lecture notes on partial di erential equations chapter …aksikas/math337-cha2.pdf1 ’ & $ %...

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Lecture Notes on Partial Differential Equations

Chapter II

First Order PDE : Method of Characteristics

Ilyasse Aksikas

Department of Mathematical and Statistical Sciences

University of Alberta

aksikas@ualberta.ca

1 THE METHOD OF CHARACTERISTICS 2'

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This chapter introduces the method of characteristics for solvingfirst-order linear PDE.

1 The Method of Characteristics

The method of characteristics is a method which can be used tosolve a general first-order (only contain first order derivatives)PDE. Consider the first-order linear PDE in two variables.

a(x, t)∂u

∂x+ b(x, t)

∂u

∂t+ c(x, t)u = 0. (1)

The goal of the method of characteristics, when applied to thisequation, is to change coordinates from (x, t) to a new coordinatesystem (x0, r) in which the PDE becomes an ordinary differentialequation along certain curves in the x − t plane.

Definition 1.1. The curves {[x(r), t(r)] : 0 ≤ r ≤ ∞}, along whichthe PDE reduces to an ODE, are called the characteristics curvesor just the characteristics.

The new variable r will vary, and the new variable x0 will be

1 THE METHOD OF CHARACTERISTICS 3'

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constant along the characteristics.

Question : How do we find the characteristic curves ?

Notice that if we choose

dx

dr= a(x, t) and

dt

dr= b(x, t) (2)

then we have

du

dr=

dx

dr

∂u

∂x+

dt

dr

∂u

∂t= a(x, t)

∂u

∂x+ b(x, t)

∂u

∂t(3)

and along the characteristic curves, the PDE becomes the followingODE

du

dr+ c(x, t)u = 0. (4)

Equations (2) will be referred to as the characteristic equations.

General Strategy The general strategy for applying the methodof characteristics to a PDE of the form (1) is :

2 APPLICATION TO THE WAVE EQUATION 4'

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Step 1 : Solve the two characteristic equations (2). Find theconstant of integration by setting x(0) = x0 and t(0) = 0. We nowhave the transformation from (x, t) to (x0, r).

Step 2 : Solve the ODE (4).

Step 3 : We now have a solution u(x0, r). Solve for r and x0 interms of x and t (using the results of step 1) and substitute thesevalues in u(x0, r) to get the solution of the original PDE as u(x, t).

2 Application to the Wave Equation

Let us consider the first-order wave equation

c∂u

∂x+

∂u

∂t= 0, (5)

which describes the movement of a wave in one direction with nochange of shape. The constant c is referred to as the wave speed orvelocity of propagation.

2 APPLICATION TO THE WAVE EQUATION 5'

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In this case the characteristic equations (2) can be written asfollows :

dx

dr= c, (6)

anddt

dr= 1. (7)

and the ODE equation (4) becomes

du

dr= 0, (8)

Now we apply the method of characteristics outlined in the 3 stepsin the previous section.

Step 1 : Solve the characteristic equations :

The solution of equation (7) is t = r + k. Using the initial conditiont(0) = 0, we determine that the constant is k = 0, and we have thatr = t.

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The second characteristic equation to solve is equation (6) with theinitial condition x(0) = x0. The solution gives the characteristiccurves

x = cr + x0 = ct + x0, since x = x0 at r = 0.

Step 2 : Solve the ODE (8) :

The solution isu = constant, (9)

along a characteristic curve but the constant may be different ondifferent characteristic curves. As x0 gives us a differentcharacteristic curve, this implies that u = F (x0).

Step 3 : x0 defines which characteristic curve you are on and from(9), u is constant on a characteristic curve that depends on x0.Hence,

u = F (x0) = F (x − ct), (10)

as shown before, where the arbitrary function, F , is determined bythe initial condition.

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3 Initial-value Problems

Here the problem is defined over an infinite range in x. The generalstatement of the problem is

c∂u

∂x+

∂u

∂t= 0, −∞ < x < ∞, t > 0. (11)

u(x, 0) = f(x). (12)

The initial value problem, IVP, defined in Equations (11) and (12),is also referred to as a Cauchy problem and the solution isdetermined uniquely by the single condition at t = 0. From (10) wehave

u(x, t) = f(x − ct), (= f(x0))

If f is continuously differentiable, then ∂u∂x = f ′(x − ct) and

∂u∂t = −cf ′(x − ct) are continuous. Thus, u = f(x − ct) is a classical

solution of (11) and (12).

If f is only piecewise continuous (derivatives are not continuous),

3 INITIAL-VALUE PROBLEMS 8'

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e.g.

u =

{

1 − (x − ct)2 0 ≤ (x − ct)2 ≤ 10 (x − ct)2 > 1,

then u = f(x − ct) is called a weak solution.

Example 3.1.

2∂u

∂x+

∂u

∂t= 0, −∞ < x < ∞, t > 0,

with the initial condition

u(x, 0) =1

1 + x2,

The characteristic curves are given by the solutions to

dx

dr= 2. ⇒ x = 2r + x0,

anddt

dr= 1, ⇒ t = r.

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Hence,x = 2t + x0, ⇒ x0 = x − 2t.

The PDE becomes

du

dr= 0, ⇒ u = u(x0).

At t = 0, x = x0 and u = 11+x2

0

. Therefore,

u(x, t) =1

1 + (x − 2t)2.

Example 3.2.

−∂u

∂x+

∂u

∂t= 0, −∞ < x < ∞, t > 0,

with the initial condition

u(x, 0) =

{

1 − |x| |x| ≤ 10 |x| > 1

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The characteristic curves are given by the solutions to

dx

dr= −1, ⇒ x = −r + x0,

anddt

dr= 1, ⇒ t = r.

Thus, eliminating r and re-writing x0 in terms of x and t, we obtain

x0 = x + t.

The PDE reduces to the ODE

du

dr= 0, ⇒ u = u(x0)

Using the initial condition, t = 0, x = x0, we have

u(x0, 0) =

{

1 − |x0| |x0| ≤ 10 |x0| > 1

and so

u(x, t) =

{

1 − |x + t| |x + t| ≤ 10 |x + t| > 1

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The solution is shown in Figure 1.

From Figure 1 it is clear that the wave moves to the left. Thisallows us to summarise the results of the first order wave equation.

c > 0 ⇒ wave moves to the right,c < 0 ⇒ wave moves to the left.

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%Fig. 1 – In this case the wave speed is negative and the wave movesto the left.

4 INITIAL-BOUNDARY-VALUE PROBLEMS 13'

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4 Initial-boundary-value Problems

Now the infinite x plane is replace by the semi-infinite plane, witha boundary condition imposed on x = 0.

c∂u∂x + ∂u

∂t = 0 x > 0, t > 0,u(x, 0) = f(x),u(0, t) = g(t).

(13)

Consider the case c > 0, so that the waves are travelling from smallx to large x. The characteristics are again given by

dx

dr= c,

dt

dr= 1, ⇒ x = ct + x0.

The characteristic through the origin, splits the x − t plane intotwo regions, as shown in Figure 2.

In region R1, x > ct,the solution is determined by the initialcondition at t = 0. In this region the characteristics originate fromthe x-axis.

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%Fig. 2 – The characteristic, x = ct, splits the x − t plane into tworegions

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In region R2, x < ct, the solution is given by the boundarycondition at x = 0. Here the characteristics originate from thet-axis. Remembering that the solution is constant along thecharacteristics, this implies that the solution is determined bywhere the characteristics come from. Thus, the solution is

u(x, t) =

{

f(x − ct) x > ct,g(t − x/c) x < ct.

(14)

Note that the characteristics are

x − ct = x0, ⇒ t − x

c= −x0

c.

It is clear that (14) satisfies the initial condition, since t = 0 impliesthat x > 0 and u(x, 0) = f(x), and the boundary condition sincex = 0 implies that t > 0 and so u(0, t) = g(t). In addition, it iseasily shown that u(x, t) also satisfies the PDE.

Example 4.1.

2∂u

∂x+

∂u

∂t= 0,

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with

u(x, 0) =

{

1 − x 0 < x ≤ 1,0 x > 1,

u(0, t) = e−t, t > 0.

The characteristics are given by

x0 = x − 2t.

Therefore the plane is split into three regions and the solution is

u(x, t) =

0 x − 2t > 1,1 − (x − 2t) 0 < x − 2t ≤ 1,e−(t−x/2) x − 2t < 0.

A little bit of thought about this last example will soon illustrate amajor problem. If the wave speed is negative, i.e. c < 0, then thereis in general no solution to the initial-boundary-value problem.This is because the characteristics intersect both boundaries and ucannot be defined by two different values (resulting from values off(x) and g(t)) on the same characteristics. This is illustrated inFigure 3.

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%Fig. 3 – The characteristic, with c < 0, intersects both the x-axisand the t-axis.

5 OTHER EXAMPLES 18'

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5 Other Examples

Example 5.1.

2xt∂u

∂x+

∂u

∂t= u, −∞ < x < ∞, (15)

with the initial condition

u(x, 0) = x,

The characteristic equations are

dt

dr= 1, ⇒ t = r,

and

dx

dr= 2xt ⇒ dx

x= 2tdt ⇒ log x = t2 + C,

and rearranging we obtain

x = Aet2 ⇒ x = x0et2 .

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Thus,

x0 = xe−t2 .

On using dudr = ∂u

∂xdxdr + ∂u

∂tdtdr , (15) is now written as

du

dr= u ⇒ u = Aer,

where the constant A is actually constant along the characteristicdefined by the value of x0. Thus, A is really a function of x0 and wewrite

u = F (x0)et,

since t = r and the function F is determined by the initialconditions. At t = 0, we have x = x0 and u(x0, 0) = x0 so thatF (x0) = x0. Replacing x0 by the above expression involving x andt, we obtain the final solution

u(x, t) = x0et = xe−t2et = xet−t2 .

Example 5.2. Consider the initial-boundary-value problem

u2 ∂u

∂x+

∂u

∂t= 0, x > 0, t > 0, (16)

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withu(x, 0) =

√x x > 0,

u(0, t) = 0 t > 0.

Characteristic equations are

dx

dr= u2 (17)

anddt

dr= 1, ⇒ t = r,

and (16) gives

du

dr= 0 ⇒ u = constant on characteristic = F (x0).

Hence,x = u2r + x0 = u2t + x0,

since u is constant along the characteristic and dx/dr is thederivative of x along the characteristic curve. Therefore, we have

x0 = x − u2t.

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Thus, substituting for x0 into F (x0) we obtain the implicit solution

u(x, t) = F (x − u2t),

where F is determined by the initial condition, namely, t = 0,x = x0 and u =

√x0. Thus, F (x0) =

√x0 and so

u =√

x − u2t, x − u2t > 0.

Squaring both sides, we can manipulate this solution into anexplicit solution for u in terms of x and t.

u2 = x − u2t,

u2(1 + t) = x,

u2 =x

1 + t,

and so the final solution is

u =

√

x

1 + t, x > 0, t > 0.

We can easily check that this is the solution to (16) by calculating

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the partial derivatives,

∂u

∂x=

1

2x−1/2(1 + t)−1/2,

∂u

∂t= −1

2x1/2(1 + t)−3/2,

Therefore,

u2 ∂u

∂x=

x

1 + t

1

2x1/2(1 + t)1/2=

1

2

x1/2

(1 + t)3/2= −∂u

∂t