7More partial differential equations

IB Methods

7.2 Method of characteristics

Curves in the plane

Suppose we have a curve on the plane x : R → R

2

given by s 7→ (x(s), y(s)).

Definition

(Tangent vector)

.

The tangent vector to a smooth curve

C

given

by x : R → R

2

with x(s) = (x(s), y(s)) is

dx

ds

,

dy

ds

.

If we have some quantity

φ

:

R

2

→ R

, then its value along the curve

C

is just

φ(x(s), y(s)).

A vector field

V

(

x, y

) :

R

2

→ R

2

defines a family of curves. To imagine this,

suppose we are living in a river. The vector field tells us the how the water flows

at each point. We can then obtain a curve by starting a point and flow along

with the water. More precisely,

Definition

(Integral curve)

.

Let

V

(

x, y

) :

R

2

→ R

2

be a vector field. The

integral curves associated to

V

are curves whose tangent

dx

ds

,

dy

ds

is just

V

(

x, y

).

For sufficiently regular vector fields, we can fill the space with different curves.

We will parametrize which curve we are on by the parameter t. More precisely,

we have a curve

B

= (

x

(

t

)

, y

(

t

)) that is transverse (i.e. nowhere parallel) to our

family of curves, and we can label the members of our family by the value of

t

at which they intersect B.

C

4

C

3

C

2

C

1

B(t)

We can thus label our family of curves by (x(s, t), y(s, t)). If the Jacobian

J =

∂x

∂s

∂y

∂t

−

∂x

∂t

∂y

∂s

6= 0,

then we can invert this to find (

s, t

) as a function of (

x, y

), i.e. at any point, we

know which curve we are on, and how far along the curve we are.

This means we now have a new coordinate system (

s, t

) for our points in

R

2

. It turns out by picking the right vector field

V

, we can make differential

equations much easier to solve in this new coordinate system.

The method of characteristics

Suppose φ : R

2

→ R obeys

a(x, y)

∂φ

∂x

+ b(x, y)

∂φ

∂y

= 0.

We can define our vector field as

V(x, y) =

a(x, y)

b(x, y)

.

Then we can write the differential equation as

V · ∇φ = 0.

Along any particular integral curve of V, we have

∂φ

∂s

=

dx(s)

ds

∂φ

∂x

+

dy(s)

ds

∂φ

∂y

= V · ∇φ,

where the integral curves of V are determined by

∂x

∂s

t

= a(x, y),

∂y

∂s

t

= b(x, y).

This is known as the characteristic equation.

Hence our partial differential equation just becomes the equation

∂φ

∂s

t

= 0.

To get ourselves a well-posed problem, we have to specify our boundary data

along a transverse curve

B

. We pick our transverse curve as

s

= 0, and we

suppose we are given the initial data

φ(0, t) = h(t).

Since φ does not vary with s, our solution automatically is

φ(s, t) = h(t).

Then φ(x, y) is given by inverting the characteristic equations to find t(x, y).

We do a few examples to make this clear.

Example (Trivial). Suppose φ obeys

∂φ

∂x

y

= 0.

We are given the boundary condition

φ(0, y) = f(y).

The solution is obvious, since the differential equation tells us the function is

just a function of

y

, and then the solution is obviously

φ

(

x, y

) =

h

(

y

). However,

we can try to use the method of characteristics to practice our skills. Our vector

field is given by

V =

1

0

=

dx

ds

dy

ds

.

Hence, we get

dx

ds

= 1,

dy

ds

= 0.

So we have

x = s + c, y = d.

Our initial curve is given by

y

=

t, x

= 0. Since we want this to be the curve

s = 0, we find our integral curves to be

x = s, y = t.

Now for each fixed t, we have to solve

∂φ

∂s

t

= 0, φ(s, t) = h(t) = f(t).

So we know that

φ(x, y) = f(y).

Example. Consider the equation

e

x

∂φ

∂x

+

∂φ

∂y

= 0.

with the boundary condition

φ(x, 0) = cosh x.

We find that the vector field is

V =

e

x

1

This gives

dx

ds

= e

x

,

dy

ds

= 1.

Thus we have

e

−x

= −s + c, y = s + d.

We pick our curve as x = t, y = 0. So our relation becomes

e

−x

= −s + e

−t

, y = s.

We thus have

∂φ

∂s

t

= 0, φ(s, t) = φ(0, t) = cosh(t) = cosh[−ln(y + e

−x

)]

So done.

Example.

Let

φ

:

R

2

→ R

solve the inhomogeneous partial differential equation

∂

x

φ + 2∂

y

φ = ye

x

with φ(x, x) = sin x.

We can still use the method of characteristics. We have

u =

1

2

.

So the characteristic curves obey

dx

ds

= 1,

dy

ds

= 2.

This gives

x = s + t, y = 2s + t

so that x = y = t at s = 0. We can invert this to obtain the relations

s = y −x, t = 2x −y.

The partial differential equation now becomes

dφ

ds

t

= u · ∇φ = ye

x

= (2s + t)e

s+t

Note that this is just an ordinary differential equation in

s

because

t

is held

constant. We have the boundary conditions

φ(s = 0, t) = sin t.

So we get

φ(x(s, t), y(s, t)) = (2 − t)e

t

(1 − e

s

) + sin t + 2se

s+t

.

Putting it in terms of x and y, we have

φ(x, y) = (2 − 2x + y)e

2x−y

+ sin(2x − y) + (y −2)e

x

.

We see that our Cauchy data

φ

(

x, x

) =

sin x

should be specified on a curve

B that intersects each characteristic curve exactly once.

If we tried to use a characteristic curve

B

that intersects the characteristics

multiple times, if we want a solution at all, we cannot specify data freely along

B

. its values at the points of intersection have to compatible with the partial

differential equation.

For example, if we have a homogeneous equation, we saw the solution will

be constant along the same characteristic. Hence if our Cauchy data requires

the solution to take different values on the same characteristic, we will have no

solution.

Moreover, even if it is compatible, if

B

does not intersect all characteristics,

we will not get a unique solution. So the solution is not fixed along such

characteristics.

On the other hand, if

B

intersects each characteristic curve transversely,

the problem is well-posed. So there exists a unique solution, at least in the

neighbourhood of

B

. Notice that data is never transferred between characteristic

curves.