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Partial derivatives & Vector calculus
Partial derivatives
Functions of several arguments (multivariate functions) such as f[x,y] can be differentiated with respect to each argument
∂f
∂x
≡ ∂x f,
∂f
∂y
≡ ∂y f,
etc. One can define higher-order derivatives with respect to the same or different variables
∂^2 f
∂x^2
≡ ∂x,x f,
∂^2 f
∂y^2
≡ ∂y,y f,
∂^2 f
∂x ∂ y
∂x
∂f
∂y
≡ ∂x,y f
For most of the functions mixed partial derivatives do not depend on the order of differentiation
∂^2 f
∂x ∂ y
∂^2 f
∂ y ∂x
This holds if the mixed derivatives are continuous at a given point. For instance,
f@x_, y_D = x y; ∂x f@x, yD ∂y f@x, yD ∂y,x f@x, yD ∂x,y f@x, yD y
x
1
1
ü "Bad" functions
Multivariate series
Taylor series can be generalized for multivariate functions and the
f@x_, y_D = Sin@x + yD; Series@f@x, yD, 8 x, 0, 3<D
Sin@yD + Cos@yD x −
Sin@yD x^2 −
Cos@yD x^3 + O@xD^4
Series@f@x, yD, 8 x, 0, 3<, 8 y, 0, 3<D
y −
y^3 6
y^2 2
y 2
y^3 12
y^2 12
or, in the symmetric form
Normal@Series@f@x, yD, 8 x, 0, 3<, 8 y, 0, 3<DD Expand@Normal@Series@f@x, yD, 8 x, 0, 3<, 8 y, 0, 3<DDD
x −
x^3 6
x^2 2
y + −
x 2
x^3 12
y^2 + −
x^2 12
y^3
x −
x^3 6
x^2 y 2
x y^2 2
x^3 y^2 12
y^3 6
x^2 y^3 12
Exercise: Find a way to sort this polynomial in increasing powers of x, y.
Vector calculus
Physics makes use of vector differential operations on functions such as gradient, divergence, curl ( rotor ), Laplacian , etc.
In the current version of Mathematica realizations of these operations are new and not included in the main body of the
software. Instead, these functions are implemented in the optional VectorAnalysis package that has to be called before
performing these operations
Needs@"VectorAnalysis`"D
Unfortunately, this package seems to be inconvenient.
ü Gradient
Gradient of a scalar funtion is a vector defined as
grad f ≡ ∇ f ≡ e x
∂f
∂x
+ e y
∂ f
∂ y
+ e z
∂f
∂z
≡ 9 ∂x f, ∂y f, ∂z f=
One can speak about the gradient operator defined as
∇ ≡ e x
∂x
+ e y
∂ y
+ e z
∂z
that acts on scalar functions of vector arguments. An example of a gradient in physics is force F that is minus gradient of the
potential energy U[x,y,z] and similar for the electric field E that is minus gradient of the electric potential f
F ≡ −∇U, E ≡ −∇ φ
Examples trying to use the Mathematica' s VectorCalculus package:
Following Mathematica help:
Clear@x, y, z, UD U = x^2 + y^2 + z^2 ; Grad@UD 8 0, 0, 0<
- a wrong output. An attempt of a standard usage
U@x_, y_, z_D = x^2 + y^2 + z^2 ; H∗ 3d oscillator ∗L
F@x_, y_, z_D = Grad@U@x, y, zDD 8 0, 0, 0<
- same wrong result. Still this command is working with a special naming choice
Divergence
Divergence of a vector is a scalar defined by
div A ≡ ∇ ⋅ A ≡
∂Ax
∂ x
∂ Ay
∂y
∂Az
∂z
divergence can be represented by the operator
∇ ≡ e x
∂x
+ e y
∂ y
+ e z
∂z
same as the gradient operator above. The only difference between them is that gradient acts on scalars and divergence acts
on vectors.
In[61]:= Div@Avec_D := ∂x Avec@@ 1 DD + ∂y Avec@@ 2 DD + ∂z Avec@@ 3 DD
Examples
Div@ 8 x, y, z<D 3
Fvec@x_, y_, z_D = 9 x^2 , y^2 , z^2 =; Div@Fvec@x, y, zDD 2 x + 2 y + 2 z
Fvec@x_, y_, z_D = 8 y, x, x y<; Div@Fvec@x, y, zDD 0
ü Laplacian
Laplacian is a second-order vector differential operation. Laplacian of a scalar f is defined as div grad f and denoted by D or
“^2
∆ f ≡ div grad f ≡ ∇ ⋅ ∇ f ≡ ∇^2 f
From this definition follows
∆ f =
∂^2 f
∂x^2
∂^2 f
∂y^2
∂^2 f
∂ z^2
The Laplace operator
∆ ≡ ∇^2 =
∂^2
∂ x^2
∂^2
∂ y^2
∂^2
∂ z^2
can be obtained by squaring the gradient / divergence operator above.
Laplace@ f _D := ∂x,x f + ∂y,y f + ∂z,z f
Example
f@x_, y_, z_D = x^2 + y^2 + z^2 ; Laplace@ f @x, y, zDD 6
Laplacian of a vector is defined in components through Laplacians of the components as
∆ A ≡ e 1 ∆A 1 + e 2 ∆A 2 + e 3 ∆A 3
In[60]:= LaplaceVec@ Avec _D^ :=^9 ∂x,x Avec @@^1 DD^ + ∂y,y Avec @@^1 DD^ + ∂z,z Avec @@^1 DD , ∂x,x Avec @@ 2 DD + ∂y,y Avec @@ 2 DD + ∂z,z Avec @@ 2 DD , ∂x,x Avec @@ 3 DD + ∂y,y Avec @@ 3 DD + ∂z,z Avec @@ 3 DD=
ü Curl (rotor)
Curl of a vector is a vector defined by
curl A ≡ detB
e x e y e z
∂x ∂y ∂z
Ax Ay Az
F ≡ ∇ A
In[62]:= Curl@Avec_D^ := 9 ∂y Avec@@ 3 DD − ∂z Avec@@ 2 DD, ∂z Avec@@ 1 DD − ∂x Avec@@ 3 DD, ∂x Avec@@ 2 DD − ∂y Avec@@ 1 DD=
Examples
Curl@ 8 x, y, z<D 8 0, 0, 0<
Curl@ 8 y, − x, 0<D 8 0, 0, − 2 <
ü Repeated first-order differential vector operations
Repeated vector differential operations satisfy some identities that are similar to repeated vector products if one uses the “
operator and considers it as a vector.
ü Double curl
Similarly to the double vector product A ×( B × C ) ª B ( A · C ) - C ( A·B ), one has
curl curl A ≡ ∇ H ∇ AL ≡ ∇ H ∇ ⋅ A L − ∇^2 A ≡ grad Hdiv A L − ∆ A
Note that here the vector A is always rightmost because the differential operators are acting on it. Check this identity
Avec@x_, y_, z_D = 8 Avecx@x, y, zD, Avecy@x, y, zD, Avecz@x, y, zD<;
Curl@Curl@Avec@x, y, zDDD
9 − AvecxH0,0,2L@x, y, zD − AvecxH0,2,0L@x, y, zD + AveczH1,0,1L@x, y, zD + AvecyH1,1,0L@x, y, zD, − AvecyH0,0,2L@x, y, zD + AveczH0,1,1L@x, y, zD + AvecxH1,1,0L@x, y, zD − AvecyH2,0,0L@x, y, zD, AvecyH0,1,1L@x, y, zD − AveczH0,2,0L@x, y, zD + AvecxH1,0,1L@x, y, zD − AveczH2,0,0L@x, y, zD=
Grad@φ@x, y, zD ψ@x, y, zDD == φ@x, y, zD Grad@ψ@x, y, zDD + Grad@φ@x, y, zDD ψ@x, y, zD True
ü Divergence and curl of a product of a vector and a scalar
div Hφ A L ≡ ∇ ⋅ Hφ A L ≡ φ ∇ ⋅ A + ∇ φ ⋅ A
curl Hφ A L ≡ ∇ Hφ A L ≡ φ ∇ A + ∇ φ A
Check these identities
ü Divergence of a vector product
div H A B L ≡ ∇ ⋅ H A B L ≡ H ∇ AL ⋅ B − A ⋅ H ∇ BL
Check this identity
Avec@x_, y_, z_D = 8 Avecx@x, y, zD, Avecy@x, y, zD, Avecz@x, y, zD<; Bvec@x_, y_, z_D = 8 Bvecx@x, y, zD, Bvecy@x, y, zD, Bvecz@x, y, zD<; Simplify@ Div@Cross@Avec@x, y, zD, Bvec@x, y, zDDD Curl@Avec@x, y, zDD.Bvec@x, y, zD − Avec@x, y, zD.Curl@Bvec@x, y, zDD D True
Potential and its gradient
Force and electric field are negative gradients of the potential energy and electric potential, respectively:
F ≡ −∇U, E ≡ −∇ φ.
Gradient is perpendicular to the equipotential lines and it shows the direction of the strongest increase of the potential. (Thus
the force shows the direction of the strongest decrease of the potential. This is natural because all systems tend to decrease
their energy and the forces are developed accordingly to this). To prove this, consider an equipotential surface
U@x, y, zD = U 0
that goes through the point r 0 = 8 x 0 , y 0 , z 0 <. Expanding the potential around r 0 up to the first order, one obtains
U@x, y, zD = U 0 + Ux^ ′^ Hx − x 0 L + Uy^ ′^ Hy − y 0 L + Uz^ ′^ Hz − z 0 L.
Together with the preceding formula this yields the equation of the plane tangential to the equipotential surface at r 0
Ux^ ′^ Hx − x 0 L + Uy^ ′^ Hy − y 0 L + Uz^ ′^ Hz − z 0 L = 0,
where Vα^ ′^ ª∑a V. Any vector r within this plane can be represented in the form
r = r 0 + ρ,
where
ρ = e x Hx − x 0 L + e y Hy − y 0 L + e z Hz − z 0 L.
One can see that r is perpendicular to the gradient because
ρ ⋅ ∇U = Ie x Hx − x 0 L + e y Hy − y 0 L + e z Hz − z 0 L M ⋅ I e x Ux^ ′^ + e y Uy^ ′^ + e z Uz^ ′^ M
= Ux^ ′^ Hx − x 0 L + Uy^ ′^ Hy − y 0 L + Uz^ ′^ Hz − z 0 L = 0.
As an example consider the electric potential created by two charges put at (0,0,0) and (a,0,0). In the (x,y) plane (z=0) the
potential has the form
In[116]:= Q 1 =^ 1;^ Q 2 = −^ 1;^ a^ =^ 1;
V@x_, y_D =
Q 1
Ix^2 + y^2 M^1 ê^2
Q 2
IHx − aL^2 + y^2 M^1 ê^2
Plot3D@V@x, yD, 8 x, − 2, 2 + a<, 8 y, − 2, 2<, PlotRange → 8 − 5, 5<D
Out[118]=
In[103]:= EquipotentialLines^ =^ ContourPlot@V@x, yD,^8 x,^ −^ 2, 2^ +^ a<,^8 y,^ −^ 2.5, 2.5<, Contours^ →^20 D
Out[103]=
The electric field in the (x,y) plane is given by.
In[44]:= EE@x_, y_D = − Grad@V@x, yDD
Out[44]= :−
− 1 + x IH− 1 + xL^2 + y^2 M^3 ê^2
x Ix^2 + y^2 M^3 ê^2
y IH− 1 + xL^2 + y^2 M^3 ê^2
y Ix^2 + y^2 M^3 ê^2
Lines of the electric field are shown in the StreamPlot where we plot only x- and y-components
