Linear algebra and differential equations (Math 54): Lecture 25 - - PowerPoint PPT Presentation

Linear algebra and differential equations (Math 54): Lecture 25

Vivek Shende April 30, 2019

Hello, and welcome to class!

Hello, and welcome to class!

Last time

Hello, and welcome to class!

Last time

We introduced the notion of Fourier series, and discussed how to expand a function into one.

Hello, and welcome to class!

Last time

We introduced the notion of Fourier series, and discussed how to expand a function into one.

Hello, and welcome to class!

Last time

We introduced the notion of Fourier series, and discussed how to expand a function into one.

This time

Hello, and welcome to class!

Last time

We introduced the notion of Fourier series, and discussed how to expand a function into one.

This time

Having developed this tool, we return to studying the heat equation.

Heat equation review

Heat equation review

The heat equation in one variable is: ∂ ∂t u(x, t) = β ∂2 ∂x2 u(x, t)

Heat equation review

The heat equation in one variable is: ∂ ∂t u(x, t) = β ∂2 ∂x2 u(x, t) We saw that some solutions are given by u(x, t) = eλt(Aλex√

λ/β + Bλe−x√ λ/β)

λ > 0 u(x, t) = Aλ+Bλx λ = 0 u(x, t) = eλt(Aλ cos(x

• −λ/β)+Bλ sin(x
• −λ/β))

λ < 0

Heat equation review

Last week, we considered a wire of length L,

Heat equation review

Last week, we considered a wire of length L, whose endpoints were kept at temperature zero.

Heat equation review

Last week, we considered a wire of length L, whose endpoints were kept at temperature zero. In other words, we imposed boundary conditions u(0, t) = 0 = u(L, t)

Heat equation review

Of our above solutions, the only ones which take this form are the eλt sin(x

• −λ/β)

when

• −λ/β = Nπ/L

Heat equation review

Of our above solutions, the only ones which take this form are the eλt sin(x

• −λ/β)

when

• −λ/β = Nπ/L

This led to a general solution of the form u(x, t) =

∞

• N=1

cNe−β( Nπ

L ) 2t sin

Nπ L x

Initial conditions and Fourier expansion

Finally suppose we are given the initial temperature in the form of some function u(x, 0).

Initial conditions and Fourier expansion

Finally suppose we are given the initial temperature in the form of some function u(x, 0). Our job now is to express u(x, 0) =

∞

• N=1

cN sin Nπ L x

Initial conditions and Fourier expansion

Finally suppose we are given the initial temperature in the form of some function u(x, 0). Our job now is to express u(x, 0) =

∞

• N=1

cN sin Nπ L x

• In other words, to find values cN making the above formula true.

Initial conditions and Fourier expansion

Finally suppose we are given the initial temperature in the form of some function u(x, 0). Our job now is to express u(x, 0) =

∞

• N=1

cN sin Nπ L x

• In other words, to find values cN making the above formula true.

Because then the solution will be given by u(x, t) =

∞

• N=1

cNe−β( Nπ

L ) 2t sin

Nπ L x

Initial conditions and Fourier expansion

The expression u(x, 0) =

∞

• N=1

cN sin Nπ L x

• looks much like a Fourier expansion.

Initial conditions and Fourier expansion

The expression u(x, 0) =

∞

• N=1

cN sin Nπ L x

• looks much like a Fourier expansion.

Two differences from last time:

Initial conditions and Fourier expansion

The expression u(x, 0) =

∞

• N=1

cN sin Nπ L x

• looks much like a Fourier expansion.

Two differences from last time: first, the function is only defined

• n the interval [0, L],

Initial conditions and Fourier expansion

The expression u(x, 0) =

∞

• N=1

cN sin Nπ L x

• looks much like a Fourier expansion.

Two differences from last time: first, the function is only defined

• n the interval [0, L], and second, we want to expand it only in sin

rather than in sin and cos.

Fourier series review

The Fourier series of a function f (x) defined on [−L, L] is a0 2 +

∞

• n=1

an cos nπx L + bn sin nπx L an = 1 L L

−L

f (x) cos nπx L dx bn = 1 L L

−L

f (x) sin nπx L dx

Initial conditions and Fourier expansion

To absorb these differences, we extend the function f to [−L, L] simply by defining f (−x) = −f (x).

Initial conditions and Fourier expansion

To absorb these differences, we extend the function f to [−L, L] simply by defining f (−x) = −f (x). This has the virtue of ensuring that the extension is an odd function,

Initial conditions and Fourier expansion

To absorb these differences, we extend the function f to [−L, L] simply by defining f (−x) = −f (x). This has the virtue of ensuring that the extension is an odd function, which therefore has a Fourier expansion consisting only of sin nπx

L

waves,

Initial conditions and Fourier expansion

To absorb these differences, we extend the function f to [−L, L] simply by defining f (−x) = −f (x). This has the virtue of ensuring that the extension is an odd function, which therefore has a Fourier expansion consisting only of sin nπx

L

waves, exactly as we wanted.

Initial conditions and Fourier expansion

To absorb these differences, we extend the function f to [−L, L] simply by defining f (−x) = −f (x). This has the virtue of ensuring that the extension is an odd function, which therefore has a Fourier expansion consisting only of sin nπx

L

waves, exactly as we wanted. In interpreting the answer, we just ignore the value of the function

• n [−L, 0].

Example

Consider a wire of length L and diffusivity β in which the initial temperature is described by the function u(x) =

• x

0 ≤ x ≤ L/2 L − x L/2 ≤ x ≤ L and in which the temperature at the endpoints is kept at zero.

Example

Consider a wire of length L and diffusivity β in which the initial temperature is described by the function u(x) =

• x

0 ≤ x ≤ L/2 L − x L/2 ≤ x ≤ L and in which the temperature at the endpoints is kept at zero. Let us determine the temperature in the wire as a function of time.

Example

Consider a wire of length L and diffusivity β in which the initial temperature is described by the function u(x) =

• x

0 ≤ x ≤ L/2 L − x L/2 ≤ x ≤ L and in which the temperature at the endpoints is kept at zero. Let us determine the temperature in the wire as a function of time. First, we should expand out u(x) into its sin Fourier series.

Example

That is, we want to compute bn = 2 L L f (x) sin nπx L

• dx

= 2 L L/2 x sin nπx L

• dx +

L

L/2

(L − x) sin nπx L

• dx
• =

2 L L nπ 2 nπ/2 u sin u du + π

nπ/2

(nπ − u) sin u du

Example

That is, we want to compute bn = 2 L L f (x) sin nπx L

• dx

= 2 L L/2 x sin nπx L

• dx +

L

L/2

(L − x) sin nπx L

• dx
• =

2 L L nπ 2 nπ/2 u sin u du + π

nπ/2

(nπ − u) sin u du

• Noting
• u sin udu = sin u − u cos u,

Example

That is, we want to compute bn = 2 L L f (x) sin nπx L

• dx

= 2 L L/2 x sin nπx L

• dx +

L

L/2

(L − x) sin nπx L

• dx
• =

2 L L nπ 2 nπ/2 u sin u du + π

nπ/2

(nπ − u) sin u du

• Noting
• u sin udu = sin u − u cos u, this is

2 L L nπ 2 sin u − u cos u

• nπ/2

−

• sin u − u cos u
• nπ

nπ/2

−

• nπ cos u
• nπ

nπ/2

Example

2 L L nπ 2 sin u − u cos u

• nπ/2

−

• sin u − u cos u
• nπ

nπ/2

−

• nπ cos u
• nπ

nπ/2

Example

2 L L nπ 2 sin u − u cos u

• nπ/2

−

• sin u − u cos u
• nπ

nπ/2

−

• nπ cos u
• nπ

nπ/2

• = 2

L L nπ 2 2 sin nπ 2

• =

4L (nπ)2 sin nπ 2

Example

2 L L nπ 2 sin u − u cos u

• nπ/2

−

• sin u − u cos u
• nπ

nπ/2

−

• nπ cos u
• nπ

nπ/2

• = 2

L L nπ 2 2 sin nπ 2

• =

4L (nπ)2 sin nπ 2 Thus the Fourier expansion of the original function u(x) is u(x) = 4L π2

• sin(x) − 1

9 sin 3x + 1 25 sin 5x − 1 49 sin 7x + · · ·

Example

And finally, having written the initial condition as u(x) = 4L π2

• sin(x) − 1

9 sin 3x + 1 25 sin 5x − 1 49 sin 7x + · · ·

Example

And finally, having written the initial condition as u(x) = 4L π2

• sin(x) − 1

9 sin 3x + 1 25 sin 5x − 1 49 sin 7x + · · ·

• we see that the time evolution is given by

u(x, t) = 4L π2

• e−β( π

L ) 2t sin(x) − 1

9e−β( 3π

L ) 2t sin 3x + 1

25e−β( π

5L) 2t sin 5x − · · ·

Example

Another sort of boundary condition we might impose is,

Example

Another sort of boundary condition we might impose is, instead of fixing the initial and final temperatures to be zero,

Example

Another sort of boundary condition we might impose is, instead of fixing the initial and final temperatures to be zero, that the wire is insulated,

Example

Another sort of boundary condition we might impose is, instead of fixing the initial and final temperatures to be zero, that the wire is insulated, i.e., the derivatives

∂ ∂x u(x, t) vanish identically at 0, L.

Example

Another sort of boundary condition we might impose is, instead of fixing the initial and final temperatures to be zero, that the wire is insulated, i.e., the derivatives

∂ ∂x u(x, t) vanish identically at 0, L.

Let’s revisit our possible solutions. u(x, t) = eλt(Aλex√

λ/β + Bλe−x√ λ/β)

λ > 0 u(x, t) = Aλ+Bλx λ = 0 u(x, t) = eλt(Aλ cos(x

• −λ/β)+Bλ sin(x
• −λ/β))

λ < 0

Example

Another sort of boundary condition we might impose is, instead of fixing the initial and final temperatures to be zero, that the wire is insulated, i.e., the derivatives

∂ ∂x u(x, t) vanish identically at 0, L.

Let’s revisit our possible solutions. u(x, t) = eλt(Aλex√

λ/β + Bλe−x√ λ/β)

λ > 0 u(x, t) = Aλ+Bλx λ = 0 u(x, t) = eλt(Aλ cos(x

• −λ/β)+Bλ sin(x
• −λ/β))

λ < 0 Try it yourself: which satisfy the boundary conditions?

Example

Example

Yet a third scenario: we could consider ask that at 0 and at L, the temperature is fixed to be some given constants U1 and U2, possibly nonzero.

Example

Yet a third scenario: we could consider ask that at 0 and at L, the temperature is fixed to be some given constants U1 and U2, possibly nonzero. Again, we should look at our possible solutions, u(x, t) = eλt(Aλex√

λ/β + Bλe−x√ λ/β)

λ > 0 u(x, t) = Aλ+Bλx λ = 0 u(x, t) = eλt(Aλ cos(x

• −λ/β)+Bλ sin(x
• −λ/β))

λ < 0

Example

Yet a third scenario: we could consider ask that at 0 and at L, the temperature is fixed to be some given constants U1 and U2, possibly nonzero. Again, we should look at our possible solutions, u(x, t) = eλt(Aλex√

λ/β + Bλe−x√ λ/β)

λ > 0 u(x, t) = Aλ+Bλx λ = 0 u(x, t) = eλt(Aλ cos(x

• −λ/β)+Bλ sin(x
• −λ/β))

λ < 0 Try it yourself: which satisfy the boundary conditions?