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[parent] Wave Mechanics Examples: Partial Derivatives for Waves (Example)

Wave Mechanics Examples: Partial Derivatives for Waves

This companion article provides exercises for WM13, wave mechanics: Partial Derivatives for Waves. All exercises are presented first. Complete worked solutions follow in Part II.

WM13 introduced the calculus of a one-dimensional wave field

u = u (x,t).
(1)

The first partial derivatives answer two different local questions:

|----------------------------------|
|      ∂u-                         |
|ux =  ∂x = spatial slope at fixed t |
-----------------------------------
(2)

and

|------------------------------------|
|u =  ∂u-=  local time  rate at fixed x.|
--t---∂t------------------------------
(3)

For transverse string displacement, ut is the transverse velocity of the material point labeled by x, while

|-------2--|
|uxx = ∂-u-|
-------∂x2--
(4)

measures spatial curvature and

|-------2--|
|u  =  ∂-u-|
--tt---∂t2-|
(5)

is local transverse acceleration [1234].

For a right-moving translating disturbance

u(x,t) = F (x − ct),
(6)

WM13 derived

|----------|
ut-=-−-c-ux-
(7)

and

|-------2----|
-utt =-cuxx.-|
(8)

The exercises below reinforce those results without treating this as the physical Newton’s-law derivation of the string wave equation. That derivation comes later in the series.

How to use this problem set

Attempt all exercises in Part I before consulting Part II. For each derivative, state explicitly which independent variable is being varied and which is being held fixed. When a derivative has physical units, include them. In chain-rule problems, identify the intermediate phase or translating coordinate before differentiating.

Part I: Exercises

Exercise 1: What is being held fixed?

For a field u(x,t), describe in words what each derivative means and state which independent variable is held fixed.

  1. ux
  2. ut
  3. uxx
  4. utt

For transverse displacement of a string, give the physical interpretation of each derivative where appropriate.

Exercise 2: Units of wave derivatives

A transverse string displacement u is measured in meters, position x is measured in meters, and time t is measured in seconds.

Determine the units of

  1. ux,
  2. ut,
  3. uxx,
  4. utt,
  5. c2u xx if c has units of meters per second.

Explain why the units in part (e) are consistent with the relation utt = c2u xx.

Exercise 3: Estimate derivatives from finite differences

At time t0, measurements near x0 give

u(x0,t0) = 4.0 mm
(9)

and

u (x0 + 0.020 m, t0) = 3.4 mm.
(10)

At the fixed position x0, measurements at nearby times give

u(x ,t +  0.010 s) = 4.8mm.
   0  0
(11)

Use forward difference quotients to estimate

  1. ux(x0,t0),
  2. ut(x0,t0).

State the sign and physical meaning of each result.

Exercise 4: Read derivative signs from a spatial profile

The following spatial snapshot marks four points on a sinusoidal profile.

PIC

Figure. A spatial snapshot used to reason about the signs of ux and uxx.

Without calculating exact numerical values, determine the sign of ux and the sign of uxx at points P, Q, R, and S.

Use the local slope for ux and the local concavity for uxx.

Exercise 5: Differentiate a sinusoidal traveling wave

Consider

u(x,t) = A cos(kx − ωt + ϕ).
(12)

Starting from the phase

𝜃 =  kx − ωt + ϕ,
(13)

derive expressions for

  1. ux,
  2. uxx,
  3. ut,
  4. utt.

Then show that

         2
uxx = − k u
(14)

and

utt = − ω2u.
(15)

Exercise 6: Numerical derivatives at one event

A traveling wave is

u (x,t) = 0.030 m cos(5x − 20t),
(16)

where x is in meters and t is in seconds.

At the event

x = 0.20 m,     t = 0.050 s,
(17)

find

  1. u,
  2. ux,
  3. ut,
  4. uxx,
  5. utt.

Interpret the signs of ux and ut physically.

Exercise 7: Verify the wave-equation relation for a sinusoid

For the wave in Exercise 6,

  1. determine the propagation speed from c = ω∕k,
  2. compute c2u xx symbolically,
  3. show that c2u xx = utt,
  4. explain why this verification is not yet the physical derivation of the string wave equation.

Exercise 8: Chain rule for an arbitrary right-moving profile

Let

u(x,t) = F (x − ct).
(18)

Define

ξ = x − ct.
(19)

Use the chain rule to derive

  1. ux,
  2. ut,
  3. uxx,
  4. utt.

Then prove

u--=-−-c-u-|
--t-------x-
(20)

and

|------------|
|       2    |
-utt =-cuxx.-
(21)

Exercise 9: Left-moving profile and the sign change

Now let

u(x, t) = G (x + ct).
(22)

  1. Derive the relationship between ut and ux.
  2. Derive the relationship between utt and uxx.
  3. Explain why the first-order relation changes sign but the second-order wave-equation relation does not.

Exercise 10: Interpret the right-moving slope-rate relation

The following figure compares the spatial slope of a translating profile with its local time rate.

PIC

Figure. For a right-moving profile F(xct), the graphs of ux and ut∕c have opposite signs.

Answer the following.

  1. If ux > 0 at some event and c > 0, what is the sign of ut?
  2. If ux < 0, what is the sign of ut?
  3. Give a geometric explanation using a right-moving pulse.
  4. What relation would replace this one for a left-moving profile?

Exercise 11: Partial derivative versus total derivative

The figure below shows three directions through the (x,t) domain.

PIC

Figure. Partial derivatives move along coordinate directions. The total derivative of u(x(t),t) follows a path through the (x,t) domain.

Suppose an observer moves according to x = x(t) through a field u(x,t).

  1. State the multivariable chain-rule formula for du∕dt.
  2. For a right-moving profile u = F(xct), substitute ut = cux into the total derivative.
  3. What value of dx∕dt makes du∕dt = 0?
  4. Interpret that result physically.

Exercise 12: Mixed partial derivatives

Consider

u(x,t) = x2t3 + 4xt.
(23)

Calculate

  1. ux,
  2. ut,
  3. uxt,
  4. utx.

Verify that uxt = utx for this smooth function.

Exercise 13: Diagnose conceptual statements

For each statement, decide whether it is correct. If incorrect, rewrite it accurately.

  1. ut is the speed at which the wave pattern propagates.”
  2. “When calculating ux, time must be held fixed.”
  3. “If uxx = 0 at a point, the displacement must also be zero there.”
  4. “For a right-moving profile F(x ct), ut and ux always have the same sign.”
  5. “Showing that a trial waveform satisfies utt = c2u xx proves that every physical medium obeys that wave equation.”

Exercise 14: Synthesis problem

A right-moving sinusoidal wave is

                    (            π )
u (x, t) = 0.015 m cos  6x − 24t + -- .
                                 3
(24)

At the event

x = 0.10 m,     t = 0.025 s,
(25)

complete the following.

  1. Identify A, k, ω, and ϕ.
  2. Find λ, T, f, and c.
  3. Find ux and ut at the event.
  4. Verify numerically that ut = cux at that event.
  5. Find uxx and utt at the event.
  6. Verify numerically that utt = c2u xx.
  7. In one sentence, distinguish the local material velocity ut from the propagation speed c.

Part II: Complete Worked Solutions

Solution 1: What is being held fixed?

  1. ux is the spatial rate of change of the field. Position varies while time is held fixed. For string displacement it is the local slope of a spatial snapshot.
  2. ut is the local time rate of change. Time varies while position is held fixed. For string displacement it is the transverse velocity of the material point labeled by x.
  3. uxx is the rate at which spatial slope changes with position while time remains fixed. In one-dimensional wave problems it measures local curvature.
  4. utt is the rate at which ut changes with time while position remains fixed. For string displacement it is local transverse acceleration.

Solution 2: Units of wave derivatives

The field u has units of meters.

  1.        m
[ux ] = --=  1.
       m
    (26)

    For displacement versus position, the slope is dimensionless.

  2.       m-
[ut] = s .
    (27)

  3.         m      1
[uxx] = --2 = --.
        m     m
    (28)

  4.        m
[utt] = -2.
       s
    (29)

  5. Since
            2
 2    m--
[c ] = s2 ,
    (30)

    we obtain

      2       m2 1    m
[c uxx] = -2--- = -2.
          s  m    s
    (31)

    This is exactly the unit of utt, so the relation utt = c2u xx is dimensionally consistent.

Solution 3: Estimate derivatives from finite differences

For the spatial derivative,

ux(x0,t0) u(x0-+-Δx,-t0) −-u(x0,t0)
           Δx (32)
= 3.4mm  −  4.0 mm
-----------------
     0.020 m (33)
= −-0.6-mm--
 0.020 m. (34)

Converting 0.6 mm = 6.0 × 104 m,

|-------------|
ux-≈--− 0.030.-
(35)

The negative sign means the profile slopes downward as x increases at that instant.

For the time derivative,

ut(x0,t0) u (x0,t0 + Δt ) − u(x0,t0)
-------------------------
           Δt (36)
= 4.8 mm  − 4.0mm
-----------------
      0.010 s (37)
= 0.8-mm-
 0.010 s (38)
= 0.080 m/s . (39)

The positive sign means the local string point is moving in the positive transverse direction at that event.

Solution 4: Read derivative signs from a spatial profile

For a cosine-shaped spatial snapshot:

  1. At P, the curve is descending, so ux < 0. The profile is concave downward there, so uxx < 0.
  2. At Q, the profile crosses zero while descending most steeply, so ux < 0. At the zero crossing of a pure cosine, the curvature is zero, so uxx = 0.
  3. At R, the profile is at a trough. The slope is zero, so ux = 0, while the curve is concave upward, so uxx > 0.
  4. At S, the profile is rising, so ux > 0. Since the displacement there is positive for a cosine and uxx = k2u, the curvature is negative: u xx < 0.

Solution 5: Differentiate a sinusoidal traveling wave

Let

𝜃 = kx −  ωt + ϕ,     u = A cos𝜃.
(40)

For the spatial derivative, hold t fixed:

ux = ∂--
∂x(A cos 𝜃) (41)
= A sin 𝜃∂𝜃-
∂x (42)
= Ak sin 𝜃. (43)

Differentiate again:

uxx = Ak cos 𝜃∂𝜃-
∂x (44)
= Ak2 cos 𝜃 (45)
= k2u. (46)

Thus

|--------------------------------|
|                             2  |
-ux =-−-Ak-sin-𝜃,-----uxx =-−-k-u.-
(47)

For the time derivative, hold x fixed:

ut = ∂--
∂t(A cos 𝜃) (48)
= A sin 𝜃∂𝜃
---
∂t (49)
= sin 𝜃. (50)

Differentiate again:

utt = cos 𝜃∂𝜃-
∂t (51)
= 2 cos 𝜃 (52)
= ω2u. (53)

Therefore

|---------------------------2--|
-ut =-Aω-sin𝜃,-----utt =-−-ω-u.-
(54)

Solution 6: Numerical derivatives at one event

The wave is

u = 0.030 cos(5x − 20t).
(55)

At

x = 0.20,     t = 0.050,
(56)

the phase is

𝜃 = 5(0.20) 20(0.050) (57)
= 1.0 1.0 (58)
= 0. (59)

Therefore

cos𝜃 =  1,    sin𝜃 = 0.
(60)

  1. |------------|
u-=--0.030m.--
    (61)

  2. ux = − Ak sin𝜃 =  0,
    (62)

    so

    |--------|
-ux-=-0.-|
    (63)

  3. ut = A ωsin𝜃 =  0,
    (64)

    so

    |------------|
-ut-=-0-m/s.-|
    (65)

  4. uxx = k2u (66)
    = 25(0.030) (67)
    = 0.750 m1 . (68)
  5. utt = ω2u (69)
    = 400(0.030) (70)
    = 12.0 m/s2 . (71)

At this event the string is at a crest of the spatial profile and also at an instantaneous turning point of the local motion. Thus both slope and transverse velocity are zero, while the curvature and acceleration are negative.

Solution 7: Verify the wave-equation relation for a sinusoid

The wave has

k =  5rad/m,      ω = 20 rad/s.
(72)

  1.     ω    20   |--------|
c = --=  ---= -4.0m/s--.
    k    5
    (73)

  2. Since
    uxx =  − k2u = − 25u,
    (74)

    and

    c2 = 16m2 ∕s2,
    (75)

    we obtain

    c2u   = 16(− 25u) = − 400u.
   xx
    (76)

  3. But
    utt = − ω2u = − 400u.
    (77)

    Therefore

    |------------|
|u  =  c2u   .|
--tt------xx-
    (78)

  4. This calculation verifies that the proposed sinusoidal traveling wave satisfies the differential relation. It does not explain why a real stretched string must obey that equation or why its speed is set by the Tension and linear mass density. That physical derivation requires force balance and Newton’s second law.

Solution 8: Chain rule for an arbitrary right-moving profile

Let

ξ = x − ct,     u = F (ξ ).
(79)

The needed derivatives of ξ are

∂ξ-=  1,     ∂ξ-= − c.
∂x           ∂t
(80)

Then

ux = F(ξ)∂ξ-
∂x (81)
= F(ξ), (82)

and

ut = F(ξ)∂ξ-
∂t (83)
= cF(ξ). (84)

Hence

|------------|
|ut = − cux. |
-------------
(85)

Differentiate again:

uxx = F ′′(ξ)
(86)

and

utt = c2F ′′(ξ).
(87)

Therefore

|------------|
|u  =  c2u   .|
--tt------xx-
(88)

The derivation does not require F to be sinusoidal. It only requires the translating form and sufficient differentiability.

Solution 9: Left-moving profile and the sign change

Let

η = x + ct,     u = G (η ).
(89)

Then

∂η-=  1,    ∂-η = c.
∂x           ∂t
(90)

Therefore

       ′
ux =  G (η)
(91)

and

u =  cG′(η).
 t
(92)

Thus

|----------|
ut-=-+c-ux--
(93)

for a left-moving profile.

The second derivatives are

uxx = G ′′(η )
(94)

and

utt = c2G ′′(η),
(95)

so once again

|------------|
|u  =  c2u   .|
--tt------xx-
(96)

The first derivative contains one factor of the sign of the propagation term. Differentiating twice produces the square of that sign, so the second-order relation is the same for both directions.

Solution 10: Interpret the right-moving slope-rate relation

For a right-moving profile,

ut = − cux,
(97)

with c > 0.

  1. If ux > 0, then
    |-------|
ut-<-0.--
    (98)

  2. If ux < 0, then
    |-------|
ut->-0.--
    (99)

  3. At a fixed position, a right-moving profile brings material from slightly to the left into the observation point as time advances. If the spatial profile rises with increasing x, the values just to the left are smaller, so the local field decreases with time. This produces the opposite sign between ut and ux.
  4. For a left-moving profile G(x + ct),
    |------------|
-ut-=-+c-ux.-|
    (100)

Solution 11: Partial derivative versus total derivative

For a moving observer x = x(t), the measured quantity is u(x(t),t). The multivariable chain rule gives

------------------
|du        dx    |
|---= ut + ---ux.|
-dt--------dt-----
(101)

For a right-moving profile,

ut = − cux.
(102)

Therefore

du
---
dt = cux + dx
---
dtux (103)
= (       )
  dx-− c
  dtux. (104)

Thus

du- = 0
 dt
(105)

when

|--------|
|dx      |
|---=  c.|
-dt------
(106)

An observer moving to the right at exactly the wave speed follows a fixed feature of the translating profile. Along that path, the value of F(x ct) remains constant.

Solution 12: Mixed partial derivatives

Given

u(x,t) = x2t3 + 4xt,
(107)

first differentiate with respect to x:

|---------------|
u  =  2xt3 + 4t.|
--x-------------
(108)

Differentiate with respect to t:

|----------------|
|ut = 3x2t2 + 4x.|
------------------
(109)

Now

uxt = ∂--
∂t(2xt3 + 4t) (110)
= 6xt2 + 4 . (111)

And

utx =  ∂
---
∂x(3x2t2 + 4x) (112)
= 6xt2 + 4 . (113)

Therefore

|----------|
-uxt =-utx.|
(114)

This agrees with equality of mixed partial derivatives for sufficiently smooth functions.

Solution 13: Diagnose conceptual statements

  1. Incorrect. ut is the local time rate of change at fixed position. For string displacement it is local transverse material velocity. The propagation speed of the wave pattern is a separate quantity such as c = ω∕k.
  2. Correct. In ux, position changes while time is held fixed.
  3. Incorrect. uxx = 0 means the local curvature measure is zero. The displacement itself can be nonzero.
  4. Incorrect. For F(x ct),
    ut = − cux,
    (115)

    so for c > 0 the two derivatives have opposite signs whenever they are nonzero.

  5. Incorrect. Showing that one waveform satisfies utt = c2u xx only verifies compatibility with that PDE. A physical derivation is required to show why a particular medium obeys that equation and what determines c.

Solution 14: Synthesis problem

The wave is

                    (            π )
u (x, t) = 0.015 m cos  6x − 24t + -- .
                                 3
(116)

  1. By comparison with A cos(kx ωt + ϕ),
    |--------------------------------------------------π--|
A  = 0.015m,   k =  6rad/m,    ω = 24 rad/s,  ϕ =  -. |
---------------------------------------------------3---
    (117)

  2. λ = 2π
---
k = 2π
---
6 = π
--
3 m 1.047 m , (118)
    T = 2π-
ω = 2π-
24 = π--
12 s 0.262 s , (119)
    f = 1-
T = 12-
π = 3.82 Hz , (120)
    c = ω
--
k = 24
---
6 = 4.0 m/s . (121)
  3. At the event x = 0.10 m and t = 0.025 s, the phase is
    𝜃 = 6(0.10) 24(0.025) + π
--
3 (122)
    = 0.6 0.6 + π
--
3 (123)
    = π-
3. (124)

    Therefore

           √ --
sin 𝜃 = --3-.
        2
    (125)

    The spatial derivative is

    ux = Ak sin 𝜃 (126)
    = (0.015)(6)√ --
--3-
 2 (127)
    ≈−0.0779 . (128)

    The time derivative is

    ut = sin 𝜃 (129)
    = (0.015)(24)√ --
  3
----
 2 (130)
    0.312 m/s . (131)
  4. Since c = 4.0 m/s,
    − c ux = − 4.0(− 0.0779 )m/s ≈ 0.312 m/s.
    (132)

    This equals ut to rounding:

    |------------|
-ut-=-−-cux.-|
    (133)

  5. First find the displacement at the event:
    u = 0.015 cos π
--
3 (134)
    = 0.015(  )
 1-
 2 (135)
    = 0.0075 m. (136)

    Then

    uxx = k2u (137)
    = 36(0.0075) (138)
    = 0.270 m1 , (139)

    and

    utt = ω2u (140)
    = 576(0.0075) (141)
    = 4.32 m/s2 . (142)
  6. c2u xx = (4.0)2(0.270) (143)
    = 16(0.270) (144)
    = 4.32 m/s2. (145)

    Therefore

    |------------|
|utt = c2uxx. |
-------------
    (146)

  7. The quantity ut is the transverse velocity of the material point at a fixed spatial label, while c is the speed at which the wave pattern propagates through the medium.

Common mistakes

  • Mistake: differentiating both x and t when taking a partial derivative. Change one independent variable at a time.
  • Mistake: confusing ut with wave propagation speed. They describe different motions.
  • Mistake: dropping the chain-rule factors k or ω when differentiating a sinusoidal phase.
  • Mistake: forgetting the second minus sign when differentiating twice, leading to an incorrect sign in uxx or utt.
  • Mistake: using ut = cux for a left-moving profile. The left-moving relation is ut = +cux.
  • Mistake: treating a PDE verification as a physical derivation of the governing wave equation.

What WM13E1 reinforces

The exercises reinforce the interpretation of partial derivatives as directional rates of change through the (x,t) domain:

|-------------------------------|
|u  = spatial slope at fixed time ,
--x-----------------------------
(147)

|------------------------------------|
ut = local time  rate at fixed position ,
--------------------------------------
(148)

|--------------------------------|
|uxx = spatial curvature measure ,
---------------------------------
(149)

and, for transverse displacement,

|----------------------------------|
|utt = local transverse acceleration.|
-----------------------------------
(150)

For translating profiles,

|------------------------|
|F(x − ct) :  ut = − cux |
--------------------------
(151)

and

|------------------------|
G (x + ct) :  u =  +c u  ,
---------------t-------x--
(152)

while both satisfy

|------------|
|       2    |
-utt =-cuxx.-
(153)

This completes the calculus preparation needed for the later physical derivation of the one-dimensional string wave equation.

References

References

[1]   Massachusetts Institute of Technology, 18.02SC Multivariable Calculus, Unit 2, “Partial Derivatives,” MIT OpenCourseWare.

[2]   William Moebs, Samuel J. Ling, and Jeff Sanny, University Physics, Volume 1, OpenStax, 2016, Section 16.2, “Mathematics of Waves,” especially partial derivatives, traveling-wave functions, and the linear wave equation.

[3]   A. P. French, Vibrations and Waves, M.I.T. Introductory Physics Series, W. W. Norton & Company, 1971.

[4]   Frank S. Crawford, Jr., Waves, Berkeley Physics Course, Volume 3, McGraw-Hill, 1968.

[5]   Massachusetts Institute of Technology, 8.03SC Physics III: Vibrations and Waves - The Physics of Waves, Fall 2016, MIT OpenCourseWare, sections introducing traveling-wave solutions and the one-dimensional wave equation.


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