math 241: solving the heat equationdeturck/m241/solving_the_heat_eqn.pdf · math 241: solving the...

Math 241: Solving the heat equation

D. DeTurck

University of Pennsylvania

September 20, 2012

D. DeTurck Math 241 002 2012C: Solving the heat equation 1 / 21

1D heat equation with Dirichlet boundary conditions

We derived the one-dimensional heat equation

ut = kuxx

and found that its reasonable to expect to be able to solve foru(x , t) (with x [a, b] and t > 0) provided we impose initialconditions:

u(x , 0) = f (x)

for x [a, b] and boundary conditions such as

u(a, t) = p(t) , u(b, t) = q(t)

for t > 0.We showed that this problem has at most one solution, now itstime to show that a solution exists.

Linearity

Well begin with a few easy observations about the heat equationut = kuxx , ignoring the initial and boundary conditions for themoment:

Since the heat equation is linear (and homogeneous), a linearcombination of two (or more) solutions is again a solution. Soif u1, u2,. . . are solutions of ut = kuxx , then so is

c1u1 + c2u2 +

for any choice of constants c1, c2, . . .. (Likewise, if u(x , t) isa solution of the heat equation that depends (in a reasonableway) on a parameter , then for any (reasonable) functionf () the function

U(x , t) =

21

f ()u(x , t) d

is also a solution.

Linearity

Well begin with a few easy observations about the heat equationut = kuxx , ignoring the initial and boundary conditions for themoment:

Since the heat equation is linear (and homogeneous), a linearcombination of two (or more) solutions is again a solution. Soif u1, u2,. . . are solutions of ut = kuxx , then so is

c1u1 + c2u2 +

for any choice of constants c1, c2, . . .. (Likewise, if u(x , t) isa solution of the heat equation that depends (in a reasonableway) on a parameter , then for any (reasonable) functionf () the function

U(x , t) =

21

f ()u(x , t) d

is also a solution.D. DeTurck Math 241 002 2012C: Solving the heat equation 3 / 21

Linearity and initial/boundary conditions

We can take advantage of linearity to address theinitial/boundary conditions one at a time.

For instance, we will spend a lot of time on initial-valueproblems with homogeneous boundary conditions:

ut = kuxx , u(x , 0) = f (x), u(a, t) = u(b, t) = 0.

Then well consider problems with zero initial conditions butnon-zero boundary values.

We can add these two kinds of solutions together to getsolutions of general problems, where both the initial andboundary values are non-zero.

Symmetry

Some more observations:

If u(x , t) is a solution, then so is u(a x , b + t) for anyconstants a and b.Note the with the x but only + with t you cantreverse time with the heat equation. This shows that theheat equation respects (or reflects) the second law ofthermodynamics (you cant unstir the cream from yourcoffee).

If u(x , t) is a solution then so is u(a2t, at) for any constant a.Well use this observation later to solve the heat equation in asurprising way, but for now well just store it in our memorybank.

Symmetry

Some more observations:

If u(x , t) is a solution, then so is u(a x , b + t) for anyconstants a and b.Note the with the x but only + with t you cantreverse time with the heat equation. This shows that theheat equation respects (or reflects) the second law ofthermodynamics (you cant unstir the cream from yourcoffee).

If u(x , t) is a solution then so is u(a2t, at) for any constant a.Well use this observation later to solve the heat equation in asurprising way, but for now well just store it in our memorybank.

Polynomial solutions

Now its time to at least find some examples of solutions tout = kuxx .

One thing we can try is polynomial solutions. Certainly anylinear function of x is a solution.

Next, taking our cue from the initial-value problem, supposeu(x , 0) = p0(x) for some polynomial p0(x), and try toconstruct a solution of the form

u(x , t) = p0(x) + tp1(x) + t2p2(x) +

We have

ut = p1(x) + 2tp2(x) + 3t2p3(x) +

anduxx = p

0 (x) + tp

1 (x) + t

2p2 (x) +

u(x , t) = p0(x) + tp1(x) + t2p2(x) +

We have

ut = p1(x) + 2tp2(x) + 3t2p3(x) +

anduxx = p

0 (x) + tp

1 (x) + t

2p2 (x) +

u(x , t) = p0(x) + tp1(x) + t2p2(x) +

We have

ut = p1(x) + 2tp2(x) + 3t2p3(x) +

anduxx = p

0 (x) + tp

1 (x) + t

2p2 (x) +

So the heat equation tells us:

p1 = kp0 , p2 =

k

2p1 =

k2

2p0 ,

p3 =k

3p2 =

k3

3!p(6)0 , . . . , pn =

kn

n!p(2n)0

This process will stop if p0 is a polynomial, and well get apolynomial solution of the heat equation whose x-degree istwice its t-degree:

u(x , t) = p0(x) +kt

1!p0 +

k2t2

2!p0 + +

kntn

n!p(2n)0 + .

But. . .

The trouble with polynomial solutions, or even with extending theidea of polynomials to power series in two variables (ick!), is that itwould be very difficult if not impossible to figure out how tochoose the coefficients of the polynomial p0 so that the boundaryvalues, even simple ones, would be matched at both ends.

There are also tricky convergence questions, etc for power series,and we dont want to get overwhelmed with these.

Separation of variables

A more fruitful strategy is to look for separated solutions ofthe heat equation, in other words, solutions of the formu(x , t) = X (x)T (t).

If we substitute X (x)T (t) for u in the heat equationut = kuxx we get:

XdT

dt= k

d2X

dx2T .

Divide both sides by kXT and get

1

kT

dT

dt=

1

X

d2X

dx2.

XdT

dt= k

d2X

dx2T .

1

kT

dT

dt=

1

X

d2X

dx2.

XdT

dt= k

d2X

dx2T .

1

kT

dT

dt=

1

X

d2X

dx2.

Separation of Variables

So if u(x , t) = X (x)T (t) thenT

kT=

X

X.

In this last equation, everything on the left side is a functionof t, and everything on the right side is a function of x . Thismeans that both sides are constant, say equal to whichgives ODEs for X and T :

X X = 0 T kT = 0

The power of this method comes in the application ofboundary conditions, which we turn to next.

kT=

X

X.

X X = 0 T kT = 0

kT=

X

X.

X X = 0 T kT = 0

Boundary conditions

ODEs for X and T :

X X = 0 T kT = 0

As an example, lets suppose the x-interval goes from 0 to L,and we have homogeneous Dirichlet conditions:u(0, t) = u(L, t) = 0 for all t.

This implies that cant be positive or zero (since solutionsto X X = 0 with 0 cant be zero twice without beingidentically zero)

So let = 2. The general solution of X + 2X = 0 is

X = c1 cosx + c2 sinx

The easiest way to satisfy the boundary conditions on u is toinsist that X (0) = X (L) = 0. What does this imply?

Boundary conditions

ODEs for X and T :

X X = 0 T kT = 0

Boundary conditions

ODEs for X and T :

X X = 0 T kT = 0

Boundary conditions

ODEs for X and T :

X X = 0 T kT = 0

Restrictions on

We have X = c1 cosx + c2 sinx , X (0) = 0, X (L) = 0.

First, note that X (0) = c1. The condition X (0) = 0 thenforces c1 = 0.

Now we need to reconcile three things: X = c2 sinx ,X (L) = 0 and we dont want X to be identically zero (i.e., wewant c2 6= 0).

This gives a condition on : sinL = 0, or L = n for someinteger n.

Since sin(x) = sin x , we need only consider positiveintegers n. Thus

=n

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

andX = c2 sin

nx

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

Restrictions on

We have X = c1 cosx + c2 sinx , X (0) = 0, X (L) = 0. First, note that X (0) = c1. The condition X (0) = 0 then

forces c1 = 0.

=n

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

andX = c2 sin

nx

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

Restrictions on

forces c1 = 0.

=n

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

andX = c2 sin

nx

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

Restrictions on

forces c1 = 0.

=n

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

andX = c2 sin

nx

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

Restrictions on

forces c1 = 0.

=n

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

andX = c2 sin

nx

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

Implications for and T

Now, recall that = 2, and the T equation wasT kT = 0.

And we know that = n/L for n = 1, 2, 3, . . ., so we have tosolve

T +n2k2

L2T = 0, n = 1, 2, 3 . . . .

The general solution of these are

T = cen2k2t/L2 , n = 1, 2, 3, . . . .

Putting this together with our X solutions, we get solutionsu(x , t) of the form:

u(x , t) = bnen2k2t/L2 sin

nx

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

T +n2k2

L2T = 0, n = 1, 2, 3 . . . .

T = cen2k2t/L2 , n = 1, 2, 3, . . . .

nx

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

T +n2k2

L2T = 0, n = 1, 2, 3 . . . .

T = cen2k2t/L2 , n = 1, 2, 3, . . . .

nx

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

T +n2k2

L2T = 0, n = 1, 2, 3 . . . .

T = cen2k2t/L2 , n = 1, 2, 3, . . . .

nx

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

Superposition, again

We can put these solutions together to get solutions of theform

u(x , t) = b1ek2t/L2 sin

x

L+ b2e

42t/L2 sin2x

L+

+ bNeN2k2t/L2 sin

Nx

L.

This function u(x , t) automatically satisfies the boundaryconditions u(0, t) = u(L, t) = 0, since all of the pieces do.

And it would be great for satisfying initial conditions given intrigonometric form for example

u(x , 0) = 3 sinx

L 2 sin 3x

L+ sin

6x

L.

Superposition, again

We can put these solutions together to get solutions of theform

u(x , t) = b1ek2t/L2 sin

x

L+ b2e

42t/L2 sin2x

L+

+ bNeN2k2t/L2 sin

Nx

L.

This function u(x , t) automatically satisfies the boundaryconditions u(0, t) = u(L, t) = 0, since all of the pieces do.

And it would be great for satisfying initial conditions given intrigonometric form for example

u(x , 0) = 3 sinx

L 2 sin 3x

L+ sin

6x

L.

Taking things to the limit

But what if the initial condition isnt trigonometric? Could weconsider adding together an infinite number of pieces? As in

u(x , t) =n=1

bnen2k2t/L2 sin

nx

L

Then the initial values would be

f (x) =n=1

bn sinnx

L.

So the question is, which functions f (x) (for x in the interval[0, L]) can be expressed as an infinite series of sines?

The (somewhat surprising) answer is, ALL OF THEM! Lets see how this might work in practice (and well take up

the question of proving this claim later).

u(x , t) =n=1

bnen2k2t/L2 sin

nx

L

f (x) =n=1

bn sinnx

L.

u(x , t) =n=1

bnen2k2t/L2 sin

nx

L

f (x) =n=1

bn sinnx

L.

u(x , t) =n=1

bnen2k2t/L2 sin

nx

L

f (x) =n=1

bn sinnx

L.

The Fourier ansatz

Just for something concrete, lets suppose we want to solvethe problem

ut =1

5uxx , u(0, t) = u(3, t) = 0, u(x , 0) = 3x x2

for t > 0 and 0 x 3.

Well assume that we can express u as

u(x , t) =n=1

bnen22t/45 sin

nx

3

(this is the ansatz) and see if we can figure out what theconstants bn should be we know that the boundaryconditions are automatically satisfied, and perhaps we canchoose the bns so that

3x x2 =n=1

bn sinnx

3.

The Fourier ansatz

Just for something concrete, lets suppose we want to solvethe problem

ut =1

5uxx , u(0, t) = u(3, t) = 0, u(x , 0) = 3x x2

for t > 0 and 0 x 3. Well assume that we can express u as

u(x , t) =n=1

bnen22t/45 sin

nx

3

(this is the ansatz) and see if we can figure out what theconstants bn should be we know that the boundaryconditions are automatically satisfied, and perhaps we canchoose the bns so that

3x x2 =n=1

bn sinnx

3.

Integrals rather than derivatives

Were trying to find bns so that

3x x2 =n=1

bn sinnx

3.

By contrast with Taylor series, where you find the coefficientsby integration rather than differentiation.

Well use two basic facts: If n 6= m then

30

sinnx

3sin

mx

3dx = 0.

If n = m then 30

sinnx

3sin

mx

3dx =

30

sin2nx

3dx =

3

2.

Integrals rather than derivatives

Were trying to find bns so that

3x x2 =n=1

bn sinnx

3.

By contrast with Taylor series, where you find the coefficientsby integration rather than differentiation.

Well use two basic facts: If n 6= m then

30

sinnx

3sin

mx

3dx = 0.

If n = m then 30

sinnx

3sin

mx

3dx =

30

sin2nx

3dx =

3

2.

Finding the coefficients

Were still trying to find bns so that

3x x2 =n=1

bn sinnx

3.

Motivated by the facts on the previous slide, we multiply bothsides by sin mx3 and integrate both sides from 0 to 3:

30

(3x x2) sin mx3

dx =

30

( n=1

bn sinnx

3

)sin

mx

3dx

=n=1

bn

30

sinnx

3sin

mx

3dx

=3bm

2

Integration by parts

Its an exercise in integration by parts to show that 30

(3x x2) sin mx3

dx =54

m33(1 (1)m)

Therefore,

bm =36

m33(1 (1)m) =

0 m even

72

m33m odd

So we arrive at a candidate for the solution:

u(x , t) =n=0

72

(2n + 1)33e(2n+1)

22t/45 sin(2n + 1)x

3

(3x x2) sin mx3

dx =54

m33(1 (1)m)

Therefore,

bm =36

m33(1 (1)m) =

0 m even

72

m33m odd

u(x , t) =n=0

72

(2n + 1)33e(2n+1)

22t/45 sin(2n + 1)x

3

(3x x2) sin mx3

dx =54

m33(1 (1)m)

Therefore,

bm =36

m33(1 (1)m) =

0 m even

72

m33m odd

u(x , t) =n=0

72

(2n + 1)33e(2n+1)

22t/45 sin(2n + 1)x

3

Validating the solution

The series

u(x , t) =n=0

72

(2n + 1)33e(2n+1)

22t/45 sin(2n + 1)x

3

converges for t 0, and certainly satisfies the boundaryconditions. What about the initial conditionu(x , 0) = 3x x2?

Well,

u(x , 0) =n=0

72

(2n + 1)23sin

(2n + 1)x

3

.

The series

u(x , t) =n=0

72

(2n + 1)33e(2n+1)

22t/45 sin(2n + 1)x

3

Well,

u(x , 0) =n=0

72

(2n + 1)23sin

(2n + 1)x

3

.

Graphical evidence

Red graph: 3x x3, Blue graph: sumOne term:

Three terms:

Plotting the solution

Here is a plot of the sum of of the first three terms of the solution:

Another example

Now, lets look at a problem with insulated ends:

ut =1

5uxx , ux(0, t) = ux(3, t) = 0, u(x , 0) = 3x x2

for t > 0 and 0 x 3.

This time, with the boundary conditions in mind, well assumethat we can express u as

u(x , t) =n=0

anen22t/45 cos

nx

3

(and see if we can figure out what the constants an should be the boundary conditions are automatically satisfied, and wewill choose the ans so that

3x x2 =n=0

an cosnx

3.

Another example

Now, lets look at a problem with insulated ends:

ut =1

5uxx , ux(0, t) = ux(3, t) = 0, u(x , 0) = 3x x2

for t > 0 and 0 x 3. This time, with the boundary conditions in mind, well assume

that we can express u as

u(x , t) =n=0

anen22t/45 cos

nx

3

(and see if we can figure out what the constants an should be the boundary conditions are automatically satisfied, and wewill choose the ans so that

3x x2 =n=0

an cosnx

3.

Useful integrals

Were trying to find ans so that

3x x2 =n=0

an cosnx

3.

Again, well use two basic facts: If n 6= m then

30

cosnx

3cos

mx

3dx = 0.

If n = m > 0 then 30

cosnx

3cos

mx

3dx =

30

cos2nx

3dx =

3

2

whereas if n = m = 0 we get 30 1

2 dx = 3.

Useful integrals

Were trying to find ans so that

3x x2 =n=0

an cosnx

3.

Again, well use two basic facts: If n 6= m then

30

cosnx

3cos

mx

3dx = 0.

If n = m > 0 then 30

cosnx

3cos

mx

3dx =

30

cos2nx

3dx =

3

2

whereas if n = m = 0 we get 30 1

2 dx = 3.

Finding the coefficients

Were still trying to find ans so that

3x x2 =n=0

bn cosnx

3.

Motivated by the facts on the previous slide, we multiply bothsides by cos mx3 and integrate both sides from 0 to 3. Form > 0 we get:

30

(3x x2) cos mx3

dx =

30

( n=0

an cosnx

3

)sin

mx

3dx

=n=0

an

30

cosnx

3cos

mx

3dx

=3am

2

and we get 3a0 for m = 0.

Its an exercise in integration by parts to show that

30

(3x x2) cos mx3

dx =

0 m odd92 m = 0 27

m22m > 0. even

u(x , t) =3

2n=1

36

(2n)22e(2n)

22t/45 cos2nx

3

Its an exercise in integration by parts to show that

30

(3x x2) cos mx3

dx =

0 m odd92 m = 0 27

m22m > 0. even

u(x , t) =3

2n=1

36

(2n)22e(2n)

22t/45 cos2nx

3

The series

u(x , t) =3

2n=1

36

(2n)22e(2n)

22t/45 cos2nx

3

Well,

u(x , 0) =3

2n=1

36

(2n)22cos

2nx

3

The series

u(x , t) =3

2n=1

36

(2n)22e(2n)

22t/45 cos2nx

3

Well,

u(x , 0) =3

2n=1

36

(2n)22cos

2nx

3

Graphical evidence

Red graph: 3x x3, Blue graph: sumOne term, two terms:

Four terms, thirteen terms:

Plotting the solution

Here is a plot of the sum of of the first three terms of the solution:

Other boundary conditions

Questions for discussion:

How would you handle boundary conditions ux(0, t) = 0,u(3, t) = 0?

What about u(0, t) = 0, ux(3, t) = 0? What about something like u(0, t) = 0, ux(3, t) + u(3, t) = 0?

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