MA 201, Mathematics III, July-November 2018, Partial Diﬀerential Equations: 1D heat ... ·...

MA 201, Mathematics III, July-November 2018,

Partial Differential Equations:

1D heat conduction equation

Lecture 14

Lecture 14 MA 201, PDE (2018) 1 / 30

Heat Conduction Problem

Lecture 14 MA 201, PDE (2018) 2 / 30

Heat conduction in a thin rod

One important problem is the heat conduction in a thin metallic rod of finite

length.

This transient problem represents a class of problem known as diffusion problems.

Just like one-dimensional wave equation, separation of variables can be appropriately

applied.

The IBVP under consideration for heat conduction in a thin metallic rod of

length L consists of:

The governing equation

ut = αuxx, 0 < x < L, t > 0, (1)

where α is the thermal diffusivity.

Lecture 14 MA 201, PDE (2018) 3 / 30

Heat conduction in a thin rod (Contd.)

Lecture 14 MA 201, PDE (2018) 4 / 30

Heat conduction in a thin rod (Contd.): Case I

Homogeneous Dirichlet boundary condition

The boundary conditions for all t > 0 due to maintaining the temperature at

both ends at zero degrees

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

u(L, t) = 0. (2b)

The initial condition for 0 ≤ x ≤ L, i.e., initial temperature distribution:

u(x, 0) = f(x). (3)

The assumptions are:

• The rod is very thin so that it can be considered as one-dimensional.

• The curved surface is insulated.

• There is no external source of heat.

Using the separation of variable technique by assuming a solution of the form

u(x, t) = X(x)T (t)

Lecture 14 MA 201, PDE (2018) 5 / 30


The governing equation is converted to the following ODEs:X ′′ − kX = 0,

T ′ − kαT = 0.

where k is a separation constant.

Observe that the boundary conditions for this IBVP are the same as those in the

one-dimensional wave equation.

We can immediately assume that a feasible nontrivial solution exists only for the

negative values of k.

Above equations ⇒X ′′ + λ2X = 0, (4)

T ′ + λ2αT = 0. (5)

Lecture 14 MA 201, PDE (2018) 6 / 30


The solutions:X(x) = A cos(λx) +B sin(λx), (6)

T (t) = Ce−αλ2t. (7)

Solution for u(x, t):

u(x, t) = [A cos(λx) +B sin(λx)]Ce−αλ2t. (8)

Using the boundary conditions (2a) and (2b)

A = 0 and λn =nπ

L, n = 1, 2, 3, . . ..

Solution corresponding to each n:

un(x, t) = Bn sin(nπx

L) e−α

n2π2

L2t.

Lecture 14 MA 201, PDE (2018) 7 / 30


The general solution:

u(x, t) =

∞∑

n=1

un(x, t)

=∞∑

n=1

Bn sin(nπx

L) e−α

n2π2

L2t (9)

where Bn is obtained (refer to the solution of the one-dimensional wave

equation) by using the initial condition (3):

Bn =2

L

∫

L

0f(x) sin(

nπx

L) dx, n = 1, 2, 3, . . . (10)

Observe that what we have solved here is a homogenous equation

subject to homogenous (zero) conditions.

Ideally the boundary conditions are very likely to be non-homogenous for diffusion

problems.

Lecture 14 MA 201, PDE (2018) 8 / 30

Heat conduction in a thin rod: Case II

Non-homogeneous Dirichlet boundary conditions

With nonhomogeneous boundary conditions

Consider the heat conduction in a thin metal rod of length L with insulated sides

with the ends x = 0 and x = L kept at temperatures u = u00C and u = uL

0C,

respectively, for all time t > 0.

Suppose that the temperature distribution at t = 0 is u(x, 0) = φ(x), 0 < x < L.

To determine the temperature distribution in the rod at some subsequent time

t > 0.

Lecture 14 MA 201, PDE (2018) 9 / 30


The IBVP under consideration consists of

The governing equation

ut = αuxx, (11)

where α is the coefficient of thermal diffusivity for the specific metal.

The boundary conditions for all t > 0

u(0, t) = u0, (12a)

u(L, t) = uL. (12b)

The initial conditions for 0 ≤ x ≤ L

u(x, 0) = φ(x). (13)

Lecture 14 MA 201, PDE (2018) 10 / 30


Due to the nonhomogeneous boundary conditions (12),

the direct application of the method of separation of variables will not work.

We intend to convert the given IBVP into two BVPs:

one resembling the problem we solved previously

the other taking care of the nonhomogeneous terms.

Seek a solution in the form:

u(x, t) = v(x, t) + h(x), (14)

where h(x) is an unknown function of x alone to take care non-homogeneous

boundary conditions such that

h(0) = u0 & h(L) = uL.

Lecture 14 MA 201, PDE (2018) 11 / 30


Using ut − αuxx = 0, we have

vt = α[vxx + h′′(x)] (15)

subject to

v(0, t) + h(0) = u0, (16a)

v(L, t) + h(L) = uL, (16b)

and

v(x, 0) + h(x) = φ(x). (17)

Now equations (15)-(17) can be conveniently split into two BVPs as follows:

Lecture 14 MA 201, PDE (2018) 12 / 30


BVP I:

αh′′(x) = 0,

h(0) = u0, h(L) = uL.

BVP II:

vt = αvxx,

v(0, t) = 0 = v(L, t),

v(x, 0) = φ(x)− h(x).

The solution of BVP II is known to us from Case I, which is

v(x, t) =

∞∑

n=1

Bn sinnπx

Le−α

n2π2

L2t, (18)

where

Bn =2

L

∫

L

0(φ(x)− h(x)) sin

nπx

Ldx, n = 1, 2, 3, . . . (19)

Lecture 14 MA 201, PDE (2018) 13 / 30


The solution for BVP II can be easily found in two steps by integrating h′′(x) and

h′(x) :

h(x) = Ax+B.

Upon using the boundary conditions, we get

B = u0 and A = uL − u0

Hence

h(x) =uL − u0

Lx+ u0. (20)

Hence the solution u(x, t) for our IBVP is given by

the sum of (18) and (20).

Lecture 14 MA 201, PDE (2018) 14 / 30

Heat conduction problem with flux and radiation conditions

Lecture 14 MA 201, PDE (2018) 15 / 30

Heat conduction problem with flux and radiation

conditions

Heat flux is defined as the rate of heat transfer per unit cross-sectional area.

Fourier’s heat flow law states flux = −αux, where α is the conductivity.

A simplest flux condition at boundary x = 0 is given by ux(0, t) = 0, t > 0.

This is also known as insulation condition, that is, no heat flow takes place across the

boundary x = 0.

A radiation condition, on the other hand, is a specification of how heat radiates

from the end of the rod, say at x = 0, into the environment, or how the end absorbs

heat from its environment.

Linear, homogeneous radiation condition takes the form

−αux(0, t) + bu(0, t) = 0, t > 0, where b is a constant.

Lecture 14 MA 201, PDE (2018) 16 / 30

Heat conduction problem with flux and radiation conditions

If b > 0, then the flux is negative

which means that heat is flowing from the rod into its surroundings (radiation)

if b < 0, then the flux is positive,

and heat is flowing into the rod (absorption).

A typical problem in heat conduction may have

a combination of Dirichlet, insulation and radiation boundary conditions.

Lecture 14 MA 201, PDE (2018) 17 / 30

Heat conduction problem in a thin rod

Consider following heat conduction equation in a rod (0, L)

ut = αuxx 0 < x < L, t > 0, (21)

The initial condition for 0 ≤ x ≤ L:

u(x, 0) = f(x). (22)

In practice, temperature u satisfies certain boundary conditions

(a) Dirichlet Condition: u(0, t) = u0, u(L, t) = uL, t > 0

(b) Neumann Condition: ux(0, t) = u0, ux(L, t) = uL, t > 0,

(c) Mixed Condition: ux(0, t) = u0, u(L, t) = uL, t > 0, Or

u(0, t) = u0, ux(L, t) = uL, t > 0,

(d) Robin Condition: ux(0, t) + a0u(0, t) = α, ux(L, t) + aLu(L, t) = β.

Lecture 14 MA 201, PDE (2018) 18 / 30

Heat conduction problems with different conditions at ends

Problem I

Governing equation: ut = αuxx

Boundary conditions: u(0, t) = 0 = u(L, t), Initial Condition: u(x, 0) = f(x)

Solution is

u(x, t) =

∞∑

n=1

An sin(nπx

L) e−α

n2π2

L2t.

with

An =2

L

∫

L

0f(x) sin(

nπx

L)dx.

Lecture 14 MA 201, PDE (2018) 19 / 30


Problem II


Boundary conditions: ux(0, t) = 0 = u(L, t), Initial Condition: u(x, 0) = f(x)

The solution is

u(x, t) =

∞∑

n=0

An exp

[

−α

(

2n + 1

2

)2 π2

L2t

]

cos((2n + 1)πx

2L).

with

An =2

L

∫

L

0f(x) cos(

(2n + 1)πx

2L)dx.

Lecture 14 MA 201, PDE (2018) 20 / 30


Problem III


Boundary conditions: u(0, t) = 0 = ux(L, t), Initial Condition: u(x, 0) = f(x)

Solution is

u(x, t) =

∞∑

n=0

An exp

[

−α

(

2n+ 1

2

)2 π2

L2t

]

sin((2n + 1)πx

2L)

with

An =2

L

∫

l

0f(x) sin(

(2n + 1)πx

2L)dx.

Lecture 14 MA 201, PDE (2018) 21 / 30


Problem IV

Governing equation ut = αuxx

Boundary conditions: ux(0, t) = 0 = ux(L, t), Initial Condition: u(x, 0) = f(x)

The solution is

u(x, t) =∞∑

n=1

An cos(nπx

L) e−α

n2π2

L2t

with

An =2

L

∫

L

0f(x) cos(

nπx

L)dx.

TRY

Problems II-IV yourself.

Lecture 14 MA 201, PDE (2018) 22 / 30

Heat conduction problems with radiation conditions at one

end

Consider a problem related to radiation condition.

Take up the following IBVP:

Governing equation

ut = αuxx, 0 < x < L, t > 0 (23)

Boundary Conditions

u(0, t) = 0, (24a)

ux(L, t) + hu(L, t) = 0, t > 0, (24b)

where h is a constant.

Initial condition

u(x, 0) = f(x), 0 ≤ x ≤ L. (25)

Lecture 14 MA 201, PDE (2018) 23 / 30

Heat conduction problems with radiation conditions at one

end

Eigen values λ’s are given by

λ/h = − tan(λL). (26)

Corresponding eigen functions are

un(x, t) = Bn sin(λnx)e−αλ2

nt. (27)

Hence the solution:

u(x, t) =

∞∑

n=1

Bn sin(λnx) e−αλ2

nt, (28)

The initial condition will give us Bn as

Bn =2

L

∫

L

0f(x) sin(λnx) dx.

Lecture 14 MA 201, PDE (2018) 24 / 30

Example

A thin metal bar of length π is placed in boiling water (temperature 1000 degrees).

After reaching 1000 throughout, the bar is removed from the boiling water and

immediately, at time t = 0, the ends are immersed and kept in a medium with constant

freezing temperature 00 degrees. Taking the thermal diffusivity of the metal to be one,

find the temperature u(x, t) for t > 0.

Solution: The IBVP will be

ut = uxx, u(0, t) = 0 = u(π, t), u(x, 0) = 100

We know

u(x, t) =∞∑

n=1

An sinnπx

Lexp (−αn2π2/L)t with

An =2

L

∫

L

0u(x, 0) sin

nπx

L.

We evaluate An as

An =2

π

∫

π

0100 sin nx dx =

200

nπ(1− cosnπ)

The solution:

u(x, t) =400

π

∞∑

n=1

1

2n− 1sin(2n − 1)x exp{−(2n − 1)2t}.

Lecture 14 MA 201, PDE (2018) 25 / 30

Non-Homogeneous Heat Equation: Duhamel’s Principle

Find temperature u such that

ut − αuxx = f(x, t), (x, t) ∈ (0, L) × (0,∞) (29)

with BCs u(0, t) = 0, u(L, t) = 0, t > 0 and (30)

with ICs u(x, 0) = 0, x ∈ [0, L]. (31)

Then the solution u is given by

u(x, t) =

∫

t

0v(x, t− τ, τ)dτ. (32)

What is v?

Lecture 14 MA 201, PDE (2018) 26 / 30

Duhamel’s Principle (Contd.)

Here, v is the solution of the problem

vt − αvxx = 0, (x, t) ∈ (0, L) × (0,∞) (33)

with BCs v(0, t) = 0, v(L, t) = 0, t > 0 and (34)

with ICs v(x, 0) = f(x, s), s > 0. (35)

for some real parameter s > 0.

Note that the solution v of the above problem depends on x, t and s.

Thus, v = v(x, t, s). Accordingly, we can modify the BCs and ICs as

with BCs v(0, t, s) = 0, v(L, t, s) = 0, t > 0, s > 0 and

with ICs v(x, 0, s) = f(x, s), s > 0.

Lecture 14 MA 201, PDE (2018) 27 / 30

Example

Solve

ut − uxx = t sinx, 0 < x < π

u(0, t) = 0, u(π, t) = 0, t > 0

u(x, 0) = 0.

Solution: We first solve the related problem for v(x, t, s)

vt − vxx = 0, 0 < x < π

v(0, t, s) = 0, v(π, t, s) = 0, t > 0, s > 0

v(x, 0, s) = f(x, s) = s sinx.

Lecture 14 MA 201, PDE (2018) 28 / 30

Example

From method of separation of variables, for fixed s, we obtain

v(x, t, s) =∞∑

n=1

Bn sinnx e−n2t

with Bn

Bn =2

L

∫

L

0f(x, s) sinnxdx

=2

π

∫

π

0s sinx sinnxdx.

Hence,

B1 = s & Bn = 0 for n 6= 1.

Lecture 14 MA 201, PDE (2018) 29 / 30

Example

For any s > 0, we obtain

v(x, t, s) = se−t sinx

Therefore

v(x, t− τ, τ) = τe−(t−τ) sinx

Hence, due to Duhamel’s principle, u(x, t) is given by

u(x, t) =

∫

t

0v(x, t− τ, τ)dτ =

∫

t

0τe−(t−τ) sinxdτ

= e−t sinx

∫

t

0τeτdτ.

Lecture 14 MA 201, PDE (2018) 30 / 30

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