ENGR3300.003 Advanced Engineering Mathematics

ENGR3300.003

Advanced Engineering Mathematics

Fall 2021

Massimo (Max) V. Fischetti [email protected]: Joy Roy, [email protected]

mailto:[email protected]

mailto:[email protected]

Outline of the class:1. Vector calculus

• Vector algebra• Scalar and vector fields• Differential calculus: Gradient, divergence, curl• Integral calculus: Line integrals, surface integrals, volume integrals• Two important theorems: Divergence, Stokes

2. Fourier analysis: Series, transforms, generalization to ‘orthogonal functions’3. Partial differential equations (PDEs)

• Wave equation• Diffusion (heat) equation• Laplace and Poisson equations• Continuity equation

4. Complex analysis

We’ll revisit later… It’s better to deal with it when we shall really need it

Partial Differential Equations (Chapter 12)1. Basic ideas and examples2. The wave equation (Sec. 12.2)

• One spatial dimension: A violin string• Separation of variables, the solution as a Fourier series (Sec. 12.3)• Example: A violin string plucked in the middle

• One-dimensional waves in free space (an infinitely-long string)• Two spatial dimensions (Sec. 12.8)

• Vibration of a square membrane (Sec. 12.9)• Laplacian in polar coordinates (Sec. 12.10)• Vibrations of a circular membrane (a drum)

• Bessel equation and Bessel functions• The Fourier-Bessel series

• Three spatial dimensions (Sec. 12.11)• Laplacian in spherical and cylindrical coordinates

Examples of partial differential equations (PDEs)𝜕𝜕2𝑢𝑢𝜕𝜕𝑡𝑡2

= 𝑐𝑐2𝜕𝜕2𝑢𝑢𝜕𝜕𝑥𝑥2

1D wave equation

𝜕𝜕𝑢𝑢𝜕𝜕𝑡𝑡

= 𝜅𝜅𝜕𝜕2𝑢𝑢𝜕𝜕𝑥𝑥2

1D diffusion equation

𝜕𝜕2𝑢𝑢𝜕𝜕𝑥𝑥2

+𝜕𝜕2𝑢𝑢𝜕𝜕𝑦𝑦2

= 0 2D Laplace equation


+𝜕𝜕2𝑢𝑢𝜕𝜕𝑦𝑦2

= 𝜌𝜌 𝑥𝑥, 𝑦𝑦 2D Poisson equation

𝜕𝜕2𝑢𝑢𝜕𝜕𝑡𝑡2


+𝜕𝜕2𝑢𝑢𝜕𝜕𝑦𝑦2

2D wave equation



+𝜕𝜕2𝑢𝑢𝜕𝜕𝑦𝑦2



+𝜕𝜕2𝑢𝑢𝜕𝜕𝑦𝑦2

+𝜕𝜕2𝑢𝑢𝜕𝜕𝑧𝑧2



+𝜕𝜕2𝑢𝑢𝜕𝜕𝑦𝑦2

+𝜕𝜕2𝑢𝑢𝜕𝜕𝑧𝑧2

= 𝜌𝜌 𝑥𝑥, 𝑦𝑦, 𝑧𝑧 3D Poisson equation



+𝜕𝜕2𝑢𝑢𝜕𝜕𝑦𝑦2

+𝜕𝜕2𝑢𝑢𝜕𝜕𝑧𝑧2

3D wave equation



+𝜕𝜕2𝑢𝑢𝜕𝜕𝑦𝑦2

+𝜕𝜕2𝑢𝑢𝜕𝜕𝑧𝑧2


In compact notation:

∇2𝑢𝑢 = 0 3D Laplace equation

∇2𝑢𝑢 = 𝜌𝜌 𝑥𝑥,𝑦𝑦, 𝑧𝑧 3D Poisson equation


= 𝑐𝑐2∇2𝑢𝑢 3D wave equation


= 𝜅𝜅 ∇2𝑢𝑢 3D diffusion/heat equation

𝜕𝜕𝜌𝜌𝜕𝜕𝑡𝑡

= − ∇ � 𝑗𝑗 3D continuity equation

Partial differential equations

• Equation in an unknown u(r) with r in a domain D• u(r) must satisfy boundary conditions (Laplace, Poisson) and/or

initial (wave, continuity, diffusion) on a ‘surface’ in the domainFor example: For a circular membrane (that is, a drum), described by its vertical displacement u from equilibrium at position (r,φ) at time t, u(r,φ,t), one must specify

u(R,φ,t) = 0 (the membrane does not move on the rim of the drum at r=R)

u(r,φ,t=0) = f(r,φ) (initial position of the membrane)∂u(r,φ,t=0)/∂t = g(r,φ) (initial velocity of the membrane)

initial condition

boundary conditions


• For a homogeneous linear problemL[u(r)] = 0

if u1(r) and u1(r) are two solutions, that is: L[u1(r)] = 0

and L[u2(r)] = 0,

then:au1(r)+b u2(r)

is also a solution, the same situation we have for ordinary differential equations

The wave equation in one spatial dimension

1. Homogeneous string (constant mass-density ρ)

2. Ignore gravity (strong tension)3. No sliding (no motion along x)

αT1

T2 β

x x+Δx0 Lx

y

Since the string does not slide, the total force along x acting on the infinitesimal element must be zero:

𝑇𝑇1 cos𝛼𝛼 = 𝑇𝑇2 cos𝛽𝛽 = 𝑇𝑇 a constant, the ′tension′ (1)The vertical acceleration must be equal to the total force along y:

𝑇𝑇2 sin𝛽𝛽 − 𝑇𝑇1 sin𝛼𝛼 = 𝜌𝜌∆𝑥𝑥𝜕𝜕2𝑢𝑢𝜕𝜕𝑡𝑡2

(2)Divide Eq. (2) by Eq. (1):

𝑇𝑇2 sin𝛽𝛽𝑇𝑇2 cos𝛽𝛽

−𝑇𝑇1 sin𝛼𝛼𝑇𝑇1 cos𝛼𝛼

= tan𝛽𝛽 − tan𝛼𝛼 = 𝜌𝜌∆𝑥𝑥𝑇𝑇𝜕𝜕2𝑢𝑢𝜕𝜕𝑡𝑡2

(3)

But:

tan𝛽𝛽 =𝜕𝜕𝑢𝑢𝜕𝜕𝑥𝑥 𝑥𝑥+∆𝑥𝑥

and tan𝛼𝛼 =𝜕𝜕𝑢𝑢𝜕𝜕𝑥𝑥 𝑥𝑥

(4)

Inserting into Eq. (3):

1∆𝑥𝑥

𝜕𝜕𝑢𝑢𝜕𝜕𝑥𝑥 𝑥𝑥+∆𝑥𝑥

−𝜕𝜕𝑢𝑢𝜕𝜕𝑥𝑥 𝑥𝑥

=𝜌𝜌𝑇𝑇𝜕𝜕2𝑢𝑢𝜕𝜕𝑡𝑡2

(5)

or:


=𝜌𝜌𝑇𝑇𝜕𝜕2𝑢𝑢𝜕𝜕𝑡𝑡2

or 𝑐𝑐2𝜕𝜕2𝑢𝑢𝜕𝜕𝑥𝑥2

=𝜕𝜕2𝑢𝑢𝜕𝜕𝑡𝑡2

with ′sound velocity′ 𝑐𝑐 =𝑇𝑇𝜌𝜌

u(x+Δx)u(x)

General form of the solutions of the wave equation in one spatial dimension:Any function of the form f(x±ct) is a solution of the wave equation.Indeed:

𝜕𝜕𝑓𝑓𝜕𝜕(𝑥𝑥±𝑐𝑐𝑡𝑡)

=𝜕𝜕𝑓𝑓𝜕𝜕𝑥𝑥

𝜕𝜕𝑥𝑥𝜕𝜕(𝑥𝑥±𝑐𝑐𝑡𝑡)

=𝜕𝜕𝑓𝑓𝜕𝜕𝑥𝑥

(1)

But also:𝜕𝜕𝑓𝑓

𝜕𝜕(𝑥𝑥±𝑐𝑐𝑡𝑡)=𝜕𝜕𝑓𝑓𝜕𝜕𝑡𝑡

𝜕𝜕𝑡𝑡𝜕𝜕(𝑥𝑥±𝑐𝑐𝑡𝑡)

= ±1𝑐𝑐𝜕𝜕𝑓𝑓𝜕𝜕𝑥𝑥

(2)

So, the 2nd derivatives are:𝜕𝜕2𝑓𝑓

𝜕𝜕(𝑥𝑥±𝑐𝑐𝑡𝑡)2=

𝜕𝜕𝜕𝜕(𝑥𝑥±𝑐𝑐𝑡𝑡)

𝜕𝜕𝑓𝑓𝜕𝜕(𝑥𝑥±𝑐𝑐𝑡𝑡)

=𝜕𝜕2𝑓𝑓𝜕𝜕𝑥𝑥2

using (1)

𝜕𝜕2𝑓𝑓𝜕𝜕(𝑥𝑥±𝑐𝑐𝑡𝑡)2

=𝜕𝜕

𝜕𝜕(𝑥𝑥±𝑐𝑐𝑡𝑡)𝜕𝜕𝑓𝑓

𝜕𝜕(𝑥𝑥±𝑐𝑐𝑡𝑡)=

1𝑐𝑐2𝜕𝜕2𝑓𝑓𝜕𝜕𝑡𝑡2

using (2)

So:𝜕𝜕2𝑓𝑓𝜕𝜕𝑥𝑥2

=1𝑐𝑐2𝜕𝜕2𝑓𝑓𝜕𝜕𝑡𝑡2

General form of the solutions of the wave equation in one spatial dimension

f(x) f(x-ct)

x

f

t=0

Separation of variablesWhenever the coefficients of the PDE (e.g., c2 in our case) do not depend on both x and t, assume:

𝑢𝑢 𝑥𝑥, 𝑡𝑡 = 𝐹𝐹 𝑥𝑥 𝐺𝐺(𝑡𝑡)

Now let us assume as boundary conditions for the violin string:

𝑢𝑢 0, 𝑡𝑡 = 𝑢𝑢 𝐿𝐿, 𝑡𝑡 = 0 for any 𝑡𝑡

We will also need two initial conditions (“two”, since the PDE is 2nd-order):

𝑢𝑢 𝑥𝑥, 0 = 𝑓𝑓 𝑥𝑥 (initial position of the string)

𝜕𝜕𝑢𝑢(𝑥𝑥, 0)𝜕𝜕𝑡𝑡

= 𝑔𝑔 𝑥𝑥 (initial velocity of the string)

This is not the most general solution. However, we shall see that the general solution will be obtained by a linear combination (superposition) of solutions of this form.

Separation of variables-I

𝜕𝜕2𝑢𝑢(𝑥𝑥, 𝑡𝑡)𝜕𝜕𝑡𝑡2

= 𝑐𝑐2𝜕𝜕2𝑢𝑢(𝑥𝑥, 𝑡𝑡)𝜕𝜕𝑥𝑥2

with b. c.′ s: 𝑢𝑢 0, 𝑡𝑡 = 𝑢𝑢 𝐿𝐿, 𝑡𝑡 = 0

Set: 𝑢𝑢 𝑥𝑥, 𝑡𝑡 = 𝐹𝐹 𝑥𝑥 𝐺𝐺 𝑡𝑡Insert into the PDE:

𝐹𝐹(𝑥𝑥)𝑑𝑑2𝐺𝐺 𝑡𝑡𝑑𝑑𝑡𝑡2

= 𝑐𝑐2𝐺𝐺(𝑡𝑡)𝑑𝑑2𝐹𝐹(𝑥𝑥)𝑑𝑑𝑥𝑥2

Divide by F(x)G(t):1

𝐺𝐺(𝑡𝑡)𝑑𝑑2𝐺𝐺 𝑡𝑡𝑑𝑑𝑡𝑡2

= 𝑐𝑐21

𝐹𝐹(𝑥𝑥)𝑑𝑑2𝐹𝐹(𝑥𝑥)𝑑𝑑𝑥𝑥2

This can be true only if both sides are equal to the same constant K:1


= 𝑐𝑐21


= 𝐾𝐾

Separation of variables-II

𝜕𝜕2𝑢𝑢(𝑥𝑥, 𝑡𝑡)𝜕𝜕𝑡𝑡2

= 𝑐𝑐2𝜕𝜕2𝑢𝑢(𝑥𝑥, 𝑡𝑡)𝜕𝜕𝑥𝑥2

with b. c.′ s: 𝑢𝑢 0, 𝑡𝑡 = 𝑢𝑢 𝐿𝐿, 𝑡𝑡 = 0

Therefore:1


= 𝐾𝐾 and1


=𝐾𝐾𝑐𝑐2

or:𝑑𝑑2𝐺𝐺 𝑡𝑡𝑑𝑑𝑡𝑡2

− 𝐾𝐾𝐺𝐺 𝑡𝑡 = 0

and 𝑑𝑑2𝐹𝐹(𝑥𝑥)𝑑𝑑𝑥𝑥2

−𝐾𝐾𝑐𝑐2𝐹𝐹 𝑥𝑥 = 0 with 𝐹𝐹 0 = 𝐹𝐹 𝐿𝐿 = 0

Separation of variables-IIIConsider:

𝑑𝑑2𝐹𝐹(𝑥𝑥)𝑑𝑑𝑥𝑥2

−𝐾𝐾𝑐𝑐2𝐹𝐹 𝑥𝑥 = 0 with 𝐹𝐹 0 = 𝐹𝐹 𝐿𝐿 = 0 .

The boundary conditions can be satisfied only if K<0. Define k2=−K/c2. Then:

𝑑𝑑2𝐹𝐹(𝑥𝑥)𝑑𝑑𝑥𝑥2

+ 𝑘𝑘2𝐹𝐹 𝑥𝑥 = 0 with 𝐹𝐹 0 = 𝐹𝐹 𝐿𝐿 = 0

implies𝐹𝐹 𝑥𝑥 = 𝐴𝐴 sin

𝑛𝑛𝜋𝜋𝐿𝐿𝑥𝑥

so that:

𝑘𝑘2 =𝑛𝑛𝜋𝜋𝐿𝐿

2→ 𝐾𝐾 = −c2

𝑛𝑛𝜋𝜋𝐿𝐿

2

Separation of variables-IVBack to the other equation:

𝑑𝑑2𝐺𝐺 𝑡𝑡𝑑𝑑𝑡𝑡2

+ c2𝑛𝑛𝜋𝜋𝐿𝐿

2𝐺𝐺 𝑡𝑡 = 0

Define 𝜔𝜔𝑛𝑛 = 𝑐𝑐 𝑛𝑛𝜋𝜋𝐿𝐿

.

Then:

𝐺𝐺 𝑡𝑡 = �𝑛𝑛

𝐴𝐴𝑛𝑛𝑒𝑒+𝑖𝑖𝜔𝜔𝑛𝑛𝑡𝑡 + 𝐵𝐵𝑛𝑛𝑒𝑒−𝑖𝑖𝜔𝜔𝑛𝑛𝑡𝑡 = �𝑛𝑛

𝐶𝐶𝑛𝑛cos(𝜔𝜔𝑛𝑛𝑡𝑡) + 𝐷𝐷𝑛𝑛sin(𝜔𝜔𝑛𝑛𝑡𝑡)

Finally, the general solution is:

𝑢𝑢 𝑥𝑥, 𝑡𝑡 = 𝐹𝐹 𝑥𝑥 𝐺𝐺 𝑡𝑡 = �𝑛𝑛

𝐴𝐴𝑛𝑛𝑒𝑒+𝑖𝑖𝜔𝜔𝑛𝑛𝑡𝑡 + 𝐵𝐵𝑛𝑛𝑒𝑒−𝑖𝑖𝜔𝜔𝑛𝑛𝑡𝑡 sin𝑛𝑛𝜋𝜋𝐿𝐿𝑥𝑥

These two forms are equivalent, since the (in principle complex) coefficients An, Bn, Cn, and Dn, are yet to be determined

Examples of partial differential equations (PDEs)𝜕𝜕2𝑢𝑢𝜕𝜕𝑡𝑡2


1D wave equation





+𝜕𝜕2𝑢𝑢𝜕𝜕𝑦𝑦2



+𝜕𝜕2𝑢𝑢𝜕𝜕𝑦𝑦2

= 𝜌𝜌 𝑥𝑥, 𝑦𝑦 2D Poisson equation



+𝜕𝜕2𝑢𝑢𝜕𝜕𝑦𝑦2

2D wave equation



+𝜕𝜕2𝑢𝑢𝜕𝜕𝑦𝑦2



+𝜕𝜕2𝑢𝑢𝜕𝜕𝑦𝑦2

+𝜕𝜕2𝑢𝑢𝜕𝜕𝑧𝑧2



+𝜕𝜕2𝑢𝑢𝜕𝜕𝑦𝑦2

+𝜕𝜕2𝑢𝑢𝜕𝜕𝑧𝑧2

= 𝜌𝜌 𝑥𝑥, 𝑦𝑦, 𝑧𝑧 3D Poisson equation



+𝜕𝜕2𝑢𝑢𝜕𝜕𝑦𝑦2

+𝜕𝜕2𝑢𝑢𝜕𝜕𝑧𝑧2

3D wave equation



+𝜕𝜕2𝑢𝑢𝜕𝜕𝑦𝑦2

+𝜕𝜕2𝑢𝑢𝜕𝜕𝑧𝑧2


In compact notation:

∇2𝑢𝑢 = 0 3D Laplace equation

∇2𝑢𝑢 = 𝜌𝜌 𝑥𝑥,𝑦𝑦, 𝑧𝑧 3D Poisson equation


= 𝑐𝑐2∇2𝑢𝑢 3D wave equation


= 𝜅𝜅 ∇2𝑢𝑢 3D diffusion/heat equation

𝜕𝜕𝜌𝜌𝜕𝜕𝑡𝑡

= − ∇ � 𝑗𝑗 3D continuity equation

We were solving this equation


• Equation in an unknown u(r) with r in a domain D• u(r) must satisfy boundary conditions (Laplace, Poisson) and/or

initial (wave, continuity, diffusion) on a ‘surface’ in the domainFor example: For a circular membrane (that is, a drum), described by its vertical displacement u from equilibrium at position (r,φ) at time t, u(r,φ,t), one must specify:

u(R,φ,t) = 0 (the membrane does not move on the rim of the drum at r=R)

u(r,φ,t=0) = f(r,φ) (initial position of the membrane)∂u(r,φ,t=0)/∂t = g(r,φ) (initial velocity of the membrane)

initial conditions

boundary conditions

Now we must satisfy the initial conditions:

Example A: initial position given, zero initial velocity:𝑢𝑢 𝑥𝑥, 𝑡𝑡 = 0 = 𝑓𝑓 𝑥𝑥 (initial position)𝜕𝜕𝑢𝑢 𝑥𝑥, 𝑡𝑡 = 0

𝜕𝜕𝑡𝑡= 0 (initial velocity)

𝑢𝑢 𝑥𝑥, 𝑡𝑡 = �𝑛𝑛

𝐴𝐴𝑛𝑛cos(𝜔𝜔𝑛𝑛𝑡𝑡) + 𝐵𝐵𝑛𝑛sin(𝜔𝜔𝑛𝑛𝑡𝑡) sin𝑛𝑛𝜋𝜋𝐿𝐿𝑥𝑥

At t=0

𝑢𝑢 𝑥𝑥, 0 = 𝑓𝑓 𝑥𝑥 = ∑𝑛𝑛 𝐴𝐴𝑛𝑛 sin 𝑛𝑛𝜋𝜋𝐿𝐿𝑥𝑥

so:

𝐴𝐴𝑛𝑛=2𝐿𝐿�0

𝐿𝐿d𝑥𝑥 𝑓𝑓 𝑥𝑥 sin

𝑛𝑛𝜋𝜋𝐿𝐿𝑥𝑥 =

1𝐿𝐿�−𝐿𝐿

𝐿𝐿d𝑥𝑥 𝑓𝑓 𝑥𝑥 sin


Complex exponentials written as trig functionsOf course, the coefficients An and Bn are different from those in a previous slide, but they are unknowns, so we may use the same symbols

All Bn are zero because of the initial condition on the velocity:At t=0, ∂u(x,0)/∂t = Σn Bn ωn sin (nπx/L) = 0

assuming a period 2Lwith and odd extension of f(x) for x<0

Example B: initial velocity given, zero initial displacement:𝑢𝑢 𝑥𝑥, 𝑡𝑡 = 0 = 0 (initial position)

𝜕𝜕𝑢𝑢(𝑥𝑥, 𝑡𝑡 = 0)𝜕𝜕𝑡𝑡

= 𝑔𝑔 𝑥𝑥 initial velocitySince

𝜕𝜕𝑢𝑢(𝑥𝑥, 𝑡𝑡)𝜕𝜕𝑡𝑡

= �𝑛𝑛

−𝐴𝐴𝑛𝑛𝜔𝜔𝑛𝑛sin(𝜔𝜔𝑛𝑛𝑡𝑡) + 𝐵𝐵𝑛𝑛𝜔𝜔𝑛𝑛cos(𝜔𝜔𝑛𝑛𝑡𝑡) sin𝑛𝑛𝜋𝜋𝐿𝐿𝑥𝑥

then, at t=0:𝜕𝜕𝑢𝑢(𝑥𝑥, 0)𝜕𝜕𝑡𝑡

= �𝑛𝑛

𝐵𝐵𝑛𝑛𝜔𝜔𝑛𝑛 sin𝑛𝑛𝜋𝜋𝐿𝐿𝑥𝑥 = 𝑔𝑔(𝑥𝑥)

which implies:

𝐵𝐵𝑛𝑛 =2𝜔𝜔𝑛𝑛𝐿𝐿

�0

𝐿𝐿d𝑥𝑥 𝑔𝑔 𝑥𝑥 sin


All An are zero because of the initial condition on the position:At t=0, u(x,0) = Σn An sin (nπx/L) = 0

Has the solution the form of a wave, f(x±ct)?Consider case A (given initial position):


𝐴𝐴𝑛𝑛 cos(𝜔𝜔𝑛𝑛𝑡𝑡) sin𝑛𝑛𝜋𝜋𝐿𝐿𝑥𝑥

Use the trigonometric identity

cos𝛼𝛼 sin𝛽𝛽 =12

sin 𝛼𝛼 + 𝛽𝛽 + sin(𝛼𝛼 − 𝛽𝛽)so:

𝑢𝑢 𝑥𝑥, 𝑡𝑡 =12�𝑛𝑛

𝐴𝐴𝑛𝑛 sin 𝜔𝜔𝑛𝑛𝑡𝑡 +𝑛𝑛𝜋𝜋𝐿𝐿𝑥𝑥 + sin 𝜔𝜔𝑛𝑛𝑡𝑡 −


Recall that 𝜔𝜔𝑛𝑛 = 𝑐𝑐𝑛𝑛𝜋𝜋𝐿𝐿

. Thus:

𝑢𝑢 𝑥𝑥, 𝑡𝑡 =12�𝑛𝑛

𝐴𝐴𝑛𝑛 sin𝑛𝑛𝜋𝜋𝐿𝐿

(𝑥𝑥 + 𝑐𝑐𝑡𝑡) − sin𝑛𝑛𝜋𝜋𝐿𝐿

(𝑥𝑥 − 𝑐𝑐𝑡𝑡)

which is indeed of the general form 𝑢𝑢 𝑥𝑥, 𝑡𝑡 = 𝑓𝑓 𝑥𝑥 ± 𝑐𝑐𝑡𝑡 .Case B is similar. Good exercise to work it out.

A summary of what we have done:Given the equation:



with boundary and initial conditions:1. Separate variables: u(x,t) = F(x) G(t)2. Expand F(x) into eigenfunctions sn(x) (sines in our case) that satisfy the boundary

conditions3. Determine Gn(t), obtaining a general solution that satisfies the boundary conditions:

u(x,t) = Σn [An cos (ωnt) + Bn sin (ωnt)] sn(x) 4. Fix the coefficients An and Bn so that the initial conditions are satisfied

Example: A violin string plucked in the middle-I


𝐴𝐴𝑛𝑛 cos(𝜔𝜔𝑛𝑛𝑡𝑡) + 𝐵𝐵𝑛𝑛 sin(𝜔𝜔𝑛𝑛𝑡𝑡) sin 𝑘𝑘𝑛𝑛𝑥𝑥

𝑘𝑘𝑛𝑛 =𝜋𝜋𝐿𝐿𝑛𝑛

𝜔𝜔𝑛𝑛 = 𝑐𝑐𝑘𝑘𝑛𝑛 = 𝑐𝑐𝜋𝜋𝐿𝐿𝑛𝑛

LL/2

dx

u

0

𝑢𝑢 𝑥𝑥, 0 =2𝑑𝑑𝐿𝐿𝑥𝑥 0 ≤ 𝑥𝑥 ≤

𝐿𝐿2

= 2𝑑𝑑 −2𝑑𝑑𝐿𝐿𝑥𝑥

𝐿𝐿2≤ 𝑥𝑥 ≤ 𝐿𝐿

Example: A violin string plucked in the middle-IIAt t=0:

𝑢𝑢 𝑥𝑥, 𝑡𝑡 = 0 = �𝑛𝑛

𝐴𝐴𝑛𝑛 sin 𝑘𝑘𝑛𝑛𝑥𝑥

So:


𝐿𝐿d𝑥𝑥 𝑢𝑢 𝑥𝑥, 0 sin


and𝐵𝐵𝑛𝑛= 0 because the initial velocity is zero, as we saw before

Example: A violin string plucked in the middle-III



𝑛𝑛𝜋𝜋𝐿𝐿𝑥𝑥 with

Set:

𝑦𝑦 =𝑛𝑛𝜋𝜋𝐿𝐿𝑥𝑥; 𝑥𝑥 =

𝐿𝐿𝑛𝑛𝜋𝜋

𝑦𝑦; d𝑥𝑥 =𝐿𝐿𝑛𝑛𝜋𝜋

d𝑦𝑦Then:




2𝐿𝐿�0

𝐿𝐿/2d𝑥𝑥

2𝑑𝑑𝐿𝐿

𝑥𝑥 sin𝑛𝑛𝜋𝜋𝐿𝐿𝑥𝑥 +

2𝐿𝐿�𝐿𝐿/2

𝐿𝐿d𝑥𝑥 2𝑑𝑑 −

2𝑑𝑑𝐿𝐿𝑥𝑥 sin


=2𝐿𝐿2𝑑𝑑𝐿𝐿�0

𝑛𝑛𝜋𝜋/2d𝑦𝑦



𝑦𝑦 sin𝑦𝑦 +2𝐿𝐿�𝑛𝑛𝜋𝜋/2

𝑛𝑛𝜋𝜋d𝑦𝑦


2𝑑𝑑 −2𝑑𝑑𝑛𝑛𝜋𝜋

𝑦𝑦 sin𝑦𝑦 =

=4𝑑𝑑𝑛𝑛2𝜋𝜋2

�0

𝑛𝑛𝜋𝜋/2d𝑦𝑦 𝑦𝑦 sin𝑦𝑦 +

4𝑑𝑑𝑛𝑛𝜋𝜋

�𝑛𝑛𝜋𝜋/2

𝑛𝑛𝜋𝜋d𝑦𝑦 1 −

1𝑛𝑛𝜋𝜋

𝑦𝑦 sin𝑦𝑦 =


�0




𝑛𝑛𝜋𝜋d𝑦𝑦 sin𝑦𝑦 −

4𝑑𝑑𝑛𝑛2𝜋𝜋2


𝑛𝑛𝜋𝜋d𝑦𝑦 𝑦𝑦 sin𝑦𝑦

𝑢𝑢 𝑥𝑥, 0 =2𝑑𝑑𝐿𝐿𝑥𝑥 0 ≤ 𝑥𝑥 ≤

𝐿𝐿2

= 2𝑑𝑑 −2𝑑𝑑𝐿𝐿𝑥𝑥

𝐿𝐿2≤ 𝑥𝑥 ≤ 𝐿𝐿

Example: A violin string plucked in the middle-IVNow, integrating by parts:

�𝑑𝑑𝑦𝑦 𝑦𝑦 sin𝑦𝑦 = −𝑦𝑦 cos 𝑦𝑦 + �𝑑𝑑𝑦𝑦 cos 𝑦𝑦 = −𝑦𝑦 cos 𝑦𝑦 + sin𝑦𝑦

So:

𝐴𝐴𝑛𝑛 =4𝑑𝑑𝑛𝑛2𝜋𝜋2

�0




𝑛𝑛𝜋𝜋d𝑦𝑦 sin𝑦𝑦 −



𝑛𝑛𝜋𝜋d𝑦𝑦 𝑦𝑦 sin𝑦𝑦 =


−𝑦𝑦 cos 𝑦𝑦 + sin𝑦𝑦 0𝑛𝑛𝜋𝜋/2 −


−𝑦𝑦 cos 𝑦𝑦 + sin𝑦𝑦 𝑛𝑛𝜋𝜋/2𝑛𝑛𝜋𝜋 −


cos 𝑦𝑦 𝑛𝑛𝜋𝜋/2𝑛𝑛𝜋𝜋 =


−𝑛𝑛𝜋𝜋2

cos𝑛𝑛𝜋𝜋2

+ sin𝑛𝑛𝜋𝜋2

−4𝑑𝑑𝑛𝑛2𝜋𝜋2

−𝑛𝑛𝜋𝜋 cos 𝑛𝑛𝜋𝜋 + sin 𝑛𝑛𝜋𝜋 +𝑛𝑛𝜋𝜋2

cos𝑛𝑛𝜋𝜋2

− sin𝑛𝑛𝜋𝜋2

−4𝑑𝑑𝑛𝑛𝜋𝜋

cos 𝑛𝑛𝜋𝜋 − cos𝑛𝑛𝜋𝜋2

Example: A violin string plucked in the middle-VI

The sound velocity (i.e., the propagation speed of waves on the string):

𝑐𝑐 =𝑇𝑇𝜌𝜌

The fundamental frequency of the string:

𝜈𝜈1 =𝜔𝜔12𝜋𝜋

=1

2𝜋𝜋𝑐𝑐𝑘𝑘1 =

12𝜋𝜋

𝑇𝑇𝜌𝜌𝜋𝜋𝐿𝐿

To tune:1. Adjust the tension T (guitar, violin, and all other string instruments)2. Adjust the mass (thickness of the string; higher mass, lower note)3. Adjust the length (organ pies: longer pipes, lower notes)

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ENGR3300.003 Advanced Engineering Mathematics

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