Definition 6.2.1. A hypersurface of class C k (1 ≤ k ≤∞) is a subset S ⊂ R n such that 

for each x 0 ∈ S there exists an open neighborhood V ⊂ R n of x 0 and a real-valued function 

ϕ ∈ C k (V ) such that ∇ϕ(x) ≠0for all x ∈ S ∩ V where 

S ∩ V = {x ∈ V : ϕ(x) =0}. 

Remark 6.2.2. Since, by definition, for each x 0 ∇ϕ(x 0 ) ≠ 0 we can apply the implicit 

function theorem (without loss of generality let us assume that ∂ xn ϕ(x 0 ) ≠0) to solve ϕ(x) = 

0 for x n = ψ(x ′ ) where x ′ =(x 1 , ··· ,x n−1 ) near x 0 . Thus a neighborhood of x 0 can be mapped 

to a piece of the hyperplane x n =0by x ↦→ (x ′ ,x n − ψ(x ′ )). The same neighborhood can be

6.2. LINEAR AND QUASILINEAR EQUATIONS OF FIRST ORDER 9 

also be represented in parametric form as the image of an open set in R n−1 (with coordinates 

x ′ ) under the map 

x ′ ↦→ (x ′ ,ψ(x ′ )). 

Thus x ′ can be thought of as giving local coordinates on S near x 0 . 

A hypersurface S is called characteristic for L at x if the normal vector ν(x) is in Char x (L) 

and S is called non-characteristic if it is not characteristic at any point. 

An important property of the characteristic variety is contained in the following: 

Let F be a smooth one-to-one mapping of Ω onto Ω ′ ⊂ R n and set y = F (x). 

Assume that the Jacobian matrix 

[ ] ∂yi 

J x = (x) 

∂x j 

is nonsingular for x ∈ Ω, so that {y 1 ,y 2 , ··· ,y n } is a coordinate system on Ω ′ . 

We have 

∂ 

n∑ ∂y i ∂ 

= 

∂x j ∂x j ∂y i 

i=1 

which we can write symbolically as ∂ x = Jx T ∂ y , where Jx 

T is the transpose of 

J x . The operator L is then transformed into 

L ′ = ∑ ( 

a α F −1 (y) ) ( α 

J T F −1 (y) y) ∂ on Ω ′ . 

|α|≤k 

When this expression is expanded out, there will be some differentiations of J T F −1 (y) but 

such derivatives are only formed by “using up” some of the ∂ y on J T F −1 (y), so they do not 

enter in the computation of the principal symbol in the y coordinates, i.e., they do not enter 

the highest order terms. We find that 

χ L (x, ξ) = ∑ 

( ) α 

a α (F −1 (y)) J T F −1 (y) ξ . 

|α|=k 

Now since F −1 (y) =x, on comparing with the expression 

χ L (x, ξ) = ∑ 

a α (x)ξ α 

|α|=k 

we see that Char x (L) is the image of Char y (L ′ ) under the linear map J T F −1 (y) .


Note that if ξ ≠ 0 is a vector in the x j -direction (i.e., ξ i = 0 for i ≠ j), then ξ ∈ Char x (L) 

if and only if the coefficient of ∂j 

k in L vanishes at x. Now, given any ξ ≠ 0, by a rotation 

of coordinates we can arrange for ξ to lie in a coordinate direction. Thus the condition 

ξ ∈ Char x (L) means that, in some sense, L fails to be “genuinely kth order” in the ξ 

direction at x. 

L is said to be elliptic at x if Char x (L) =∅ and elliptic on Ω if it elliptic at each x ∈ Ω. 

Elliptic operators exert control on all derivatives of all order. 

Example 6.2.3. The first three examples are in R 2 as discussed above. 

1. L = ∂ 1 : Char x (L) ={ξ ≠0:ξ 1 =0}. 

2. L = ∂ 1 ∂ 2 : Char x (L) ={ξ ≠0:ξ 1 =0 or ξ 2 =0}. 

3. L = 1 2 (∂ 1 + i∂ 2 ): L is elliptic on R 2 . 

4. L = 

n∑ 

∂j 2 (Laplace Operator): L is elliptic on R n . 

j=1 

n∑ 

5. L = ∂ 1 − ∂j 2 (Heat Operator): Char x (L) ={ξ ≠0:ξ j =0, for j ≥ 2}. 

j=2 

6. L = ∂ 2 1 − 

n∑ 

∂j 

2 

j=2 

(Wave Operator): Char x (L) ={ξ ≠0:ξ 2 j = ∑ n 

j=2 ξ2 j }. 

Remark 6.2.4. In the notation introduced in Definition 6.2.1 we say that a surface S is 

oriented if for each s ∈ S we have made a choice of a vector ν(x) which is orthogonal to S 

and is a continuously varying function of x. Such a vector is called a normal vector to S at 

x. OnS ∩ V = {x : ϕ(x) =0} we have 

ν(x) =± ∇ϕ(x) 

|∇ϕ(x)| . 

Thus ν(x) is a C k−1 function on S. If S is the boundary of a domain Ω then we usually 

choose the orientation so that ν points out of Ω. 

At this point we can also define the normal derivative by 

. 

∂ ν u = ν ·∇u.


Definition 6.2.5. A hypersurface S is called characteristic for L at x ∈ S if the normal 

vector ν(x) to S at x is in Char x (L), and S is called non-characteristic if it is not charateristic 

at any x in S. 

We now turn to the development for real first order systems. First recall the basic problem 

in ODEs is the IVP: 

Given a function F on say R 3 and (t 0 ,u 0 ) ∈ R 2 , find a function u(t) defined 

in a neighborhood of t 0 such that F (t, u, u ′ )=0and u(t 0 )=u 0 . 

In this disscussion we will consider the analog of this which is the initial value problem 

for a first order partial differential equation. We will focus on the linear and quasi-linear 

cases. 

Let us first consider the linear equation 

n∑ 

a j ∂ j u + bu = f(x). (6.2.1) 

j=1 

where a j , b and f are assumed to be C 1 functions of x. If we denote by A the vector field 

then we have 

A(x) =(a 1 (x), ··· ,a n (x), 

Char x (L) ={ξ ≠0:A(x) · ξ =0}. 

That is, Char x (L) ∪{0} is the hyperplane orthogonal to A(x). From this we see that: 

A hypersurface S is characteristic at x if and only if A(x) is tangent to S at x. 

INITIAL VALUE PROBLEM: Find a solution u to (6.2.1) with given initial values 

u = ϕ on a given hypersurface S. 

If S is characteristic at a point x 0 , then the quantity ∑ a j (x 0 )∂ j u(x 0 ) is completely 

determined as a certain directional derivative of ϕ along S at x 0 . For this reason it may 

not be possible to make it equal to f(x 0 ) − b(x 0 )u(x 0 ). As an example, if the equation is 

∂ 1 u = 0 and S is the hyperplane x n = 0, we cannot have u = ϕ on S unless ∂ 1 ϕ =0. 

Namely, consider the case of R 2 . The general solution is given by u(x 1 ,x 2 )=f(x 2 ) where f 

is arbitrary. But if S corresponds to x 2 = 0 then the solution must satisfy u(x 1 , 0) = φ(x 1 ) 

and the only choice is that ϕ ≡ 0. 

Thus to make the initial value problem well behaved, we must assume that S is noncharacteristic. 

It turns out that to solve for u it is useful to compute the integral curves of 

the vector field A(x).


Definition 6.2.6. The integral curves of the vector field A(x) are, by definition, the parameterized 

curves x(t) that satisfy the system of ODEs 

dx 

dt = A(x), i.e., dx j 

dt = a j(x), j =1, 2, ··· ,n. (6.2.2) 

Along such a curve a soution u of the equation (6.2.1) must satisfy 

du 

dt = 

n∑ 

j=1 

∂u 

∂x j 

dx j 

dt = ∑ a j ∂ j u = f − bu. (6.2.3) 

That is, along such a curve a solution u of the equation (6.2.1) will satisfy the ODE 

du 

= f − bu. (6.2.4) 

dt 

By the fundamental existence uniqueness theorem from ODEs, through each point x 0 of S 

there passes a unique integral curve x(t) ofA, namely the solution of (6.2.2) with x(0) = 

x 0 . Along this curve the solution u of (6.2.1) must also be a solution of the ODE (6.2.4) 

with u(0) = ϕ(x 0 ). Moreover, since A is non-characteristic, x(t) ∉ S (at least for |t| ≠0 

sufficiently small) and the curves x(t) fill out a neighborhood of S. The same result as stated 

in Theorem 6.2.7 is given in the simpler case of R 2 in Subsection 6.2.1 (see, in particular, 

Theorem 6.2.14). 

Theorem 6.2.7. Assume that S is a hypersurface of class C 1 which is non-characteristic 

for (6.2.1), and that the functions a j , b, f, and ϕ are C 1 and real-valued. Then for any 

sufficiently small neighborhood Ω of S in R n there is a unique solution u ∈ C 1 of (6.2.1) that 

satisfies u = ϕ on S. 

This theorem is a special case of the corresponding result for quasi-linear equations so 

we will defer the proof of this result to the proof of the following more general result (see 

Theorem 6.2.7). 

Consider a first order quasi-linear equation 

n∑ 

a j (x, u)∂ j u = b(x, u). (6.2.5) 

j=1 

In this case, we consider variables (x 1 , ··· ,x n ,u) ∈ R n+1 and note that if u is a function 

of x, then the normal to the graph of u (i.e., (x, u(x)) ∈ R n+1 )inR n+1 is proportional to 

⃗v =(∂ 1 u, ···∂ n u, −1). So (6.2.5) says that 

A(x, u) =(a 1 (x, u), ··· ,a n (x, u),b(x, u))


is tangent to the graph of y = u(x) at any point (since it is orthogonal to ⃗v). 

This suggests that we look at the integral curves of the vector field A(x, u) inR n+1 given 

by solving the equations 

dx j 

dt = a j(x, y), j=1, ··· ,n, 

dy 

dt 

= b(x, y). (6.2.6) 

As you will see, any graph y = u(x) inR n+1 which is the union of an (n − 1)-parameter 

family of these integral curves will define a solution of (6.2.5). Conversely, suppose that u is 

a solution of (6.2.5). If we solve the equations 

dx j 

dt = a j(x, u(x)), 

x j (0)=(x 0 ) j 

to obtain a curve x(t) passing through x 0 , and then set y = u(x(t)), we have 

dy 

dt = 

n∑ 

j=1 

∂ j u dx j 

dt = 

n∑ 

a j (x, u)∂ j u = b(x, u). 

j=1 

Thus if the graph y = u(x) intersects an integral curve of A in one point (x 0 ,u(x 0 )), it 

contains the whole curve. 

Suppose we are given intial data u = ϕ on a hypersurface S in R n . If we form the 

submanifold 

S ∗ = {(x, ϕ(x)) : x ∈ S} 

of R n+1 , the graph of the solution should be the hypersurface (in R n+1 ) generated by the integral 

curves of A passing through S ∗ . Again, we need to assume that S is non-characteristic 

in some sense. This is more complicated than the linear case because a j depend on u as well 

as x. We need the following geometric interpretation: 

For x ∈ S, the vector field A(x, ϕ(x)) = ( a 1 (x, ϕ(x)), ··· ,a n (x, ϕ(x)) ) should not 

be tangent to S at x. Note that this condition involves ϕ as well as S. 

If S is represented parametrically by a mapping g : R n−1 → R n and we take coordinates 

s =(s 1 , ··· ,s n−1 ) ∈ R n−1 , so that g(s) =(g 1 (s), ··· ,g n (s)) , then the above condition can 

be expressed as 

⎛ 

⎞ 

∂g 1 ∂g 1 

··· a 1 (g(s),φ(g(s))) 

∂s 1 ∂s n−1 

det 

⎜ . . 

. 

⎟ ≠0. (6.2.7) 

⎝∂g n ∂g n 

⎠ 

··· a n (g(s),φ(g(s))) 

∂s 1 

∂s n−1


Theorem 6.2.8. Suppose that S is a C 1 hypersurface and a j , b and φ are C 1 real valued 

functions. Suppose that the vector field V =(a 1 (x, φ(x)), ··· ,a n (x, φ(x))) is not tangent to 

S at any x ∈ S (this is the noncharacteristic condition). Then there is a unique solution 

u ∈ C 1 of 

n∑ 

a j (x, u)∂ j u = b(x, u) 

j=1 

in a neighborhood Ω of S such that u = φ on S. 

The proof of this result is constructive. 

Proof. The uniqueness follows from the discussion above which states that u must be the 

union of integral curves of A in Ω passing through S ∗ . 

Now any hypersurface S can be covered by by open sets on which it admits a parametric 

representation x = g(s), with s ∈ R n−1 . If we solve the problem on each such open set, 

by uniqueness the local solutions must agree on the overlap of the open sets and hence 

patch together to give a solution for all of S. It therefore suffices to assume that S is given 

parametrically by x = g(s) with s ∈ R n−1 . 

For each s ∈ R n−1 , consider the initial value problem 

∂x j 

∂t (s, t) =a j(x, y), j=1, ··· ,n, 

∂y 

(s, t) =b(x, y), (6.2.8) 

∂t 

x j (s, 0) = g j (s), 

y(s, 0) = ϕ(g(s)). 

Here s is a parameter vector, so we have a system of ODEs in t. By the fundamental existence 

uniqueness theorem for ODEs, there exists a unique solution (x, y) defined for small t, and 

(x, y) isaC 1 function of s and t jointly. By the non-characteristic condition (6.2.7) and the 

inverse mapping theorem, the mapping (s, t) ↦→ (x, t) is invertible on some neighborhood Ω 

of S, yielding s and t as C 1 functions of x on Ω such that t(x) =0andg(s(x)) = x when 

x ∈ S. Now set 

u(x) =y(s(x),t(x)). 

We have u = ϕ on S, and we claim that u satisfies (6.2.5). By the chain rule and the


fact that ∂s k 

∂t = 0, since s k and t are functionally independent, and ∂t 

∂t 

( 

n∑ ∂u 

n∑ n−1 

) 

∑ ∂u ∂s k 

a j = a j + ∂u ∂t 

∂x 

j=1 j ∂s 

j=1 

k ∂x j ∂t ∂x j 

k=1 

This completes the proof. 

= 

= 

= 

∑n−1 

k=1 

∑n−1 

k=1 

∑n−1 

k=1 

∂u 

∂s k 

∂u 

∂s k 

n∑ 

j=1 

n∑ 

j=1 

∂u 

∂s k 

∂s k 

=0+ ∂u 

∂t = b. 

∂s k 

a j + ∂u 

∂x j ∂t 

∂s k ∂x j 

∂x j ∂t + ∂u 

∂t 

∂t + ∂u ∂t 

∂t ∂t 

n∑ 

j=1 

a j 

∂t 

∂x j 

n∑ 

j=1 

∂x j 

∂t 

∂t 

∂x j 

= 1 we have 

Example 6.2.9. In R 3 solve the linear initial value problem Lu = x 1 ∂ 1 u+2x 2 ∂ 2 +∂ 3 u =3u 

with u = ϕ(x 1 ,x 2 ) on the plane x 3 =0. 

Solvability: We could observe that the solvability condition is satisfied by considering The 

vector field A(x) =(x 1 , 2x 2 , 1) and the characteristic manifold 

Char x (L) ={ξ =(ξ 1 ,ξ 2 ,ξ 3 ) ≠0:A(x) · ξ =0}. 

We note that the initial surface S in this case is x 3 = 0 with constant normal vector ν = 

(0, 0, 1). In the present case we see that 

A(x) · ν =1≠0 

so the surface is non-characteristic. 

We could also note that with s =(s 1 ,s 2 ) ∈ R 3−1 and 

g(s) =(g 1 (s),g 2 (s),g 3 (s))=(s 1 ,s 2 , 0) 

paramterizing the surface S we have for (6.2.7) 

⎛ 

1 0 s 1 

⎞ 

det ⎝0 1 2s 2 

⎠ = ϕ(s 1 ,s 2 ). 

0 0 ϕ(s 1 ,s 2 )


So under the assumption that ϕ(s 1 ,s 2 ) ≠ 0 we can solve the problem. 

Solution:In the present case the initial value problem (6.2.2) (see also (6.2.6) and (6.2.8)) 

gives 

dx 1 

dt = x dx 2 

1, 

dt =2x dx 3 

2, 

dt =1, du 

dt =3u, 

with initial conditions 

(x 1 ,x 2 ,x 3 ,u) ∣ t=0 

=(s 1 ,s 2 , 0,φ(s 1 ,s 2 )). 

Solving this system of ODEs yields 

x 1 = s 1 e t ,x 2 = s 2 e 2t ,x 3 = t, u = ϕ(s 1 ,s 2 )e 3t . 

The next step is to invert the first three equations to obtain s 1 , s 2 and t 

t = x 3 , s 1 = x 1 e −x 3 

, s 2 = x 2 e −2x 3 

. 

Thus we find that 

u = ϕ(x 1 e −x 3 

,x 2 e −2x 3 

) e 3x 3 

. 

Example 6.2.10. Let us now consider a quasi-linear example in R 2 and we want to solve 

u∂ 1 u + ∂ 2 u = 1 with u = 1 2 s on the segment x 1 = x 2 = s, 0


6.2.1 Special Case: Quasilinear Equations in R 2 

As a special case let us consider n = 2, i.e., let us consider first order quasi-linear (or linear) 

equations in two variables in the form 

a(x, y, u) ∂u + b(x, y, u)∂u 

∂x ∂y 

= c(x, y, u). (6.2.9) 

For this example we adhere to common practice and refer to the variables as x and y or x 

and t, depending on the given problem, instead of x 1 and x 2 . Also in this case it is often 

useful to use the notations 

u x = ∂u 

∂x ,u y = ∂u 

∂y ,u t = ∂u 

∂t , etc. 

for the various partial derivatives. 

Let u = u(x, y) be a solution (also refered to as a solution surface. The vector (u x ,u y , −1) 

is normal to the surface u(x, y) − u = 0 and the equation (6.2.9) expresses the orthogonality 

condition 

A(x, y, u) · (u x ,u y , −1)=0, A(x, y, u) ≡ (a(x, y, u),b(x, y, u),c(x, y, u)). (6.2.10) 

u 

(u x ,u y , −1) 

(a, b, c) 

u = u(x, y) 

y 

x 

Solution Surface 

To solve (6.2.9) amounts to constructing a surface such that the normal to the surface 

satisfys the orthogonality condition (6.2.10). This is the same as saying that we seek a 

surface u = u(x, y) such that the tangent plane to the surface at (x, y, u) contains the 

vector (a(x, y, u),b(x, y, u),c(x, y, u)). More specifically, since we are really interested in the


initial value problem, we seek a solution of (6.2.9) that contains a given curve, say C 0 given 

parametrically by ( x 0 (s),y 0 (s),u 0 (s) ) . Recall that a three dimensional surface (in our case 

the solution surface u = u(x, y)) can be represented (at least locally) parametrically as a two 

parameter family 

( 

x(s, t),y(s, t),u(s, t) 

) 

. 

In this way we see that a solution surface is built-up from a two parameter family of curves 

( 

x(s, t),y(s, t),u(s, t) 

) 

. And , in order that the initial values be achieved we need 

( 

x(s, 0),y(s, 0),u(s, 0) 

) 

= 

( 

x0 (s),y 0 (s),u 0 (s) ) . 

If these curves lie on a solution surface, then tangent vectors to the curves must lie in the 

tangent plane to the surface at the point. This means that for a fixed s the curves (in t) 

must satisfy the equations 

dx 

(s, t) =a(x, y, u), 

dt x(s, 0) = x 0(s), 

dy 

dt (s, t) =b(x, y, u), y(s, 0) = y 0(s), (6.2.11) 

du 

(s, t) =c(x, y, u), 

dt u(s, 0) = u 0(s). 

( 

x0 (s),y 0 (s),u 0 (s) ) 

u 

(u x ,u y , −1) 

( x(s, τ),y(s, τ ),u(s, τ ) 

) 

(a, b, c) 

y 

x 

characteristic 

( 

x0 (s),y 0 (s) ) 

Solution Surface and Characteristic 

If the vector field A were tangent to the curve C 0 , then a solution of (6.2.11) would 

coincide with C 0 and our method for constructing a surface would fail. We will see that this 

problem can be avoided by not perscribing data on the curve with 

dx 

dt = a, 

dy 

dt = b.


Such a curve is called a characteristic curve. To construct a solution surface we first find 

x = x(s, t), y = y(s, t), u = u(s, t) 

and then solve the two equations x = x(s, t) and y = y(s, t) for s and t in terms of x and y. 

In order to guarantee that this can be done requires a result from advanced calculus – the 

inverse function theorem which states: 

If x = x(s, t) and y = y(s, t) are C 1 maps in a neighborhood of a point (s 0 ,t 0 ), 

the Jacobian 

⎛ ⎞ 

∂x ∂x 

⎜ ⎟∣ 

|J| = det ⎜ ∂t 

⎝∂y 

∂t 

∂s⎟ 

∂y⎠ 

∣ 

∂s 

∣ 

(s0 ,t 0 ) 

≠0. 

and, in addition, x 0 = x(s 0 ,t 0 ) and y 0 = y(s 0 ,t 0 ), then there exists a neighborhood 

R of (s 0 ,t 0 ) and there exists unique C 1 mappings 

and 

s = s(x, y), t = t(x, y) 

s = s(x 0 (s),y 0 (s)), 

0=t(x 0 (s),y 0 (s)) 

With this we can construct our solution surface as 

and 

u = u(s, t) =u(s(x, y),t(x, y)) = u(x, y), 

u 0 (s) =u(s, 0) = u(s(x 0 (s),y 0 (s)),t(x 0 (s),y 0 (s))) = u(x 0 ,y 0 ). 

Example 6.2.11. Solve 

We first seek solutions of 

xu x +(x + y)u y = u +1, u(x, 0) = x 2 . 

dx 

dt = x, 

In this case we can parameterize C 0 by 

dy 

dt = x + y, 

du 

dt = u +1. 

x 0 (s) =s, y 0 (s) =0, u 0 (s) =s 2


and the initial conditions are 

x(s, 0) = x 0 (s) =0,y(s, 0) = y 0 (s) =0,u(s, 0) = u 0 (s) =s 2 . 

Thus we solve 

x ′ = x 

x(s, 0) = s 

} 

⇒ 

x(s, t) =se t 

y ′ − y = se t 

y(s, 0)=0 

} 

⇒ 

y(s, t) =ste t 

du 

u +1 = dt 

u(s, 0) = s 2 ⎫ 

⎬ 

⎭ ⇒ u(s, t) =(1+s2 )e t − 1. 

Since x = se t and y = ste t we see that t = y x and hence s = xe−y/x .Thus 

u(x, y) =(1+x 2 e −2y/x )e y/x − 1=e y/x + x 2 e −y/x − 1, for x ≠0. 

Example 6.2.12. (Unidirectional Wave Motion) In this linear example we seek a function 

u = u(x, t) such that 

∂u 

∂t + c∂u =0, (6.2.12) 

∂x 

u(x, 0) = F (x). (6.2.13) 


dx 

dτ = c, 

dt 

dτ =1, 

du 

dτ =0, 

subject to 

x(s, 0) = s, t(s, 0)=0, u(s, 0) = F (s). 

Thus we obtain 

x = cτ + s, t = τ, u = F (s). 

Solving for s and τ in terms of x and t we have 

We then get 

s = x − ct, 

τ = t 

u(x, t) =u(s(x, t),τ(x, t)) = F (x − ct).


Example 6.2.13. Let us now consider an example in which data is prescribed on characteristics. 

In this linear example we seek a function u = u(x, y) such that 


subject to 

Thus we obtain 

dx 

dt = x, 

x ∂u 

∂x + y ∂u 

∂y = u + 1 (6.2.14) 

u(x, x) =x 2 . (6.2.15) 

dy 

dt = y, 

du 

dt = u +1, 

x(s, 0) = s, y(s, 0) = s, u(s, 0) = s 2 . 

x = se t , y = se t , u = s 2 e t + e t − 1. 

In this example it is not possible to solve for s and t in terms of x and y. Not that A(x, y) = 

(x, y) and the characteristic manifold 

Char (x,y) (L) ={ξ =(ξ 1 ,ξ 2 ) ≠0:A(x, y) · ξ =0}. 

We note that the initial curve S in this case is x = y with constant normal vector ν =(1, −1). 

In the present case we see that 

so the curve is a characteristic. 

A(x, x) · ν = x − x =0 

Within the context of these examples in R 2 

solvability condition (6.2.7) as 

we can restate Theorem 6.2.8 with the 

Theorem 6.2.14. Suppose a(x, y, u), b(x, y, u), c(x, y, u) are C 1 in Ω ⊂ R 3 , C 0 is C 1 initial 

curve given by (x 0 (s),y 0 (s),u 0 (s)) ⊂ Ω and 

( ) 

a(x0 (s),y 

det 

0 (s),u 0 (s)) x ′ 0(s) 

b(x 0 (s),y 0 (s),u 0 (s)) y 0(s) 

′ ≠0. 

Then there exists a unique C 1 solution of 

a(x, y, u)u x + b(x, y, u)u y = c(x, y, u), 

with 

u(x 0 (s),y 0 (s)) = u 0 (s).


Proof. Our regularity assumptions guarantee that the initial value problem 

dx 

dt 

dy 

dt 

du 

dt 

= a(x, y, u), x(s, 

= b(x, y, u), y(s, 

= c(x, y, u), u(s, 

0) = x 0(s) 

0) = y 0(s) 

0) = u 0(s) 

has a unique local solution that is C 1 in t and s. By our hypotheses 

⎛ ⎞ 

∂x ∂x 

|J| = det ⎜ ∂t ∂s 

⎟ 

⎝∂y 

∂y⎠ 

= 

∣ a(x 0(s),y 0 (s),u 0 (s)) x ′ 0(s) 

b(x 0 (s),y 0 (s),u 0 (s)) y 0(s) 

′ 

∣ 

∂t ∂s t=0 

∣ ≠0 

Thus |J| ≠ 0 in a neighborhood of the initial curve C 0 and by the inverse function theorem 

we can solve for 

s = s(x, y), t = t(x, y) 

and s = s(x 0 (s),y 0 (s)), 

0=t(x 0 (s),y 0 (s)), so we can define 

u = u(x, y) =u(s(x, y),t(x, y)). 

We first note that u satisfies the initial conditions: 

u(x 0 (s),y 0 (s)) = u(s, 0) = u 0 (s). 

Furthermore, by the chain rule 

( 

) 

∂s 

au x + bu y = a u s 

∂x + u ∂t 

t 

∂x 

( 

∂s 

+ b u s 

=(as x + bs y )u s +(at x + bt y ))u t . 

) 

∂y + u ∂t 

t 

∂y 

Now by our construction of x, y, and since x t = a, y t = b, wehave 

and 

as x + bs y = s x x t + s y y t = ∂s 

∂t =0, 

at x + bt y = s x x t + t y y t = ∂t 

∂t =1


and so 

au x + bu y = c. 

Recall that in the case of a differential equation in R 2 a solution surface is a surface in 

R 3 which (at least locally) can be parameterized by a two parameter family. Thus what 

we have shown is that the collection of all characteristic curves thorugh C 0 gives a solution 

surface. Uniqueness will follow by arguing that any solution surface is essentially a collection 

of characteristic curves. In particular let u(x, y) be a solution of the equation and fix a point 

P 0 (x 0 ,y 0 ,z 0 ) on the surface. Let γ :(x(t),y(t),z(t)) be the curve through P 0 determined by 

dx 

dt = a(x, y, u(x, y)), x(0) = x 0 

dy 

dt = b(x, y, u(x, y)), y(0) = y 0 

z(t) =u(x, y), z(0) = z 0 . 

Then along this curve 

dz 

dt = u dx 

x 

dt + u dy 

y 

dt = au x + bu y = c 

since u is a solution. Thus we see that γ is the characteristic curve through P 0 . In other 

words, a solution is always a union of characteristic curves. Through any point on a solution 

surface there is a unique characteristic curve. 

Therefore if C 0 is not a characteristic curve, there is a unique solution surface that 

contains it. If, on the other hand, C 0 is a characteristic curve then 

x ′ 0(s) =a(x 0 (s),y 0 (s),ϕ ( s)) 

y 0(s) ′ =b(x 0 (s),y 0 (s),ϕ ( s)) 

which contradicts |J| ≠0. 

Remark 6.2.15. The above discussion shows that if C 0 were a characteristic curve we could 

construct infinitely many solutions containing C 0 . Namely, take any curve C 1 that meets C 0 

in a point P 0 and such that |J| ≠0on C 1 . Then construct the solution surface through C 1 . 

As discussed above, this solution surface must contain the characteristic curve C 0 .


P 0 

C 0 

C 1 

Example 6.2.16. Recall the Example 6.2.12 

Solution Surface for Each Curve C 1 

u t + cu x =0, 

u(x, 0) = F (x) 

with solution given by 

u(x, t) =F (x + ct). 

We note that the initial surface is t = 0 and the characteristics are determined by 

dx 

dτ = c, x(s, 0) = s, dt 

=1,t(s, 0)=0, ⇒ t(s, τ) =τ, x(s, τ) =cτ + s, ⇒ x − ct = s. 

dτ 

Thus we see that the solution is constant along characterisitics, i.e., (x, t) on a characteristic 

means that x − ct = s = constant and the solution is given by u(x, t) =F () and u is 

constant on a characteristic, i.e., a line x − ct = s for s ∈ R. 

A generalization of this problem is the case in which c = c(x, t). Then the characteristics 

are determined by 

and along this curve 

dx 

dτ 

= c(x, t), x(s, 

0) = s, dt 

dτ 

=1,t(s, 0)=0, 

du 

dt (x(t),t)=u dx 

x 

dt + u dt 

t 

dt = u xc(x(t),t)+u t ≡ 0.


Hence u is constant on a characteristic. 

Consider, for example, 

The characteristics are determined by 

which yields the parabolas 

u t +2tu x =0, u(x, 0) = e −x2 . 

dx 

dt =2t 

x = t 2 + k, k constant . 

The characteristic through a point (ξ,0) is x = t 2 + ξ. Since u is constant on this curve we 

have 

u(x, t) = exp(−ξ 2 )=e −(x−t2 ) 2 . 

Example 6.2.17. In the study of fluid flow an important physical characteristic is the 

formation of shock waves. The simplest example of the formation of shocks can be witnessed 

in the study of certain quasi-linear equations in R 2 called hyperbolic conservation laws. 

These are equations of the form 

subject to initial conditions 

The characteristics are determined by 

Along this curve 

u t + c(u)u x =0, x ∈ R, t>0 (6.2.16) 

u(x, 0) = ϕ(x), x ∈ R. (6.2.17) 

dx 

dt = c(u) 

du 

dt (x(t),t)=u x(x(t),t) dx 

dt + u t(x(t),t) dt 

dt = u xc(u)+u t ≡ 0. 

Hence u is constant on a characteristic. The characteristics are straight lines since 

d 2 x 

dt 2 

= d dt 

( ) dx 

= d dt dt c(u) =c′ (u) du 

dt =0,


which implies that x = C 1 t + C 2 . Now since u is constant on a characteristic, then on a 

characteristic, say the characteristic passing through (ξ,0), 

dx 

dt 

= c(u(x(t),t)) = c(u(ξ,0)) = c(ϕ(ξ)). 

Thus we see that the slope of the characteristics depend on c(u) and the initial data. The 

equation for the characteristic passing through (ξ,0) is given by 

and the solution on this line is given by 

where x = c ( ϕ(ξ) ) t + ξ. 

x = c ( ϕ(ξ) ) t + ξ, 

u(x, t) =ϕ(ξ) =ϕ(x − c(ϕ(ξ))t) 

t 

(x, t) 

(ξ,0) 

x 

Characteristic through (ξ,0) 

Example 6.2.18. One particularly famous example of a hyperbolic conservation law which 

is often used as a one dimensional model for the Navier-Stokes equations is the Burgers’ 

equation given by 

Let us consider (6.2.18) subject to initial conditions 

u t + uu x =0, x ∈ R, t>0. (6.2.18) 

⎧ 

⎨ 2, x < 0 

u(x, 0) = ϕ(x) = 2 − x, 0 ≤ x ≤ 1 . (6.2.19) 

⎩ 

1, x > 1 

For x1 the slope is 1.


Let us demonstrate that u − ϕ(x − c(u)t) = 0 implicitly defines a solution u(x, t) ofthe 

equation. First, differentiating with respect to x, wehave 

u x = ϕ ′ (x − c(u)t) [ x − c(u)t ] = x ϕ′ (x − c(u)t) [ 1 − c ′ (u)u x t ] 

and we can solve for u x 

u x = 

ϕ ′ (x − c(u)t) 

1+tc ′ (u)ϕ ′ (x − c(u)t) . (6.2.20) 

Now we differentiate with respect to t, to obtain 

u t = ϕ ′ (x − c(u)t) [ x − c(u)t ] = t −ϕ′ (x − c(u)t) [ tc ′ (u)u t + c(u) ] 

and we can solve for u t 

u t = −c(u)ϕ′ (x − c(u)t) 

1+tc ′ (u)ϕ ′ (x − c(u)t) . (6.2.21) 

So, combining (6.2.20) and (6.2.21), we have 

u t + c(u)u x = 

−c(u)ϕ′ (x − c(u)t) 

1+tc ′ (u)ϕ ′ (x − c(u)t) + c(u) ϕ ′ (x − c(u)t) 

1+tc ′ (u)ϕ ′ (x − c(u)t) 

= −c(u)ϕ′ (x − c(u)t)+c(u)ϕ ′ (x − c(u)t) 

=0. 

1+tc ′ (u)ϕ ′ (x − c(u)t) 

and 

u(x, 0) = ϕ(x − 0) = ϕ(x). 

t 

t =1 

(ξ,0) 

x 

Solution Exists Only for t1 since the characteristics 

cross beyond that line and the values on u on the intersecting characteristics are different –


thus, as a function, u is not well defined. More specifically, recall that our solution is defined 

by 

u(x, t) =ϕ(ξ) where x = ϕ(ξ)t + ξ. 

The characteristics are described by 

⎧ 

⎨ 2t + ξ, ξ < 0 

x = (2 − ξ)t + ξ, 0 ≤ ξ ≤ 1 . 

⎩ 

t + ξ, ξ > 1 

We can compute that the characteristics intersect at 

((2 − ξ)t + ξ) ∣ ∣ 

ξ=0 

=(t + ξ) ∣ ∣ 

ξ=1 

, 

or 2t =1+t, i.e., t =1. Fort1) 

t 

t =1 

x = t +1 

x =2t 

u =2 

(ξ,0) 

u =1 

x 

Solution 

If 0 ≤ ξ ≤ 1 the characteristic passing through (ξ,0) is x =(2− ξ)t + ξ which implies 

ξ =(x − 2t)/(1 − t) and so 

( ) x − 2t 

u(x, t) =2− = 2 − x , 2t ≤ x ≤ t +1, t < 1. 

1 − t 1 − t 

u 

t =1 

u =2 

t 

u =1 

x


Solution 

At t = 1 refered to as the “breaking time,” a “shock” develops. As an exercise you will 

show that for general c(u), the breaking time is given by 

t b = min 

ξ 

−1 

ϕ ′ (ξ)c ′ (ϕ(ξ)) ,t b > 0, 

at which 

1+c ′ (ϕ(ξ))ϕ ′ (ξ)t =0.


Exercise Set 1: First Order Linear and Quasi-Linear Equations 

1. Solve the initial value problems 

(a) yz x − xz y =2xyz for z = t 2 on C: x = t, y = t, t>0. 

(b) zz x + yz y = x with z =2t on C: x = t, y = t, t ∈ R. 

2. Solve ∂ x u + ∂ y u = u with u = cos(x) when y =0. 

3. Solve x 2 ∂ x u + y 2 ∂ y u = u 2 with u = 1 when y =2x. 

4. Solve u∂ x u + ∂ y u = 1 with u = 0 when y = x. What happens in this problem if we 

replace u =0byu =1 

5. Solve u x + u y = u 2 with the initial condition u(x, 0) = h(x). 

6. Show that for the initial value problem, 

u t + c(u)u x =0,x∈ R, t > 0, 

u(x, 0) = φ(x) 

the breaking time will occur at the minimum value of t for which 

−1 

t b = min 

φ ′ (ξ)c ′ (φ(ξ)) 

for which 

1+φ ′ (ξ)c ′ (φ(ξ))t =0. 

7. (a) Solve the initial value problem 

uu x + u t =0, u(x, 0) = f(x). 

(b) If f(x) =x, show that the solution exists for all t>0. 

(c) If f(x) =−x, show that a shock develops, that is, the solution blows up in finite 

time. 

⎧ 

8. Let D be a constant and H the Heaviside function. Solve 

⎪⎨ u t + cu x =0,x∈ R,t>0, 

⎪⎩ 

u(x, 0) = −H(−x)xe x/D 

if (c 0 is a constant) 

a) c(x) =c 0 (1 − x/L) 

b) c(x, t) =c 0 (1 − x/L − t/T ).

6.3. CHARACTERISTICS AND HIGHER ORDER EQUATIONS 31 

6.3 Characteristics and Higher Order Equations 

6.3.1 Characteristics and Classification of 2nd Order Equations 

L = 

n∑ ∂ 2 

a ij (x) 

(6.3.1) 

∂ xi ∂ xj 

i,j=1 

where a ij are real valued functions in Ω ⊂ R n and a ij = a ji . Fix a point x 0 ∈ Ω. The 

characteristic polynomial is given by 

We say that the operator L is: 

σ x0 (L, ξ) = 

n∑ 

a ij (x 0 )ξ i ξ j (6.3.2) 

i,j=1 

1. Elliptic at x 0 if the quadratic form (6.3.2) is non-singular and definite, i.e., can be 

reduced by a real linear transformation to the form 

n∑ 

ã i ξi 2 + l. o. t. 

i=1 

2. Hyperbolic at x 0 if the quadratic form (6.3.2) is non-singular and indefinite and can be 

reduced by a real linear transformation to a sum of n squares, (n − 1) of the same sign, 

i.e., to the form 

n∑ 

ξ1 2 − ã i ξi 2 + l. o. t. 

i=2 

3. Ultra-Hyperbolic at x 0 if the quadratic form (6.3.2) is non-singular and indefinite and 

can be reduced by a real linear transformation to a sum of n squares, (n ≥ 4) with 

more than one terms of either sign. 

4. Parabolic at x 0 if the quadratic form (6.3.2) is singular, i.e., can be reduced by a real 

linear transformation to a sum of fewer than n squares, (not necessarily of the same 

sign). 

It can be shown that in the constant coefficient case a reduction to one of these forms is 

always possible with a simple constant matrix transformation of coordinates. 

The case of two independent variables and non-constant coefficients can also be analyzed. 

a ∂2 u 

∂x 2 +2b ∂2 u 

∂x∂y + c∂2 u 

∂y 2 + F (x, y, u, u x,u y ) = 0 (6.3.3)


where a = a(x, y), b = b(x, y), c = c(x, y) are C 2 real valued functions in Ω ⊂ R 2 and (a, b, c) 

doesn’t vanish at any point. Let us restrict to the linear case 

a ∂2 u 

∂x +2b ∂2 u 

2 ∂x∂y + u 

c∂2 ∂y + d∂u 

2 ∂x + e∂u + fu = g (6.3.4) 

∂y 

Example 6.3.1. Consider the one dimensional wave equation 

Let 

u xx − u yy =0. 

α = ϕ(x, y) =x + y, β = ψ(x, y) =x − y. 

So we have 

(α + β) 

(α − β) 

x =Φ(α, β) = , y =Ψ(α, β) = . 

2 

2 

We can express the solutions in terms of the (x, y) or(α, β) coordinates 

By the chain rule we have 

Thus we have 

u(x, y) =u(Φ(α, β), Ψ(α, β)) = U(α, β). 

u x = U α ϕ x + U β ψ x = U α + U β , (6.3.5) 

u xx = U αα +2U αβ + U ββ , (6.3.6) 

u y = U α ϕ y + U β ψ y = U α − U β , (6.3.7) 

u yy = U αα − 2U αβ + U ββ . (6.3.8) 

0=u xx − u yy =4U αβ , or U αβ =0. 

The general solution of this equation is given by 

which, in turn, implies 

U(α, β) =G(α)+F (β) 

u(x, y) =G(x + y)+F (x − y). 

This solution represents a pair of waves traveling at the same speed but in opposite directions. 

Returning to (6.3.4), if we introduce a nonsingular C 1 coordinate transformation 

{ α = ϕ(x, y) 

β = ψ(x, y),


i.e., the Jacobian is not zero 

|J| = det 

⎡ 

⎢ 

⎣ 

∂α 

∂x 

∂β 

∂x 

⎤ 

∂α 

∂y 

⎥ 

∂β ⎦ ≠0. 

∂y 

Applying the chain rule we have 

u x = u α ϕ x + u β ψ x , u y = u α ϕ y + u β ψ y , 

u xx = u αα ϕ 2 x +2u αβ ϕ x ψ x + u ββ ψx 2 + u α ϕ xx + u β ψ xx , 

u yy = u αα ϕ 2 y +2u αβ ϕ y ψ y + u ββ ψy 2 + u α ϕ yy + u β ψ yy , 

u xy = u αα ϕ x ϕ y + u αβ (ϕ x ψ y + ϕ y ψ x )+u ββ ψ x ψ y + u α ϕ xy + u β ψ xy . 

Then the equation (6.3.4) is transformed into a new equation of exactly the same form 

where 

ã ∂2 u 

∂α +2˜b ∂2 u 

2 ∂α∂β + ˜c ∂2 u ∂u 

+ ˜d 

∂β2 ∂α + ẽ ∂u 

∂β + ˜fu = ˜g (6.3.9) 

1. ã = aα 2 x +2bα x α y + cα 2 y 

2. ˜b = aα x β x + b(α x β y + α y β x )+cα y β y 

3. ˜c = aβx 2 +2bβ x β y + cβy 

2 (6.3.10) 

4. ˜d = aαxx +2bα xy + cα yy + dα x + eα y 

5. ẽ = aβ xx +2bβ xy + cβ yy + dβ x + eβ y 

6. ˜f = f and ˜g = g 

Furthermore, we have the following extremely important invariance 

˜D =(˜b 2 − ã˜c) =(b 2 − ac)J 2 = DJ 2 . 

With this we then obtain the following theorem. The proof is constructive. 

Theorem 6.3.2. D is positive, negative or zero if and only if ˜D is. Furthermore, we have 

1. D>0 implies Hyperbolic


2. D =0implies Parabolic 

3. D0 implies ã = ˜c =0 

2. D =0implies ã = ˜b =0 

3. D 0. If a = c = 0 we are done so we assume 

without loss of generality that a ≠ 0. We seek α = ϕ(x, y) and β = ψ(x, y) so that ã 

and ˜c are zero. We first note that the calculations (6.3.10) suggest that we consider an 

expression of the form 

av 2 x +2bv x v y + cv 2 y = 0 (6.3.11)


where v could represent either ϕ or ψ. We note that (6.3.11) can be factored into 

[ ( √ ) ][ ( √ ) ] 

−b + b2 − ac 

−b − b2 − ac 

a v x − 

v y v x − 

v y . (6.3.12) 

a 

a 

Thus from (6.3.12) we seek ϕ and ψ so that, for example, 

[ ( √ ) ] 

−b + b2 − ac 

ϕ x − 

ϕ y =0, (6.3.13) 

a 

and 

[ 

ψ x − 

( √ ) ] 

−b − b2 − ac 

ψ y =0. (6.3.14) 

a 

With these choices (6.3.11) is satisfied with v given by ϕ and ψ. In addition we must 

impose a noncharacteristic solvability condition 

∣ 

∂(ϕ, ψ) ∣∣∣∣ 

∂(x, y) = ϕ x 

ψ x 

ϕ y 

ψ y 

∣ ∣∣∣∣ 

( √ ) 

−b + b2 − ac 

= 

ϕ y ψ y − 

a 

( −b − 

√ 

b2 − ac 

a 

) 

ϕ y ψ y 

= 2 a 

√ 

b2 − ac ϕ y ψ y ≠0. (6.3.15) 

A solution of (6.3.13) is found by solving 

dx 

dt =1, 

( √ ) 

dy −b + 

dt = − b2 − ac 

, 

a 

dϕ 

dt =0. 

For this problem we take the initial conditions 

x 0 (s) =x 0 ,y 0 (s) =y 0 + s, 

ϕ 0 (s) =s. 

With this choice we find (in our earlier notation we used a and b which are not the 

same as those occuring in the present problem) 

∣ ∣∣∣∣∣∣ a x ′ 1 0 

0(s) 

∣b 

y 0(s) 

′ ∣ = ( √ ) 

b − b2 − ac 

≠0. (6.3.16) 

1 

a ∣


Moreover, 

and so 

ϕ(s, t) =s 

ϕ s =1=ϕ x x s + ϕ y y s . 

If ϕ y = 0, then it follows from (6.3.13) that ϕ x = 0, a contradiction. Now, as we have 

learned in our earlier work, the condition (6.3.16) guarantees that we can solve for 

(s, t) in terms of (x, y) to obtain ϕ(x, y). 

Similarly, we can obtain ψ(x, y) by solving 

dx 

dt =1, 

( √ ) 

dy b + 

dt = b2 − ac 

, 

a 

For this problem we take the initial conditions 

dψ 

dt =0. 

x 0 (s) =x 0 ,y 0 (s) =y 0 + s, 

ψ 0 (s) =s. 

Again, ψ(s, t) =s and as above ψ y ≠0. ThusJ ( (ϕ, ψ)/(x, y) ) ≠0. 

In summary we have found a change of coordinates α = ϕ(x, y), β = ψ(x, y) that 

reduces the equation to 

2˜bu αβ + l.o.t. = 0. 

It is easy to show that 

˜b =2 

( ac − b 

2 

a 

) 

ϕ y ψ y (6.3.17) 

and so ˜b ≠ 0 and we can divide by it to put (6.3.17) into the normal form 

Remark 6.3.4. 

u αβ = F (α, β, u, u α ,u β ). 

(a) The reduction to normal form is a local result. 

(b) The special initial curve can be replaced by any curve (x 0 (s),y 0 (s),ϕ 0 (s)) with 

ϕ 0 (s) =s and (x 0 (s),y 0 (s)) that specifies a curve in the xy-plane where the PDE 

is hyperbolic. 

(c) The curves ϕ(x, y) = constant, ψ(x, y) = constant are called characteristics of 

(6.3.4). Further, equation (6.3.11) is called the characteristic equation for the 

PDE.


(d) Note that the characteristics for the first order PDE’s for ϕ and ψ are determined 

by 

( √ ) 

dx 

dt =1, dy b ± 

dt = b2 − ac 

a 

or 

( √ ) 

dy b ± 

dx = b2 − ac 

. 

a 

But if ϕ(x, y) = constant and ϕ solves (6.3.13), then 

and so 

dy 

dx = −ϕ x 

ϕ y 

= 

ϕ x + ϕ y 

dy 

dx =0 

( b − 

√ 

b2 − ac 

a 

) 

. (6.3.18) 

Hence the characteristics for a 2nd order PDE (6.3.4) coincide with the characteristics 

of the associated 1st order PDE (6.3.13) (Similarly for ψ). 

(e) In order to obtain the other hyperbolic form we set 

ξ = α + β, 

η = α − β 

so that 

and 

Thus we have 

α = ξ + η 

2 , β = ξ − η 

2 , 

u α = u η η α + u ξ ξ α = u η + u ξ , u β = u η η β + u ξ ξ β = −u η + u ξ . 

u αβ =(−u ηη + u ηξ )+(−u ξη + u ξξ )=u ξξ − u ηη . 

Example 6.3.5. Consider the equation y 2 u xx − x 2 u yy = 0, x > 0, y > 0. Here 

b 2 − ac = x 2 y 2 > 0. The characteristics are given by ( see (6.3.18)) 

( √ ) 

dy b + 

dx = b2 − ac 

= x a y , 

and 

( √ ) 

dy b − 

dx = b2 − ac 

= − x a 

y .


Thus the characteristic curves are given by 

y 2 − x 2 = constant, 

y 2 + x 2 = constant. 

We take 

α = ϕ(x, y) =x 2 + y 2 , β = ψ(x, y) =x 2 − y 2 . 

We solve for x, y to obtain 

If we use (6.3.10) we find 

√ √ 

α + β α − β 

x = , y = . 

2 

2 

˜b =8x 2 y 2 =2(α 2 − β 2 ) 

˜d =2(y 2 − x 2 )=−2β 

ẽ =2(y 2 + x 2 )=2α 

and also 

u αβ = 

2β 

4(α 2 − β 2 ) u 2β 

α − 

4(α 2 − β 2 ) u β 

= βu a − αu β 

2(α 2 − β 2 ) . 

Example 6.3.6. Suppose that a, b, c, d, e, f in (6.3.4) are constants. Then the characteristics 

are given by 

dy 

dx = b ± √ b 2 − ac 

≡ ν ± . 

a 

Thus the characteristics are straight lines 

α = ϕ(x, y) =y − ν + x, 

β = ψ(x, y) =y − ν − x. 

In this case we get ã = ˜c = 0 and 

˜b = aϕx ψ x + b(ϕ x ψ y + ϕ y ψ x )+cϕ y ψ x 

= aν + ν − − b(⃗u + + ν − )+c 

= 2(ac − b2 ) 

. 

a


For the first order terms we have 

˜d = aϕ xx +2bϕ xy + cϕ yy + dϕ x + eϕ y = dν + + e ≡ d ′ , 

ẽ = aψ xx +2bψ xy + cψ yy + dψ x + eψ y = dν − + e ≡ e ′ . 

Recalling that there is a 2 timdes the ˜b term and multiplying by K, the transformed 

equation is 

u αβ + Kd ′ u α + Ke ′ u β + Kfu = Kg 

where 

K = 

a 

4(ac − b 2 ) . 

We note that a further reduction is always possible in this case. Namely we can remove 

the first order terms. Let 

u = e λα+µβ v. 

Then we have 

u α = e λα+µβ (λv + v α ) 

u β = e λα+µβ (µv + v β ) 

u αα = e λα+µβ (v αα +2λv α + λ 2 v) 

u ββ = e λα+µβ (v ββ +2µv β + µ 2 v) 

u αβ = e λα+µβ (v αβ + λv β + µv a + λµv). 

If we choose µ = −d ′ K, λ = −e ′ K and define 

f 1 = Kf + K 2 d ′ e ′ , 

g 1 = e −(λα+µβ) Kg, 

then the equation can be written in terms of v as 

v αβ + f 1 v = g 1 . 

II. Parabolic Case: Assume that D = b 2 − ac = 0. In this case, if a = 0, then b = 0 and 

we are done since c ≠ 0 (otherwise the equation is not truely second order). Thus we 

assume that a ≠ 0 and the characteristic equation (6.3.11) factors as 

( 

a v x + b 2 

y) 

a v =0. (6.3.19)


Thus we need to find a solution ϕ(x, y) of (6.3.19) and then select ψ(x, y) so that 

∣ 

∂(ϕ, ψ) ∣∣∣∣ 

∂(x, y) = ϕ x 

ψ x 

ϕ y 

ψ y 

∣ ∣∣∣∣ 

= ϕ x ψ y − ψ x ϕ y 

= − b a ϕ yψ y − ψ x ϕ y 

= −ϕ y 

(ψ x + b ) 

a ψ y ≠0. 

We may, for example, take ψ = x and prescribe inital conditions for (6.3.18) so that 

ϕ y ≠ 0. With these choices, ã = 0 (since ϕ satisfies (6.3.18)) and from (6.3.10) we find 

˜b = aϕx ψ x + b(ϕ x ψ y + ϕ y ψ x )+cϕ y ψ y 

( 

= a − b ) [( 

a ϕ y ψ x + b − b ) ] 

a ϕ y ψ y + ϕ y ψ x + cϕ y ψ x 

[ 

] 

= ϕ y −bψ x − b2 

a ψ y + bψ x + cψ y 

[ ] 

= −ϕ y ψ y − b2 

a + c =0. 

Thus with α = ϕ(x, y), β = ψ(x, y), (6.3.4) becomes 

˜cu ββ + ˜du α + ẽu β + ˜fu = ˜g. 

Furthermore, since b 2 = ac, and with the choice ψ = x we have 

( 

˜c = aψx 2 +2bψ x ψ y + cψy 2 = a ψ x + b 2 

y) 

a ψ = a ≠0. 

Therefore we can divide by ˜c = a to get the desired form. 

If ϕ(x, y) is a solution of (6.3.19) and ϕ(x, y) = constant, then it follows from (6.3.18) 

that the characteristics could be obtained from 

dy 

dx = b a . 

Example 6.3.7. Let us transform the equation 

u xx − 2xu xy + x 2 u yy − 2u y =0


into canonical form. Here we have b 2 −ac = x 2 −x 2 = 0, so the equation is of parabolic 

type and the characteristics are found from 

or 

Take 

dy 

dx = b a = −x, 

y = − x2 

2 + c. 

ϕ(x, y) =y + x2 

2 , 

ψ(x, y) =x. 

Then 

˜d = aϕ xx +2bϕ xy + cϕ yy + dϕ x + eϕ y = (1)(1)+0+0+0+(−2)(1) = −1, 

ẽ = aψ xx +2bψ xy + cψ yy + dψ x + eψ y =0, 

and so 

u ββ − u α =0. 

III. Elliptic Case: Assume that D = b 2 − ac < 0. For this case we choose to proceed 

in a purely formal fashion. This problem turns out to be rather messy and lengthy 

to carry out in detail. For a detailed treatment we refer to the text by Garabedian 

[6]. In particular, we will assume that a, b and c are analytic functions near the 

point in question (x 0 ,y 0 ). This can be avoided but not without great difficulty. The 

characteristic equation (6.3.18) factors in this case into 

a 

( 

v x − 

[ −b + i 

√ 

ac − b 

2 

a 

] 

v y 

)( 

v x − 

[ √ ] ) 

−b − i ac − b 

2 

v y =0. 

a 

We seek a solution of the characteristic equation solving, for example, 

( [ √ ] ) 

−b + i ac − b 

2 

v x − 

v y =0. 

a 

Let us suppose that we find z(x, y) =ϕ(x, y)+iψ(x, y) so that 

Note that in the present case we must have ac < 0 which means that a and c have the 

same sign and are not zero. We seek a holomorphic function ϕ = ϕ 1 + iϕ 2 such that 

aϕ x + 

(b + i √ ) 

ac − b 2 ϕ y =0,


or 

a (ϕ 1x + iϕ 2x )+ 

( 

b + √ ) 

ac − b 2 )(ϕ 1y + iϕ 2y ) =0. 

Note that, since we have assumed that a, b, c are real the solution ψ of 

aψ x + 

(b − i √ ) 

ac − b 2 ψ y =0, 

is given by ψ = ϕ = ϕ 1 − iϕ 2 . 

To argue that these complex equations are solvable we collecting real and imaginary 

parts in the equation for ϕ to obtain a first order linear system of partial differential 

equations 

or 

aϕ 1x + bϕ 1y − √ (ac − b 2 ) ϕ 2y =0 

aϕ 2x + bϕ 2y + √ (ac − b 2 ) ϕ 1y =0. 

( ) 

∂ ϕ1 

= 1 ( 

−b 

∂x ϕ 2 a − √ (ac − b 2 ) 

If we prescribe any Cauchy initial data 

( 

ϕ1 

ϕ 2 

) 

(0,y)= 

√ ) (ac − b2 ) ∂ 

−b ∂y 

( ) 

g1 (y) 

g 2 (y) 

( 

ϕ1 

ϕ 2 

) 

. 

this system has a unique solution by the Cauchy-Kovalevski Theorem. 

We now introduce the real transformations 

and note that 

ξ = ϕ 1 =Reϕ, 

η = ϕ 2 =Imϕ 

( ) 

ϕ1x ϕ 

det J = det 

1y 

ϕ 2x ϕ 2y 

⎛ 

= 1 a det ⎝ −bϕ 1y + √ ⎞ 

(ac − b 2 )ϕ 2y ϕ 1y 

2 

− √ ⎠ 

(ac − b 2 )ϕ 1y − bϕ 2y ϕ 2y 

( ) 

ϕ1x ϕ 

= det 

1y 

ϕ 2x ϕ 2y 

(√ ) 2 ( ) 

(ac − b2 ) ϕ2y ϕ 

= 

det 

1y 

a 

−ϕ 1y ϕ 2y 

(√ ) 2 

(ac − b2 ) (ϕ ) 

2 

= 

a 

2y + ϕ 2 1y ≠0.


Thus, by our construction of ϕ, wehave 

which implies 

aϕ 2 x +2bϕ x ϕ y + cϕ 2 y =0 

a (ϕ 1x + iϕ 2x ) 2 +2b (ϕ 1x + iϕ 2x )(ϕ 1y + iϕ 2y )+c (ϕ 1y + iϕ 2y ) 2 =0. 

Collecting the real parts we have 

a ( ϕ 2 1x − ϕ 2 2x) 

+2b (ϕ1x ϕ 1y − ϕ 2x ϕ 2y )+c ( ϕ 2 1y − ϕ 2 2y) 

=0. 

We now collect the terms involving ξ = ϕ 1 on the left and those involving η = ϕ 2 on 

the right to get 

aξ 2 x +2bξ x ξ y + cξ 2 y = aη 2 x +2bη x η y + cη 2 y 

which from (6.3.10) gives 

ã = ˜c. 

Next we collect the imaginary parts to get 

2 

(aϕ 1x ϕ 2x + b ( ) 

) 

ϕ 2x ϕ 1y + ϕ 2y ϕ 1x + cϕ1y ϕ 2y =0, 

which implies 

˜b = aξx η x + b ( ξ y η x + ξ x η y 

) 

+ cξy η y =0. 

Also, since ˜b = 0, we know from invariance that 

−ã˜c = ˜b 2 − ã˜c 0, y > 0, 

into canonical form. 

In this case we have 

dy 

dx = iy x = b + i√ ac − b 2 

a


or 

We take 

Then 

− ln(x) =i ln(y)+K, K a constant. 

α = ϕ 1 (x, y) = ln(x), β = ϕ 2 (x, y) = ln(y). 

ã = x 2 ( 1 

x 2 ) 

+2b 1 x (0)+0=1=˜c, 

˜b =0, 

( 

˜d = x 2 − 1 ) 

+0+y 2 (0) = −1, 

x 2 ( ) −1 

ẽ = x 2 (0)+0+y 2 = −1. 

y 2 

Thus we have 

u αα + u ββ − u α − u β =0. 

We conclude this discussion by considering another example. The main point of this 

example is to illustrate that the classification of a differential equation is a local result. 

Example 6.3.9. The Tricomi Equation. 

u yy − yu xx =0. 

For this equation, D = b 2 − ac = y. So when y0 

the equation is hyperbolic and when y = 0 the equation is parabolic. 

For this example a = −y, b = 0 and c = 1 so the characteristic equation 

dy 

dx = b ± √ b 2 − ac 

a 

reduces to 

√ 

dy −ac 

dx = ± a 

= ± 1 √ y 

. 

Since 

For y>0 the equation is hyperbolic and the characteristic curves are are given by 

3x ± 2y 3/2 = constant


The transformations 

ξ =3x − 2y 3/2 , η =3x +2y 3/2 , 

reduce the equation to the normal form 

u ξη − 1 u ξ − u η 

6 ξ − η =0. 

y 

hyperbolic 

0 

elliptic 

Characteristics for Tricomi Equation 

x 

parabolic 

Return to Characteristics for Higher Order Equations 

Let us return to the general notation for linear differential operators and characteristics 

introduced at the begining of the chapter. Namely, a kth order linear equation on Ω in R n 

is given by 

L = ∑ 

a α (x)∂ α . 

|α|≤k 

The principal part of the equation is consists of all kth order terms 

P = ∑ 

a α (x)∂ α . 

|α|=k 

Let us consider some examples For the Laplacian in two variables we have equation and 

principal part are the same and given by 

L = P = ∂ (2,0) 

1 + ∂ (0,2) 

2 = ∂2 

+ ∂2 

. 

∂x 2 1 ∂x 2 2


For the wave equation in R 2 we again have the equation and principal part are the same 

L = P = ∂ (2,0) 

1 − ∂ (0,2) 

2 = ∂2 

− ∂2 

. 

∂x 2 1 ∂x 2 2 

For the heat operator in R 2 we have the equation and principal part which in this case 

are not the same 

L = ∂ (2,0) 

1 − ∂ (0,1) 

2 = ∂2 

− 

∂ , 

∂x 2 1 ∂x 2 

and 

P = ∂ (2,0) 

1 = ∂2 

. 

∂x 2 1 

The characteristic form (or principal symbol) atx ∈ Ω is the homogeneous polynomial 

of degree k defined for ξ ∈ R n by 

χ L (x, ξ) = ∑ 

a α (x)ξ α . 

|α|=k 

A vector ξ is characteristic or is called a characteristic direction, for L at x if 

χ L (x, ξ) =0. 

The characteristic variety is the set of all characteristic vectors ξ, i.e., 

Char x (L) ={ξ ≠0:χ L (x, ξ) =0}. 

As we already have defined, a surface (or curve) S is said to be characteristic with respect 

to L at x 0 if the normal vector ν(x 0 )atx 0 defines a characteristic directions. 

Example 6.3.10. Consider 

L = a(x, y) ∂ 

∂x + b(x, y) ∂ ∂y = a (1,0)∂ (1,0) + a (0,1) ∂ (0,1) . 

The characteristic form is 

χ L (x, ξ) =a(x, y)ξ 1 + b(x, y)ξ 2 . 

Let C be a characteristic curve given parametrically by (x(t),y(t)). The tangent to this 

curve is (x ′ (t),y ′ (t)) and the normal is (y ′ (t), −x ′ (t)). Since C is a characteristic curve we 

must have 

a(x(t),y(t))y ′ (t) − b(x(t),y(t))x ′ (t) =0. 

Thus the characteristic curve can be found from 

dx 

dt = a(x(t),y(t)), dy 

dt = b(x(t),y(t)) 

just as we defined them earlier.


Example 6.3.11. Reconsider the wave equation in R 2 

Lu = ∂ (2,0) 

1 u − ∂ (0,2) 

2 u = ∂2 u 

− ∂2 u 

=0. 

∂x 2 1 ∂x 2 2 

If ξ =(ξ 1 ,ξ 2 ) defines a characteristic direction, then we must have 

ξ 2 1 − ξ 2 2 =0. 

Now if C : (x(t),y(t)) is a characteristic curve, let 

ξ 1 = dy 

dt , 

ξ 2 = − dx 

dt 

denote a normal to C. Then we have 

( ) 2 ( ) 2 ( dy dx dy 

− = 

dt dt dt − dx )( dy 

dt dt + dx ) 

=0 

dt 

This equation is satisfied if we take 

or 

dy 

dx = ±1, 

y − x = c, y + x = c, 

and we see that the characteristic curves are straight lines (just as we saw earlier). 

Example 6.3.12. For the equation (6.3.4), the principal part is 

A characteristic curve (x(t),y(t)) has normal 

Pu = a ∂2 u 

∂x 2 +2b ∂2 u 

∂x∂y + c∂2 u 

∂y 2 . 

that satisfies 

or 

or 

a 

( ) 2 dy 

− 2b 

dt 

ξ 1 = dy 

dt , 

ξ 2 = − dx 

dt 

aξ 2 1 +2bξ 1 ξ 2 + cξ 2 2 =0, 

( dy 

dt 

)( ) dx 

+ c 

dt 

ady 2 − 2bdydx + cdx 2 =0, 

( ) 2 dx 

=0, 

dt


or 

[ ( √ ) ][ ( √ ) ] 

b + b2 − ac 

b − b2 − ac 

a dy − 

dx dy − 

dx =0. 

a 

a 

Hence the characteristic curves can be found by solving 

( √ ) 

dy b ± 

dx = b2 − ac 

, 

a 

just as we saw earlier. 

In an attempt to further illustrate the significance of characteristics consider the following 

example. 

Example 6.3.13. Consider the Cauchy problem that we looked at earlier 

Lu = ∂u 

∂x =0, u(x, 0) = f(x), (x, y) ∈ R2 . 

From the PDE we must have u(x, y) =F (y) and the initial condition gives u(x, 0) = 

f(x) =F (0) so that f must be a constant or the problem has no solution. If f(x) =c is a 

constant then we can take any F (y) such that F (0) = c so we get infinitely many solutions. 

Note that the characteristics for this example are determined by 

dy 

dx 

=0, ⇒ y = constant. 

So we have prescribed our initial data along a characteristic curve. Note that y is constant 

on a characteristic. 

6.3.2 The Cauchy Problem for Higher Order Equations 

Let S be a hypersurface of class C k .Ifu ∈ C k−1 defined near S, then the Cauchy data of u 

on S is the set 

{u, ∂ ν u, ··· ,∂ν 

k−1 u}. 

We observe that we can consider the so-called Cauchy problem in a neighborhood of a 

single point and then introduced a change of variables to transform the problem so that S 

contains the origin and near the origin S coincides with the hyperplane x n =0. Thusit 

is often the case when treating higher order initial value problems that we distinguish the 

variable x n and denote it by t and then define x =(x 1 , ··· ,x n−1 ) ∈ R n−1 . Also we then 

consider ∂ t and ∂ x and multi-indices α =(α 1 , ··· ,α n−1 ). With this the Cauchy Problem is:


Given functions {φ j (x)} k−1 

j=0 solve 

F ( x, t, (∂ α x ∂ j t u) |α|+j≤k 

) 

=0 

with 

∂ j t u(x, 0) = φ j (x), 0 ≤ j ≤ (k − 1). 

This problem is far to general since the Cauchy data determine many derivatives on S. 

Indeed, up to order k the only unknown is ∂t k u. For the Cauchy problem to be well-posed it 

must be assumed that F = 0 can be solved for ∂t k . This is a characteristic condition on S. 

Numerous examples were given to illustrate this situation. 

We could embark on an in-depth treatment of the Quasi-linear case. In particular, we 

could consider the notion of noncharacteristic surface for the Cauchy problem: An initial 

value problem 

∑ ( ) 

a α,j x, t, ((∂ 

β 

x ∂tu) i |β|+i≤k−1 ∂ 

α 

x ∂ j t u = b ( ) 

x, t, (∂x β ∂tu) i |β|+i≤k−1 

|α|+j=k 

∂ j t u(x, 0) = φ j (x), 0 ≤ j ≤ (k − 1) 

is called non-characteristic if 

a 0,k 

( 

x, 0, (∂ 

α 

x ∂ j t u) |α|+j≤k 

) 

≠0 

With this we could be in a position to prove the main fundamental existence and uniqueness 

result (such as it is) for PDE’s with analytic data. Rather that take such a deep 

excursion we will simple state the main result – the Cauchy-Kovalevski Theorem. 

Theorem 6.3.14. If G, {φ j } k−1 

j=0 are analytic near the origin, the Cauchy problem 

⎧ ( 

) 

⎨ ∂t k u = G x, t, (∂x α ∂ j t u) |α|+j≤k 

j


Theorem 6.3.15. Suppose that a α , f, ϕ 1 , ··· ,ϕ k and the surface S are analytic (note that 

to say S is analytic means that if S is given by F (x 1 , ··· ,x n )=0then F is an analytic 

function). Suppose further that the initial surface S is not characteristic at x 0 ∈ S, i.e., 

∑ 

a α (x 0 ) [ ν(x 0 ) ] α 

≠0, 

|α|≤k 

where ν(x 0 ) is the normal to S at x 0 . Then the Cauchy problem has a solution u(x) in a 

neighborhood of x 0 . This solution is unique in the class of analytic solutions. 

Remark 6.3.16. 1. The Cauchy-Kovalevski theorem says that under certain conditions 

we can seek a solution in the form of a power series. While sometimes useful, this is 

not a very practical method of solution. 

2. The Cauchy-Kovalevski theorem guarantees that the Cauchy problem 

∆u(x, y) =0, |x| < ∞, y > 0, 

u(x, 0) = f(x), |x| < ∞, 

∂u 

(x, 0) = g(x), |x| < ∞ 

∂x 

has a unique solution for every analytic f and g. It does not, however, say that the 

problem is well posed. Recall the Hadamard example we discussed earlier. 

At this point we mention that there are numerous related topics that should be considered 

if we had the time. 

Additional topics in this area include 

1. Holmgren uniqueness theorem 

2. Non-continuous dependence example of Hadamard 

3. Classical Lewy example of an equation with no solution of class C 1 for any C ∞ (not 

analytic) right hand side 

4. Malgrange-Ehrenpreis theorem on local solvability for constant coefficient operators. 

Unfortunately, with our time constraints the best we can do is to state some of these 

results for completeness. 

The first is the Holgren Uniqueness Theorem which says that analyticity is not required 

in the Cauchy-Kovalevski theorem to guarantee uniqueness. 

Theorem 6.3.17 (Holmgren’s Uniqueness Theorem). Under the same assumptions as 

the Cauchy-Kovalevski theorem, any two C k solutions in a neighborhood of x 0 must coincide 

in a perhaps smaller neighborhood of x 0 .


The next theorem due to Malgrange and Ehrenpreis gives an existence theorem for general 

constant coefficient equations. 

Theorem 6.3.18 (Malgrange-Ehrenpreis Theorem). If 

L = ∑ 

a α ∂ α , 

|α|≤k 

is a kth order linear differential equation with constant coefficients on R n and f ∈ 

C0 ∞ (R n ) (the space of infinitely differentiable functions with compact support), then there 

exists a u ∈ C ∞ (R n ) such that Lu = f. 

On the flip side we should also mention the classical nonexistence example due to Hans 

Lewy (1957). This result shattered all hopes of generalizing the Cauchy-Kovalevski theorem 

to the non-analytic case. 

Consider the differential operator L defined in R 3 with coordinates (x, y, t) given by 

L = ∂ 

∂x + i ∂ ∂y − 2i(x + iy) ∂ ∂t . 

Theorem 6.3.19 (Hans Lewy Theorem). Let f be a continuous real-valued function depending 

only on t. If there is a C 1 function u of (x, y, t) satisfying Lu = f in some neighborhood 

of the origin, then f must actually be analytic. 

Once this result is known, it is possible to show that there are C ∞ functions g on R 3 

such that the equation Lu = g has no solution u ∈ C 1+α (α>0) in any neighborhood of 

any point.


Exercise Set 2: Classification and Canonical Forms 

1. Classify the operators and find the characteristics (if any) through the point (0, 1): 

(a) u tt + u xt + u xx =0 

(b) u tt +4u xt +4u xx =0 

(c) u tt − 4u xt + u xx =0 

2. Find where the operators are hyperbolic, parabolic and elliptic: 

(a) Lu = u tt + tu xx + xu x 

(b) Lu = x 2 u tt − u xx + u 

(c) Lu = tu tt +2u xt + xu xx + u x 

3. Transform the equation to canonical form: 2u xx − 4u xy − 6u yy + u x =0. 

4. Transform the equations to canonical form: 

(a) e 2x u xx +2e x+y u xy + e 2y u yy + u x =0. 

(b) x 2 u xx + y 2 u yy = 0 for x>0, y>0. 

5. Transform the equations to canonical form: 

(a) u xx +2u xy +17u yy =0. 

(b) 4u xx +12u xy +9u yy − 2u x + u =0. 

6. Find the characteristics of Lu = u tt − tu xx through the point (0, 1). 

7. Determine where the following equation is hyperbolic or parabolic. 

yu xx +(x + y)u xy + xu yy =0 

Where the equation is hyperbolic, show that the general solution may be written as 

u(x, y) = 1 ∫ 

f(β) dβ + g(y − x). 

y − x 

where β = y 2 − x 2 

8. Classify the following equation and find the canonical form, u xx − 2u xt + u tt =0. Show 

that the general solution is given by u(x, t) =tf(x + t)+g(x + t). 

9. Show that (1 + x 2 )u xx +(1+y 2 )u yy + xu x + yu y = 0 is elliptic and find the canonical 

form of the equation.

Bibliography 

[1] R. Dautray and J.L. Lions, Mathematical analysis and numerical methods for science 

and technology, 

[2] G. Folland, Introduction to partial differential equations, 

[3] A. Friedman, Generalized functions and partial differential equations, 

[4] A. Friedman, Partial differential equations, 

[5] K.E. Gustafson Partial differential equations, 

[6] P.R. Garabedian, Partial Differential Equations, New York, John Wiley & Sons, 1964. 

[7] I.M. Gel’fand and G.E. Shilov, Generalized functions, v. 1, 

[8] G. Hellwig, Partial differential equations, New York, Blaisdell Publ. Co. 1964. 

[9] L. Hörmander, Linear partial differential operators, 

[10] F. John, Partial differential equations, 

[11] J. Kevorkian, Partial differential equations, 

[12] P. Lax, Hyperbolic systems of conservation laws and the mathematical theory of shocks, 

[13] I.G. Petrovsky, Lectures on partial differential equations, Philadelphia, W.B. Saunders 

Co. 1967. 

[14] M. Pinsky, Introduction to partial differential equations with applications, 

[15] W. Rudin, Functional analysis, 

[16] F. Treves, Basic Partial Differential Equations, New York, Academic Press, 1975. 

[17] F. Treves, Linear Partial Differential Equations with Constant Coefficients, New York, 

Gordon & Breach, 1966. 

53

54 BIBLIOGRAPHY 

[18] K. Yosida, Functional analysis, 

[19] Generalized functions in mathematical physics, V.S. Vladimirov 

[20] H.F. Weinberger, Partial differential equations, Waltham, Mass., Blaisdel Publ. Co., 

1965. 

[21] E.C. Zachmanoglou and D.W. Thoe, Introduction to partial differential equations with 

applications, 

[22] A.H. Zemanian, Distribution theory and transform analysis,

Chapter 6 Partial Differential Equations

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