Today

• Diffusion equation -

• derivation (transport eqns in general)

• initial conditions, boundary conditions

• steady state

• separation of variables

Conservation equations

Qab(t) =Z b

ac(x, t) dx

a b

c(x,t) is linear mass density of ink in a long narrow tube.

Conservation equations

Qab(t) =Z b

ac(x, t) dx

a b

c(x,t) is linear mass density of ink in a long narrow tube.

Conservation equations

Qab(t) =Z b

ac(x, t) dx

a b

c(x,t) is linear mass density of ink in a long narrow tube.

dQab

dt

(t) =d

dt

Z b

ac(x, t) dx =

Z b

a

�

�tc(x, t) dx

Define the flux Ja to be the amount of mass crossing the line x=a per unit of time (particles moving right count as positive flux) .

In that case, the change of Q inside the a-b box can also be counted watching flux, that is, flux at a - flux at b:

dQab

dt(t) = �Jb + Ja

JbJa

Conservation equations - Transport equation

Qab(t) =Z b

ac(x, t) dx

a bdQab

dt

(t) =d

dt

Z b

ac(x, t) dx =

Z b

a

�

�tc(x, t) dx

dQab

dt(t) = �Jb + Ja

Conservation equations - Transport equation

Qab(t) =Z b

ac(x, t) dx

a bdQab

dt

(t) =d

dt

Z b

ac(x, t) dx =

Z b

a

�

�tc(x, t) dx

dQab

dt(t) = �Jb + Ja

Need a model for flux. Let’s consider simpler case first (not diffusion yet!)

If fluid in pipe is moving with velocity v, flux is vc:

dQab

dt(t) = �Jb + Ja

Z b

a

�

�tc(x, t) dx )

Ja = vc(a, t)

= �vc(b, t) + vc(a, t) = � vc(x, t)����b

a

= �Z b

av

�c

�xdx

�c

�t= �v

�c

�xCalled Transport equation.

= �Z b

av

�c

�xdx

Conservation equations - Diffusion equation

Qab(t) =Z b

ac(x, t) dx

a bdQab

dt

(t) =d

dt

Z b

ac(x, t) dx =

Z b

a

�

�tc(x, t) dx

Ja

= �D�c

�x

����x=a

Now lets consider diffusion. We can derive this from chemical potential but

it also makes sense that for diffusion:

dQab

dt(t) = �Jb + Ja

Conservation equations - Diffusion equation

Qab(t) =Z b

ac(x, t) dx

a bdQab

dt

(t) =d

dt

Z b

ac(x, t) dx =

Z b

a

�

�tc(x, t) dx

Ja

= �D�c

�x

����x=a

Now lets consider diffusion. We can derive this from chemical potential but

it also makes sense that for diffusion:

dQab

dt(t) = �Jb + Ja

dQab

dt(t) = �Jb + Ja = D

�c

�x

����x=b

�D�c

�x

����x=a

= D�c

�x

����b

a

=Z b

aD

�2c

�x2dx

Z b

a

�

�tc(x, t) dx

�

�tc(x, t) = D

�2

�x2c(x, t))

Conservation equations - Diffusion equation

Qab(t) =Z b

ac(x, t) dx

a bdQab

dt

(t) =d

dt

Z b

ac(x, t) dx =

Z b

a

�

�tc(x, t) dx

Ja

= �D�c

�x

����x=a

Now lets consider diffusion. We can derive this from chemical potential but

it also makes sense that for diffusion:

dQab

dt(t) = �Jb + Ja

dQab

dt(t) = �Jb + Ja = D

�c

�x

����x=b

�D�c

�x

����x=a

= D�c

�x

����b

a

=Z b

aD

�2c

�x2dx

Z b

a

�

�tc(x, t) dx

�

�tc(x, t) = D

�2

�x2c(x, t))

The Diffusion Equation

dc

dt

= D

d

2c

dx

2

Initial and boundary conditions

Initial and boundary conditions

• One derivative in time requires an initial condition in t.

Initial and boundary conditions

• One derivative in time requires an initial condition in t.

• Two derivatives in space require two “initial conditions” in x (i.e. one at x=0 and one at x=L). Called boundary conditions (BCs).

Initial and boundary conditions

• One derivative in time requires an initial condition in t.

• Two derivatives in space require two “initial conditions” in x (i.e. one at x=0 and one at x=L). Called boundary conditions (BCs).

• Initial condition: c(x,0) = f(x) where f(x) gives initial concentration profile.

Initial and boundary conditions

• One derivative in time requires an initial condition in t.

• Two derivatives in space require two “initial conditions” in x (i.e. one at x=0 and one at x=L). Called boundary conditions (BCs).

• Initial condition: c(x,0) = f(x) where f(x) gives initial concentration profile.

• Boundary conditions:

Initial and boundary conditions

• One derivative in time requires an initial condition in t.

• Two derivatives in space require two “initial conditions” in x (i.e. one at x=0 and one at x=L). Called boundary conditions (BCs).

• Initial condition: c(x,0) = f(x) where f(x) gives initial concentration profile.

• Boundary conditions:

• c(0,t) = c0 and c(L,t) = cL Dirichlet conditions

Initial and boundary conditions

• One derivative in time requires an initial condition in t.

• Two derivatives in space require two “initial conditions” in x (i.e. one at x=0 and one at x=L). Called boundary conditions (BCs).

• Initial condition: c(x,0) = f(x) where f(x) gives initial concentration profile.

• Boundary conditions:

• c(0,t) = c0 and c(L,t) = cL

• anddc

dx

(0, t) = m0dc

dx

(L, t) = mL

Dirichlet conditions

Neumann conditions

Initial and boundary conditions

• One derivative in time requires an initial condition in t.

• Two derivatives in space require two “initial conditions” in x (i.e. one at x=0 and one at x=L). Called boundary conditions (BCs).

• Initial condition: c(x,0) = f(x) where f(x) gives initial concentration profile.

• Boundary conditions:

• c(0,t) = c0 and c(L,t) = cL

• anddc

dx

(0, t) = m0dc

dx

(L, t) = mL

Dirichlet conditions

Neumann conditions

Ja = �D

dc

dx

(a, t)Recall flux:Neumann conditions also called flux conditions (no-flux when m0 = mL = 0)

Initial and boundary conditions

• One derivative in time requires an initial condition in t.

• Two derivatives in space require two “initial conditions” in x (i.e. one at x=0 and one at x=L). Called boundary conditions (BCs).

• Initial condition: c(x,0) = f(x) where f(x) gives initial concentration profile.

• Boundary conditions:

• c(0,t) = c0 and c(L,t) = cL

• and

• c(0,t) = c0 and

dc

dx

(0, t) = m0dc

dx

(L, t) = mL

Dirichlet conditions

Neumann conditions

Mixed conditionsdc

dx

(L, t) = mL

(no-flux conditions)

Initial and boundary conditions

• One derivative in time requires an initial condition in t.

• Two derivatives in space require two “initial conditions” in x (i.e. one at x=0 and one at x=L). Called boundary conditions (BCs).

• Initial condition: c(x,0) = f(x) where f(x) gives initial concentration profile.

• Boundary conditions:

• c(0,t) = c0 and c(L,t) = cL

• and

• c(0,t) = c0 and

•

dc

dx

(0, t) = m0dc

dx

(L, t) = mL

Dirichlet conditions

Neumann conditions

Mixed conditionsdc

dx

(L, t) = mL

a

dc

dx

(0, t) + bc(0, t) = m0 Robin conditions

(no-flux conditions)

The Diffusion equation

• What does a steady state of the Diffusion equation look like?

The Diffusion Equation

dc

dt

= D

d

2c

dx

2

The Diffusion equation

• What does a steady state of the Diffusion equation look like?

The Diffusion Equation

dc

dt

= D

d

2c

dx

2

0 = D

d

2c

dx

2

The Diffusion equation

• What does a steady state of the Diffusion equation look like?

The Diffusion Equation

dc

dt

= D

d

2c

dx

2

0 = D

d

2c

dx

2

css(x) = Ax + B

The Diffusion equation

• What does a steady state of the Diffusion equation look like?

• A and B can be determined using the BCs. Getting A from Neumann conditions requires using the IC as well (total mass conservation).

The Diffusion Equation

dc

dt

= D

d

2c

dx

2

0 = D

d

2c

dx

2

css(x) = Ax + B

Separation of variables

Separation of variables

• Doc cam

This pdf is also posted on the lecture slides page.

• Find the Fourier series for f(x) = 2u0(x)-1 on the interval [-1,1].

fFS(x) = A0 + a1 cos

⇣�x

L

⌘+ a2 cos

✓2�x

L

◆+ · · ·

+b1 sin

⇣�x

L

⌘+ c2 sin

✓2�x

L

◆+ · · ·b2

Fourier series

• Find the Fourier series for f(x) = 2u0(x)-1 on the interval [-1,1].

• Our hope is that f(x) = fFS(x) so we calculate coefficients as if they were equal:

fFS(x) = A0 + a1 cos

⇣�x

L

⌘+ a2 cos

✓2�x

L

◆+ · · ·

+b1 sin

⇣�x

L

⌘+ c2 sin

✓2�x

L

◆+ · · ·b2

Fourier series

• Find the Fourier series for f(x) = 2u0(x)-1 on the interval [-1,1].

• Our hope is that f(x) = fFS(x) so we calculate coefficients as if they were equal:

fFS(x) = A0 + a1 cos

⇣�x

L

⌘+ a2 cos

✓2�x

L

◆+ · · ·

+b1 sin

⇣�x

L

⌘+ c2 sin

✓2�x

L

◆+ · · ·

bn =1L

Z L

�Lf(x) sin

⇣n�x

L

⌘dx

an =

1

L

Z L

�Lf(x) cos

⇣n�x

L

⌘dx

A0 =1

2L

Z L

�Lf(x) dx

b2

Fourier series

• Find the Fourier series for f(x) = 2u0(x)-1 on the interval [-1,1].

• Our hope is that f(x) = fFS(x) so we calculate coefficients as if they were equal:

fFS(x) = A0 + a1 cos

⇣�x

L

⌘+ a2 cos

✓2�x

L

◆+ · · ·

+b1 sin

⇣�x

L

⌘+ c2 sin

✓2�x

L

◆+ · · ·

bn =1L

Z L

�Lf(x) sin

⇣n�x

L

⌘dx

an =

1

L

Z L

�Lf(x) cos

⇣n�x

L

⌘dx

A0 =1

2L

Z L

�Lf(x) dx

• To simplify formulas, usually define a0 = 2A0 =

1L

Z L

�Lf(x) dx

b2

Fourier series

• Find the Fourier series for f(x) = 2u0(x)-1 on the interval [-1,1].

• Our hope is that f(x) = fFS(x) so we calculate coefficients as if they were equal:

fFS(x) = A0 + a1 cos

⇣�x

L

⌘+ a2 cos

✓2�x

L

◆+ · · ·

+b1 sin

⇣�x

L

⌘+ c2 sin

✓2�x

L

◆+ · · ·

bn =1L

Z L

�Lf(x) sin

⇣n�x

L

⌘dx

an =

1

L

Z L

�Lf(x) cos

⇣n�x

L

⌘dx

A0 =1

2L

Z L

�Lf(x) dx

• To simplify formulas, usually define a0 = 2A0 =

1L

Z L

�Lf(x) dx

b2

Fourier series

• Find the Fourier series for f(x) = 2u0(x)-1 on the interval [-1,1].

• Our hope is that f(x) = fFS(x) so we calculate coefficients as if they were equal:

bn =1L

Z L

�Lf(x) sin

⇣n�x

L

⌘dx

an =

1

L

Z L

�Lf(x) cos

⇣n�x

L

⌘dx

A0 =1

2L

Z L

�Lf(x) dx

• To simplify formulas, usually define a0 = 2A0 =

1L

Z L

�Lf(x) dx

fFS(x) =

a0

2

+ a1 cos

⇣�x

L

⌘+ a2 cos

✓2�x

L

◆+ · · ·

+b1 sin

⇣�x

L

⌘+ c2 sin

✓2�x

L

◆+ · · ·b2

Fourier series

• Find the Fourier series for f(x) = 2u0(x)-1 on the interval [-1,1].

• Our hope is that f(x) = fFS(x) so we calculate coefficients as if they were equal:

bn =1L

Z L

�Lf(x) sin

⇣n�x

L

⌘dx

an =

1

L

Z L

�Lf(x) cos

⇣n�x

L

⌘dx

A0 =1

2L

Z L

�Lf(x) dx

• To simplify formulas, usually define a0 = 2A0 =

1L

Z L

�Lf(x) dx

fFS(x) =

a0

2

+ a1 cos

⇣�x

L

⌘+ a2 cos

✓2�x

L

◆+ · · ·

+b1 sin

⇣�x

L

⌘+ c2 sin

✓2�x

L

◆+ · · ·b2

A0 is the average value of f(x)!

Fourier series

• Calculate the coefficients.

https://www.desmos.com/calculator/tlvtikmi0y Does f(x) = fFS(x) for all x?Problems at jumps! x=-1, 0, 1

Fourier series

fFS(x) =

a0

2

+ a1 cos

⇣�x

L

⌘+ a2 cos

✓2�x

L

◆+ · · ·

+b1 sin

⇣�x

L

⌘+ c2 sin

✓2�x

L

◆+ · · ·b2

bn =1L

Z L

�Lf(x) sin

⇣n�x

L

⌘dx

an =

1

L

Z L

�Lf(x) cos

⇣n�x

L

⌘dx

A0 =1

2L

Z L

�Lf(x) dx

a0 = 2A0 =1L

Z L

�Lf(x) dxa0 = 2A0 =

1L

Z L

�Lf(x) dx

an = 0

a0 = 0

bn = 2 (1- (-1)n) / nπ

fFS(x) =4

⇡

sin⇣⇡x

L

⌘+

4

3⇡sin

✓3⇡x

L

◆+

4

5⇡sin

✓5⇡x

L

◆

Fourier series

• Theorem Suppose f and f’ are piecewise continuous on [-L,L] and periodic beyond that interval. Then f(x) = fFS(x) at all points at which f is continuous. Furthermore, at points of discontinuity, fFS(x) takes the value of the midpoint of the jump. That is,

fFS(x) =f(x+) + f(x�)

2

Heat/Diffusion equation - example

• Find the solution to the heat/diffusion equation

• subject to BCs

• and with IC

• The “warming up the milk bottle” example.

ut

= 7uxx

u(0, t) = 0 = u(4, t)

u(x, 0) =

⇢1, 0 x 20, 2 < x 4

https://www.desmos.com/calculator/zvowjmu30g

Using Fourier Series to solve the Diffusion Equation

u(0, t) = u(2, t) = 0

u(x, 0) = x

ut

= 4uxx

https://www.desmos.com/calculator/yt7kztckeuhttps://www.desmos.com/calculator/wcdvgrveez

Using Fourier Series to solve the Diffusion Equation

u(0, t) = u(2, t) = 0

u(x, 0) = x

ut

= 4uxx

bn =(�1)n+14

n�

u(x, t) =

a0

2

+

1X

n=1

ane�n2�2tcos

n�x

2

u(x, t) =1X

n=1

bne�n2�2t sinn�x

2

(0 for n odd)

a0 = 1, an = � 8

n2�2for n even

(A)

(B)

https://www.desmos.com/calculator/yt7kztckeuhttps://www.desmos.com/calculator/wcdvgrveez

Using Fourier Series to solve the Diffusion Equation

u(0, t) = u(2, t) = 0

u(x, 0) = x

ut

= 4uxx

bn =(�1)n+14

n�

u(x, t) =

a0

2

+

1X

n=1

ane�n2�2tcos

n�x

2

u(x, t) =1X

n=1

bne�n2�2t sinn�x

2

(0 for n odd)

a0 = 1, an = � 8

n2�2for n even

(A)

(B)

https://www.desmos.com/calculator/yt7kztckeuhttps://www.desmos.com/calculator/wcdvgrveez

Using Fourier Series to solve the Diffusion Equation

u(0, t) = u(2, t) = 0

u(x, 0) = x

ut

= 4uxx

bn =(�1)n+14

n�

u(x, t) =

a0

2

+

1X

n=1

ane�n2�2tcos

n�x

2

u(x, t) =1X

n=1

bne�n2�2t sinn�x

2

(0 for n odd)

a0 = 1, an = � 8

n2�2for n even

(A)

(B)

https://www.desmos.com/calculator/yt7kztckeuhttps://www.desmos.com/calculator/wcdvgrveez

Using Fourier Series to solve the Diffusion Equation

u(0, t) = u(2, t) = 0

u(x, 0) = x

ut

= 4uxx

bn =(�1)n+14

n�

u(x, t) =

a0

2

+

1X

n=1

ane�n2�2tcos

n�x

2

u(x, t) =1X

n=1

bne�n2�2t sinn�x

2

(0 for n odd)

a0 = 1, an = � 8

n2�2for n even

(A)

(B)

https://www.desmos.com/calculator/yt7kztckeuhttps://www.desmos.com/calculator/wcdvgrveez

Using Fourier Series to solve the Diffusion Equation

u(0, t) = u(2, t) = 0

u(x, 0) = x

ut

= 4uxx

bn =(�1)n+14

n�

u(x, t) =

a0

2

+

1X

n=1

ane�n2�2tcos

n�x

2

u(x, t) =1X

n=1

bne�n2�2t sinn�x

2

(0 for n odd)

a0 = 1, an = � 8

n2�2for n even

(A)

(B)

• Show Desmos movies.https://www.desmos.com/calculator/yt7kztckeuhttps://www.desmos.com/calculator/wcdvgrveez

Using Fourier Series to solve the Diffusion Equation

ut

= 4uxx

du

dx

����x=0,2

= 0

u(x, 0) = cos

3�x

2

Using Fourier Series to solve the Diffusion Equation

ut

= 4uxx

du

dx

����x=0,2

= 0

u(x, 0) = cos

3�x

2

The IC is an eigenvector! Note that it satisfies the BCs.

Using Fourier Series to solve the Diffusion Equation

ut

= 4uxx

du

dx

����x=0,2

= 0

u(x, 0) = cos

3�x

2

v3(x) = cos

3�x

2

The IC is an eigenvector! Note that it satisfies the BCs.

Using Fourier Series to solve the Diffusion Equation

ut

= 4uxx

du

dx

����x=0,2

= 0

u(x, 0) = cos

3�x

2

vn(x) = cos

n�x

2

v3(x) = cos

3�x

2

The IC is an eigenvector! Note that it satisfies the BCs.

Using Fourier Series to solve the Diffusion Equation

ut

= 4uxx

du

dx

����x=0,2

= 0

u(x, 0) = cos

3�x

2

vn(x) = cos

n�x

2

v3(x) = cos

3�x

2

un(x, t) = e�ntcos

n�x

2

The IC is an eigenvector! Note that it satisfies the BCs.

Using Fourier Series to solve the Diffusion Equation

ut

= 4uxx

du

dx

����x=0,2

= 0

u(x, 0) = cos

3�x

2

vn(x) = cos

n�x

2

v3(x) = cos

3�x

2

un(x, t) = e�ntcos

n�x

2

The IC is an eigenvector! Note that it satisfies the BCs.

⇤

⇤tun(x, t) = �ne�nt

cos

n⇥x

2

Using Fourier Series to solve the Diffusion Equation

ut

= 4uxx

du

dx

����x=0,2

= 0

u(x, 0) = cos

3�x

2

vn(x) = cos

n�x

2

v3(x) = cos

3�x

2

un(x, t) = e�ntcos

n�x

2

The IC is an eigenvector! Note that it satisfies the BCs.

⇤

⇤tun(x, t) = �ne�nt

cos

n⇥x

2

⇥2

⇥x2un(x, t) = �n2�2

4

e�ntcos

n�x

2

Using Fourier Series to solve the Diffusion Equation

ut

= 4uxx

du

dx

����x=0,2

= 0

u(x, 0) = cos

3�x

2

vn(x) = cos

n�x

2

v3(x) = cos

3�x

2

un(x, t) = e�ntcos

n�x

2

The IC is an eigenvector! Note that it satisfies the BCs.

⇤

⇤tun(x, t) = �ne�nt

cos

n⇥x

2

⇥2

⇥x2un(x, t) = �n2�2

4

e�ntcos

n�x

2

44

Using Fourier Series to solve the Diffusion Equation

ut

= 4uxx

du

dx

����x=0,2

= 0

u(x, 0) = cos

3�x

2

vn(x) = cos

n�x

2

v3(x) = cos

3�x

2

un(x, t) = e�ntcos

n�x

2

The IC is an eigenvector! Note that it satisfies the BCs.

⇤

⇤tun(x, t) = �ne�nt

cos

n⇥x

2

⇥2

⇥x2un(x, t) = �n2�2

4

e�ntcos

n�x

2

44

Using Fourier Series to solve the Diffusion Equation

ut

= 4uxx

du

dx

����x=0,2

= 0

u(x, 0) = cos

3�x

2

vn(x) = cos

n�x

2

v3(x) = cos

3�x

2

un(x, t) = e�ntcos

n�x

2

The IC is an eigenvector! Note that it satisfies the BCs.

⇤

⇤tun(x, t) = �ne�nt

cos

n⇥x

2

⇥2

⇥x2un(x, t) = �n2�2

4

e�ntcos

n�x

2

44

Using Fourier Series to solve the Diffusion Equation

ut

= 4uxx

du

dx

����x=0,2

= 0

u(x, 0) = cos

3�x

2

vn(x) = cos

n�x

2

v3(x) = cos

3�x

2

un(x, t) = e�ntcos

n�x

2

The IC is an eigenvector! Note that it satisfies the BCs.

⇤

⇤tun(x, t) = �ne�nt

cos

n⇥x

2

⇥2

⇥x2un(x, t) = �n2�2

4

e�ntcos

n�x

2

�n = �n2⇥2

44

Using Fourier Series to solve the Diffusion Equation

ut

= 4uxx

du

dx

����x=0,2

= 0

u(x, 0) = cos

3�x

2

vn(x) = cos

n�x

2

v3(x) = cos

3�x

2

un(x, t) = e�ntcos

n�x

2

The IC is an eigenvector! Note that it satisfies the BCs.

⇤

⇤tun(x, t) = �ne�nt

cos

n⇥x

2

⇥2

⇥x2un(x, t) = �n2�2

4

e�ntcos

n�x

2

�n = �n2⇥2

44

u(x, t) = e�9�2tcos

3�x

2

So the solution is