C ha pter 2

L inear First-Or der PDEs

The general ﬁrst-order partial diﬀerential equation (PDE) for a two-variable function, denoted as

u=u(x; y), can be expressed in the form:

F (x; y; u; u

; u

) = 0:

Here, u

and u

repre sent the partial derivatives of u with respe ct to x and y, respectively. The

function F establishe s a functional relationship between the function u, its partial derivatives, a nd

the independent variables x and y. The general ﬁrst-order partial diﬀerential equation f or a function

u=u(x₁; : : : ; xₙ) o f n independent variables, denoted as x₁; : : : ; xₙ, can be represented as:

F (x; u; ru)=0;

Here, ru is a vector denoted as ru= (@₁u; : : : ; @ₙu), which comprises the partial derivatives of u

with respect to each independent variable x₁; : : : ; xₙ.

Deﬁnition 2.1. A classical solution of the equation F (x; u; ru)=0, for x2Rⁿ, ru= (@₁u; :::; @ₙu),

is a smooth function u=u(x) deﬁned on an open set Ω⊂Rⁿ such that F (x; u(x); ru(x))=0, is an

identity for all x 2Ω.

For example, it is possible to verify that functions of the form u =h(x

+ y

) for arbitrary smooth

functions h is the classical solution of the equation

¡xu

=0:

For example, the function u = x

+ y

is a classical solution to the equation for all (x; y) 2R

, while

u = x

+ y

is a solution only on R

¡f(0; 0)g. The gr aph of these two solution shown below.

Observe that the graph of a classica l solution of a ﬁrst-order PDE in two variable x; y is a smooth

surface:

2.1 Cl assiﬁcation of ﬁrst-order PDEs

In this book, we will exclusively focus on the study of ﬁrst-order PDEs falling within the categories

of linear, semi-linear, and quasi-linear. The subject of fully nonline ar equations will be introduced

in a separate book dedicated to that topic.

Linear equations. The general form of a linear ﬁrst-order PDE for a function u=u(x), x =

; : : : ; x

) is given by:

j=1

(x) @

u(x) + v

(x) u = r(x);

for some (usually) continuous functions vⱼ(x) and r(x).

Semi-linear equations. A semi-linear equation is characterized by the general form:

j=1

(x) @

u(x) = r(x; u);

The diﬀerence between a linear and semi-linear equation is that a semi-linear equation can

be nonlinear with respect to u (and not with the partial deriva tives u

and u

Quasi-linear equations. A quasi-linear equation assumes the general form:

j=1

(x; u) @

u(x) = r(x; u);

The diﬀerence between a quasi-linear and semi-linear equation is that in the former case, the

coeﬃcients of partial derivatives are function of u a s well.

Fully-nonlinear equations. A fully nonlinear equation is an equation where one or all of the

partial derivatives are nonlinear. For example, the equation:

j² + ju

j² =1;

is a fully nonlinear ﬁrst-order equatio n for u=u(x; y). We wont study this type of equation

in this book.

Exercise 2.1. Classify the following ﬁrst-order eq uations

a) u

+ u

= e

b) xu

+ yu

= e

c) u

+ (u

)

= 1 + u

d) u

+ uu

= 1

Exercise 2.2. Consider t he linear equation

+ u

= ¡u:

Verify that every fun ction of the forms u = f(y ¡ x) e

¡x

, u = f (y ¡x) e

¡y

satisﬁes the equation, where f is a

smooth arbitrary function.

Exercise 2.3. Consider t he following quas i- linear equation

+ uu

= 0:

a) Verify tha t every function of the implicit form u = f(y ¡ux) solves the equation.

b) What is t he exp licit solution if we know u satis ﬁes the con dition u = 2y + 1 along the y-axis ?

Exercise 2.4. Verify that the function u = 1 ¡ x

+ y

is a solution t o the fully nonlinear equation

+ ju

= 1:

What is the domain of u if it considered as a c lassical solution? The solu tion satisﬁes th e auxiliary condition

u = 1 on the unit circle x

+ y

= 1. Note that u = x

+ y

¡1 solves the PDE and the auxiliary condition too.

2 Li near First-Order PDEs

2.2 Characteristic method and OD Es along cu rves

The characteristic method is a powerful technique for solving ﬁrst-order partial diﬀerential equa-

tions, and it is especially useful for semi-linear and quasi-linear equations. By using this method,

one can derive the general solution to such equations. Furthermore, the characteristic method has

a geometric interpretation that can be illustrated through the Cauchy problem.

2.2.1 Introductory remark: ODE along a curve

In our study of ordinary diﬀerential equations (ODEs), we explore d equations of the form:

= f (x; u):

Here, u = u(x) represents a single-variable function. Geometrically, we interpret the x-variable as the

x-axis in the standard direction. The solution to this equation consists of a one-parameter family of

functions u = u(x; c), where c 2R, such that for any x within the domain of u, the following relation

holds:

u(x; c) = f(x; u(x; c)):

Now, let's shift our focus to a parametric curve γ(t) in the xy-plane. An ordinary diﬀerential

equation (ﬁrst-order) along γ(t) takes the form:

du ◦ γ

= f (t; u ◦ γ):

Here, u◦γ is deﬁned at any t as (u◦γ)(t)= u(γ(t)). If we denote w as u◦γ, we arrive at the equation:

= f (t; w):

γ(t) = (x(t); y(t))

= f(t; w)

For instance, consider γ(t) given by γ = (t; t

), a parabola in the xy-plane along which the

diﬀerential equati on:

= w;

is deﬁned. Suppose u at the point (1 ; 0) is 1, corresponding to γ(0). Then, we obtain w as:

w(t) = u(γ(t)) = e

2.2 Characteristic method and ODEs along curves 3

Remark 2.2. Solving diﬀerential equations along curves can sometimes result in non-valid solu-

tions. For instance , let's consider the circle C represented by the parametric curve:

γ(t) = (cos(t); sin(t));

for t in the interval [0; 2π), along with the initial value problem:

= w; w(0) = 1;

deﬁned for w(t) = u(γ(t)). The solution to this equation is w(t) = e

for t 2[0; 2π). However, the

function u is not continuous on the circle because:

u(γ(0)) = u(1; 0) = 1;

and



lim

t!2π

γ(t)



= u(γ(0)) = 1 =/ lim

t!2π

u(γ(t))

Exercise 2.5. Consider the famil y of د fy = x

+ c; c 2Rg for parameter c 2R. Consider th e ord inary diﬀerential

eq uation

= 0;

alo ng all curves i n this family. Show that u on the xy-plane can be descri bed by u = h(y ¡ x

) where h is an

ar bitrary smooth function.

2.2.2 A simple type of e quations

Let's begin with the following simple equation:

+ v(x; y) u

= 0; (2.1)

where u is a smooth two-variable function, u = u(x; y). We'll relate the independent variable y to

x through the equation:

= v(x; y);

and assume that the solution to this equation is expressed as y = Y (x; c), where c is a para meter

of the solution t o this ordinary diﬀerential equation (ODE). This family of curves is known as the

characteristic curves of the given partial diﬀerential equation (PDE). The reason is that along each

curve y = Y (x; c), the PDE reads as an ODE:

u(x; Y (x; c)) = 0: (2.2)

Note that, by the chain rule, we have:

u(x; Y (x; c)) = u

(x; y) +

(x; y) = u

(x; y) + v(x; y) u

(x; y):

4 Li near First-Order PDEs

Equation (2.2) can be simply solved for the constant function:

u(x; Y (x; c)) = C ;

where C is constant along the characteristic curve y = Y (x; c) for a ﬁxed c. Therefore, C is a f u nction

of c, written as C = h(c), where h is an arbitrary function.

u = C

y = Y (x; c

)

y = Y (x; c

)

u = C

y = Y (x; c

)

u = C

(x; Y (x; c)) = 0

Let's assume that the equation y = y(x; c) can be solved for c as c = g(x; y). Then, we can

express u(x; y) as

u(x; y) = h(g(x; y)); (2.3)

for an arbitrary smooth function h. Now, let's verify that the solution (2.3) satisﬁes equation (2.1):

(

= h

(g(x; y)) g

= h

(g(x; y)) g

This implies:

+ v(x; y) u

= h

(g(x; y))(g

+ v(x; y) g

)

Utilizing the relation c = g(x; y), we have

0 = g

dx + g

dy;

which leads t o:

= 0:

Substituting this into our previous equation:

(h(g(x; y)))

+ v(x; y)(h(g(x; y)))

= 0:

This conﬁrms that the solution (2.3) satisﬁes equation (2.1).

Example 2.3. Consider the partial diﬀerential equation:

+ u

= 0:

To apply the characteristic method, we begin by ﬁnding the characteristic equation:

= 1:

This equation has the solution y = x + c, where c is a parameter. Now, let's consider the characteristic

family, denoted as fc = y ¡x; c 2Rg. Along each characteristic curve, u remains constant due to

the equation:

= 0. Thus, we can express u as u = C along each line c = y ¡x, where C depends

on c. Consequently, we have:

u(x; y) = h(y ¡x):

2.2 Characteristic method and ODEs along curves 5

Here, h is an arbitrary smooth function. The characteristic curves in this case are straight lines

with a slope of 1.

(x; y)

¡1

u = C

= h(c

)u = C

= h(c

)

u = C

¡1

= h(c

¡1

)

x ¡ y

Remark 2.4. In the example we solved earlier, we obtained the solution u in terms of an arbitrary

function h(y ¡x). Consequently, any function of the form u =sin(y ¡x), u = e

¡(y ¡x)

, u = (y ¡x)

x ¡ y, and so on, satisﬁe s the given partial diﬀerential equation u

= u

= 0. This type of solution

is known as a general solution.

The concept of a general solution here is akin to the concept of the general solution for a ﬁrst-

order ordinar y diﬀerential equation that typically contains a constant parameter rather than an

arbitrary function. In subsequent discussions, we will explore how to determine the speciﬁc form of

the arbitrary function h with the help of auxiliary conditions for the problem.

Exercise 2.6. Find the general solution of the equation

+ xu

= u:

2.2.3 Characteristic method for semi-linear PDEs

Let's consider the following equation:

(x; y)u

(x; y; u); (2.4)

where v

; v

, and v

are smooth functions. The objective is to transform this partial diﬀere ntial

equation into a set of ﬁrst-order ordinary diﬀerential equations along characteristic curves.

Recall the diﬀerential of a two-variable function u = u(x; y) as du = u

dx + u

dy. Comparing

the expression of du with equation (2.4) implies the following system

(x; y; u)

(x; y)

: (2.5)

By relating x to y through the characteristic equation:

= v(x; y); (2.6)

where v =

, we obtain a family of curves γ

: c = g(x; y), where c is an arbitrary constant. The

given PDE reduces to t he following equation along each γ

: (2.7)

Suppose this equation is solved for u = U (x; c; C), where C depends on γ

and thus can be expressed

as C = h(c) for an arbitrary smooth function h. Hence, the general solution can be expressed as:

u = U (x; g(x; y); h(g(x; y))):

6 Li near First-Order PDEs

Example 2.5. Let's solve the following partial diﬀerential equation:

+ yu

= xy:

The characteristic equation is given by:

;

which we can solve t o obtain y = cx. The ordinary diﬀerential equat ion for u becomes:

= y:

Substituting y = cx into this equation, we have :

= cx:

This ordinary diﬀerential equation can be solved to ﬁnd:

u =

+ C:

Here, C can be expressed as an arbitrary smooth function in terms of c. Therefore, the gene ral

solution can be written as:

u(x; y) =

+ h





Theorem 2.6. The general solution of equation ( 2.4) is

u = U (x; g(x; y); h(g(x; y))); (2.8)

where h is an arbitrary smooth function, c = g(x; y) is the equation of characteristic curves solution

of the equation ( 2.6), and U is the solution of the equation ( 2.7).

Proof. We have

(

= U

+ U

+ u

(g) g

= U

+ u

(g) g

Multiplying the ﬁrst equation by v

and the second one by v

, we obtain

+ v

= v

+ U

+ v

) + u

(g) (v

+ v

By the equation c = g(x; y), we have

0 = g

dx + g

dy;

and by the equality

;

we obtain:

+ v

= 0:

Hence, we obtain the equality:

+ v

= v

On the other hand, from the equat ion u = U (x; c; C), we have du = U

dx. Utilizing the equation

;

2.2 Characteristic method and ODEs along curves 7

yields v

= v

, that proves the equation v

+ v

= v

. 

Deﬁnition 2.7. Given a ﬁxed value of c in the real numbers, t he curve γ

, which is the solut ion

of equation ( 2.6), is referred to as a characteristic curve of the diﬀerential equation ( 2.4). Since

equation ( 2.4) reduces to an ordinary diﬀerential equation when evaluated along γ

for any ﬁxed

c, we obtain an inﬁnite system of ordinary diﬀerential equations for the family of characteristic

curves fγ

; c 2Rg. The system ( 2.5) is known as the characteristic system of the associated partial

diﬀerential equation.

Example 2.8. The existence of a general solution, even for linear ﬁrst-order PDEs, is not always

a trivial question. Consider the equation:

+ yu

= αu;

where α is a constant. We will examine three cases: α=0; ¡1, and 1.

For α=0, the characteristic system is:

;

and the equation for the characteristic curves in the xy-plane is:

ydx+xdy=0:

The general solution of this simple ODE is y =cx, which is shown in the following ﬁgure:

0°

30°

60°

90°

120°

150°

180°

210°

240°

270°

300°

330°

Along the characteristic line γ

, we have

=0, and thus u is constant along γ

. Furthermore,

all characteristic lines intersect at the origin, so γ

carries the information of u at the origin. This

implies that u(x; y)=u(0; 0)=C, a constant for all ( x; y). There is no other solution of the equation

in this case.

Now, consider α =¡1. We will show that the only possible solution is u≡0. In this case, the

solution u along γ

satisﬁes the ODE:

= ¡

and therefore, u=

along γ

: y =cx with respect to x. On the other hand, the PDE implies u(0;

0) =0, and thus C =0, implying that u(x; y) is identically zero in this c ase.

For α=1, the solution u satisﬁes the ODE:

with the solution u=Cx, for a constant C. The general solution in this case is:

u(x; y) = f





8 Li near First-Order PDEs

The form of the solution imposes a restriction on the form of the function f if u is assumed to be

smooth inside the unit disk. For example, for f (z)=z

, the solution is not even continuous at the

orig in.

Exercise 2.7. Consider t he following equation

2 x

+ u

= 0:

a) Draw the characteristic curves of the equation.

b) Show that u(4; 3) = u(9; 4).

c) Write down the general solution of the equation. Note that th e solution is not diﬀerentiable with re spect

to x at x = 0.

Exercise 2.8. Consider t he following equation

+ y

= αu; y > 0

where α is a con stant.

a) Fi rst assume α =0. Draw the character istic curve in the xy-plane and use the relation

= 0 to ﬁnd u(1; 1).

if u =

1 + x

on the x-axis.

b) Find the general soluti on of the equati on fo r α = ¡1. What is u(1; 1) if u =

1 + x

on the x-axis?

Exercise 2.9. Consider t he equ ation

+ u

= 0:

Show th at i f u is the solution of the equation, then it is impossible that u = x o n the unit circle x

+ y

= 1.

Exercise 2.10. Consider the following function

+ xu

= 0:

a) What are the characteristic curves of the equation in the xy-plane?

b) Find the general soluti on of the equati on and draw some integral surfaces of the equation.

Exercise 2.11. Find the general solution of the following equations

a) xu

+ yu

= xyu:

b) ¡yu

+ xu

= 2xyu:

c) u

+ u

= x(y ¡x) u

d) xu

+ yu

= x(y + u)

Exercise 2.12. Consider the following equation

+ 3x

= y

a(x) + y

b(x);

where a; b are smooth functions. Show that the gener al solutio n of the equatio n is

u(x; y) = f(y ¡x

) A(x) + g(y ¡x

) B(x);

where A(x); B(x) are respectively anti-derivatives of xa(x) and x

b(x).

Exercise 2.13. Find the general solution of the following equation

+ u

= u + xu

Exercise 2.14. Consider the following equation

+ yu

= f (u):

If f(0) =/ 0 , show that the equati on can not have a smooth solution inside a unit disk.

Exercise 2.15. Consider the following equation

¡yu

+ xu

= u:

If u(x; y) is a smooth solution in the closure of t he unit ball in the xy-plane , show that u = 0.

2.2 Characteristic method and ODEs along curves 9

Hint: Since u is smooth in the closure of the unit ball B, it has an absolute max and min in cl(B). If the

max and min occurs inside B, then u = 0. Let max and min occur on the boundary. Conclude the relation

¡yu

+ xu

= 0 on the boundary. Use this relation and conclude u = 0.

Exercise 2.16. Consider the following equation

+ u

= u:

Take y as the independent variable for the characteristic curves x = x(y), and ﬁnd the general solution of the

eq uation.

2.2.4 Semi-linear equation in general dimension

Now, we generalize the method for semi-linear PDEs in general dimension. Consider the following

equation:

j=1

(x) @

u = r(x; u); x 2R

; (2.9)

where v

(x) are smooth functions of x = (x

; : : : ; x

). To solve this equation, we ﬁrst consider the

characteristic system:

(x)

= ··· =

(x)

r(x)

The characteristic equation in the space (x

; : : : ; x

) is

(x)

= ··· =

(x)

By taking x

as the independent variable, we can write the system as:

(x)

: (2.10)

We can solve system (2.10) for a given parameter c=(c

; : : : ; c

) a s the implicit system:

= g

(x); : : : ; c

(x):

We assume that the ab ove implicit solutions are solvable for x in terms of x

and c, so we can write:

; c); : : : ; x

; c):

The intersection of surfaces

= g

(x); : : : ; c

(x);

reduces to a smooth curve γ

, c=(c

; : : : ; c

), which is the same as the parametric curves:

: fx

; c); : : : ; x

; c)g:

Next, we consider the derivative of u along the curve γ

, which is given by:

r(x; u)

(x)

: (2.11)

By solving this equation, we obtain a function u = U (x

; c

; : : : ; c

; C). Therefore, taking C =h(c),

for an arbitrary smooth function f , the general solution to equation (2.9) can be expressed as:

u = U (x

; g

(x); : : : ; g

(x); h(g

(x); : : : ; g

(x))):

10 Li near First-Order PDEs

Example 2.9. Let's solve the partial diﬀerential equation:

+ u

= 0:

The characteristic equations in the space (x; y; z) is given by:

= 1;

which is solved for c

= y ¡x, c

= z ¡x. The characteristic curves parameterized by x is

= (x; x + c

; x + c

)

for c = (c

; c

). The given PDE along γ

b ecomes:

= 0, which implies u=C. Replacing C by

h(c

; c

) for an arbitrary smooth function h, we obtain the general solution

u(x; y; z)=h(y ¡x; z ¡x):

Exercise 2.17. Consider the following equation

¡zu

+ yu

= αu;

where α is a con stant.

a) Show that the characteristic curves of the equation have the following equat ions

(x) = (x; c

cos(x) ¡c

sin(x); c

sin(x) + c

cos(x)):

What is the shape of the characteristic curves?

b) The characteristic curves passes through the plane x = 0. Show that for α = 0, we have

u(0; 1; 1) = u



; ¡1; 1



c) Show that the solution of the given PDE along γ

(x) is of the form

u(x; c

; c

) = h(c

; c

) e

αx

;

for arbitrary function h. Find t he general solution.

Exercise 2.18. Consider the following equation

+ 2 y

+ 2 z

= 0;

where y ≥0 and z ≥0. Find the general solution of the equation. Veri fy the equ ality u(2; 4; 4) = u(1; 1; 1).

Exercise 2.19. The singular equation

+ yu

+ zu

= 0:

If u(x; y ; z) is a smooth solution inside the unit ball in R

, show that u(x; y; z) = C a constant for all x; y ; z.

2.2.5 Characteristic method for quasi-linear equat ions

The method of characteristics for quasi-linear partial diﬀerential equations may lead to the emer-

gence of new phenomena, such as shocks, which we will discuss later. Let's consider the following

equation

(x; y; u) u

+ v

(x; y; u) u

= v

(x; y; u): (2.12)

The coeﬃcients of the quasi-linear equations depend on u. The characteristic system of the PDE is:

: (2.13)

2.2 Characteristic method and ODEs along curves 11

Let c

= φ(x; y; u) and c

= (x; y; u) be implicit solutions of the characteristic system. We have

the following theorem:

Theorem 2.10. Let v

, and v

be smooth functions in equation ( 2.12), and let c

=φ(x; y;u) and

= (x; y; u) be implicit solutions of its associated characteristic system. Then, the general solution

of the equation in implicit form is given by f(φ; )=0 for any smooth function f satisfying f

=/ 0.

Example 2.11. Let's consider the following equation called the Burger's equation:

+ uu

= 0:

The characteristic system is given by:

;

By taking x as the independent variable and setting y = y(x), we obtain:

= u;

= 0:

This system is solved for c

= u and c

= y ¡xu. The general implicit solution of the equation is

derived by f(c

; c

) = 0, for arbitrary smooth function f, or equivalently f(u; y ¡xu) = 0. Alterna-

tive ly, we can write the solution as u = g(y ¡xu) for arbitrary smooth function g. Note that u in

both cases is in the implicit form.

Example 2.12. Consider the equation

+ yuu

+ zuu

= 0:

To solve this equation, we need to determine the characteristic system, which is given by:

= yu;

= zu;

= 0:

Solving this system of diﬀerential equations, we obtain e

¡x

y = c

u, e

¡x

z =c

u, and u=c, where c;

and c

are arbitrary constants. Substituting these values into the equation, we get the general

implicit solution as

f(u; e

¡x

¡1

; e

¡x

¡1

) = 0:

Alternatively, the solution can be expre ssed as u = g



¡x

;

¡x



Proof. (of the theorem) Since f (φ(x; y; u(x; y)); (x; y; u(x; y))) is identically zero, taking

derivatives with respect to x and y gives:

(

(φ

+ u

) + f

(

+ u

) = 0

(φ

+ u

) + f

(

+ u

) = 0

Note that f

=/ 0 implies that f

and f

cannot both be identically zero. This in turn implies that

the following determinant is zero:



+ u



= 0:

Expanding the determinant and simplifying, we get:

(φ

+ (φ

¡ φ

) u

= φ

¡ φ

: (2.14)

12 Li near First-Order PDEs

On the other hand, we have:

(

dφ = v

+ v

d = v

+ v

= 0

;

that gives

¡ φ

¡φ

: (2.15)

Matching relations (2.14) and (2.15) gives v

+ v

= v

, which completes the proof. 

Exercise 2.20. Find the general solution of the following equations

a) uu

+ u

= u

b) sinuu

+ u

= 1 (hint: you can take y a s the independent variable just for the simplicity)

c) xu

+ (1 + u) u

= u

d) u

+ uu

¡uu

= 1

Exercise 2.21. Consider the following equation

+ uu

= 0:

a) Fi nd the general sol ution in the implicit fo rm. Then for the auxiliary condition u(0; y) = y, ﬁnd an explici t

solution. Determine the dom ain (x; y) where th e solution exists smoothly.

b) In general, the im plicit solutions can not be transformed to an explicit form, although, th e exis tence of

such explicit forms is guaranteed by the implicit function th eorem. Let u(0; y) = e

. Obtain the implicit

solution of the equation and use the theorem to show the existence of the expl icit solution.

Exercise 2.22. Consider the following equation

(

+ uu

= 0

u(0; y; z) = f (y; z)

a) Fi nd the solution of the equat ion in the implicit form.

b) Find the explicit solution if f(y; z) = y + z.

Exercise 2.23. Consider the following quasi-linear equation

u +

j=2

(x; u) @

u = r(x; u); x 2R

The cha racteristic systems is

= v

; j = 2; : : : ; n

= r

Let the system sis solved fo r

c = (x; u); c

= φ

(x; u); j = 2; : : : ; n:

Show th at the function f (φ

(x; u); (x; u)) = 0 solves the given partial diﬀerential equa tion.

2.3 Theoretical aspects

2.3.1 The geometrical interpretation of a ﬁrst-order PDE

Consider the partial diﬀerential equation:

(x; y; u) u

+ v

(x; y; u) u

= v

(x; y; u): (2.16)

2.3 Theoretical aspects 13

Here, v

; v

, and v

are functions of x; y, and u. Let z = u(x; y) is a solution to this equation. This

function deﬁnes a surface on the space (x; y; z):

S: (x; y; u(x; y)):

Recall that the vector n~ : (¡u

; ¡u

; 1) is perpendicular on S at any point (x; y; z) on S. Now,

assume that the vector ﬁeld V

: (x; y; z) 7!(v

; v

) is given on the space. The partial diﬀerential

equation (2.16) then can be stated from the geometrical point of view as follow:

Geometrical interpretation of a ﬁrst-order PDE: Given a vector ﬁeld V

: (x; y; z)!(v

;

; v

), the equation of a tangent surface z = u(x; y) to V

is given by the partial diﬀerential equation

(x; y; u) u

+ v

(x; y; u) u

= v

(x; y; u):

Conversely, the solution of the above partial diﬀerential equation deﬁnes a surface z = u(x; y) that

is locally tangent to the vector ﬁeld V

For example, consider the equation: u

+ u

= 0. The vector ﬁeld is

: (x; y; z) !(1; 1; 0)

The ﬁgure below shows two diﬀerent surfaces that are tangent to this vector ﬁeld at all point of the

surfaces

In fact, it can be seen that the all surface generated by the function u = f (y ¡x) for arbitrary

smooth f has this tangent property as

·n~ = (1; 1; 0) ·(f

; ¡f

; 1) = 0 :

Problem 2.1. Let V = (v

(x; y; z); v

(x; y; z)) be a given vector ﬁeld in space.

a) Show that the existence of a surface φ(x; y; z) = 0 which is perpendi cular to the vector ﬁelds at each point

on the surface satisﬁes the following equation

dx + v

dy + v

dz = 0:

b) The exi stence of the solu tion to the a bove ODE is not a trivia l questi on. Show that if r×V = 0, then there

is a sur face φ(x; y; z) = 0 that sa tisﬁes the above ODE. Hint: Note that if r×V = 0, then V is potential.

c) As an examp le, consider the vector ﬁeld V = (y; x; z). Find a surface which is perpendicular to V at each

point on the surface. The ﬁgure below depicts one of such surfaces .

14 Li near First-Order PDEs

Problem 2.2. Consider the vector ﬁeld V = (¡y; x; 0). This ﬁeld does not sati sfy the condition r × V = 0,

however, a su rface still exists which is per pendicular to V . Show that the surfa ce expressed pa ra metrically as

Σ(x; z) = (x; αx; ¡xz) has such a property.

Problem 2.3. Show that there is no surface φ(x; y; z) = 0 with the normal vectors (x; y; xy).

2.3.2 Parametric solution surfaces

Now, consider a curve γ(t) on the solution surface (yet unknown) S. This curve is tangent to the

vector ﬁeld V

at any point on the curve. Therefore, the equation of γ(t) is determined by the

following ordinary diﬀerential equation

= v

(x(t); y(t); z(t))

= v

(x(t); y(t); z(t))

= v

(x(t); y(t); z(t))

: (2.17)

This ﬁrst-order system of ordinary diﬀerential equation has a unique solution if an initial condition

is set for the system. Let's assume that we know a point p

on S. We can set the initial condition as:

γ(0) = p

x(0) = x

y(0) = y

z(0) = z

With this initial condition, a curve γ

(t) is obtained on S. In this way, to determine S, we need

to a family of curves fγ

(t)g where p lies on a curve on S as shown below:

(t) = Σ(t; s)

z = u(x; y)

p(s)

For example, assume the curve p(s) = (s; 0; s sin(s)) lies on S o f the solution of the equation

+ u

= 0. The solution surface is spanned by the curves det ermined by the following system of

ODEs

= 1

= 0

;

accompanied with the conditions x(0) = s; y(0) = 0; z( 0) = s sin(s). This system is solved as

x(t; s) = t + s; y(t; s) = t; z(t; s) = s sin(s):

2.3 Theoretical aspects 15

In this way, we obtain a parametric surface

Σ(t; s) = (t + s; t; s sin(s)):

It is simply seen that this parametric surface is algebraically represented as

u(x; y) = (y ¡x) sin(y ¡x):

The ﬁgu re below depicts the “data line” p(s) in black and the space characteristic curves γ

(t) in red.

Remark. The advantage of using parametric form for representing solutions is that it can represent

very complicated surfaces, whereas explicit functions z = u(x; y) represent only restricted classes of

surfaces. The example below furt her clariﬁes this point .

Example 2.13. Consider the following problem

(

0.2xu

¡uu

= y

fx=y

Here u is give n along the curve x = y

in the xy-plane. This data can be parameterized in terms of

s as: p(s)=(s

; s; s). The system of characteristic equations is:

= 0.2x

= ¡u

= y

;

with the initial con ditions x

= s

; y

= s and z

= s. The parametric representation of the integral

surface is obtained as:

Σ(t; s) = (s

0.2t

; s(cost ¡sint); s(sint + cost)):

As shown in ﬁgure below, the solution of the equation represents a complicated surface which can

not be expressed by an explicit function z = u(x; y).

16 Li near First-Order PDEs

Even though this surface is not associated with an explicit function, it is possible to deﬁne

an explicit function that locally coincides with this surface. Thus, classical solutions of partial

diﬀerential equati ons c an only be deﬁned locally in a neighborhood of the data curve.

Exercise 2.24. Use the parametrization and ﬁnd the solu tion of the following Ca uchy problems. Use Matlab

and draw the integral surface

a) uu

+ yu

= x, u = sin(x) on y = 1.

b) (y ¡u) u

+ (u ¡x) u

= u, u = x

on y = 0.

The following code generates the integral surfaces of t he given equation in part b).

s = lins pace(0, 2, 100); % d efine the range of s

t = lins pace(0, 10, 100); % define the range of t

[S, T] = meshgrid(s, t); % create a grid of s and t values

X = S .* cos(T); % compute the x-coordinates

Y = S .* sin(T); % compute the y-coordinates

Z = S .^ 2 .* exp(0.2 * T);% compute the z-coordinates

surf(X, Y, Z) % plot the surface

shading interp; camlight

xlabel('x');

ylabel('y');

zlabel('z');

title('Surface plot of x = s*cos(t), y = s*sin(t), z = s^2*exp(0.2*t)');

2.3.3 Cauchy problem

In this section, we demonstrate how to obtain the particular solution o f a given ﬁrst-order PDE

from the general solution, using an auxiliary condition or additional information about the solution.

This process is similar to deriving particular solutions from the general solutions of ODEs, using

initial conditions.

Deﬁnition 2.14. A problem of the form

(

(x; y; u) u

+ v

(x; y; u) u

= v

(x; y; u)

(2.18)

where C is a curve in an open set Ω ⊂R

in the xy-plane, is called a Cauchy problem.

Let's consider the following problem:

+ u

= 0

fy=0g

1 + 2x

The general solution to the equation is u = h(y ¡x) for an arbitrary smooth function h. Utilizing

the initial condition u(x; 0) =

1 + 2x

= h( ¡x);

and thus h(y ¡x) =

1 + 2(y ¡x)

, and thus t he particular solution to the given Cauchy problem is

given by:

u(x; y) =

1 + 2(y ¡x)

2.3 Theoretical aspects 17

Exercise 2.25. Find the partic ular solution of the followi ng Cauchy problem

(

+ (x + y)u

= u

x=0

=sin(y)

Exercise 2.26. Find the partic ular solution of the followi ng Cauchy problem

(

+ u

= 2x(y ¡x)u

x=0

Exercise 2.27. The Cauchy problem can be gener alized to higher dimensi ons. Fi nd the solution of the following

eq uation for u = u(x; y; z):

(

+ yu

+ z u

= 0

fz=1g

=xy

Examples below highlight some of the issues that can arise when attempting to extract a par-

ticular solution from a general solution.

Example 2.15. Consider the Cauchy problem:

(

¡2x u

= 0

fy=x

The general solution of the PDE is u = f (y + x

) for an arbitrary smooth function f. We can use the

auxiliary condition u = x on the line C: y = x

to determine f , which gives us x = f(2x

). However,

this equation cannot be solved as f ( 2 ) can have two possible values, ±1, leading to a contradiction.

-2 -1 1 2

-4

-2

This problem occurs because the characteristic curves of the PDE, y =¡x

+c, intersect the data

curve C: y = x

at more than one point, resu lting in multiple possible values of u. To resolve this

issue, we can consider only one branch of the data curve, such as y = x

; x ≥0, to obtain a unique

solution u(x; y) =

y + x

for y ≥¡x

-2 -1 1 2

-4

-2

However, this solution is still not unique in the region y < ¡x

, where the characteristic curves

do not intersect the data curve. The value of u along these curves can be chosen arbitra rily.

18 Li near First-Order PDEs

-2 -1 1 2

-4

-2

If we change the data curve to y = ¡x

, which is also a characteristic curve, there is no unique

solution in the region y > ¡x

and y < ¡x

. In this case, x = f (0), which i s not solvable from the

general solution u = f (y + x

) and the data curve y = ¡x

-2 -1 1 2

-4

-2

Example 2.16. Let us consi der the following equation:

(

¡yu

+ xu

= u

;

where C is the x-axis f or x ≥0. To apply the method of characteristics, we ﬁrst need to ﬁnd the

equation of the characteristic curves. Using the characteristic equation

¡y

;

we get x

+ y

= c; where c is a positive constant. Therefore, the characteristic curves are circles

centered at the origin. In one of previous exercise, we asked the read to show that this equation does

not have any smooth solution inside a disk. We can see this fact by trying to solve this equation

explicitly. Taking x as the independent variable, we can rewrite the PDE as

¡1

To solve this equation, we need to express y as a function of x. Howe ver, this cannot be done

through the obtained implicit function x

+ y

= c. One way to overcome this diﬃculty is to use

the parametric representation of the characterist ic curves. Let t be a parameter and consider the

characteristic system:

= ¡y

= x

The solution to this system is

(t) = (s cos(t); s sin(t));

2.3 Theoretical aspects 19

where s is a non-negative parameter. Note that we used the data curve C to write the initial point

of the cha racteristic curve as γ

(0) = (s; 0). Now, we can express u in terms of t as f ollows:

= u;

which is a separable ODE with solution

u(γ

(t)) = u(γ

(0)) e

= u(s) e

= se

To determin e the domain o f t, note that γ

(0) = γ

(2π), which implies

u(γ

(2π)) = u(γ

(0)) = s:

However, we also have

u(γ

(2π)) = se

2π

;

so we conclude that the domain of t can not be [0; 2π]. Note that for x = s cost; y = s sint and u = se

we obtain the integral surface

Σ(t; s) = (s cost; s sint; se

This solution can be put in the algebraic form as

u(x; y) = x

+ y

atan



Exercise 2.28. Find the partic ular solution of the followi ng Cauchy problem

(

+ (x + y)u

= u

x=0

=sin(y)

Exercise 2.29. Find the partic ular solution of the followi ng Cauchy problem

(

+ u

= 2x(y ¡x)u

x=0

Exercise 2.30. The Cauchy problem can be gener alized to higher dimensi ons. Fi nd the solution of the following

eq uation for u = u(x; y; z):

(

+ yu

+ z u

= 0

fz=1g

=xy

Exercise 2.31. Consider the following Cauchy problem

(

¡yu

+ !

= αu

;

where C is the line y = x for x ≥0, and !

is a constant. Use parametric representation of characteristic curves in

ter ms of the parameter t and ﬁnd the solution for α = 0. If α =/ 0 , ﬁnd the domain of t and indica te the d omain

of the soluti on in the xy-plane.

Exercise 2.32. Consider the following Cauchy problem

(

(¡y ¡αx) u

+ (x ¡αy) u

= u

= f (x)

;

where C is the x-axis for x ≥0, and α > 0. Show that the problem is solvable in classical sen se only if f(x) = 0.

2.3.4 Well-posedness and existence of i ntegral surfaces

As we observed in previous examples, if the data curve of a Cauchy problem is not a characteristic

curve, then there is a solution that can be extended locally. The following ﬁgure shows this situation

schematically.

20 Li near First-Order PDEs

(t)

C(s)

C(s

)

Here, the data cur ve C is parametrized by s as C = C(s), and the planar characteristic curves

γ are parameterized by t. Note tha t γ

(t) is a characteristic curve passing throu gh C(s) at t = 0.

Let s

b e a point on C(s) in its domain. If γ

(0) and C

) are non-parallel, then there is a t

> 0

such that γ

(t) exists for 0 ≤t < t

Theorem 2.17. Consider the Cauchy problem

(

(x; y) u

+ v

(x; y) u

= v

(x; y; u)

where v₁; v₂, and v₃ are smooth functions, and C is a smooth curve in the xy-plane. Assume there

exists (x₀; y₀)2C such that

; y

) jj (v

; y

); v

; y

)): (2.19)

Then there exists an open neighborhood Ω of (x₀; y₀) and a smooth function u = u(x; y) on Ω that

solves the given Cauchy problem.

The p roof of the theorem is based on a standard theorem on the existence and uniqueness of

the solution to ordinary diﬀerential equations. Note that if condition (2.19) holds, then due to the

continuity of v

; v

, and C

at (x

; y

), the condition holds for an open neighborhood of (x

; y

Then, the existence of a domain Ω for the solution u(x; y) is reduced to the existence and uniqueness

of the ordinary diﬀerential equation

= v

(x(t); y(t); u):

However, it is important to note that the theorem only provides a suﬃcient condition for the

existence of a solution, and there may be cases where the condition is not satisﬁed but a solution

still exists. For instance, consider the problem:

(

+ y

= 0

u(x; 0) = f(x)

This problem has the solution u(x; y) = f(x ¡2 y

) which is deﬁned for y ≥0, even though it i s not

generally diﬀerentiable on the x-axis.

The existence of a parametric surfac e for a quasi-linear ﬁrst-order PDE is established in a similar

manner. We have the following theorem.

Theorem 2.18. Consider the Cauchy problem

(

(x; y; u) u

+ v

(x; y; u) u

= v

(x; y; u)

;

2.3 Theoretical aspects 21

where v

; v

are smooth functions and C is a smooth and non-characteristic curve in xy-plane.

Let ¡(s) be the parametrized space data curve (C(s); f(s)). Fix s

in the domain of ¡(s) and let

; y

; u

) = ¡(s

). If

) ×V (x

; y

; u

) =/ 0;

where V = (v

; v

), then there exist α; β >0 such that the given problem has an integral surface

Σ(t; s) for t 2(¡β; β) and s 2(s

¡α; s

+ α).

2.4 Time de pendent functions and ﬂuid ﬂow

Consider a ﬂuid distributed along the x-axis at time t, with u(x; t) representing its density function.

Let x

b e a ﬁxed point on the x-axis. The rate of change in u(x

; t) as time progresses is given

by the partial derivative u

; t). Meanwhile, u

(x; t

) measures the rate of diﬀerence in density

between x and its neighboring points.

; t

)

; t

)

; t

)

; t

)

Now, consider a ﬂuid with density function u(x; t) ﬂowing with velocity function V (x). This

veloc ity function can be thought of as a one-dimensional vector ﬁeld: x!V (x). The properties of

this ﬁeld aﬀect the density function u. For example, suppose we have a control volume [¡1; 1 ] moving

with velocity V (x)=x. The total mass within this volume at time t = 0 is given by the integral

M =

¡1

u(x; 0) dx:

What can be said about the mass of this packet at time t = 1? Let x

b e a ﬁxed point on the x-axis.

According to the velocity ﬁeld, the p osition of x

at time t can be deter mined by the diﬀerential

equation

=V . For V (x)=x, we obtain x(t)=x

. Therefore, the control volume [¡1; 1] expands

to the volume [¡e; e] at time t = 1. Assuming there is no sink or source of mass production, the

total mass is conserved, and thus

M =

¡e

u(x; 1) dx:

Since the volume of the control volume has increased, the density u(x; 1) must have decreased to

maintain the same mass. The main objective of this section is to address how the density fu nction

u of a n-dimensional ﬂuid ﬂow changes over time.

Exercise 2.33. Consider the following density distr ibution

u(x; t) = e

(x¡ t)

It is simp ly seen that the total mass al ong the x-axis for any t is equal to

¡1

u(x; t) dx = 2π

;

22 Li near First-Order PDEs

and thus we can write that

lim

t!1

¡1

u(x; t) dx = 2π

Find the steady state distribution when t !1 and calculate

¡1

lim

t!1

u(x; t) dx:

This shows that the foll owing eq uality does not hold always

lim

t!1

¡1

u(x; t) dx =

¡1

lim

t!1

u(x; t) dx:

2.4.1 Change of density along ﬂow lines

Consider a velocity ﬁel d V = (v

(x); : : : ; v

(x)) of a ﬂuid ﬂow, where x = (x

; : : : ; x

). Let x

b e a

ﬁxed point in R

, and let γ

(t) be the ﬂow line of x

, determined by the system of ODEs

= v

(x); j = 1; : : : ; n;

with initial condition x(0)=x

. Our goal is to ﬁnd the rate of change of the density function u along

the ﬂow line γ

(t). This rate of change is given by the time derivative:

(γ

; t) = u

(γ

; t) + ru(γ

; t) ·

dγ

;

where ru(t; x) represents the gradient of u. This can be expressed in coordinate form as:

(x; t) = u

(x; t) +

j=1

(x) @

u(x; t):

The derived formula provides us with the change of u along the ﬂow lines. Therefore, the question

arises: can we determine u(x; t) if we know the initial density u(x; 0)?

Exercise 2.34. Verify that the above formula is a straightforward applica tion of the chain rule.

Example 2.19. Consider a ﬂuid ﬂow in the xy-plane, where the density at t = 0 is given by

u(x; y; 0) = f(x; y) = x

¡x

¡y

, and the velocity ﬁeld V is

(x; y) 7!(¡αx ¡y; x ¡αy);

with α ≥0 being a constant. The trajectories of the particles are determined by the following system

of diﬀerential equations:

= ¡αx ¡ y

= x ¡αy

For the ﬁrst case, let α = 0, that leads to the trajectory

)

(t) = (x

cos(t) ¡ y

sin(t); x

sin(t) + y

cos(t));

where (x

; y

) is the initial position at t = 0. In matrix form, the ﬂow line passing through (x

; y

)

is given by:



x(t)

y(t)





cos(t) ¡sin(t)

sin(t) cos(t)





2.4 Time dependent functions and fluid flow 23

Note that the coeﬃcient matrix is the rotation matrix R

. Consider a control volume D centered

at (0.5; 0.5). The following graph illustrates u(x(t); y( t); t) for t = 0 :

: 2π. Observe how the control

volume centered at (0.5; 0.5) rotates around the origin while also rotating around itself. The velocity

ﬁeld is responsible for this rotation, as it is a rotational ﬁeld with a non-zero curl, denoted as

r×V = 2.

The following ﬁgure shows the density at diﬀerent time slices, speciﬁcally for times t = 0 ;

, and

π. As shown in the ﬁgure, the density rotates in a circular pattern around t he origin as it is clear

from the diﬀerential equations of the trajectory

= ¡y;

= x. Furthermore, we observe that the

density along the trajectory of the ﬂuid particles remains constant, which means the density remains

constant a long the trajectory. This type of ﬂuid ﬂow, where the density remains constant along the

trajectory of the ﬂuid particles, is called an incompressible ﬂow.

Now, let α = 0.1. The trajectory of the control volume is an inward spiral towards the origin,

given by



x(t)

y(t)



= e

¡0.1t



cos(t) ¡sin(t)

sin(t) cos(t)





The following graph illustrates this scenario, showing how the control volume centered at (0.5; 0.5)

forms an inward spiral while also rotating around itself. Observe how the control volume gradu ally

becomes smaller and by the conservation law, the density in D increases to balance the decrea se in

the volume.

24 Li near First-Order PDEs

The above example illustrate how the velocity ﬁled V ( x) aﬀects the density of particles along

the ﬂow lines. The following theorem make a relationship between the divergence r·V (x) and the

change of density along the ﬂow line.

Theorem 2.20. Consider a control volume D in a ﬂuid ﬂow that moves according to the velocity

ﬁeld V. If the divergence of V, denoted by r·V (γ

(t)), is positive, then the volume of D increases

along the trajectory γ

(t). On the other hand, if r·V (γ

(t)) is negative, the volume of D decreases

along the trajectory, and if r·V (γ

(t)) is zero, the volume of D remains constant.

Corollary 2.21. If r·V = 0, the density of a ﬂuid-ﬂow remains constant along its trajectory in

the absence of any material sink or source.

In the ﬁrst scenario of the example, the velocity ﬁeld V = (¡y; x) had a divergence of r·V = 0,

which implies that the volume of the control volume remained constant along γ

(t). As a result,

the density u remained unchanged along the ﬂow lines. In the second scenario, the velocity ﬁeld

was V = (¡y ¡αx; x ¡αy), with a divergence of r·V = ¡2α < 0 for α > 0. This indicates that the

volume of the control volume decreased with time, leading to an increase in the density u to balance

the volume decrease.

2.4.2 Continuity equation

Given a region Ω in a ﬂuid ﬂow, the total mass of the ﬂuid in Ω at time t, denoted by M

Ω

(t), changes

due to two factors. The ﬁrst factor is the net ﬂow rate of ﬂuid passing through the boundary bnd(Ω)

per unit time. This can be expressed mathematically as the surface integral

Φ(t) := ¡



bnd(Ω)

u(x; t)V (x) ·νdV :

where V is the velocity vector of the ﬂuid and ν is the unit outward normal vector to the boundary.

The second factor is the presence of a source or sink of material in the ﬂuid, which can be represented

mathemat ically as the volume integral

Q(t) =

ZZZ

Ω

f(x; t) dV ;

2.4 Time dependent functions and fluid flow 25

where f (x; t) is the rate of material added or removed per unit volume at point x at time t.

Therefore, we can write the rate of change of M

Ω

(t) as:

Ω

(t)

= ¡



bnd(Ω)

u(x; t)V (x) ·νdV +

ZZZ

Ω

f(x; t) dV :

This equation is known as the mass conservation equation, or the continuity equation, and it plays

a fundamental role in ﬂuid dynamics.

On the other hand, the left hand side of the above equation is

Ω

(t)

ZZZ

Ω

ρ(x; t) dV =

ZZZ

Ω

(x; t) dV ;

and by equating the two expressions for

Ω

, we arrive at the continuity equation for the ﬂuid ﬂow

in Ω:

ZZZ

Ω

(x; t) dV = ¡



bnd(Ω)

u(x; t)V (x) ·νdS+

ZZZ

Ω

f(x; t) dV :

The material ﬂux J(x; t) = u(x; t)V (x) measures the rate at which ﬂuid passes through a unit area

per unit time. When considering the ﬂux at a particular point x on the boundary bnd(Ω) of the

bounded, open set Ω in R

, the unit normal vector ν(x) is needed to determine the ﬂux through the

surface. This is because the ﬂux J(x; t) can be decomposed into two components: 1) a component in

the direction of the unit normal vector ν(x), and 2) a tangential component that does not leave the

closure of Ω, denoted by Ω

. Only the component of J(x; t) that is perpendicul ar to the boundary,

i.e., J(x; t) ·ν(x), contributes to the ﬂow leaving Ω through bnd(Ω). The ﬁgure below illustrates

this concept.

unit area

Ω

The diﬀerential form of the continuity equation can be derived by applying Gau ss's theorem to

the divergence of the material ﬂux density. Speciﬁcally, the theorem yields:

ZZZ

Ω

(x; t) dV = ¡

ZZZ

Ω

r·[u(x; t)V (x)]dV +

ZZZ

Ω

f(x; t) dV ;

which simpliﬁes to:

ZZZ

Ω

(x; t) + r·[u(x; t) V (x)] ¡ f (x; t)gdV = 0:

Since the integral holds for every arbitrary subdomai n of Ω, we conclude that the integrand must

be identically zero, yielding the continuity equation:

(x; t) + r·[u(x; t) V (x)] = f ( x; t): (2.20)

This equation expresse s the conse rvation of mass, where f(x; t) represents the source or sink of

material i n the ﬂuid.

26 Li near First-Order PDEs

Exercise 2.35. Show the relation

div [u(x; t) V (x)] = ru ·V (x) + u(x; t) div [V (x)];

and conclude that t he conservation equation can be written as follows

+ V ·ru = ¡u div [V ]:

Assuming f is identically zero in equation (2.20), we can rewrite the equation using the result

from the previous exercise as follows:

+ V ·ru = ¡ur·V : (2.21)

Let γ

(t) be the trajectory of a particle at z which is determined by the equatio n

= V (z). In this

case, the left hand side of the above equation reads

(t; γ

(t)), and then we have

(γ

(t); t) = ¡ur·V (γ

(t)):

If r·V (γ

(t)) = 0, then

(γ

(t); t) and this implies that the density of x remains the same a t its

trajectory, an therefore, equation ( 2.21) leads to the solution

u(γ

(t); t) = u(γ

(0); 0);

and by the fact γ

(0) = z, we obtain

u(γ

(t); t) = u(z; 0):

If we denote γ

(t) by x and then z = γ

(¡t), we derive the solution

u(x; t) = u(γ

(¡t); t):

Deﬁnition 2.22. A ﬂow moving along the velocity ﬁled V (x) is called incompressible if div (V ).

Exercise 2.36. We aim to show th at the total mass within a moving c ontrol volume D, deﬁned by the trajectory

(t), remains constant when there are no external sources or sinks. To begin, we ex press the total mass M

D(t)

as the integral of the ﬂui d density u(x; t) over the volume D(t) at time t:

D(t)

ZZZ

D (t)

u(x; t) dx:

Taking the time derivative of the above equation, we get:

D (t)

ZZZ

D(t)

u(x; t) dx:

Using the Leibniz integral rule, we can write:

ZZZ

D (t)

u(x; t) dx =

ZZZ

D(t)

(x; t) dx +



bnd(D(t))

uV (x) ·ν dS:

Use the divergence theorem (Gauss theorem) and show

ZZZ

D (t)

u(x; t) dx = 0;

and conclude that M

D (t)

remains constant along γ

(t).

2.5 Linear an d semi-linear transport equations

The general form of a semi-linear transport equation in R

is given by

+V (x) ·ru= f (x; t);

2.5 Li near and semi-linear transp ort equatio ns 27

where V (x) is the velocity ﬁeld. However, in some cases, the velocity ﬁeld may also depend on the

density u in addition to x, leading to the quasi-linear equation:

+V ( x; u) ·ru= f(x; t):

In general, the independent variable for transport equations is time t. For semi-linear equations,

the characteristic system is given by

=V (x);

while for quasi-linear equations, the characteristic system becomes:

= V (x; u)

= f(x; t)

Example 2.23. The simplest form of the transport equation is:



+ c u

= 0

u(x; 0) = u

(x)

This equation is also known as the unilateral wave equation due to t he directionality of the wave

propagation. Since V (x)=c is constant, the ﬂow is incompressible, and the trajectory x(t)=ct+x

yields the ordinary diﬀerential equation

= 0, which has the solution:

u(x(t); t)=u(x

; 0)=u

) = u

(x ¡ct):

This solution corresponds to a traveling wave propagating to the right (if c>0) or to the left (if c<0 ).

The solution to the transport equation in h igher dimensions is similar. Consider the equation:



+ C ·ru = 0

u(x; 0) = u

(x)

;

where C = (c

; :::; c

) and x = (x

; : : : ; x

). The ﬂow lines are:

(t) = c

t + x

;

where j =1; :::; n and x

= x

(0). Thus, we have u(t; x(t))= f(x

), and for x

= x ¡Ct, the solution is:

u(x; t) = u

¡c

t; : : : ; x

¡c

t):

Example 2.24. Consider the following equation

+ cxu

= αu

u(x; 0) =

1 + x

;

Here, c is a constant. When α = ¡c, the equation is conservative, and can be written as:

+ r·[cxu] = 0:

Let's ﬁrst solve the equation for α = ¡c, where c > 0. We can derive the ﬂow lines by solving:

= cx;

28 Li near First-Order PDEs

which gives us γ

: x(t) = x

. The PDE along γ

can then be expressed as:

= ¡cu;

which can be solved to give:

u(x(t); t) = u(x

; 0) e

¡ct

1 + x

Replacing x

by xe

¡ct

, we obtain

u(x; t) =

¡ct

1 + x

¡2ct

As expected, the total mass at t = 0 is conserved for t > 0. In fact, we have:

¡1

u(x; t) dx =

¡1

¡ct

1 + x

¡2ct

z=xe

¡ct

¡1

1 + z

= π:

This is the same as the total mass at t = 0. Therefore, we can write:

¡1

u(x; t) dx =

¡1

u(x; 0) dx:

Note that since V (x) = cx, t he particles move with velocity proportional to x. Consequently, the

initial proﬁle becomes fatter in time as the total mass remains the same.

-4 -2 0 2 4

0.2

0.4

0.6

0.8

For negative constant c, c < 0, the solution has the same structure. However, u(x; t) becomes

mor e concentrated a round x = 0 in this case. It is seen that for t !1, u(x; t) approaches the Dirac

delta function.

-4 -2 0 2 4

0.5

1.5

2.5

If α = 0, the equation is no longer conservative, and the solution is:

u(x; t) =

1 + x

¡2ct

This this case, we have

¡1

u(x; t) dx =

¡1

1 + x

¡2ct

= πe

¡1

u(x; 0) dx:

2.5 Li near and semi-linear transp ort equatio ns 29

Exercise 2.37. Consider the following equation

+ e

= 0;

for t ≥0. Draw th e characte ristic curves an d ﬁnd the general solution of the equatio n.

Exercise 2.38. Consider the following equation

+ xu

¡ yu

= αu

u(x; y; 0) =

(

1 x

+ y

≤ 1

0 otherwise

a) For what value of α the equation is con servative?

b) For th is α, ﬁ nd u(x; y; t).

c) General ize the above results in R

j=1

(¡1)

u = αu

u(x; y; 0) =

(

j=1

≤1

0 otherwise

Problem 2.4. In the ﬂ ow of a inco mpressible ﬂuid in a long pipeline, the velocity depends to the area intersectio n;

se e the following.

Use the conservation of mass principle, and show that

Problem 2.5. Suppose th e density function of a ﬂuid is given by u(t; x; y) = e

¡t

+ y

). Find the total mass

in the unit ball B = f(x; y); x

+ y

< 1g. Wha t is the rate of change of the total mass in B at time t = 1?

Problem 2.6. Suppose the initia l density function for a 1D ﬂow is u

(x) = 1 + x. If the velocity ﬁeld for the

ﬂow is v = t + x, we know that u(x; t) = e

¡2t

(1 + t + x) for t > 0. Verify the conservation law for this ﬂow in an

ar bitrary segment [x

; x

Problem 2.7. For a 2D ﬂow, suppose V = (x; y). If the init ial density fun ction is u

(x; y) = x

+ y

, then the

density function at t > 0 is u(t; x; y) = (x

+ y

) e

¡4t

. Verify directly the conservati on law for thi s ﬂow.

Problem 2.8. Assume tha t the initial density of a matter is given by u

(x; y) = e

¡(x

)

kg/m

. If the mass

moves w ith the constant speed V = 1m / s in the direction of x-axis, ﬁnd the total mass in the unite disk

(x ¡1)

+ y

< 1 the time t = 1.

Problem 2.9. Verify that the functi on u(x; y; t) = xy is a solution to the problem

(

+ xu

¡ yu

= 0

u(x; y; 0) = xy

Verify the conservation law

u(x; y; t) dS =

bnd(B)

·νd`;

where B is the unit bal l in R

Problem 2.10. Write down the co ntinuity equation for function u for the velocit ies given below, and determine

if they are incompressible

a) V = ¡x

b) V = u

30 Li near First-Order PDEs

c) V = (y; x)

d) V = (1; 0; u)

Problem 2.11. Consider the veloci ty ﬁeld V = (¡y; x) in R

. If the initial density is u

(x; y) =

1 + x

, ﬁnd

the density of the point (¡3; 0) at t ime t = π without solving the tran sport equation. Find the total mass in the

unit disk at time t = 1.

Problem 2.12. Suppose the ini tial density fu nction of a 2D ﬂuid is given by u

(x; y) = e

¡x

¡y

. If th e ﬂuid moves

with the velocity V = (¡y; 10x ¡2y), ﬁnd the parametric equations of a particle initially loca ted at (x

; y

) =(0; 3).

Find the de nsity of the point (x; y) = (0; 3e

¡π/3

) at time t =

Problem 2.13. Su ppose the velocity ﬁl ed of a 2D ﬂuid is given by V = (1 + x

; 1). Find the ﬂow line passi ng

through the origin and show that the density is decreasing along this line.

Problem 2.1 4. Write the continuity equation for a ﬂuid movi ng with the velocity V = (xy; y ¡x). Is the ﬂow

incompressible? Repeat the problem for the vel oc ity ﬁle d V = (αx + y; e

¡ y) and ﬁnd α such th at the ﬂow is

incompressible.

Problem 2.15. Assume that the density function of a ﬂuid (in absence of any source term) is u(x; y; t) = e

¡t

). If we know t he divergence of the velocity ﬁeld V = (v

; v

) i s equal to 1, show the r elation xv

+ yv

= 0.

Problem 2.16. Assume that the initial density of a matter is given by u

(x) =

π(1 + x

)

kg/m. If the mass moves

with constant velocity V = 1m/s along x-a xis, ﬁ nd the total mass in th e segment [0; 2] at t = 1.

Problem 2.1 7. Assume that the velocity of some ﬂuid is gi ve n by V = x. Write down the continuity equation

for the ﬂow and obtain the density at t = 1 if the initial density is u

(x) =

¡2

1 + x

Problem 2.18. Dete rmine the characte ris tic line along which th e solution to the problem

(

+ x u

= 0

u(x; 0) = e

¡jxj

;

is u = e

¡1

Problem 2.19. Draw the characteris tic lines of the equation

¡x u

= 0:

If the initial data is u(x; 0) = tan

¡1

x +

, ﬁnd u(x; t).

Problem 2.20. Consider the equation

+ v(x)u

= 0;

where v(x) is the following function

v(x) =

¡x + 1 0 < x ≤1

¡x ¡1 ¡1 < x < 0

0 jxj≥1

If the initial data f is the function

f(x) =



jxj jxj≤1

1 jxj≥1

Problem 2.21. So lve the fo llowing linear probl em

(

+ (1 + x

) u

= t

u(x; 0) = x

Problem 2.22. Find the solution to the following



+ xu

= u

u(x; 0) = sin(x)

Problem 2.23. So lve the fo llowing equation

(

+ xu

= xe

¡t

u(x; 0) = sin(x)

2.5 Li near and semi-linear transp ort equatio ns 31

Problem 2.24. Consider the equation

2/3

+ u

= 0;

for the function u = u(x; t).

a) Write th e e quati on of the characteris tic line passi ng through the point t = 1; x = 2.

b) Find the value u(1; 2) ¡u(4; 3) (you should provide the rea son for the value)

c) Find u(1; 2) if u(x; 0) = e

¡x

Problem 2.25. Consider the foll owing se mi-linear problem

(

+ u

= 2te

¡u

u(x; 0) = ln(1 + x

)

Find region in the plane (x; t) where u < 1.

Problem 2.26. Assume that the velocity ﬁel d of a matter is give as V = (¡y; x). Find the density of the point

(1; 0) at time t = 1 if the plan (x; y) is initially ﬁlle d with a matter of density

i. u

= x

+ y

ii. u

= xy

Problem 2.27. So lve the fo llowing problem

(

+ u

+ yu

= xy

u(x; y; 0) = sin(xy)

Problem 2.28. So lve the fo llowing problem

(

+ u

= u

u(x; y; 0) = u

(x; y)

Problem 2.29. Consider the equation

+ yu

¡xu

= 0:

a) Show that the equa tions of characteristic curves are



x = x

cos(t) + y

sin(t)

y = ¡x

sin(t) + y

cos(t)

b) Find the solution if u(x; y; 0) = xy.

c) Now solve the following equation

(

+ yu

¡xu

= u

u(x; y; 0) = xy

2.6 Qua si- linear equations and shockwave

In general, V in the continuity equation u

+ r·[uV ] = 0 can be a function of x and u, such that

V = V (x; u). One example of this is in modeling traﬃc ﬂow on highways, where the velocity of cars

depends inversely on the density of cars u sing the h ighway, describe d by the function

V (u) = V

max



1 ¡



;

where u

is the density of c ars at maximum capacity, V

max

is the maximum speed of cars, and

0 ≤u ≤u

repre sents the density of traﬃc on the highway. In this case, the continuity equation

becomes

+ V

max



1 ¡



= 0: (2.22)

32 Li near First-Order PDEs

Anot her important equation with V as a function of u is the B urgers' equation, where V =

giving the PDE





= 0;

or equivalently, u

+ uu

= 0.

More generally, we can consider problems of the form:



+div [g(u)] = 0

u(x; 0) = u

(x)

;

which can be written equivalently as:



+ g

(u) u

= 0

u(x; 0) = u

(x)

The characteristic system for this problem is given by:

= g

(u)

= 0

which leads t o the following implicit solution: u = u

(x ¡ g(u)t).

Example 2.25. Let's consider the continuity equation problem given by

+uu

=0;

with the initial condition u(0; x)=x. Although the initial condition u(0; x) for x<0 is not physically

meaningful a s i t leads to negative density, we retain this example to highlight an important feature

of the continuity equation. By solving the characteristic system of equations

= u;

= 0;

we obtain the solution x=ut+x

and u(t; x(t))=u(0; x(0))=x

, where x

is replaced by x ¡ut.

This yields the solution

u(x; t)=

1 + t

;

which has a domain of [0; 1) for t>0.

Suppose we now modify the init ial condition to ¡x. The resul ting solution is

u(x; t)=

t ¡1

;

which has a domain of [0; 1). As shown in the ﬁgure, the characteristic lines in this case collid e at

t=1, leading to a shock in the solution.

-1 -0.5 0 0.5 1

0.2

0.4

0.6

0.8

2.6 Quasi-linear equations and shoc kwave 33

The value of u(1; 0) can take any value in the range (¡1; 1), as particles with diﬀerent densities

converge at x=0 at t=1 and the density becomes discontinuous at this point. The ﬁgure below

shows this phenomena schematically. This concentration changes the initial densi ty of x=0 to a

discontinuous value.

x = 0

u(1; x)

t = 1 u

)

t = 1

(¡x

)

¡x

Exercise 2.39. Sol ve the following quasi-linear equation and determine the domain of the solution



+ uu

= 0

u(x; 0) = jxj

Remark 2.26. What can we say about the solution u(x; t) for t > 1? As we have seen, the solution

is discontinuous at t = 1, which means that the diﬀerential equation fails to hold for t ≥1 as well.

However, this does not mean that we cannot study the propagation of the shock beyond t = 1 from

a physical point of view. In fact, the development of a shock is a physical p henomenon that can b e

observed and studied experimentally.

One possible mathematical way to extend the discontinuous or shock solutions beyond the shock

time t = 1 is by using the concept of shockwave solutions. In this approach, we assume that the

discontinuous solution propagates like a wave in time. Shockwave solutions are a common way of

extending the solutions of partial d iﬀerential equations beyond t he point of discontinu ity.

2.6.1 Riemann problem

In this section, we consider the initial value problem given by the scalar conservatio n law:

+ uu

= 0

u(x; 0) = u

(x): =

1 x < 0

1 ¡x 0 < x < 1

0 x > 1

: (2.23)

To solve this problem, we ﬁrst determine the characteristic system of equations:

= u

= 0

;

which leads to the characteristic lines x(t)=ut+x

. Along each charact eristic line, the solution u

remains constant. Thus, we can write:

u(t; x(t)) = u

);

where x

is the initial position of the characteristic line. Using the initial condition u

(x), we can

ﬁnd the value of u along e ach characteristic line.

34 Li near First-Order PDEs

Next, we express x

in terms of x and t as x

=x¡ut. Substituting this into the expression for

u(t; x(t)), we obtain the solution u as a function of x; u and t:

u(x; t) =

1 x ¡ut < 0

1 ¡(x ¡ut) 0 < x ¡ut < 1

0 x ¡ut > 1

Simplifying the expression algebraically, we get:

u(x; t) =

1 x < t

1 ¡x

1 ¡t

t < x < 1

0 x > 1

: (2.24)

The ﬁgure below shows the solution u(x; t) at several instances of time. Note how a shock (discon-

tinuity) is developed at t=1.

t = 0

t = 0.5

t = 0.75

t = 1

discontinuity at t=1

Let's analyze the Riemann problem again using equation (2.23). In this equation, the function u

has a dual role, serving both as the density function and the material velocity V , where V =

= u.

Now, let's examine the initial condition shown below at t = 0:

t = 0

• Particles initially located at x

≤0 move to the right with velocity V =1, as their density

u =1 is conserved. For example, a p article initially located at x

=0 will reach x=1 at t = 1,

as evident from the characteristic equation x(t)=ut+x

. When x

= 0 and u=1, x(1)=1.

• Particles located initially at x 2(0; 1) move with the veloc ity V = 1 ¡x while they preserving

their densities. For example, a particle at x

has density u=

and moves with speed

V =

to the right, reaching x=0 at t=1.

• All particles initially located at x ≥1 have zero velocity and remain stationary on the x-axis

for t >0.

2.6 Quasi-linear equations and shoc kwave 35

The above observations imply that the density of point x=1 at time t = 1 can take any value between

0 and 1, as particles with diﬀerent densities between 0 and 1 reach this point at t=1. Therefore,

we can write the density u(x; 1) as

u(x; 1) =



1 x < 1

0 x > 1

;

while u(1; 1) is not uniquely deﬁned.

t = 1

2.6.2 Extension of the solution

We will delve into the details of this problem in the second volume of the book. However, in this

context, we aim to understand the elements of shock propag ation through the conservation law. Our

primary concern is to determine the density function u(x; t) for t>1. It should be noted that the

partial diﬀerential equation is not valid at t = 1, and hence, we cannot use the solution (2.24) for

t > 1. In fact, using this solution for t > 1 results in the existence of multiple non-physical solutions,

as shown in the ﬁgure below.

t = 2

t = 1.5

To unde rstand the p ropagation of shocks better, let us ﬁrst consider the following one-dimen-

siona l wave equation:



+ cu

= 0

u(x; 0) = u

(x)

;

where u

is the following discontinuous function

(x) =



1 x < 0

0 x > 0

As expected, the wave equation carries the initial condition with speed c to the right or left depending

on the sign of c in t he equation. Hence, the solution can be expressed as:

u(x; t) = u

(x ¡ct):

The following ﬁgure shows the solution of the wave equation for t = 1 and t = 2. Note that the

extension of the initial condition is plausible, and the solution is discontinuous only at one point

for each time instance.

t = 1

t = 2

1 + c1

36 Li near First-Order PDEs

Let's return to the Riemann problem:

+ uu

= 0

u(x; 1) =



1 x < 1

0 x > 1

Since u can only take the values 0 or 1, except at the discontinuity point, it's reasonable to assume

that this equ ation behaves like a wave equation. If we make this assumption, then the question

becomes: at what speed will the initial condition propagate?

To answer this, we can use the conservation law and write the PDE as an integral equation:









dx = 0:

for any interval (x

; x

) containing the shock position x at time t. Integrating this equation over

the interval (x ¡"; x + "), where " is a small positive constant, we obtain

x¡"

x+"

(x; t) dx +

= 0; (2.25)

where u

and u

are the values of u to the left and right of the shock, respectively. On the other

hand, we can write the integral in the above equation as

x¡"

x+"

(x; t) dx = lim

δt!0

δt



x¡"

x+"

u(x; t + δt) dx ¡

x¡"

x+"

u(x; t) dx



The ﬁrst integral in the right-hand side of the above equation is equal to

x¡"

x+"

u(x; t + δt) dx = u

(" + δx) + u

(" ¡δx);

and the second one is

x¡"

x+"

u(x; t) dx = u

" + u

The scenario is shown in the ﬁgure below.

x + δx

x ¡" x + "

x + "

x ¡"

t + δt

Hence, we can write

x¡"

x+"

(x; t) dx = v(u

+ u

);

where v is the ve locity of the shock point x. Substituting the result into equation (2.25) yields

v =

+ u

For our example with u

= 1 and u

= 0, we obtain the shock velocity v =

. The solution for t = 1;

2 is illustrated in the following ﬁgure:

2.6 Quasi-linear equations and shoc kwave 37

t = 1

t = 2

The solution u(x; t) can be obtained using the shockwave solution of the Riemann problem. The

formula for the solutio n is given as:

u(x; t) =



1 x < 1 + v(t ¡1)

0 x > 1 + v(t ¡1)

;

where v =

is the shock velocity for the g iven example.

This solution is called the shockwave solution be cause it has a sharp transition between the two

states, which moves with constant velocity v. The following ﬁgure illustrates the characteristic lines

with the shock l ine.

∗

= 1

u = 1

∗

= 1

u = 0

t = 2x ¡1

Example 2.27. This example illustrate the fact that t he graph of shockwave in xt-plane can be

a curve. This means that the ve locity of shock propagation can vary with time. Let us solve the

following problem

+ uu

= 0

u(x; 0) =



1 + x ¡1 < x < 0

1 ¡x 0 < x < 1

The equation of characteristic lines are x(t) = u

) t + x

where u

) =



1 + x

¡1 < x

< 0

1 ¡x

0 < x

< 1

. Sub-

stituting u

) into the equation of characteristic line yields

x(t) = x



(1 + x

)t ¡1 < x

< 0

(1 ¡x

)t 0 < x

< 1

A simple algebraic calculation helps to solve x

in terms of x; t as follows

x ¡t

1 + t

¡1 < x < t

x ¡t

1 ¡t

t < x < 1

Accordingly, the solution u(x; t) is obtained as

u(x; t) =

1 + x

1 + t

¡1 < x < t

1 ¡x

1 ¡t

t < x < 1

38 Li near First-Order PDEs

The solution for t =

;

and t = 1 are shown below.

−

0 0

5 1

= 0

−

= 0

−

1 1

= 1

t is observed that at t = 1, a discontinuity is formed at x

∗

= 1. For t > 1, the solution extends as

a sho ckwave with speed v that needs to be determined. Similar to the argument above, it can be

justiﬁed that the shockwave propagates with speed v =

at t = 1. However, what about v for t > 1?

It is important to note that according to the conservation law, the total mass of the system

remains conserved. This can be expressed as:

¡1

u(x; t) dx =

¡1

u(x; 0) dx = 1:

If the sho ckwave were to move with a constant velocity v =

for all t > 1, then the area under the

curve u(x; t) would increase with time, which would violate the conservation of mass. In order to

maintain the constant mass equal to 1, the height of the triangle, h, must decrease with time. This

is depicted in the following ﬁgure.

¡1 1.45

2.74

t = 2

t = 1

t = 6

h = 2/ 14

h = 1

h = 2/ 6

It is important to note that the velocity of the shockwave is not constant in the case where the

initial data has a jump discont inuity. Since the left value of the solution is h and the right value is

0, we can derive the shock velocity as v =

h, which in turn gives v =

1 + x

, where x is the distance

from the discontinuity at x = 1. Therefore, the shock velo city varies with respect to x.

To ﬁnd the shock velocity in terms of t, we can integrate the relation v =

1 + x

with resp ect

to t. This gives us x(t) = 2t + 2

¡1, which implies that the shock velocity at time t is v =

2t + 2

for t ≥1. The characteristic li nes in the (x; t)-plane can be plotted to visualize the behavior of the

solution.

−

1 0 1 2

shock

2.6 Quasi-linear equations and shoc kwave 39

Exercise 2.40. An alternative method to ﬁnd the shockwave velocity for the above problem is to use the following

formula

(x; t) =

x + 1

t + 1

;

and to solve the equa tion

v =

x + 1

2(t + 1)

Carry out the calculati on and conclude v =

2t + 2

Exercise 2.41. Let us solve the foll owing problem



+ uu

= 0

u(x; 0) = u

(x)

;

where u

is as follows

(x) =



3 x < 0

1 x > 0

Draw the ch ar ac teristic lines in the (x; t)-plan e and ﬁnd the shockwave solution.

Exercise 2.42. Consider the following problem



+ uu

= 0

u(x; 0) = u

(x)

;

where u

is a continuously diﬀerentiable fun ction.

a) Show that if u

< 0, the equation will develop a shock.

b) show that the ﬁrst time of the appearance of the shock is derive d by the following formula

∗

¡1

min

(ξ)

Hint : The shock is formed i f two character istic li nes collide

+ u

)t = x

+ u

)t:

This gives t =

¡ x

) ¡ u

)

. Minimize this and conclude the above rela tion.

Exercise 2.43. Consider the following equation



+ g

(u) u

= 0

u(x; 0) = u

(x)

If the probl em devel ops a shock, show that the velocity of the shockwave is

v =

g(u

) ¡ g(u

)

+ u

;

where u

; u

ar e the left and right limit of u at the shock point x respectively.

Exercise 2.44. Consider the following problem

+ uu

= 0

u(x; 0) =

0 x ≤0

0 ≤ x ≤ "

1 x ≥"

;

for " > 0. The problem does not devel op shock.

a) Draw the initial condition and obtain the solution u

(x; t).

b) Let " !0 and ﬁnd the solution of the follow ing equation

+ uu

= 0

u(x; 0) =

0 x < 0

1 x > 0

The solution of this equation is called the rarefaction solution.

40 Li near First-Order PDEs

Problem 2.30. Find an explicit solution of the followi ng equation



+ uu

= x

u(x; 0) = 0

Problem 2.31. Find the explicit solution to the followin g problem

(

+ e

u¡x

= 0

u(x; 0) = x

Problem 2.32. So lve the fo llowing problem



+ uu

= u

u(x; 0) = 1 + x

Problem 2.33. Find an implicit soluti on to the following problem



+ xuu

= 0

u(x; 0) = x

Use the implicit function theorem to justify that the solution exists in some interval of t 2[0; T ).

Problem 2.34. Find an implicit soluti on to the following problem



u + sin(u) @

u = 1

u(x; 0) = x

Use the implicit function theorem to justify that the solution exists in some interval of t.

Problem 2.35. Draw the solution to the following problems for t = 0; 1; 2



u + uu

= 0

u(x; 0) = u

(x)

;

where

(x) =

0 x ≤0

x 0 ≤x ≤1

1 x ≥1

Problem 2.36. Consider the foll owing problem

+ uu

= 0

u(x; 0) =

2 x ≤0

2 ¡x 0 ≤x ≤ 1

1 x ≥1

i. Draw t he characteri stic lines and determine t

∗

ii. Find the shock wave speed.

iii. Draw the solution u(x; 0 .5), u(x; 1), u(x; 2) and u(x; 4).

Problem 2.37. Draw the solution u(1; x) and u(2; x) to the following problem

u + uu

= 0

u(x; 0) =



2 x ≤0

1 x ≥0

Problem 2.38. So lve the fo llowing problem for t < t

∗

and draw the solutio n for few values o f t.

+ uu

= 0

u(x; 0) =

(

1 + x ¡1 < x < 0

¡x

0 ≤ x

Problem 2.39. For th e following equa tion, draw the charac teristics and ﬁnd the shock wave speed. Draw the

sol ut ion u(1; x) and u(2; x):

+ uu

= 0

u(x; 0) =



2 ¡jxj ¡1 < x ≤ 1

0 otherwise

2.6 Quasi-linear equations and shoc kwave 41

Problem 2.40. The traﬃc ﬂow in highways is usually modeled by the fol lowing equation

+ (1 ¡u)u

= 0;

where u is the density of cars (number of cars in a unit length). Assume that the initial proﬁle is as follows

u(x; 0) =

1 x ≤0

x + 1 0 ≤x ≤1

2 x ≥1

a) Draw the characteristic lines of the eq uation and ﬁnd the collision time t

b) Dr aw u(x; t) where T = 2.

Problem 2.41. For the fol lowing equation, ﬁnd the collision t ime t

∗

. Use a software to ﬁne u(x; 0.5) and u(x; 1).

(

+ 2uu

= 0

u(x; 0) = e

¡jxj

Problem 2.42. So lve the fo llowing problem

+ uu

= 0

u(x; 0) =



1 x < 0

2 x > 0

Problem 2.43. So lve the fo llowing problem

+ uu

= 0

u(x; 0) =

0 x < 0

1 0 < x < 1

0 x > 1

42 Li near First-Order PDEs

Appendix A

C urves and Surface

A.1 Curves and surfaces

The techniques outlined in this chapter for solving ﬁrst-order PDEs are primarily based on the

concept of the derivative of a scalar function along a smooth curve. To apply these techniques, it is

necessary to have an understanding of how smooth curves and surfaces are represente d in the plane

or in R

for n > 2.

A.1.1 Curves and derivative along a curve

Curves in the plane R

can be represented in several ways: through an explicit funct ion as the graph

f(x; f(x)); x 2D

g where f is a smooth function, an implicit func tion φ(x; y)=c, or a parametric

form γ(t)=(x(t); y(t)), where t ranges over an open interval I ⊂R. An implicit representation φ(x;

y)=c o f a curve can be transformed into an explicit one, y = f (x), under certain conditions using

the implicit function theorem; see the appendix of the book. Note that the graph of an explicit

function y = f(x) is a special case of a parameterized curve where the parameter is chosen to be x,

so that γ(x)=(x; f(x)).

While explicit and implicit representations of curves focus on the curves as a geometrical object,

or a set of points in the xy-plane, parametrized c urves can be considered as the path or trajectory

of a moving particle in the plane. Accordingly, the parameter t represents time, and the function

γ(t) gives the position of the particle at each time t. This interpreta tion enables us to determine the

veloc ity and acceleration vectors of the particle along its path. Moreover, the concept of parame-

trization can be generalized to higher dimensions, allowing us to represent more complicate d curves

and surfaces in R

A curve in R

parametrized by a parameter t is deﬁne d as a map γ: I ⊂R !R

given by

γ(t)=(x₁(t); x₂(t); :::; xₙ(t)). Here, x₁; x₂; :::; xₙ are scalar functions of t that describe the coordinate s

of points on the curve, and t is a parameter that varies over an open interval I ⊂R. The tangent

vector at a speciﬁc point γ(t) is deﬁned as the derivative of γ(t) with respect to t, written as

(t)=(x

(t); x

(t); :::; x

(t)). The magnitude of this vector is d enoted by jγ

(t)j, and in the c ontext

of physics, it repr esents the speed of a moving particle along γ(t).

The arc length s( t) of a curve is deﬁned as the integral of the speed jγ

(t)j over the interval [0; t],

that is,

s(t)=

jγ

(τ)jdτ :

By the fundamental theorem of calculus, the derivative of s(t) with respect to t is given by

= jγ

(t)j;

which is the speed of a moving particle along the curve γ(t).

The equation of a straight line γ = (x

(t); : : : ; x

(t)) in R

passing through c = (c

; : : : ; c

) is

simply derived by the diﬀerential equations

dγ

= a;

where a = (a

; : : : ; a

) is a constant. In the coordinate system, we have

= a

for j = 1; : : : ; n,

which is solve for x

= a

t + c

, or in the algebrai c form:

¡c

= ··· =

¡c

The vector a determines the direction of the line and is tange nt to γ as γ

(t) = a.

For a general curve γ(t) whose direction changes continuously according to the vector function

V (x) = (v

(x); : : : ; v

(x)), where x =(x

; :::; x

), the equation is derived by solving the following

diﬀerential equati on:

dγ

= V (γ(t));

or in the coordinate system,

= V (x

(t); : : : ; x

(t)); for j =1; :::; n, where γ(t)=(x

(t); :::; x

(t)).

If γ(t) is considered as the trajectory of a particle in R

, the vector V (γ(t)) is the velocity of the

particle at the point γ(t), and the vector function V = (v

;:::; v

) is said to be the velocity vector ﬁeld.

Example A.1. Consider a partic le moving in the xy-plane with the directional vector function V (x;

y)=(¡y; x). The trajectory γ(t) of the particle can be derived by solving the system of diﬀerential

equations:

= ¡y;

= x:

This system is equivalent to the exact equation xdx + ydy = 0 in the xy-plane, which has the implicit

solution x

+ y

= c. Alternatively, the system can be converted to a second-order equation for x(t)

as x

+ x = 0, with solution

x(t) = x

cos(t) ¡ y

sin(t);

where x

is the initial position of the particle (i.e., x(0)=x

), and y

= ¡x

(0). Using x(t) and the

second diﬀerential equation, we can ﬁnd

y(t) = x

sin(t) + y

cos(t);

and thus the equation of the trajectory is obtained as

γ(t) = (x

cos(t) ¡y

sin(t); x

sin(t) + y

cos(t)):

In particular, if x

= 1 and y

= 0, we obtain the trajectory

(1;0)

(t) = (cos(t); ¡sin(t)):

As we can see, the trajectory of the particle is the unit circle in the xy-plane. We can alternatively

solve the system by putting it in the matrix form







0 ¡1

1 0





;

44 Curves and Surface

with the fu ndamental matrix Φ(t) given by:

Φ(t) =



cos(t) ¡sin(t)

sin(t) cos(t)



Note that:

Φ(t)





= γ

(1;0)

(t):

The velocity vector of the particle with the trajectory γ

(1;0)

(t) is

(1;0)

(t) = (¡sin(t); cos(t));

and throu gh the relation

(1;0)

(t) · γ

(1;0)

(t) = 0;

we observe that γ

(1;0)

(t) is perpendicular to the traj ectory of the particle.

Let γ(t) be a parametric curve in R

, and let u = u(x

; : : : ; x

) be a continuously diﬀerentiable

function. The derivative of u along γ(t) is deﬁned as:

j=1

u(γ(t))

= ru(γ(t)) · γ

(t);

where ru denotes the gradient of u, given by the vector:

ru =

The operator r is also known as nabla and plays an important role in the context of multivariable

functions. Fo r example, if u represents the density f unction in R

, and γ(t) is a smooth curve, then

(γ(t)) represents the rate of change of the density when a control volume of mass is moving along

γ.

Example A.2. The above explanation highlights the relationship between curves in R

and systems

of ﬁrst-order ordinary diﬀerential equations. Speciﬁcally, a curve in R

can be obtained by solving

a system of ﬁrst-order ordinary diﬀerential equations, and conversely, the solution of a system of

ordinary diﬀerential equations corresponds to a curve. This c onnection between curves and diﬀer-

ential equations is fundamental to many areas of mathematics and has wide-ranging applications

in physics, engineering, and other ﬁelds.

Consider the system

= ¡y ¡αx

= x ¡αy

;

where α is a constant. The trajectory associated with α = 0; 0.3; ¡0.3, are shown b elow

A.1 Curv es and surfaces 45

In the xy-plane, the equation the equation can be written as

(x ¡αy) dx + (y + αx)dy = 0;

which is not exact if α =/ 0, meaning that energy is not conserved along the paths of mo tion. The

equivalent se cond-order equation of the system is:

+ 2αx

+ x = 0;

which can which can be further transformed to:





= ¡2α jx

Integrating this equation yields:

E(t) = E(0) ¡2 α

(s)j

ds:

If α > 0, the particles move down spirally to the origin along V (x; y), while if α < 0, they gain energy

and move outward spirally.

Exercise A.1. Find the force ﬁeld associated to th e velocity ﬁled given in the example above using t he relation

F =mγ

(t), and show that this force ﬁeld is centrifugal. Consider the t rajectory sta rting at (1; 0) for s implicity.

Additionally, demonstrate that the energy of the particle is conserved along its trajectory for the energy given by

E(t) = mjγ(t)j

mjγ

(t)j

Exercise A.2. Co nsider the vector ﬁeld V = (¡y; x; 1) and suppose a par ticle located initially at (1; 0; 0) moves

according to this vector ﬁeld. Find the trajectory of the particle an d show the relation γ

(t) · γ

(t) = 0 for all t.

Thus, conclude that the frame



(t)

jγ

(t)j

;

(t)

jγ

(t)j

;

(t) × γ

(t)

jγ

(t) × γ

(t)j



deﬁnes a local coordinate system for the particle.

46 Curves and Surface

Exercise A.3. With a curve mapping, we can determine the length of a curve. For a given explicit curve y = f (x)

or (x; f(x)) for x 2(x

; x

) the diﬀerential length is

ds = 1 + jf

dx;

and thus the arc length i s

s =

1 + jf

dx:

Simi larly, for a general parametriza tion γ(t), the diﬀerential le ngth is

ds = jγ

(t)jdt = jx

(t)j

+ jy

(t)j

dt;

and thus

s =

(t)j

+ jy

(t)j

dt:

The interesting point is that the arc l ength formula is independent of the parametriza tion. Let C ⊂ R

be a

geometric curve and let γ

: (a; b) !C and γ

: (c; d) !C be two smooth parametrization of C. Show

jγ

(t)jdt =

jγ

(t)jdt:

Exercise A.4. Consider the temperature dist ribution in the xy-plane given by T (x; y)=10 exp(x

¡y

). Suppose

a runner is moving along the curve γ(t)=(cos(t); sin(t)), and wearing a hand clock m arke d from 0 to 60. What is

the angular velocity of the hand of the runner's watch as they move alon g the curve? Find the angular ve locity

of the hand clock if the runn er runs given by the following function γ(t) = (cos(!

t); sin(!

t)) for some !

> 0.

A.1.2 Surfaces in space

There are several ways to represent a surface in three-dimensional space. One common approach

is to express the surface as the graph of an explicit function of the form z = f(x; y), where z is the

dependent variable and x; y are the independent variables. Alternatively, the surface can be deﬁned

implicitly through an equation of the form φ(x; y; z)=c.

The ﬁrst approach is a convenient way to visualize and manipulate the surface. However, this

representation is only possible for surfaces that can be written as a function of x and y, such as

paraboloids or planes. The second approach is more general and can represent surfaces that cannot

be written explicitly as a function of x and y, such as spheres or tori. The implicit function theorem

provides conditions for the existence and diﬀerentiability of a smooth function z = f(x; y) that

satisﬁes the equation φ(x; y; f (x; y)) = c. In particular, the the orem requires the existence of a

point (x

; y

; z

) such that φ(x

; y

; z

) = c and the partial derivative

@φ

; y

; z

) is nonzero. This

condition guarantees that the equation φ(x; y; z)=c deﬁnes the surface locally as a graph over the

xy-plane, and the function f (x; y) can be obtained by solving for z in terms of x and y.

Finally, a surface can be parameterized using a set of e quations that describe how x; y, a nd z

vary with two independent variables, typically denoted by t and s. In particular, a parameterized

surface in three-dimensional space is given by the equation

Σ(t; s)=(x(t; s); y(t; s); z(t; s));

where x; y, and z are functions o f the independent variables t and s, and (t; s) belongs to an open

set D in the plane. This method allows for the representation of surfaces with complex shapes

and topologies, such as the torus or the Mobius strip, and can also be used to represent higher-

dimensional surfaces. For instance, an m-dimensional hypersurface in R

where m < n can be

represented by a vector function of the f orm

Σ(t

; : : : ; t

) = (x

; : : : ; t

); : : : ; x

; : : : ; t

)):

A.1 Curv es and surfaces 47

Moreover, the parametric representation of a surface reveals the relationship between curves and

surfaces. For example, a surface Σ( t; s) in R

contains the lines γ

(t) = (x(t; s

); y(t; s

); z( t; s

)),

where t ranges over an interval for each s

, and the curves γ

(s) = (x(t

; s); y(t

; s); z(t

; s)),

where s ranges over an interval for each t

. For instance, let D =[0; π]×[¡π; π], and consider the

parametrization given by:

Σ(t; s)=(sint coss; sint sins; cost);

for (t; s) in D. This parametrization maps the rectangle D onto the surface of the unit sphere . Each

line fs

g×[0; π], for s2[¡π; π], is mappe d to the vertical curve Σ(t; s

), and each line [¡π; π]×ft

to the horizontal curves Σ(s; t

). In the spherical coordinate system, t is usually denoted by θ, the

angle with the z-axis, and s by φ, the ang le with the x-axis.

It is worth noting that the graph of an explicit fu nction z =f (x; y) is actually a special case of

a parameterized surface, where the surface is deﬁned by the equation Σ(x; y)=(x; y; f(x; y)). The

tangent and normal vectors to a surface Σ can be found as follows. Let (t

; s

) be a ﬁxed point on

Σ. The curve map γ

(t)=(x(t; s

); y(t; s

); z(t; s

)) lies on Σ, and its tangent vector γ

) is equal

to Σ

; s

). Similarly, the curve map γ

(s)=(x(t

; s); y(t

; s); z(t

; s)) lies on Σ, and its tangent

vector is Σ

; s

). The space spanned by Σ

; s

) and Σ

; s

) is the tangent space to the surface

Σ at ( t

; s

), which is just the tangent plane to Σ at that point. A surface Σ is said to be smooth if

(t; s) and Σ

(t; s) exist and are linearl y independent for all (t; s) in the domain of Σ. The normal

vector to Σ is then given by the cross product

ν =

×Σ

jΣ

×Σ

Note that if Σ

; s

) and Σ

; s

) are linearly independent, then Σ

; s

) ×Σ

; s

)=/ 0, and

the normal vector is well-deﬁned. For example, the normal vector to the unit sphere at the point

Σ(π /2; 0)=(1; 0; 0) with the parametrizati on given above can be calculated as follows: Σ

(π /2;

0)=(0; 0; ¡1) and Σ

(π /2; 0)=(0; 1; 0). Therefore, Σ

(π /2; 0)×Σ

(π /2; 0)=(1; 0; 0) a nd jΣ

(π /2;

0)×Σ

(π/2; 0)j=1, so the normal vector is ν =

×Σ

jΣ

×Σ

= (1; 0; 0), which is clearly perpendicular to

the tangent plane of the sphere at the point Σ(π/2; 0).

Exercise A.5. Th e are a diﬀerential dS is equal to jΣ

×Σ

jdtds, and thus the area of the surface Σ is equal to

S =

jΣ

×Σ

jdtds:

In particular, dS = 1 + jruj

dxdy is the diﬀerential area of an explicit sur face z = u(x; y). For example, if Σ

represent s a mat erial surface o f density ρ(x; y; z), th en the t otal mass on Σ is equal to

M =

ρ(Σ(t; s))jΣ

×Σ

jdtds:

48 Curves and Surface

Use the spherical coordinate and and calculate the area of the pot ion 0 ≤θ ≤ θ

of the unit sphe re for θ

Calc ulate the total mass on the surfac e if ρ = (x

+ y

)z.

Exercise A.6. Write down the equation of th e tangent plane to the smooth surface φ(x; y; z) = 0 at (x

; y

; z

)

on the surface.

Exercise A.7. Let φ(x; y; z) = 0 denote a surfac e Σ. Show that rφ is perpendicular to all tange nt vectors on

the surface Σ. Hint: Let γ = (x(t); y(t); z(t)) be arbitrary smooth curve lying on Σ .

Let's now consider the equations of surfaces in R

. Suppo se we have a plane passing through

a point p

= (x

; y

; z

) and perpendicular to a constant vector n~ = (a; b; c). Its equation can be

expressed as a

a(x ¡x

)+b(y ¡ y

)+c(z ¡z

)=0;

which is the expansion of the dot product (p ¡p

) ·n~ = 0, for p = (x; y; z) on the plane. Equivalently,

if we are given two linearly independent vectors V

= (a

; b

; c

) and V

= (a

; b

; c

), we can write

the parametric equation of the plane containing these two vectors as

Σ(t; s)=(ta

+sa

; tb

+sb

; tc

+sc

Note both vectors lie in the obtained plane. If we were provided with only one vector, the equation

of the plane could not be uniquely determined.

A.1 Curv es and surfaces 49