Integration of Dirac Delta Function and Vector Calculus, Exercises of Physics

Solutions to various integration problems involving dirac delta functions in one and three dimensions, as well as computations of gradient and divergence of a vector field. It also includes an optional exercise on the general rule for integrating delta functions.

Typology: Exercises

2019/2020

Uploaded on 09/24/2020

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1) Compute the following integrals involving the Dirac delta function
1.1) R5
2δ(x)f(x)dx
1.2) R5
1δ(x3)f(x)dx
1.3) R4
1δ(2x3)f(x)dx
1.4) R3
2δ(x21)f(x)dx
1.5) R1
5δ(x24)f(x)dx
2) Compute the following integrals involving the 3dDirac delta function
2.1) R2
2dx R1
3dy R5
0dzδ(~x x+ z))f(x, y, z)
2.2) R2
2dx R1
3dy R5
0dzδ(~x x+ 2 ˆy+ z))f(x, y, z )
2.3) R2
2dx R3
3dy R5
0dzδ(~x x+ ˆy+ z))f(x, y, z )
3) Consider the vector field ~
F= 3xˆx+ (2x+z)ˆy+ 3yˆz
3.1) Compute ~
· ~
F
3.2) Compute ~
× ~
F
3.3) Can you split ~
Finto ~
F=~
F1+~
F2such that ~
× ~
F1=~
0 and ~
· ~
F2= 0?
3.4) Find a scalar Vsuch that ~
V=~
F1
3.5) Find a vector ~
Asuch that ~
× ~
A=~
F2
4) Optional Exercise: There is an identity for the most general delta function (in one
dimension) one can write down
Zx2
x1
δ(g(x))dx =X
iZx2
x1
δ(xxi)
|g0(xi)|dx
where the sum above is over all the points where g(x) is zero in the interval [x1, x2] i.e.
g(xi) = 0 i.
4.1) Split the integral over a sum over the zeros of the function g(x)
Zx2
x1
δ(g(x))dx =X
iZxi+
xi
δ(g(x))dx
In the above you are only integrating over small intervals. Why can you discard the rest?
Can you give an idea on what size the intervals should be so as not to run into trouble?
4.2) In the region around each xiyou can write g(x)(xxi)g0(xi). Why? You thus have
Zx2
x1
δ(g(x)) = X
iZxi+
xi
δ(xg0(xi)g0(xi)xi)dx
4.3) Prove that Rx2
x1δ(axxi)dx =1
|a|R
1
|a|x2
1
|a|x1δ(xxi
|a|)dx. Why is the absolute value important?
1
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  1. Compute the following integrals involving the Dirac delta function

5

− 2

δ(x)f (x)dx

5

− 1

δ(x − 3)f (x)dx

4

− 1

δ(2x − 3)f (x)dx

3

− 2

δ(x

2 − 1)f (x)dx

1

− 5

δ(x

2 − 4)f (x)dx

  1. Compute the following integrals involving the 3d Dirac delta function

2

− 2

dx

1

− 3

dy

5

0

dzδ(~x − (ˆx + 2ˆz))f (x, y, z)

2

− 2

dx

1

− 3

dy

5

0

dzδ(~x − (ˆx + 2ˆy + 2ˆz))f (x, y, z)

2

− 2

dx

3

− 3

dy

5

0

dzδ(~x − (ˆx + ˆy + 2ˆz))f (x, y, z)

  1. Consider the vector field

F = 3xxˆ + (2x + z)ˆy + 3yzˆ

3.1) Compute

F

3.2) Compute

∇ ×

F

3.3) Can you split

F into

F =

F

1

F

2

such that

∇ ×

F

1

0 and

F

2

3.4) Find a scalar V such that

∇V =

F

1

3.5) Find a vector

A such that

∇ ×

A =

F

2

  1. Optional Exercise: There is an identity for the most general delta function (in one

dimension) one can write down

x 2

x 1

δ(g(x))dx =

i

x 2

x 1

δ(x − xi)

|g

′ (x i

dx

where the sum above is over all the points where g(x) is zero in the interval [x 1 , x 2 ] i.e.

g(x i

) = 0 ∀i.

4.1) Split the integral over a sum over the zeros of the function g(x)

x 2

x 1

δ(g(x))dx =

i

xi+

xi−

δ(g(x))dx

In the above you are only integrating over small intervals. Why can you discard the rest?

Can you give an idea on what size the  intervals should be so as not to run into trouble?

4.2) In the region around each x i

you can write g(x) ≈ (x − x i

)g

′ (x i

). Why? You thus have

x 2

x 1

δ(g(x)) =

i

xi+

xi−

δ(xg

(x i

) − g

(x i

)x i

)dx

4.3) Prove that

x 2

x 1

δ(ax−x i

)dx =

1

|a|

1

|a|

x 2

1

|a|

x 1

δ(x−

xi

|a|

)dx. Why is the absolute value important?

This completes the proof.

4.4) Show that this general rule is consistent with the rule for δ(x

2

− α)