TP1 01 Vector Calculus 03 Laplacian
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Transcript of TP1 01 Vector Calculus 03 Laplacian
Basic Mathematics
The Laplacian
R Horan & M Lavelle
The aim of this package is to provide a short selfassessment programme for students who want toapply the Laplacian operator.
Copyright cΒ© 2005 [email protected] , [email protected] Revision Date: April 4, 2005 Version 1.0
Table of Contents
1. Introduction (Grad, Div, Curl)2. The Laplacian3. The Laplacian of a Product of Fields4. The Laplacian and Vector Fields5. Final Quiz
Solutions to ExercisesSolutions to Quizzes
The full range of these packages and some instructions,should they be required, can be obtained from our webpage Mathematics Support Materials.
Section 1: Introduction (Grad, Div, Curl) 3
1. Introduction (Grad, Div, Curl)The vector differential operator β , called βdelβ or βnablaβ, is definedin three dimensions to be:
β =β
βxi +
β
βyj +
β
βzk
The result of applying this vector operator to a scalar field is calledthe gradient of the scalar field:
gradf(x, y, z) = βf(x, y, z) =βf
βxi +
βf
βyj +
βf
βzk .
(See the package on Gradients and Directional Derivatives.)
The scalar product of this vector operator with a vector field F (x, y, z)is called the divergence of the vector field:
divF (x, y, z) = β Β· F =βF1
βx+
βF2
βy+
βF3
βz.
Section 1: Introduction (Grad, Div, Curl) 4
The vector product of the vector β with a vector field F (x, y, z) is thecurl of the vector field. It is written as curlF (x, y, z) = βΓ F
βΓ F (x, y, z) = (βF3
βyβ βF2
βz)iβ (
βF3
βxβ βF1
βz)j + (
βF2
βxβ βF1
βy)k
=
β£β£β£β£β£β£β£i j kβ
βx
β
βy
β
βzF1 F2 F3
β£β£β£β£β£β£β£ ,
where the last line is a formal representation of the line above. (Seealso the package on Divergence and Curl.)
Here are some revision exercises.
Exercise 1. For f = x2y β z and F = xiβ xyj + z2k calculate thefollowing (click on the green letters for the solutions).
(a) βf (b) β Β· F
(c) βΓ F (d) βf ββΓ F
Section 2: The Laplacian 5
2. The LaplacianThe Laplacian operator is defined as:
β2 =β2
βx2+
β2
βy2+
β2
βz2.
The Laplacian is a scalar operator. If it is applied to a scalar field, itgenerates a scalar field.
Example 1 The Laplacian of the scalar field f(x, y, z) = xy2 + z3 is:
β2f(x, y, z) =β2f
βx2+
β2f
βy2+
β2f
βz2
=β2
βx2(xy2 + z3) +
β2
βy2(xy2 + z3) +
β2
βz2(xy2 + z3)
=β
βx(y2 + 0) +
β
βy(2xy + 0) +
β
βz(0 + 3z2)
= 0 + 2x + 6z = 2x + 6z
Section 2: The Laplacian 6
Exercise 2. Calculate the Laplacian of the following scalar fields:(click on the green letters for the solutions).(a) f(x, y, z) = 3x3y2z3 (b) f(x, y, z) =
βxz + y
(c) f(x, y, z) =β
x2 + y2 + z2 (d) f(x, y, z) =1β
x2 + y2 + z2
Quiz Choose the Laplacian of f(r) =1rn
where r =β
x2 + y2 + z2.
(a) β 1rn+2
(b)n
rn+2
(c)n(nβ 1)
rn+2(d)
n(n + 5)rn+2
The equation β2f = 0 is called Laplaceβs equation. This is an impor-tant equation in science. From the above exercises and quiz we see
that f =1r
is a solution of Laplaceβs equation except at r = 0.
Section 2: The Laplacian 7
The Laplacian of a scalar field can also be written as follows:
β2f = β Β·βf
i.e., as the divergence of the gradient of f . To see this consider
β Β·βf = β Β· (βf
βxi +
βf
βyj +
βf
βzk)
=β
βx(βf
βx) +
β
βy(βf
βy) +
β
βz(βf
βz)
=β2f
βx2+
β2f
βy2+
β2f
βz2= β2f
Exercise 3. Find the Laplacian of the scalar fields f whose gradientsβf are given below (click on the green letters for the solutions).
(a) βf = 2xzi + x2k (b) βf =1yz
iβ x
y2zj β x
yz2k
(c) βf = ezi + yj + xezk (d) βf =1x
i +1yj +
1zk
Section 3: The Laplacian of a Product of Fields 8
3. The Laplacian of a Product of FieldsIf a field may be written as a product of two functions, then:
β2(uv) = (β2u)v + uβ2v + 2(βu) Β· (βv)
A proof of this is given at the end of this section.Example 2 The Laplacian of f(x, y, z) = (x + y + z)(xβ 2z) may bedirectly calculated from the above rule
β2f(x, y, z) = (β2(x + y + z))(xβ 2z) + (x + y + z)β2(xβ 2z)+2β(x + y + z) Β·β(xβ 2z)
Now β2(x + y + z) = 0 and β2(x β 2z) = 0 so the first line on theright hand side vanishes.To calculate the second line we note that
β(x+y+z) =β(x + y + z)
βxi+
β(x + y + z)βy
j+β(x + y + z)
βzk = i+j+k
Section 3: The Laplacian of a Product of Fields 9
and
β(xβ 2z) =β(xβ 2z)
βxi +
β(xβ 2z)βy
j +β(xβ 2z)
βzk = iβ 2k
and taking their scalar product we obtain
β2f(x, y, z) = 0 + 2β(x + y + z) Β·β(xβ 2z)= 2(i + j + k) Β· (iβ 2k) = 2(1 + 0β 2)= β2.
This example may be checked by expanding (x + y + z)(xβ 2z) anddirectly calculating the Laplacian.
Exercise 4. Use this rule to calculate the Laplacian of the scalarfields given below (click on the green letters for the solutions).
(a) (2xβ 5y + z)(xβ 3y + z) (b) (x2 β y)(x + z)
(c) (y β z)(x2 + y2 + z2) (d) xβ
x2 + y2 + z2
Section 3: The Laplacian of a Product of Fields 10
Proof that β2(uv) = (β2u)v + uβ2v + 2(βu) Β· (βv) .
By definition β2(uv) =β2
βx2(uv) +
β2
βy2(uv) +
β2
βz2(uv) . Consider
therefore:β2
βx2(uv) =
β
βx
(β
βx(uv)
)=
β
βx
(βu
βxv + u
βv
βx
)=
β2u
βx2v +
βu
βx
βv
βx+
βu
βx
βv
βx+ u
β2v
βx2
=β2u
βx2v + 2
βu
βx
βv
βx+ u
β2v
βx2
where the product rule was repeatedly used.
By symmetry we also have:β2
βy2(uv) =
β2u
βy2v + 2
βu
βy
βv
βy+ u
β2v
βy2and
β2
βz2(uv) =
β2u
βz2v + 2
βu
βz
βv
βz+ u
β2v
βz2. Adding these results yields the
desired result.
Section 4: The Laplacian and Vector Fields 11
4. The Laplacian and Vector FieldsIf the scalar Laplacian operator is applied to a vector field, it acts oneach component in turn and generates a vector field.
Example 3 The Laplacian of F (x, y, z) = 3z2i + xyzj + x2z2k is:
β2F (x, y, z) = β2(3z2)i + β2(xyz)j + β2(x2z2)k
Calculating the components in turn we find:
β2(3z2) =β2
βx2(3z2) +
β2
βy2(3z2) +
β2
βz2(3z2) = 0 + 0 + 6 = 6
β2(xyz) =β2
βx2(xyz) +
β2
βy2(xyz) +
β2
βz2(xyz) = 0 + 0 + 0 = 0
β2(x2z2) =β2
βx2(x2z2) +
β2
βy2(x2z2) +
β2
βz2(x2z2) = 2z2 + 0 + 2x2
So the Laplacian of F is:
β2F = 6i + 0j + (2z2 + 2x2)k = 6i + 2(x2 + z2)k
Section 4: The Laplacian and Vector Fields 12
Quiz Select from the answers below the Laplacian of the vector fieldF = x3yi + ln(z)j + ln(xy)k .
(a) 6xyi (b) 6xyiβ 1z2
j β x2 + y2
x2y2k
(c) 3xi +1z2
j β(
1x2β 1
y2
)k (d) y3i + x2j β y2 β x2
z2k
Quiz Choose the Laplacian of F = 3x2ziβ sin(Οy)j +ln(2x3)k at thepoint (1,β2, 1)?
(a) 0 (b) 6iβ 3k (c) 3iβ 3k (d) 6iβ Ο2j β 6k
Quiz Which of the following choices is the Laplacian of the vector fieldF = ln(y)i + z2j β sin(2Οx)k at (1, 1, Ο)?
(a) βi + 2j (b) 0 (c) 4j + 4Ο2k (d) βi + j + k
Section 5: Final Quiz 13
5. Final QuizBegin Quiz Choose the solutions from the options given.
1. Choose the Laplacian of f(r) = 5x3y4z2.
(a) 30xy4z2 +60x3y2z2 +10x3y4 (b) 30x + 20y2 + 10
(c) 30xy4z2 +75x3y2z2 +15x3y4 (d) 30xy4z2 +12x3y2z2 +15x3y4
2. Choose the Laplacian of f(x, y, z) = ln(r) where r =β
x2 + y2 + z2.
(a) 0 (b)2r2
(c)1
2r2(d)
1r2
3. Select the Laplacian of F = x3i + 7yj β 3 sin(2y)k.
(a) 6xi (b) 6xi + 12 sin(2y)k(c) 6xi + 3 sin(2y)k (d) 6xiβ 12 sin(2y)k
End Quiz
Solutions to Exercises 14
Solutions to ExercisesExercise 1(a)To find the gradient of the scalar field f = x2y β z, we need thepartial derivatives:
βf
βx=
β
βx(x2y β z) = 2x2β1 Γ y = 2xy ,
βf
βy=
β
βy(x2y β z) = x2 Γ y1β1 = x2 ,
βf
βz=
β
βz(x2y β z) = 0β z1β1 = β1 .
Therefore the gradient of f = x2y β z is
βf(x, y, z) =βf
βxi +
βf
βyj +
βf
βzk
= 2xyi + x2j β k .
Click on the green square to return
Solutions to Exercises 15
Exercise 1(b)To find the divergence of the vector field F = xi β xyj + z2k, werecognise that its components are
F1 = x , F2 = βxy , F3 = z2 ,
So the divergence is the scalar expression
β Β· F =βF1
βx+
βF2
βy+
βF3
βz
=β
βx(x) +
β
βy(βxy) +
β
βz(z2)
= x1β1 β xΓ y1β1 + 2Γ z2β1
= 1β x + 2z .
Click on the green square to return
Solutions to Exercises 16
Exercise 1(c)The curl of the vector field F whose components are
F1 = x , F2 = βxy , F3 = z2 ,
is given by the vector expression:
βΓ F =(
βF3
βyβ βF2
βz
)iβ
(βF3
βxβ βF1
βz
)j +
(βF2
βxβ βF1
βy
)k
=(
β
βy(z2)β β
βz(βxy)
)iβ
(β
βx(z2)β β
βz(x)
)j
+(
β
βx(βxy)β β
βy(x)
)k
= (0 + 0) iβ (0β 0) j + (βy β 0) k
= βyk .
Click on the green square to return
Solutions to Exercises 17
Exercise 1(d)To subtract the curl of the vector F = xiβxyj+z2k from the gradientof the scalar field f = x2y β z
βf ββΓ F .
we use the results of Exercise 1a and Exercise 1c, where it was foundthat
βf = 2xyi + x2j β k and βΓ F = βyk .
Therefore the difference of these two vectors is
βf ββΓ F = 2xyi + x2j β k β (βy)k= 2xyi + x2j β (1β y)k .
Click on the green square to return
Solutions to Exercises 18
Exercise 2(a)The Laplacian of the scalar field f = 3x3y2z3 is:
β2f =β2f
βx2+
β2f
βy2+
β2f
βz2
=β2
βx2(3x3y2z3) +
β2
βy2(3x3y2z3) +
β2
βz2(3x3y2z3)
=β
βx(9x2y2z3) +
β
βy(6x3yz3) +
β
βz(9x3y2z2)
= 18xy2z3 + 6x3z3 + 18x3y2z .
Extracting common factors, the scalar β2f can also be written as
β2f = 6xz(3y2z2 + x2z2 + 3x2y2
)= 6xz
(3y2(z2 + x2) + x2z2
).
Click on the green square to return
Solutions to Exercises 19
Exercise 2(b)The Laplacian of the scalar field f =
βxz + y = x1/2z1/2 + y is:
β2f =β2
βx2
(x
12 z
12 + y
)+
β2
βy2
(x
12 z
12 + y
)+
β2
βz2
(x
12 z
12 + y
)=
β
βx
(12x( 1
2β1)z12
)+
β
βy(1) +
β
βz
(12x
12 z( 1
2β1)
)=
β
βx
(12xβ
12 z
12
)+ 0 +
β
βz
(12x
12 zβ
12
)=
12(β1
2)x(β 1
2β1)z12 +
12(β1
2)x
12 z(β 1
2β1)
= β14xβ
32 z
12 β 1
4x
12 zβ
32 .
This result may be rewritten as
β2f = β14x
12 z
12
(xβ2 + zβ2
)= β1
4β
xz
(1x2
+1z2
).
Click on the green square to return
Solutions to Exercises 20
Exercise 2(c) To calculate β2β
x2 + y2 + z2 , define u = x2+y2+z2,so f = u1/2. From the chain rule we have
βf
βx=
βu12
βuΓ βu
βx=
12u( 1
2β1) Γ 2x = xuβ12 .
Therefore the second derivative is (from the product and chain rules):
β2f
βx2=
β
βx(x)Γ uβ
12 + xΓ βuβ
12
βuΓ βu
βx= uβ
12 β x2uβ
32 .
Since f and u are symmetric in x, y and z we have
β2f
βy2= uβ
12 β y2uβ
32 and
β2f
βz2= uβ
12 β z2uβ
32 .
Adding these results and using x2 + y2 + z2 = u yields
β2f =(uβ
12 β x2uβ
32
)+
(uβ
12 β y2uβ
32
)+
(uβ
12 β z2uβ
32
)= 3uβ
12 β (x2 + y2 + z2)uβ
32 = 2uβ
12 =
2βx2 + y2 + z2
.
Click on the green square to return
Solutions to Exercises 21
Exercise 2(d) To find the Laplacian of f = (x2 + y2 + z2)β12 , we
again define u = x2 + y2 + z2. From the chain rule
βf
βx=
βuβ12
βuΓ βu
βx= β1
2u(β 1
2β1) Γ 2x = βxuβ32 .
Thus the second order derivative is:β2f
βx2= β β
βx(x)Γ uβ
32 β xΓ βuβ
32
βuΓ βu
βx= βuβ
32 + 3x2uβ
52 .
Due to the symmetry under interchange of x, y and z :
β2f
βy2= βuβ
32 + 3y2uβ
52 ,
β2f
βz2= βuβ
32 + 3z2uβ
52 .
Therefore we find that the Laplacian of f vanishes:
β2f =(βuβ
32 + 3x2uβ
52
)+
(βuβ
32 + y2uβ
52
)+
(βuβ
32 + 3z2uβ
52
)= β3uβ
32 + 3(x2 + y2 + z2)uβ
52 = β3uβ
32 + 3uβ
32 = 0 .
Click on the green square to return
Solutions to Exercises 22
Exercise 3(a)If the gradient of the scalar function f is βf = 2xzi + x2k, then theLaplacian of f is given by the divergence of this vector
β2f = div (βf) .
Therefore the Laplacian of f is
β2f = div(2xzi + x2k
)=
β
βx(2xz) +
β
βy(0) +
β
βz(x2)
= 2z + 0 + 0= 2z .
Click on the green square to return
Solutions to Exercises 23
Exercise 3(b)
If a scalar field f has gradient βf =1yz
iβ x
y2zjβ x
yz2k , its Laplacian
is given by the divergence
β2f = div (βf) .
Using this formula we can find β2f as follows
β2f = div(
1yz
iβ x
y2zj β x
yz2k
)=
β
βx
(1yz
)+
β
βy
(β x
y2z
)+
β
βz
(β x
yz2
)= (0)β (β2)Γ x
y3zβ (β2)Γ x
yz3
= 2xy2 + z2
y3z3.
Click on the green square to return
Solutions to Exercises 24
Exercise 3(c)As above if the gradient of f is
βf = ezi + yj + xezk
its Laplacian is the scalar divergence of the vector βf :
β2f = div (βf) .
Therefore
β2f = div (ezi + yj + xezk)
=β
βx(ez) +
β
βy(y) +
β
βz(xez)
= (0) + 1 + xez
= 1 + xez .
Click on the green square to return
Solutions to Exercises 25
Exercise 3(d)
If βf =1x
i +1yj +
1zk , then the Laplacian of f is the divergence of
the gradient of f :β2f = div (βf) .
Therefore
β2f = div(
1x
i +1yj +
1zk
)=
β
βx
(1x
)+
β
βy
(1y
)+
β
βz
(1z
)= β 1
x2β 1
y2β 1
z2.
This may also be written as follows
β2f = βx2y2 + z2x2 + y2z2
x2y2z2.
Click on the green square to return
Solutions to Exercises 26
Exercise 4(a)To find the Laplacian of f = (2xβ 5y + z)(xβ 3y + z), we use
β2(uv) = (β2u)v + uβ2v + 2(βu) Β· (βv)
with u = 2xβ 5y + z and v = xβ 3y + z. This implies
βu = 2iβ 5j + k
βv = iβ 3j + k
and soβ2u = β2v = 0
The above rule therefore gives:
β2f = 2βu Β·βv = 0 + 0 + 2(2iβ 5j + k) Β· (iβ 3j + k)= 2(2 + 15 + 1)= 36 .
Click on the green square to return
Solutions to Exercises 27
Exercise 4(b)To find the Laplacian of f = (x2 β y)(x + z), we use
β2(uv) = (β2u)v + uβ2v + 2(βu) Β· (βv)
with u = x2 β y and v = x + z. This implies
βu = 2xiβ j
βv = i + k
and so
β2u = β Β· 2xi =β(2x)βx
= 2 and β2v = 0
The above rule therefore gives:
β2f = 2(x + z) + 0 + 2(2iβ j) Β· (i + k)= 2(x + z) + 2(2)= 2(x + z + 2) .
Click on the green square to return
Solutions to Exercises 28
Exercise 4(c)To find the Laplacian of f = (y β z)(x2 + y2 + z2), we use
β2(uv) = (β2u)v + uβ2v + 2(βu) Β· (βv)
with u = y β z and v = x2 + y2 + z2. This implies
βu = j β k and β2u = 0
while
βv = 2xi + 2yj + 2zk
β2u = 2 + 2 + 2 = 6
The above rule therefore gives:
β2f = 0 + (y β z)Γ 6 + 2(j β k) Β· (2i + 2j + 2k)= 6 + 2(0 + 2β 2)= 6 .
Click on the green square to return
Solutions to Exercises 29
Exercise 4(d) To find the Laplacian of f = xβ
x2 + y2 + z2, weuse
β2(uv) = (β2u)v + uβ2v + 2(βu) Β· (βv)
with u = x and v =β
x2 + y2 + z2. Therefore βu = i and β2u = 0 .
In Exercise 2(c) we saw β2v = β2/β
x2 + y2 + z2. To find βv use
the chain rule:β
βx(x2 + y2 + z2)β
12 = 2xΓ (β 1
2 )Γ (x2 + y2 + z2)β32
and similarly for the other partial derivatives. Thus
βv = β 1(x2 + y2 + z2)
32(xi + yj + zk)
The above rule therefore gives:
β2f = 0 + xΓ β2βx2 + y2 + z2
+ 2i Β· β1(x2 + y2 + z2)
32(xi + yj + zk)
= β 2xβx2 + y2 + z2
β 2x2
(x2 + y2 + z2)32
.
Click on the green square to return
Solutions to Quizzes 30
Solutions to QuizzesSolution to Quiz: The first and the second order partial derivativesof f = rβn , r = (x2 + y2 + z2)
12 with respect to the variable x read:
βf
βx=
βrβn
βrΓ βr
βx= βnrβnβ1 Γ x
r= βn
x
rn+2.
β2f
βx2= βn
1rn+2
βx
βxβ xnΓ β
βr
(1
rn+2
)Γ βr
βx
= βn1
rn+2+ n(n + 2)
x2
rn+4.
Due to the symmetry under interchange of x, y and z we have also
β2f
βy2= βn
1rn+2
+ n(n + 2)y2
rn+4,
β2f
βz2= βn
1rn+2
+ n(n + 2)z2
rn+4.
Adding these second order derivatives yields the Laplacian:
β2f = β 3n
rn+2+ n(n + 2)
x2 + y2 + z2
rn+4=
n(nβ 1)rn+2
.
End Quiz
Solutions to Quizzes 31
Solution to Quiz:The Laplacian of the vector field F = x3yi + ln(z)j + ln(xy)k is avector β2F whose i , j components are correspondingly
β2(x3y) =β2
βx2(x3y) +
β2
βy2(x3y) +
β2
βz2(x3y) = 6xy ,
β2(ln(z)) =β2
βz2(ln(z)) =
β
βz
(1z
)= β 1
z2,
while the k component is
β2(ln(xy)) = β2(ln(x) + ln(y)) = β 1x2β 1
y2= βx2 + y2
x2y2.
So the Laplacian of F is:
β2F = 6xyiβ 1z2
j β x2 + y2
x2y2k .
End Quiz
Solutions to Quizzes 32
Solution to Quiz:The Laplacian of the vector field F = 3x2ziβ sin(Οy)j + ln(2x3)k isa vector β2F whose i , j and k components are correspondingly
β2(3x2z
)=
β2
βx2(3x2z) +
β2
βy2(3x2z) +
β2
βz2(3x2z) = 6z ,
β2 (β sin(Οy)) = β β2
βy2(sin(Οy)) = β β
βy(Ο cos(Οy)) = Ο2 sin(Οy) ,
β2(ln(2x3)
)=
β2
βx2(ln(2x3)) =
β2
βx2(ln(2) + 3 ln(x))) = β3
1x2
,
Therefore the Laplacian of F is
β2F = 6zi + Ο2 sin(Οy)j β 31x2
k ,
and evaluating it at the point (1,β2, 1) we get
β2F = 6i + Ο2 sin(β2Ο)j β 3k = 6iβ 3k .
End Quiz
Solutions to Quizzes 33
Solution to Quiz:The Laplacian of F = ln(y)i+z2jβsin(2Οx)k at (1, 1, Ο) is the vectorwhose i, j and k components are in turn given by:
β2 ln(y) =β2 ln(y)
βx2+
β2 ln(y)βy2
+β2 ln(y)
βz2= 0β 1
y2+ 0 ,
β2z2 =β2z2
βx2+
β2z2
βy2+
β2z2
βz2= 0 + 0 + 2 ,
β2 sin(2Οx) =β2 sin(2Οx)
βx2+
β2 sin(2Οx)βy2
+β2 sin(2Οx)
βz2
= β4Ο2 sin(2Οx) + 0 + 0 ,
Therefore we have
β2F = β 1y2
i + 2j β 4Ο2 sin(2Οx)k
Since sin(2Ο) = 0, we find at (1, 1, Ο) that β2F = βi + 2j .End Quiz