The main definitions, formulas and conditions of the multivariable and vector calculus course on one page, unit by unit, with links to the lessons that explain each topic in full.
This sheet follows the six units of the course in order. Each unit starts with the big idea in a few sentences, then gives the definitions, formulas and theorems you will look up most often, with the conditions they need. When you want the full explanation, a proof or more examples, follow the lesson links under each topic.
Coordinates in space are right-handed and angles are in radians. Spherical coordinates are written (ρ,θ,ϕ), with the azimuth θ measured in the xy-plane from the positive x-axis and the polar angle ϕ measured from the positive z-axis. A formula holds wherever both sides are defined unless a condition says otherwise.
Vectors and the Geometry of Space
Multivariable calculus happens in space, so it starts with a precise way to describe points, directions and flat or curved surfaces. This unit sets up coordinates and vectors, the dot and cross products that measure angles, areas and volumes, the matrices that describe linear maps, and the equations of lines, planes, quadric surfaces and round coordinate systems.
Points and vectors
The axes are right-handed. Curling the fingers of the right hand from the positive x-axis toward the positive y-axis leaves the thumb along the positive z-axis. With Δx=x2−x1, Δy=y2−y1 and Δz=z2−z1, the distance between P1(x1,y1,z1) and P2(x2,y2,z2) is
d=(Δx)2+(Δy)2+(Δz)2,
and the sphere with center (h,k,l) and radius r>0 is
(x−h)2+(y−k)2+(z−l)2=r2.
An equation x2+y2+z2+Dx+Ey+Fz+G=0 becomes this form after completing the square in each variable, and it is a sphere, a single point or empty as the right side comes out positive, zero or negative. For example x2+y2+z2−4x+6y−2z=2 is the sphere with center (2,−3,1) and radius 4.
The vector from A(x1,y1,z1) to B(x2,y2,z2) is AB=⟨x2−x1,y2−y1,z2−z1⟩, and a vector a=⟨a1,a2,a3⟩ equals a1i+a2j+a3k in the standard basis.
Quantity
Formula
Conditions
Magnitude
∣a∣=a12+a22+a32
zero only for a=0
Sum
a+b=⟨a1+b1,a2+b2,a3+b3⟩
tail of b at the head of a
Scalar multiple
ca=⟨ca1,ca2,ca3⟩ with length ∣c∣∣a∣
reversed direction when c<0
Unit vector
∣a∣a
a=0
Parallel
b=ca for a scalar c
a and b nonzero
Forces add as vectors, and an object is in equilibrium when the resultant of all forces on it is 0.
For a=⟨a1,a2,a3⟩ and b=⟨b1,b2,b3⟩, and with θ in [0,π] the angle between them when both are nonzero,
a⋅b=a1b1+a2b2+a3b3=∣a∣∣b∣cosθ.
The product is commutative and distributive, and a⋅a=∣a∣2.
Quantity
Formula
Conditions
Angle
cosθ=∣a∣∣b∣a⋅b
a, b nonzero, 0≤θ≤π
Orthogonal
a⋅b=0
0 is orthogonal to every vector
Direction cosines
cosα=∣a∣a1, cosβ=∣a∣a2, cosγ=∣a∣a3
a=0, squares add to 1
Scalar projection
compab=∣a∣a⋅b
a=0, negative for an obtuse angle
Vector projection
projab=∣a∣2a⋅ba
a=0
Work
W=F⋅PQ
constant force, straight motion from P to Q
The remainder b−projab is orthogonal to a. Projecting b=⟨1,1,2⟩ onto a=⟨1,0,1⟩ gives ⟨23,0,23⟩, and the remainder has length 26, the distance from (1,1,2) to the line through the origin along a.
The Cauchy-Schwarz inequality∣a⋅b∣≤∣a∣∣b∣ holds for all vectors, with equality exactly when one vector is a scalar multiple of the other. It implies the triangle inequality ∣a+b∣≤∣a∣+∣b∣.
For vectors in space, the cross product is the vector with the components listed below as a column,
a×b=a2b3−a3b2a3b1−a1b3a1b2−a2b1.
It is perpendicular to both factors, b×a=−(a×b), it is distributive, and it is not associative. In right-handed coordinates i×j=k, j×k=i and k×i=j. For nonzero a and b with angle θ,
∣a×b∣=∣a∣∣b∣sinθ,
and when they are not parallel the product points where the right thumb points as the fingers curl from a toward b. Two nonzero vectors are parallel exactly when their cross product is 0.
Quantity
Formula
Conditions
Parallelogram area
∣a×b∣
adjacent sides a and b
Triangle area
21∣a×b∣
sides a and b from one vertex
Torque about O
τ=r×F
r from O to where F acts
Scalar triple product
a⋅(b×c)=det of the matrix with rows a, b, c
unchanged by a cyclic shift, sign change on a swap
Parallelepiped volume
∣a⋅(b×c)∣
edges a, b, c from one vertex
Tetrahedron volume
61∣a⋅(b×c)∣
the same three edges
The triple product is zero exactly when the three vectors are coplanar, positive when (a,b,c) is a right-handed triple and negative when it is left-handed. For the triangle with vertices P(1,0,0), Q(0,2,0) and R(0,0,3), the product PQ×PR=⟨6,3,2⟩ has length 7, so the area is 27 and the plane through the three points is 6x+3y+2z=6.
An m×n matrix A with columns a1,…,an acts on x in Rn by
Ax=x1a1+⋯+xnan.
Every linear map is multiplication by exactly one matrix, whose jth column is the image of ej. The product AB is defined when A has as many columns as B has rows, it applies B first and then A, so (AB)x=A(Bx), and in general AB=BA. A square matrix is invertible exactly when detA=0, and then Ax=b has the single solution x=A−1b. For ad−bc=0,
The rules det(AB)=detAdetB and detAT=detA hold for square matrices of one size. Swapping two rows changes the sign, scaling one row by k scales the determinant by k, and adding a multiple of one row to another changes nothing. The same holds for columns.
Area and volume. A 2×2 matrix multiplies every area by ∣detA∣, and a 3×3 matrix multiplies every volume by ∣detA∣.
Orientation. An invertible map preserves orientation when detA>0 and reverses it when detA<0. An ordered basis (v1,v2,v3) is right-handed when v1⋅(v2×v3)>0.
Change of basis. If the columns of an invertible P form a basis B, the coordinates of x are [x]B=P−1x, and the map with standard matrix A has matrix P−1AP in those coordinates.
For example A with rows ⟨2,1⟩ and ⟨1,2⟩ and P with columns ⟨1,1⟩ and ⟨1,−1⟩ give P−1AP=diag(3,1), so the map stretches by 3 along ⟨1,1⟩ and fixes ⟨1,−1⟩.
The line through the point with position vector r0=⟨x0,y0,z0⟩ with direction v=⟨a,b,c⟩=0 is
r(t)=r0+tv,t∈R.
When a, b and c are all nonzero it has the symmetric form
ax−x0=by−y0=cz−z0.
A zero component drops its fraction, so b=0 with a,c=0 gives y=y0 together with ax−x0=cz−z0. The plane through r0 with normal n=⟨a,b,c⟩=0 is
n⋅(r−r0)=0,that isax+by+cz=d
with d=n⋅r0. Three points P, Q, R not on one line lie in exactly one plane, with normal PQ×PR.
Question
Formula
Conditions
Angle between lines
cosθ=∣v1∣∣v2∣∣v1⋅v2∣
0≤θ≤2π, the same with normals for planes
Angle between a line and a plane
sinφ=∣v∣∣n∣∣v⋅n∣
0≤φ≤2π
Line meets the plane n⋅r=d
t=n⋅vd−n⋅r0
n⋅v=0, otherwise parallel
Two planes meet in a line
direction n1×n2
normals not parallel
Point to plane
D=a2+b2+c2∣ax1+by1+cz1−d∣
point (x1,y1,z1)
Skew lines
D=∣v1×v2∣∣(r2−r1)⋅(v1×v2)∣
v1×v2=0
Two lines are skew when they are not parallel and do not meet, which happens exactly when v1×v2=0 and (r2−r1)⋅(v1×v2)=0. For example the point (1,2,3) lies at distance 3∣2−2+6−1∣=35 from the plane 2x−y+2z=1.
A trace is the intersection of a surface with a plane, found by fixing one coordinate. An equation that does not contain z is a cylinder with rulings parallel to the z-axis, and the same holds for the other variables. With a,b,c>0, the standard quadric surfaces with axis along the z-axis are
Surface
Equation
Traces in z=k
Ellipsoid
a2x2+b2y2+c2z2=1
ellipses for ∣k∣<c, points at k=±c
Elliptic paraboloid
cz=a2x2+b2y2
ellipses where ck>0, here with any c=0
Hyperbolic paraboloid
cz=a2x2−b2y2
hyperbolas for k=0, crossing lines for k=0, here with any c=0
Elliptic cone
a2x2+b2y2=c2z2
ellipses for k=0, the vertex for k=0
Hyperboloid of one sheet
a2x2+b2y2−c2z2=1
ellipses for every k
Hyperboloid of two sheets
c2z2−a2x2−b2y2=1
ellipses for ∣k∣>c, empty for ∣k∣<c
To identify an equation without cross terms, first look for a missing variable, which means a cylinder. Then complete the square in every variable that appears both squared and to the first power. With all three variables squared and the right side scaled to 1, no minus sign on the left gives an ellipsoid, one gives a hyperboloid of one sheet and two give a hyperboloid of two sheets, and the axis belongs to the variable whose sign differs. With right side 0 and mixed signs it is a cone. Two squared variables and one to the first power give a paraboloid, elliptic for equal signs and hyperbolic for opposite signs. For example x2−2x+y2−z2=3 becomes (x−1)2+y2−z2=4, a hyperboloid of one sheet whose axis is the line x=1, y=0.
Cylindrical coordinates (r,θ,z) use polar coordinates in the xy-plane and keep the height,
x=rcosθ,y=rsinθ,z=z,
with r≥0 the distance to the z-axis. Spherical coordinates (ρ,θ,ϕ) use the distance ρ to the origin, the same azimuth θ, and the angle ϕ from the positive z-axis,
xyz=ρsinϕcosθ,=ρsinϕsinθ,=ρcosϕ,
with ρ≥0, 0≤ϕ≤π and usually 0≤θ<2π. The two systems are linked by r=ρsinϕ and z=ρcosϕ. Many physics books swap the letters θ and ϕ, so check which angle a source means.
Conversion
Formula
Conditions
r and ρ
r=x2+y2, ρ=x2+y2+z2
always
Azimuth
cosθ=rx and sinθ=ry
r>0
Polar angle
ϕ=arccosρz
ρ>0
The formula θ=arctanxy gives the right direction only for x>0, and a negative value then needs 2π added, so choose θ in the quadrant given by the signs of x and y. The point (−1,1,6) has r=2, θ=43π, ρ=22 and ϕ=6π, while arctan(−1)=−4π would point the wrong way.
Equation
Surface
r=c with c>0
cylinder of radius c about the z-axis
θ=β
half-plane with edge on the z-axis
ρ=c with c>0
sphere of radius c about the origin
ϕ=α with 0<α<2π or 2π<α<π
upper or lower nappe of a cone with vertex at the origin
ϕ=2π
the xy-plane
Off the z-axis the coordinates in these ranges are unique. On the z-axis the angle θ is arbitrary, and at the origin ϕ is arbitrary too.
A vector function traces a curve in space as its parameter runs, the way a moving object traces its path. Its calculus works one component at a time. From the derivatives of the position vector come tangent lines, the length of the curve, how sharply it bends and twists, and the velocity and acceleration of the motion.
Vector functions and their derivatives
A vector function r(t)=⟨f(t),g(t),h(t)⟩ has the intersection of the domains of its components as its natural domain. Limits, continuity, derivatives and integrals are all taken component by component, so
r′(t)=h→0limhr(t+h)−r(t)=⟨f′(t),g′(t),h′(t)⟩
exists exactly when all three component derivatives exist. A continuous r on an interval traces a space curve, oriented in the direction of increasing t. The circular helix
r(t)=⟨acost,asint,bt⟩
with a>0 and b=0 lies on the cylinder x2+y2=a2 and rises 2πb per turn. For b>0 it turns counterclockwise as seen from above while it rises.
A parametrization on an interval I is regular when it is C1 on I, with one-sided derivatives at endpoints that belong to I, and r′(t)=0 for every t in I. Where r′(t0)=0, the tangent line and the unit tangent are
L(s)=r(t0)+sr′(t0),T(t)=∣r′(t)∣r′(t).
For example r(t)=⟨cost,sint,t⟩ has r′(0)=⟨0,1,1⟩, so its tangent line at (1,0,0) is ⟨1,s,s⟩. For differentiable u, v and f,
with the order of the factors kept in the cross product rule. The chain rule reads dsdu(f(s))=f′(s)u′(f(s)). If r is differentiable and ∣r(t)∣ is constant on an interval, then r′(t) is perpendicular to r(t). Integrals also go component by component, and for r continuous on [a,b] with antiderivative R,
For r of class C1 on [a,b], the speed is ∣r′(t)∣ and the length is
L=∫ab∣r′(t)∣dt.
For a piecewise C1 parametrization, add the lengths of the pieces. The length counts multiplicity, so a curve traced twice has twice the length. The arc length function from a is s(t)=∫at∣r′(u)∣du, with dtds=∣r′(t)∣.
For a regular r the function s is strictly increasing with a C1 inverse, and R(s)=r(t(s)) has ∣R′(s)∣=1, the parametrization by arc length. A reparametrization r(φ(u)) with φ of class C1 and φ′=0 keeps a regular curve regular and leaves its length unchanged. It preserves the orientation when φ′>0 and reverses it when φ′<0, which replaces T by −T.
For example the helix ⟨3cost,3sint,4t⟩ has speed 5, length 10π over 0≤t≤2π, and arc length parametrization
Each object needs its own hypotheses, and each row adds its condition to the rows above it.
Object
Needs
T and the normal plane
C1 and r′=0
κ
also C2
N, B, osculating plane and circle
also κ=0
τ and the Frenet-Serret formulas
also C3
The curvature measures turning per unit length,
κ=dsdT=∣r′∣3∣r′×r′′∣.
For the graph y=f(x) of a C2 function this becomes κ=(1+f′2)3/2∣f′′∣. Where κ=0, the principal unit normal and the binormal are
N=∣T′∣T′,B=T×N=∣r′×r′′∣r′×r′′.
The frame T,N,B is right-handed and orthonormal. The osculating plane is spanned by T and N, and the osculating circle lies in it with radius κ1 and center r+κ1N. The torsion is the signed number
τ=∣r′×r′′∣2(r′×r′′)⋅r′′′,
and with this sign a helix that turns counterclockwise as it rises in right-handed coordinates has τ>0. The frame moves by the Frenet-Serret formulas
dsdTdsdNdsdB=κN,=−κT+τB,=−τN.
A curve with κ=0 on an interval lies in one plane exactly when τ=0 there. The helix ⟨acost,asint,bt⟩ has
For a position r(t) the velocity is v=r′, the speed is v=∣v∣ and the acceleration is a=r′′. For r of class C2 with v=0,
a=aTT+aNN,aT=dtdv=∣v∣v⋅a,aN=κv2=∣v∣∣v×a∣,
where the N term appears only when κ=0. Also aT2+aN2=∣a∣2, and the acceleration has no binormal part.
A projectile under gravity alone, with no air resistance, launched from height h≥0 at speed v0 and angle α above the horizontal, has
x(t)y(t)=v0tcosα,=h+v0tsinα−21gt2
until it lands. For h=0 and 0<α<2π, the flight time, range and maximum height are
Tf=g2v0sinα,D=gv02sin2α,H=2g(v0sinα)2.
With a launch speed of 20 meters per second, α=6π and g=9.8 meters per second squared, these are about 2.04 seconds, 35.3 meters and 5.10 meters.
For a particle of constant mass m the angular momentum L=r×mv has derivative r×F. A central force, parallel to r, keeps L constant, and if L=0 the motion stays in one plane and sweeps area at the rate 2m∣L∣. For a planet moving only under the inverse-square gravity of a sun of mass M fixed at the origin, with L=0, Kepler’s laws hold. The orbit is a conic with a focus at the sun, an ellipse when bounded, equal areas are swept in equal times, and an elliptical orbit with semimajor axis a has period T with
A function of several variables can change in infinitely many directions at once. This unit defines limits and derivatives that account for every direction, organizes the derivative as a gradient or a Jacobian matrix, and uses it to approximate functions, to solve equations locally and to find largest and smallest values.
Functions, domains and level sets
When only a formula is given, the domain is the largest set where it gives a real number. The graph of f(x,y) is the surface z=f(x,y), the level curve at level c is the set where f(x,y)=c, and for three variables the level surface is the set where F(x,y,z)=k. A contour map draws level curves for equally spaced values. For example f(x,y)=9−x2−y2 has the closed disk x2+y2≤9 as its domain, the range [0,3], and circles centered at the origin as level curves for 0≤c<3, while the level curve for c=3 is the origin alone.
A point P is an interior point of a set S when some open ball about P lies in S, and a boundary point when every open ball about P meets both S and its complement. The set is open when all its points are interior, closed when it contains all its boundary points, bounded when it lies in some ball, and compact when it is closed and bounded. Many sets are neither open nor closed.
Let (a,b) be a limit point of the domain D of f. Then f(x,y)→L as (x,y)→(a,b) when for every ε>0 there is δ>0 with ∣f(x,y)−L∣<ε for every (x,y) in D with
0<(x−a)2+(y−b)2<δ.
Paths can disprove a limit. Two paths in D toward (a,b) with different limits show that the limit does not exist. Along y=mx the function x2+y2xy tends to 1+m2m, which depends on m.
Paths cannot prove a limit. The function x4+y2x2y tends to 0 along every line through the origin, but equals 21 along y=x2, so it has no limit at the origin.
Squeeze by distance. If ∣f(x,y)−L∣≤g(x,y) near (a,b) and g→0, then f→L. Since x2≤x2+y2, the bound x2+y2x2y≤∣y∣≤r shows that this function tends to 0.
The function f is continuous at a point (a,b) of its domain when the limit there equals f(a,b), with only points of D tested. Sums, products, quotients with nonzero denominator and compositions of continuous functions are continuous, so polynomials are continuous everywhere and rational functions on their domains.
They are the slopes of the traces in the planes y=b and x=a. In the subscript notation the order of differentiation reads from left to right, and in the ∂ notation from right to left,
fxy=(fx)y=∂y∂x∂2f.
Clairaut’s theorem. If fxy and fyx exist and are continuous on an open disk containing (a,b), then fxy(a,b)=fyx(a,b). More generally, for f of class Ck on an open set, every partial derivative of order at most k is independent of the order of differentiation. For example f=x3y2+exy has fx=3x2y2+yexy and fxy=fyx=6x2y+(1+xy)exy. The existence of both partial derivatives at a point does not even make f continuous there.
The classical partial differential equations, with constants α=0 and c=0, are Laplace’s equation uxx+uyy=0, the heat equation ut=α2uxx and the wave equation utt=c2uxx.
Let U be open in Rn, f map U to Rm and a lie in U. The function is differentiable at a when there is a linear map L, the total derivative Df(a), with
h→0lim∥h∥∥f(a+h)−f(a)−L(h)∥=0.
Then every partial derivative exists at a, Df(a) is multiplication by the m×n Jacobian matrix [∂xj∂fi(a)], with row i for the component fi, and f is continuous at a. The converse fails. Partial derivatives can exist where f is not differentiable, but if all partial derivatives exist near a and are continuous at a, then f is differentiable at a. In particular every C1 function on an open set is differentiable there.
For f(x,y) differentiable at (a,b), the linearization, whose graph is the tangent plane, is
L(x,y)=f(a,b)+fx(a,b)(x−a)+fy(a,b)(y−b),
with normal vector ⟨fx(a,b),fy(a,b),−1⟩. The total differential dz=fxdx+fydy approximates the actual change Δz with an error that is small compared to dx2+dy2. For example f=x2+3xy at (1,2) has f=7, fx=8 and fy=3, so f(1.1,1.9)≈7+0.8−0.3=7.5, against the exact value 7.48. The function equal to x2+y2xy away from the origin and 0 at the origin has fx(0,0)=fy(0,0)=0 but is not continuous there, so it has no tangent plane at the origin.
with the partial derivatives of f evaluated at the point (x,y) that the inner functions reach. In a dependency diagram, add over every route from the top variable down to the independent variable the product of the derivatives along the route. Differentiability of f is needed, since for the function x2+y2xy above and g(t)=⟨t,t⟩ the partial derivatives would predict 0 while f(g(t))=21 for t=0. For T(x,y,t) along a path, the total derivative
dtdT=∂t∂T+∂x∂Tdtdx+∂y∂Tdtdy
differs in general from the partial derivative ∂t∂T. For example z=x2y with x=cost and y=sint gives dtdz=−2costsin2t+cos3t.
If f is differentiable at a, then Duf(a)=∇f(a)⋅u for every unit u. Without differentiability this formula can fail even when every directional derivative exists. When ∇f(a)=0,
the largest rate of increase is ∥∇f(a)∥, in the direction of ∇f(a),
the smallest is −∥∇f(a)∥, in the opposite direction,
the rate is 0 exactly in the directions perpendicular to ∇f(a).
If f is C1 on an open set and ∇f(a)=0, the level set through a is a smooth curve or surface near a with normal ∇f(a), and its tangent line or plane is ∇f(a)⋅(x−a)=0. At a critical point, where ∇f(a)=0, the gradient gives no direction and no normal. For example f=x2y at (1,2) has ∇f=⟨4,1⟩, so in the direction of ⟨3,4⟩ the rate is 512+4=516, and the largest rate is 17.
Implicit function theorem. Let F be C1 on an open set W in Rn+m with values in Rm, and let F(a,b)=0. If the m×m matrix DyF of partial derivatives in y1,…,ym has detDyF(a,b)=0, then near (a,b) the solutions of F(x,y)=0 are exactly the points y=g(x) for a unique C1 function g with g(a)=b, and
Dg(x)=−[DyF]−1DxF,
with both matrices evaluated at (x,g(x)). For one equation F(x,y,z)=0 solved for z this gives
∂x∂z=−FzFx,∂y∂z=−FzFy,Fz=0.
If F is C1 and ∇F(p)=0, the level surface F=c through p=(x0,y0,z0) has the tangent plane
Fx(p)(x−x0)+Fy(p)(y−y0)+Fz(p)(z−z0)=0.
For the sphere x2+y2+z2=9 at (1,2,2) the partial derivatives of z are −21 and −1, and the tangent plane is x+2y+2z=9.
Inverse function theorem. If f maps an open set of Rn to Rn, is C1, and has detDf(a)=0, then f is one-to-one from a neighborhood of a onto a neighborhood of f(a), with a C1 inverse and
D(f−1)(f(a))=[Df(a)]−1.
Both theorems are local. The map (u,v)↦(eucosv,eusinv) has Jacobian determinant e2u=0 everywhere, yet it is not one-to-one, since v and v+2π give the same point.
The Hessian of f at a is the matrix Hf(a)=[fxixj(a)], symmetric when f is C2. For f of class C2 on an open set containing a and h=x−a, the second-order Taylor polynomial is
T2(x)=f(a)+∇f(a)⋅h+21hTHf(a)h.
In two variables the quadratic part is 21(fxxh12+2fxyh1h2+fyyh22), with all derivatives at a. For example excosy at the origin has T2=1+x+21(x2−y2). In multi-index notation, for f of class Ck,
Tk(a+h)=∣α∣≤k∑α!Dαf(a)hα.
If f is Ck+1 on an open set containing the segment from a to a+h, the remainder Rk(h)=f(a+h)−Tk(a+h) is the sum of the terms of order k+1 with every derivative taken at a+θh, for some 0<θ<1. If also every partial derivative of order k+1 is at most M in size on that segment, then
A critical point is an interior point of the domain where ∇f=0 or some first partial derivative does not exist. Every local extremum at an interior point is a critical point, but not every critical point is an extremum. A saddle point is a critical point that is neither a local maximum nor a local minimum.
Second derivative test. Let f be C2 on an open set containing a with ∇f(a)=0. A positive definite Hessian gives a strict local minimum, a negative definite one a strict local maximum, and an indefinite one a saddle point. In two variables, with
D=fxx(a)fyy(a)−fxy(a)2,
Condition
Conclusion
D>0 and fxx(a)>0
strict local minimum
D>0 and fxx(a)<0
strict local maximum
D<0
saddle point
D=0
no information
For example f=x3−3x+y2 has critical points (1,0), with D=12 and fxx=6, a local minimum, and (−1,0), with D=−12, a saddle. When D=0, as for both x4+y4 and x4−y4 at the origin, classify by other means, such as the sign of f(x)−f(a) or restrictions to lines and curves. In three or more variables a degenerate Hessian that is still indefinite gives a saddle, while a semidefinite one gives no conclusion.
Extreme value theorem. A function continuous on a nonempty closed and bounded set takes an absolute maximum and an absolute minimum there. The closed region method finds them for a region whose boundary is made of finitely many smooth curves.
Evaluate f at every critical point inside the region.
On each boundary curve ri(t) with αi≤t≤βi, find the extreme values of gi(t)=f(ri(t)), including the endpoints.
Evaluate f at every corner.
The largest value is the absolute maximum and the smallest is the absolute minimum. For f=x2+y2−2x on the disk x2+y2≤4, the only interior critical point gives f(1,0)=−1, the boundary gives 4−4cost between 0 and 8, so the maximum is 8 at (−2,0) and the minimum is −1 at (1,0). On a set that is not closed or not bounded an extremum may fail to exist and needs its own argument. For example a continuous f on a closed nonempty set with f→∞ as x2+y2→∞ has an absolute minimum.
Let f and g be C1 on an open set in Rn. If f restricted to g=c has a local extremum at a and ∇g(a)=0, then for some number λ
∇f(a)=λ∇g(a).
With m<n constraints g1=c1,…,gm=cm whose gradients are linearly independent at a, the condition becomes
∇f(a)=λ1∇g1(a)+⋯+λm∇gm(a).
The conditions are necessary, not sufficient. Collect every candidate, meaning the Lagrange points, the singular points of the constraint where the gradients are dependent, the points where f or g is not C1, and any interior critical points, corners and endpoints when there are inequality constraints. On a closed and bounded feasible set, comparing the values of f at the candidates finds the absolute extrema, and otherwise existence needs a separate argument. For example f=x+2y on the circle x2+y2=5 gives 1=2λx and 2=2λy, so the candidates are (1,2) and (−1,−2), where the maximum 5 and the minimum −5 occur.
A multiple integral adds up a quantity spread over a region of the plane or a solid in space, as a limit of sums of small pieces. Fubini’s theorem turns it into repeated single integrals, and a good choice of coordinates, whether polar, cylindrical, spherical or a custom change of variables, makes the bounds simple.
Double integrals and Fubini’s theorem
On a rectangle R the double integral is the limit of Riemann sums ∑f(xij∗,yij∗)ΔAij as the largest subrectangle shrinks. A bounded function whose discontinuities lie on finitely many smooth curves is integrable, and so is every continuous function. When f≥0 the integral is the volume under the graph, and in general a signed volume. The average value over a region of positive area A(D) is
favg=A(D)1∬DfdA.
Fubini’s theorem. If f is continuous on R=[a,b]×[c,d], then
∬RfdA=∫ab∫cdf(x,y)dydx=∫cd∫abf(x,y)dxdy.
A region of type I is a≤x≤b, g1(x)≤y≤g2(x), and a region of type II is c≤y≤d, h1(y)≤x≤h2(y), with continuous boundary functions in order. For f continuous on them,
type Itype II∫ab∫g1(x)g2(x)fdydx,∫cd∫h1(y)h2(y)fdxdy.
To reverse the order, sketch the region from the given bounds and describe it again with the other variable outside, so the new outer bounds are constants. For example
∫01∫x1ey2dydx=∫01∫0yey2dxdy=2e−1.
Integrals add over regions that overlap only along their boundaries.
With x=rcosθ, y=rsinθ and r≥0, the area element is dA=rdrdθ. For a region α≤θ≤β, h1(θ)≤r≤h2(θ), with 0≤β−α≤2π and continuous 0≤h1≤h2, and f continuous on it,
∬DfdA=∫αβ∫h1(θ)h2(θ)frdrdθ,
with f evaluated at (rcosθ,rsinθ). The bounds must cover each point once, so the angle interval has length at most 2π and a curve that retraces itself is used over one tracing only. For example the circle x2+y2=2y is r=2sinθ for 0≤θ≤π, and
For a lamina on D with continuous density σ(x,y)≥0,
Quantity
Formula
Conditions
Area
A=∬D1dA
boundary of zero area
Mass
m=∬DσdA
Moments
Mx=∬DyσdA and My=∬DxσdA
Center of mass
xˉ=mMy and yˉ=mMx
m>0, the centroid when σ is constant
Moments of inertia
Ix=∬Dy2σdA, Iy=∬Dx2σdA, I0=Ix+Iy
Radius of gyration
R=mI
m>0, about the same axis as I
For example the half disk x2+y2≤a2, y≥0, has its centroid at (0,3π4a).
A joint density f≥0 of (X,Y) has total integral 1 over the plane, and P((X,Y)∈D)=∬DfdA. The marginal density of X is fX(x)=∫−∞∞f(x,y)dy, and X and Y are independent exactly when f(x,y)=fX(x)fY(y) outside a set that does not affect any double integral. Expected values are E[g(X,Y)]=∬gfdA when ∬∣g∣fdA is finite, and when E[X2] and E[Y2] are finite
Cov(X,Y)=E[XY]−E[X]E[Y].
Independent variables have covariance 0, but covariance 0 does not imply independence. For the density x+y on the unit square, E[X]=E[Y]=127, E[XY]=31 and the covariance is −1441.
Triple integrals are limits of Riemann sums over boxes, and a continuous function on a box can be integrated in any of the six orders. A solid of type 1 lies over a region D of the xy-plane between continuous surfaces u1≤u2, and for f continuous on it
∭EfdV=∬D[∫u1(x,y)u2(x,y)fdz]dA.
The volume is ∭E1dV, the mass is m=∭EδdV for a continuous density δ≥0, and for m>0 the center of mass has coordinates m1∭ExδdV and so on for y and z. The moment of inertia about the z-axis is Iz=∭E(x2+y2)δdV. Round solids are simpler in other coordinates.
Coordinates
Point
Volume element
Rectangular
(x,y,z)
dV=dxdydz
Cylindrical
(rcosθ,rsinθ,z)
dV=rdzdrdθ
Spherical
(ρsinϕcosθ,ρsinϕsinθ,ρcosϕ)
dV=ρ2sinϕdρdϕdθ
The bounds need r≥0, ρ≥0, 0≤ϕ≤π and an angle interval for θ of length at most 2π, and each point of the solid must be covered once. Useful descriptions are
the cone ϕ=c with 0<c<2π, which opens upward around the positive z-axis,
the plane z=h with h>0, which is ρ=hsecϕ for 0≤ϕ<2π,
the ball x2+y2+(z−a)2≤a2 with a>0, which is 0≤ρ≤2acosϕ for 0≤ϕ≤2π.
For example the solid between the paraboloid z=x2+y2 and the plane z=4 has volume
∫02π∫02(4−r2)rdrdθ=8π,
and the part of the ball ρ≤2 inside the cone ϕ≤3π has volume
∫02π∫0π/3∫02ρ2sinϕdρdϕdθ=38π.
A solid ball of radius a and density 1 has Iz=158πa5. Symmetry helps too. An integrand with f(−x,y,z)=−f(x,y,z) over a solid unchanged by x↦−x integrates to 0.
and its absolute value is the local scaling factor for area, while its sign only records orientation. Where it is nonzero, the map has a local C1 inverse whose Jacobian is the reciprocal.
Change of variables theorem. Let S be a closed bounded region with piecewise smooth boundary, T a C1 map on an open set containing S that is one-to-one with nonzero Jacobian on the interior of S, and R=T(S). For f continuous on R,
∬Rf(x,y)dA=∬Sf(T(u,v))∂(u,v)∂(x,y)dudv.
Failures of these conditions on the boundary of S do no harm. In space the same holds with dV=∂(u,v,w)∂(x,y,z)dudvdw, which gives the factors r and ρ2sinϕ of cylindrical and spherical coordinates. For example the region with 0≤x−y≤2 and 0≤x+y≤4 becomes a rectangle under u=x−y and v=x+y, with ∂(u,v)∂(x,y)=21, so
Over an unbounded region, or near points where f is unbounded, the integral is the limit over a regular exhaustion, closed bounded regions K1⊆K2⊆⋯ inside D with piecewise smooth boundaries that eventually contain every closed bounded subset of D. For a continuous f≥0 the limit lies in [0,∞] and is the same for every exhaustion. A continuous f of any sign is absolutely integrable when ∬D∣f∣dA is finite.
Tonelli. For continuous f≥0, the double integral and both iterated improper integrals are equal, all finite or all infinite.
Fubini. For continuous f of any sign with ∬D∣f∣dA<∞, both iterated integrals converge and equal the double integral. Check this with Tonelli applied to ∣f∣.
The course states both with inner integrals that are finite and continuous in the outer variable. Without absolute integrability the two orders can disagree. For (x2+y2)2x2−y2 on (0,1]×(0,1] the order dydx gives 4π and the order dxdy gives −4π. The Gaussian integral follows from Tonelli and polar coordinates,
Vector calculus integrates fields along curves and across surfaces, then ties those integrals to derivatives inside the region through the theorems of Green, Stokes and Gauss. Each theorem needs its orientation and its smoothness and domain conditions, and the formulas can fail when one of them is dropped.
Vector fields and potentials
A vector field assigns a vector F=⟨P,Q,R⟩ to each point of its domain, and it is continuous or C1 when its components are. A velocity field moves particles along paths with r′(t)=F(r(t)), which are its field lines. For a C1 field exactly one maximal field line passes through each point, so different field lines never cross. The gravitational field of a mass at the origin is
F(x)=−∣x∣3kx,x=0,
with k=GMm>0, an inverse-square law. A field F is conservative on an open set D when F=∇f there for a potential function f. On a connected open set two potentials differ by a constant, and physicists often use the potential energy V=−f. Every statement about potentials refers to a specific domain, and a singular point, where the field is undefined, can change the answer.
A curve is regular when it has a regular parametrization r(t), a≤t≤b, that traces it once, and piecewise regular when it is a chain of finitely many regular curves joined end to end. For f and F continuous on C,
For the vector integral the parametrization must move in the chosen direction of C. It equals ∫CF⋅Tds and, in coordinates, ∫CPdx+Qdy+Rdz. Over a piecewise regular curve, add the pieces.
Property
Scalar ∫Cfds
Vector ∫CF⋅dr
Another parametrization tracing C once
same value
same value if the direction is kept
Reversed direction
same value
changes sign
Meaning
mass of a wire with density f, length when f=1
work done by a force field, circulation when C is closed
A wire with continuous density δ≥0 has mass m=∫Cδds and, when m>0, center of mass with coordinates m1∫Cxδds and so on. Along the helix ⟨cost,sint,t⟩ for 0≤t≤2π, with speed 2,
∫Czds=∫02πt2dt=22π2,∫C⟨−y,x,1⟩⋅dr=4π.
A simple closed curve in the plane is positively oriented when it runs counterclockwise, with the enclosed region on the left.
Fundamental theorem for line integrals. If f is C1 on an open set D and C is an oriented piecewise regular curve in D from A to B, then
∫C∇f⋅dr=f(B)−f(A).
For F continuous on a connected open set D, three statements are equivalent. The field is conservative on D, its line integrals are path independent in D, and its integral around every closed piecewise regular curve in D is 0.
Cross-partial test. If F is C1 and conservative on an open set, then Py=Qx in the plane, and in space
Ry=Qz,Pz=Rx,Qx=Py,
that is ∇×F=0. The converse holds when the open set is simply connected, which means connected with every closed curve shrinkable to a point inside it. In the plane that means no holes. Space with one point removed is simply connected, while space with the z-axis removed is not. The test alone is not enough on other sets. The field
F=⟨x2+y2−y,x2+y2x⟩
has Py=Qx on the plane without the origin, yet its circulation counterclockwise around the unit circle is 2π, so it is not conservative there.
To find a potential on a simply connected set, integrate P in x with a constant g(y,z), match fy with Q to find g up to h(z), match fz with R, and check that ∇f=F. For example F=⟨2xy+z,x2,x⟩ has the potential f=x2y+xz, so its integral along any path from the origin to (1,2,3) is f(1,2,3)−f(0,0,0)=5.
A plane field ⟨P,Q⟩ has the scalar curl Qx−Py. For a C1 field the divergence is the outward flux per unit volume through small spheres about a point, and the component (∇×F)⋅u is the circulation per unit area around small circles perpendicular to the unit vector u, counterclockwise as seen from its tip. The Laplacian is ∇2f=∇⋅∇f=fxx+fyy+fzz, and a C2 function with ∇2f=0 on an open set is harmonic there.
Identity
Conditions
∇×(∇f)=0
f of class C2 on an open set
∇⋅(∇×F)=0
F of class C2 on an open set
Both rest on Clairaut’s theorem and say nothing at points where the field is undefined. For example F=⟨xy,yz,zx⟩ has divergence x+y+z and curl ⟨−y,−z,−x⟩.
Let C be a positively oriented, piecewise smooth, simple closed curve in the plane, D the closed region made of C and its inside, and P and Q of class C1 on an open set containing D. Then
∮CPdx+Qdy=∬D(∂x∂Q−∂y∂P)dA.
With the outward unit normal n, the flux form is
∮CF⋅nds=∮CPdy−Qdx=∬D(Px+Qy)dA.
A single point of D where P or Q is undefined breaks the hypothesis. For a region with holes the boundary runs with D on the left, so the outer curve goes counterclockwise and each inner curve clockwise, and the field may be undefined inside the holes. Taking Qx−Py=1 gives the area
area(D)=∮Cxdy=−∮Cydx=21∮Cxdy−ydx,
which gives πab for the ellipse x=acost, y=bsint with a,b>0. Around the unit square counterclockwise, ∮C(x2−y)dx+(x+y2)dy=∬D2dA=2.
A parametrization r(u,v) over a closed bounded region D with piecewise smooth boundary is regular when r is C1 on an open set containing D, one-to-one on the interior of D, and
ru×rv=0
on the interior of D. Seams and edges that collapse to a point may sit on the boundary of D. The vectors ru and rv are tangent to the grid curves, ru×rv is normal to the tangent plane, and the surface area is
A(S)=∬D∣ru×rv∣dA.
For the graph z=g(x,y) of a C1 function, rx×ry=⟨−gx,−gy,1⟩ and
A=∬D1+gx2+gy2dA.
The sphere of radius a has the parametrization ⟨asinϕcosθ,asinϕsinθ,acosϕ⟩ over 0≤ϕ≤π, 0≤θ≤2π, with rϕ×rθ pointing outward and of length a2sinϕ, so its area is 4πa2.
For a regular parametrization and g continuous on S,
∬SgdS=∬Dg(r)∣ru×rv∣dudv,
which does not depend on the regular parametrization chosen. With g=1 this is the area, and with a density g=δ it is the mass of a thin sheet. An orientation is a continuous choice of unit normal n. A connected orientable surface has exactly two, and a Möbius band has none, so flux through it is not defined. The flux of a continuous F across an oriented surface is
∬SF⋅dS=∬SF⋅ndS=∬DF(r)⋅(ru×rv)dudv,
where the last form needs ru×rv to point the same way as n. The graph z=f(x,y) has the upward unit normal
n=1+fx2+fy2⟨−fx,−fy,1⟩.
Reversing the orientation changes the sign of the flux, while scalar surface integrals do not change. A closed surface, the boundary of a bounded solid, is oriented by the outward normal unless stated otherwise. For example the outward flux of ⟨x,y,z⟩ through the sphere of radius a is 4πa3, and the upward flux of ⟨0,0,z⟩ through z=1−x2−y2 over the unit disk is 2π.
Let S be a bounded oriented piecewise smooth surface whose boundary ∂S is made of finitely many simple closed piecewise smooth curves with the induced orientation, and let F be C1 on an open set containing S. Then
∮∂SF⋅dr=∬S(∇×F)⋅ndS.
The induced orientation is the one in which you walk along ∂S with your head along n and the surface on your left, so n×T points into S. Reversing only the normal or only the direction breaks the formula. With several boundary curves, add the circulations around all of them. Two oriented surfaces with the same boundary and the same induced orientation have the same curl flux whenever F is C1 on an open set containing both, so a flat surface often replaces a curved one. For F=⟨−y,x,0⟩, whose curl is ⟨0,0,2⟩, the upper unit hemisphere with upward normal and the unit disk both give 2π, the circulation counterclockwise around the unit circle seen from above.
Let E be a bounded solid whose boundary ∂E is a finite union of closed piecewise smooth surfaces with the outward unit normal n, and let F be C1 on an open set containing E and its boundary. Then
∬∂EF⋅ndS=∭E∇⋅FdV.
Every piece of the boundary counts, including the inner surface around a cavity, whose outward normal points into the cavity. For F=⟨x3,y3,z3⟩ and the ball of radius a, the divergence is 3ρ2 and both sides equal 512πa5.
If F is C1 except at one interior point P0 and ∇⋅F=0 away from it, remove a small ball about P0. The flux through ∂E then equals the flux outward from P0 through any small sphere about it. The field ∣r∣3r has divergence 0 away from the origin and flux 4π through every sphere about the origin. In the same way Gauss’s law says that a point charge q at the origin, with field E=4πε0q∣r∣3r, has outward flux ε0q through a closed surface around it and 0 through one that leaves it outside.
All the main theorems integrate a derivative over a region and get the original object on the boundary.
This unit continues the differential equations of single-variable calculus. Exact equations reuse potentials from vector calculus, complex exponentials reduce linear equations with constant coefficients to algebra, and power series, the Laplace transform and matrices handle variable coefficients, switched inputs and systems of equations.
Exact equations
The equation Mdx+Ndy=0, with M and N continuous on an open set D, is exact on D when it has a potential F there, a C1 function with Fx=M and Fy=N. Its solutions then lie on the level curves F(x,y)=C.
Test for exactness. Let M and N be C1 on D. Exactness implies My=Nx on D, and when D is open and simply connected, My=Nx on D implies exactness. On an open rectangle containing (x0,y0) a potential is
F(x,y)=∫x0xM(s,y0)ds+∫y0yN(x,s)ds.
For example (2xy+3)dx+(x2+4y)dy=0 has My=Nx=2x and the solutions x2y+3x+2y2=C.
Integrating factor
Works when
Conditions
μ(x)=e∫h(x)dx
My−Nx=h(x)N
h continuous
μ(y)=e∫k(y)dy
Nx−My=k(y)M
k continuous
Both need D open and simply connected and M, N of class C1. A factor that is never zero keeps the same solutions, while one that is zero or undefined somewhere can add or lose solutions there. For example (3xy+y2)dx+(x2+xy)dy=0 has h(x)=x1, and μ=x makes it exact with potential x3y+21x2y2 on a region where x=0.
A complex number z=a+bi with i2=−1 has real part a, imaginary part b, conjugate z=a−bi and modulus ∣z∣=a2+b2=zz. To divide, multiply the top and bottom by the conjugate of the denominator, so 1−2i3+4i=−1+2i. Every z=0 has polar form r(cosθ+isinθ) with r=∣z∣ and an argument θ, unique up to multiples of 2π. Moduli multiply and arguments add, and for a positive integer n
zn=rn(cosnθ+isinnθ).
A number R(cosφ+isinφ) with R>0 has exactly n distinct nth roots,
R1/n(cosnφ+2πk+isinnφ+2πk),k=0,1,…,n−1.
For example the cube roots of −8 are 1+i3, −2 and 1−i3. The exponential series gives Euler’s formula for real θ and the complex exponential for real a and b,
eiθ=cosθ+isinθ,ea+bi=ea(cosb+isinb).
So ez is never 0, ez+w=ezew, and dtdert=rert for a complex constant r. If p and q are real, the real and imaginary parts of a complex solution of y′′+py′+qy=0 are real solutions.
On an interval where P(t)=0, the equation Py′′+Qy′+Ry=G has the normalized form
y′′+p(t)y′+q(t)y=g(t).
If p, q and g are continuous on an open interval I containing t0, the initial value problem with y(t0)=y0 and y′(t0)=y1 has exactly one solution, and it exists on all of I. Solutions of the homogeneous equation can be added and scaled. Two solutions y1 and y2 form a fundamental set, and every solution is c1y1+c2y2, exactly when their Wronskian
W=y1y2′−y1′y2
is nonzero. By Abel’s formula W(t)=W(t0)e−∫t0tp(s)ds, so for two solutions W is zero everywhere on I or nowhere. For ay′′+by′+cy=0 with real constants and a=0, the roots of the characteristic equation ar2+br+c=0 decide the general solution.
Roots
General solution
real r1=r2, when b2−4ac>0
c1er1t+c2er2t
double root r=−2ab, when b2−4ac=0
(c1+c2t)ert
α±iβ with β>0, when b2−4ac<0
eαt(c1cosβt+c2sinβt)
For example y′′+2y′+5y=0 with y(0)=1 and y′(0)=1 has roots −1±2i and the solution y=e−t(cos2t+sin2t). A boundary value problem, with values at two different points, may have one solution, none or infinitely many.
With p, q and g continuous on I and a fundamental set y1, y2, every solution of y′′+py′+qy=g is
y=c1y1+c2y2+yp
for one particular solution yp and exactly one pair of constants.
Undetermined coefficients. For ay′′+by′+cy=g with constant coefficients and a term Pm(t)eαtcosβt or Pm(t)eαtsinβt, where Pm has degree m, try
yp=tseαt(A(t)cosβt+B(t)sinβt)
with polynomials A and B of degree m, both needed even when g has only a cosine or only a sine. The power s is the multiplicity of α+iβ as a root of the characteristic equation, 0 when it is not a root. For a sum of such terms, add the particular solutions. For example y′′−y=et has 1 as a simple root, so yp=Atet, which gives A=21.
Variation of parameters. For the normalized equation with any continuous g and W=y1y2′−y1′y2=0 on I,
yp=−y1∫Wy2gdt+y2∫Wy1gdt.
The leading coefficient must be 1, so for Py′′+Qy′+Ry=G the forcing is g=PG. For example y′′+y=sect on (−2π,2π) has yp=costln(cost)+tsint.
A mass m>0 on a spring with constant k>0, damping c≥0 and applied force F(t), with x measured from equilibrium, satisfies
mx′′+cx′+kx=F(t),ω0=mk.
Without damping or force the motion is x=Rcos(ω0t−δ) with period ω02π. With c>0 and no force every solution tends to 0, in one of three ways.
Case
Solution
Behavior
Underdamped, c2<4mk
e−ct/(2m)(c1cosμt+c2sinμt) with μ=2m4mk−c2
oscillates with quasi period μ2π
Critically damped, c2=4mk
(c1+c2t)e−ct/(2m)
crosses equilibrium at most once
Overdamped, c2>4mk
c1er1t+c2er2t with r2<r1<0
crosses equilibrium at most once
For a force F0cosωt without damping, ω=ω0 gives the bounded response m(ω02−ω2)F0cosωt, while ω=ω0 gives pure resonance,
xp=2mω0F0tsinω0t,
which grows without bound. With c>0 and F0>0 every solution stays bounded, and the steady periodic response A(ω)cos(ωt−γ) has amplitude
A(ω)=(k−mω2)2+c2ω2F0.
When c2<2mk the amplitude peaks at the practical resonance frequency
ωr=mk−2m2c2<ω0,
and when c2≥2mk it decreases for all ω>0. For example m=1, c=2 and k=10 give ωr=22 and the peak amplitude 6F0. A series circuit obeys Lq′′+Rq′+C1q=E(t), the same equation with L, R and C1 in place of m, c and k.
Write y′′+p(x)y′+q(x)y=0 with p=PQ and q=PR. A point x0 is ordinary when p and q are analytic there, and then every solution is a power series ∑an(x−x0)n that converges at least where the series of p and q both do. For polynomial coefficients with no common factor, that radius is at least the distance from x0 to the nearest complex zero of P. Substitute the series, shift indices so every sum shows the same power, and set each coefficient to 0. For example y′′=xy gives a2=0 and
an+2=(n+2)(n+1)an−1,n≥1.
A singular point x0 is regular when (x−x0)p and (x−x0)2q are analytic there. At a regular singular point 0, with xp(x)=∑pnxn and x2q(x)=∑qnxn, the Frobenius series xr∑anxn with a0=0 needs a root of the indicial equation
r(r−1)+p0r+q0=0.
For real roots r1≥r2 there is always a solution for r1. A second one is another Frobenius series when r1−r2 is not an integer, it contains y1lnx when r1=r2, and it may contain y1lnx when r1−r2 is a positive integer. The Bessel equation of order ν≥0,
x2y′′+xy′+(x2−ν2)y=0,
has indicial roots ±ν and the general solution c1Jν(x)+c2Yν(x) for x>0, where
For f defined for t≥0 and integrable on every [0,T],
F(s)=L{f}(s)=∫0∞e−stf(t)dt
for the real s where the improper integral converges. A function is of exponential order a when ∣f(t)∣≤Meat for all t≥T0. If f is piecewise continuous on every [0,T] and of exponential order a, the integral converges absolutely for every s>a and F(s)→0 as s→∞.
f(t)
F(s)
Valid for
1
s1
s>0
tn, with n=0,1,2,…
sn+1n!
s>0
eat
s−a1
s>a
sinbt
s2+b2b
s>0
cosbt
s2+b2s
s>0
eatf(t)
F(s−a)
s−a in the range for F
uc(t)f(t−c) with c≥0
e−csF(s)
the range for F
δ(t−c) with c>0
e−cs
all s
(f∗g)(t)
F(s)G(s)
both of exponential order a, s>a
Here uc is 0 for t<c and 1 for t≥c, and the convolution is (f∗g)(t)=∫0tf(τ)g(t−τ)dτ. The impulse δ(t−c) is not a function but the limit of pulses of area 1 on [c,c+ε), with ∫0∞δ(t−c)g(t)dt=g(c) for continuous g.
Derivatives. If f is continuous on [0,∞) and of exponential order a and f′ is piecewise continuous on every [0,T], then for s>a
L{f′}=sF(s)−f(0).
If f′ also satisfies these conditions, then
L{f′′}=s2F(s)−sf(0)−f′(0).
Two piecewise continuous functions of exponential order with the same Laplace transform agree wherever both are continuous, so the inverse Laplace transform is read off the table after partial fractions. For ay′′+by′+cy=g with y(0)=y0 and y′(0)=y1,
(as2+bs+c)Y(s)=G(s)+a(sy0+y1)+by0.
For example y′′+3y′+2y=2 with zero initial values gives
Y(s)=s(s+1)(s+2)2=s1−s+12+s+21,
so y=1−2e−t+e−2t. An impulse kδ(t−t0) with t0>0 leaves y continuous and makes y′ jump by ak at t0.
The system x′=A(t)x+g(t) with entries continuous on an open interval I has exactly one solution through each initial value x(t0)=x0, defined on all of I. For a constant A, an eigenvalue λ with eigenvector v=0, a root of det(A−λI)=0, gives the solution eλtv. For a 2×2 matrix with trace T and determinant D the characteristic equation is λ2−Tλ+D=0.
Diagonalizable. With n independent eigenvectors, which is the case for n distinct eigenvalues, the general solution is c1eλ1tv1+⋯+cneλntvn.
Complex pair. For real A and λ=α+iβ with β=0, the real and imaginary parts of eλtv are two independent real solutions.
Defective double eigenvalue. For a 2×2 matrix A=λI with one independent eigenvector v, choose w with (A−λI)w=v, and then eλtv and eλt(tv+w) form a fundamental set.
The matrix exponential eAt=∑k≥0k!(At)k converges for every A and gives the solution x(t)=eAtx0 of x′=Ax with x(0)=x0. The rule eA+B=eAeB holds when AB=BA and can fail otherwise. For a real 2×2 system with detA=0, the eigenvalues classify the origin.
Eigenvalues
Origin
real with opposite signs
saddle
real, distinct, same sign
node
real and equal with A=λI
proper node
real and equal with A=λI
degenerate node
α±iβ with α=0, β=0
spiral
±iβ with β=0
center
The origin is asymptotically stable when every eigenvalue has negative real part and unstable when one has positive real part. When every real part is at most 0 and some is 0, it is stable exactly when each eigenvalue with zero real part has as many independent eigenvectors as its multiplicity. For example the matrix with rows ⟨1,2⟩ and ⟨2,1⟩ has eigenvalues 3 and −1, with eigenvectors ⟨1,1⟩ and ⟨1,−1⟩, so the origin is a saddle and
For an autonomous system x′=f(x,y), y′=g(x,y) with f and g of class C1 on an open set, the equilibria are the points where f=g=0, the crossings of the nullclines. The linearization at an equilibrium uses the Jacobian
J=[fxgxfygy](x∗,y∗).
An equilibrium is hyperbolic when no eigenvalue of J has zero real part.
Hartman-Grobman theorem. If F=⟨f,g⟩ is C1 on an open set containing a hyperbolic equilibrium, a homeomorphism of a neighborhood of the equilibrium onto a neighborhood of 0 carries the trajectories of the nonlinear system onto those of the linearization and keeps the direction of time. So all eigenvalues with negative real part give an asymptotically stable sink, all with positive real part a source, and real eigenvalues of opposite signs an unstable saddle. If F is C2, real distinct eigenvalues of one sign give a node and α±iβ with α=0 and β=0 give a spiral. The conclusion is local, in a neighborhood whose size the theorem does not give.
For example x′=y, y′=−sinx−21y has a stable spiral at the origin, where the eigenvalues have real part −41, and a saddle at (π,0), where detJ=−1. A nonhyperbolic equilibrium needs other tools, such as a conserved quantity or a radial equation. The system x′=−y−x(x2+y2), y′=x−y(x2+y2) has a linear center at the origin, yet r′=−r3, so every trajectory spirals into it.
The Lotka-Volterra model with a,b,c,d>0,
x′y′=ax−bxy,=−cy+dxy,
has a saddle at (0,0) and a linear center at (dc,ba) with eigenvalues ±iac. In the open first quadrant the quantity
H=dx−clnx+by−alny
is constant along trajectories, so every trajectory there other than the equilibrium is a closed curve and the populations are periodic.
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