Arfken Mathematical Methods 7e: Section 9.6 - Exercise 9.6.3 Page 1 of 14

Exercise 9.6.3

Solve the wave equation, Eq. (9.89), subject to the indicated conditions.

Determine ψ(x, t) given that at t = 0 ψ0(x) is a single square-wave pulse as defined below, andthe initial time derivative of ψ is zero.

ψ0(x) = 0, |x| > a/2, ψ0(x) = 1/a, |x| < a/2.

Solution

The initial value problem to solve is as follows.

ψtt = c2ψxx, −∞ < x

Arfken Mathematical Methods 7e: Section 9.6 - Exercise 9.6.3 Page 2 of 14

Divide both sides by dt to obtain the fundamental relationship between the total derivative of hand the partial derivatives of h.

dh

dt=∂h

∂t+dx

dt

∂h

∂x

In light of this, the PDE for u reduces to the ODE,

du

dt= 0, (1)

along the characteristic curves in the xt-plane that satisfy

dx

dt= c, x(ξ, 0) = ξ, (2)

where ξ is a characteristic coordinate. Integrate both sides of equation (2) with respect to t tosolve for x(ξ, t).

x = ct+ ξ

Now integrate both sides of equation (1) with respect to t.

u(x, ξ) = f(ξ)

f is an arbitrary function of the characteristic coordinate ξ. Eliminate ξ in favor of x and t.

u(x, t) = f(x− ct)

Consequently, the PDE for ψ becomes

∂ψ

∂t− c∂ψ

∂x= f(x− ct).

It reduces todψ

dt= f(x− ct) (3)

along the characteristic curves in the xt-plane that satisfy

dx

dt= −c, x(η, 0) = η, (4)

where η is another characteristic coordinate. Integrate both sides of equation (4) with respect to tto solve for x(η, t).

x = −ct+ η

Now integrate both sides of equation (3) with respect to t.

ψ(x, η) =

ˆ tf(x− cs) ds+G(η)

G is an arbitrary function of the characteristic coordinate η. Make the substitution r = x− cs inthe integral.

ψ(x, η) =

ˆ x−ctf(r)

(−drc

)+G(η)

= F (x− ct) +G(η)

www.stemjock.com

Arfken Mathematical Methods 7e: Section 9.6 - Exercise 9.6.3 Page 3 of 14

F is the integral of −f/c, another arbitrary function. Therefore, since η = x+ ct,

ψ(x, t) = F (x− ct) +G(x+ ct).

This is the general solution of the wave equation. Now apply the initial conditions to determine Fand G.

ψ(x, 0) = F (x) +G(x) = ψ0(x)

ψt(x, 0) = −cF ′(x) + cG′(x) = 0

Differentiate both sides of the first equation with respect to x and multiply both sides of it by c.

cF ′(x) + cG′(x) = cψ′0(x)

−cF ′(x) + cG′(x) = 0

Add both sides of each equation to eliminate F ′.

2cG′(x) = cψ′0(x)

Divide both sides by 2c.

G′(x) =1

2ψ′0(x)

Integrate both sides with respect to x, setting the constant of integration to zero.

G(x) =1

2ψ0(x)

So then

F (x) +G(x) = ψ0(x) → F (x) +1

2ψ0(x) = ψ0(x) → F (x) =

1

2ψ0(x).

What we have actually solved for are F (w) and G(w), where w is any expression we choose.

F (x− ct) = 12ψ0(x− ct)

G(x+ ct) =1

2ψ0(x+ ct)

As a result,

ψ(x, t) = F (x− ct) +G(x+ ct)

=1

2ψ0(x− ct) +

1

2ψ0(x+ ct)

=1

2[ψ0(x− ct) + ψ0(x+ ct)].

www.stemjock.com

Arfken Mathematical Methods 7e: Section 9.6 - Exercise 9.6.3 Page 4 of 14

Note that

ψ0(x− ct) =

1

a|x− ct| < a

2

0 |x− ct| > a2

=

1

a−a2< x− ct < a

2

0 x− ct < −a2

0 x− ct > a2

and

ψ0(x+ ct) =

1

a|x+ ct| < a

2

0 |x+ ct| > a2

=

1

a−a2< x+ ct <

a

2

0 x+ ct < −a2

0 x+ ct >a

2

.

Depending what region in the xt-plane the point (x, t) is chosen, ψ(x, t) will be different. Theseregions are obtained by drawing characteristic lines with slopes ±c through

(−a2 , 0

)and

(a2 , 0),

the boundaries of where the initial condition is nonzero.

www.stemjock.com

Arfken Mathematical Methods 7e: Section 9.6 - Exercise 9.6.3 Page 5 of 14

Region I

Suppose the point (x, t) is chosen in region I.

In this case x− ct < −a2 and x+ ct >a2 , so

ψ(x, t) =1

2[ψ0(x− ct) + ψ0(x+ ct)]

=1

2(0 + 0)

= 0.

This formula is valid for |x| < ct− a2 .

www.stemjock.com

Arfken Mathematical Methods 7e: Section 9.6 - Exercise 9.6.3 Page 6 of 14

Region II

Suppose the point (x, t) is chosen in region II.

In this case x− ct < −a2 and −a2 < x+ ct <

a2 , so

ψ(x, t) =1

2[ψ0(x− ct) + ψ0(x+ ct)]

=1

2

(0 +

1

a

)=

1

2a.

This formula is valid for |a2 − ct| < −x <a2 + ct.

www.stemjock.com

Arfken Mathematical Methods 7e: Section 9.6 - Exercise 9.6.3 Page 7 of 14

Region III

Suppose the point (x, t) is chosen in region III.

In this case −a2 < x− ct <a2 and x+ ct >

a2 , so

ψ(x, t) =1

2[ψ0(x− ct) + ψ0(x+ ct)]

=1

2

(1

a+ 0

)=

1

2a.

This formula is valid for |a2 − ct| < x <a2 + ct.

www.stemjock.com

Arfken Mathematical Methods 7e: Section 9.6 - Exercise 9.6.3 Page 8 of 14

Region IV

Suppose the point (x, t) is chosen in region IV.

In this case x− ct < −a2 and x+ ct < −a2 , so

ψ(x, t) =1

2[ψ0(x− ct) + ψ0(x+ ct)]

=1

2(0 + 0)

= 0.

This formula is valid for −x > a2 + c|t|.

www.stemjock.com

Arfken Mathematical Methods 7e: Section 9.6 - Exercise 9.6.3 Page 9 of 14

Region V

Suppose the point (x, t) is chosen in region V.

In this case −a2 < x− ct <a2 and −

a2 < x+ ct <

a2 , so

ψ(x, t) =1

2[ψ0(x− ct) + ψ0(x+ ct)]

=1

2

(1

a+

1

a

)=

1

a.

This formula is valid for |x| < a2 − c|t|.

www.stemjock.com

Arfken Mathematical Methods 7e: Section 9.6 - Exercise 9.6.3 Page 10 of 14

Region VI

Suppose the point (x, t) is chosen in region VI.

In this case x− ct > a2 and x+ ct >a2 , so

ψ(x, t) =1

2[ψ0(x− ct) + ψ0(x+ ct)]

=1

2(0 + 0)

= 0.

This formula is valid for x > a2 + c|t|.

www.stemjock.com

Arfken Mathematical Methods 7e: Section 9.6 - Exercise 9.6.3 Page 11 of 14

Region VII

Suppose the point (x, t) is chosen in region VII.

In this case −a2 < x− ct <a2 and x+ ct < −

a2 , so

ψ(x, t) =1

2[ψ0(x− ct) + ψ0(x+ ct)]

=1

2

(1

a+ 0

)=

1

2a.

This formula is valid for |a2 + ct| < −x <a2 − ct.

www.stemjock.com

Arfken Mathematical Methods 7e: Section 9.6 - Exercise 9.6.3 Page 12 of 14

Region VIII

Suppose the point (x, t) is chosen in region VIII.

In this case x− ct > a2 and −a2 < x+ ct <

a2 , so

ψ(x, t) =1

2[ψ0(x− ct) + ψ0(x+ ct)]

=1

2

(0 +

1

a

)=

1

2a.

This formula is valid for |a2 + ct| < x <a2 − ct.

www.stemjock.com

Arfken Mathematical Methods 7e: Section 9.6 - Exercise 9.6.3 Page 13 of 14

Region IX

Suppose the point (x, t) is chosen in region IX.

In this case x− ct > a2 and x+ ct < −a2 , so

ψ(x, t) =1

2[ψ0(x− ct) + ψ0(x+ ct)]

=1

2(0 + 0)

= 0.

This formula is valid for |x| < −ct− a2 .

Some of the formulas we found can be combined by forming unions of the regions and usingabsolute value signs. The union of regions I and IX, for example, results in the following formulafor ψ(x, t).

ψ(x, t) = 0 if |x| < c|t| − a2

The union of regions IV and VI results in the following formula for ψ(x, t).

ψ(x, t) = 0 if |x| > a2+ c|t|

The union of regions II, III, VII, and VIII results in the following formula for ψ(x, t).

ψ(x, t) =1

2aif∣∣∣a2− c|t|

∣∣∣ < |x| < a2+ c|t|

www.stemjock.com

Arfken Mathematical Methods 7e: Section 9.6 - Exercise 9.6.3 Page 14 of 14

Therefore,

ψ(x, t) =

1

aif |x| < a

2− c|t|

0 if |x| < c|t| − a2

0 if |x| > a2+ c|t|

1

2aif∣∣∣a2− c|t|

∣∣∣ < |x| < a2+ c|t|

.

The solution in each part of the xt-plane is labelled by color.

www.stemjock.com