5Distributions

IB Methods

5.3 Green’s functions for initial value problems

Consider

Ly

=

f

(

t

) subject to

y

(

t

=

t

0

) = 0 and

y

0

(

t

=

t

0

) = 0. So instead of

having two boundaries, we have one boundary and restrict both the value and

the derivative at this boundary.

As before, let

y

1

(

t

)

, y

2

(

t

) be any basis of solutions to

Ly

= 0. The Green’s

function obeys

L(G) = δ(t − τ ).

We can write our Green’s function as

G(t, τ) =

(

A(τ)y

1

(t) + B(τ)y

2

(t) t

0

≤ t < τ

C(τ )y

1

(t) + D(τ)y

2

(t) t > τ.

Our initial conditions require that

y

1

(t

0

) y

2

(t

0

)

y

0

1

(t

0

) y

0

2

(t

0

)

A

B

=

0

0

However, we know that the matrix is non-singular (by definition of

y

1

, y

2

being a

basis). So we must have

A, B

= 0 (which makes sense, since

G

= 0 is obviously

a solution for t

0

≤ t < τ).

Our continuity and jump conditions now require

0 = C(τ )t

1

(t) + D(τ)y

2

(t)

1

α(τ)

= C(τ )y

0

1

(τ) + D(τ )y

0

2

(τ).

We can solve this to get C(τ) and D(τ). Then we get

y(t) =

Z

∞

t

0

G(t, τ)f(τ) dτ =

Z

t

t

0

G(t, τ)f(τ) dτ,

since we know that when

τ > t

, the Green’s function

G

(

t, τ

) = 0. Thus the

solution

y

(

t

) depends on the forcing term term

f

only for times

< t

. This

expresses causality!

Example.

For example, suppose we have

¨y

+

y

=

f

(

t

) with

f

(0) =

˙y

(0) = 0.

Then we have

G(t, τ) = Θ(t − τ )[C(τ ) cos(t − τ) + D(τ ) sin(t − τ)]

for some

C

(

τ

)

, D

(

τ

). The continuity and jump conditions gives

D

(

τ

) = 1

, C

(

τ

) =

0. So we get

G(t, τ) = Θ(t − τ ) sin t(τ).

So we get

y(t) =

Z

t

0

sin(t − τ)f(τ) dτ.