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Math 240 Summer 2019 Notes
Yao-Rui
Lecture notes might contain typos.
1 Lecture 1 – Introduction
1.1 Brief Overview of the Course
This course is an invitation to differential equations. What is a differential equation? Broadly speaking, it is an equation that relates functions to their derivatives. A lot of the differential equations that we study comes from physics, and our main techniques to understand these equations are linear algebra and multivariable calculus.
Ordinary Differential Equations
A standard topic in introductory physics is radioactive decay, where the associated differential equation is dN dt = −kN
for some constant k. The solution is well-known; it is
N (t) = e−ktN 0.
There are a lot more physical solutions of this form. For example, the change in density as height varies is given by dρ dy
gM RT
ρ.
All these differential equations are of first degree. One can ask for the solution to higher-degree equations. For example, Hooke’s Law for an isolated frictionless body can be described by
d^2 x dt^2 = −ω^2 x.
Here the solution is well-known too; it is
x(t) = A cos(ωt + φ 0 ).
A large part of the course is to study these kinds of equations, called ordinary differential equations (ODEs). We will start with ODEs having constant coefficients. In fact, we generalize and study the higher dimensional version of it, for it is unreasonable for us to only work in one dimension in physics. One obvious example is modeling the oscillations of a bridge. To do this we will need to apply techniques from linear algebra, most notably the concept of eigenvalues and eigenvectors. This will be our starting point after some warmup with one-dimensional ODEs. There are also numerous famous ODEs that are not linear. For example, vertical upward movement with resistance in liquids can be modeled by
dv dt = μg
c μgm v^2
and the solution to this equation is
v(t) =
μgm c
tanh
μgc m
t.
Another example is Newton’s Law of Gravitation:
x^ #′′(t) = − GM ‖ x#(t)‖^3
#x(t).
Yet another example is the predator-prey model, given by
dx dt
= αx − βxy, dy dt = δxy − γy.
We will study how to solve these kinds of ODEs, or quantitatively sketch solutions to them if an explicit solutions cannot be easily found (for example, how to analyze equilibrium points and linearize the ODEs near these points). In fact, understanding the mathematics behind Newton’s Law of Gravitation is the extra credit assignment; see subsection 21.11.
Example 1.1. Let us illustrate what we mean by “solving ODEs” and “sketching solutions quan- titatively” using the following system of differential equations
x′(t) = −y(t), y′(t) = x(t), x(t 0 ) = x 0 , y(t 0 ) = y 0.
We will develop techniques to solve these kinds of equations; right now this example is just to let you have a feel for the kind of stuff we will do. Let us solve this equation. Writing x#(t) = (x(t), y(t)), one can rewrite our equations as
#x′(t) =
[
]
#x(t), #x(t 0 ) = (x 0 , y 0 ).
Letting A be the matrix above, the solution to this equation is
#x(t) = etA #x(t 0 ) =
[
cos(t) − sin(t) sin(t) cos(t)
] [
x 0 y 0
]
[
x 0 cos(t) − y 0 sin(t) x 0 sin(t) + y 0 cos(t)
]
In this example it is not hard to observe that #x(t) is actually a circle: writing r^2 = x^20 + y 02 , one sees that (x(t))^2 + (y(t))^2 = r^2.
Of course, it is not this easy to understand the behavior of most solutions to our differential equations by staring at the explicit solution in this course, so we need to develop another technique. For this we write down the vector field
#v (x, y) = (−y, x)
associated to our system of differential equations, and observe that #v ⊥(x, y) = (x, y) is conservative with associated function f (x, y) = (x^2 + y^2 )/2 satisfying
∇f (x, y) = #v ⊥(x, y), d dt
f (x(t), y(t)) = ∇f (x, y) · #v (x, y) = 0.
Hence the unique solution x#(t) must satisfy f ( #x(t)) = c for some constant c, and substituting the initial condition tells us c = (x^20 + y^20 )/2.
It is hard to solve ODEs, and we really only have a good general theory for linear ODEs (which is the main topic for this course). In real life it is not realistic to assume that our coefficients in our ODEs will always be constant. For example, the spring constant may decrease with time as the spring weakens. We will also study ODEs with varying coefficients, paying special attention to the case of second order ODEs. We resist giving an example in this introduction.
Example 1.2. Solving t + xx′^ = 0
tells us x(t) = ±
2 c − t^2 , −
2 c ≤ t ≤
2 c.
Giving us initial conditions will tell us how to choose the unique solution, if possible.
Example 1.3. Solving x′^ = et+x
tells us x(t) = − ln(−c − et), t < ln(−c)
Again giving us initial conditions will tell us how to choose the unique solution, if possible. Let us observe for this example that lim t→ln(−c)−^
x(t) = ∞.
1.3 First Order Linear ODEs
This is a differential equation of the form
x′(t) = p(t)x(t) + q(t), x(t 0 ) = x 0.
In order to solve for x(t) one simply consider the integrating factor e−^
∫ (^) p(t) dt
. Multiplying this gives us d dt
x(t)e−^
∫ (^) p(t) dt) = q(t)e−^
∫ (^) p(t) dt .
Integrating tells us
x(t) = eP^ (t)−P^ (t^0 )x 0 + eP^ (t)
∫ (^) t
t 0
e−P^ (s)q(s) ds,
where P (t) is an antiderivative of p(t).
Remark. The condition x(t 0 ) = x 0 is called the initial condition of the ODE (left ambiguous in the above examples). In general, if we are given a differential equation without conditions, then the solution we get is called the general solution since it depends on undetermined constants. Initial conditions allow us to pinpoint which solution we pick among the family of general solution.
Example 1.4. Consider the ODE
tx′^ = −x + t^2 , x(1) = 1
on the interval t > 0. We divide throughout by t to convert it into the form above:
x′^ = − x t
The integrating factor in this case is eln^ t^ = t, and we need to solve
d dt
(tx) = t^2.
After integration, the general solution is
x(t) = t^3 3
c t
If we substitute in x(1) = 1, the solution is
x(t) = t^3 3
3 t
1.4 Bernoulli ODEs
A Bernoulli ODE can be seen as a generalization of a first order linear ODE, and is a differential equation of the form x′(t) = p(t)x(t) + q(t)xn(t), n 6 = 0, 1.
(If n = 0, 1 then this ODE reduces to a first order linear ODE.) To solve for x(t) one considers the change of variables y = x^1 −n,
transforming this ODE into
y′(t) = (1 − n)p(t)y(t) + (1 − n)q(t).
We can then solve for the general solution y(t) by an integrating factor, and x(t) = 1 −n
y(t). If an initial condition is given, then we can find the particular solution.
1.5 Ricatti ODEs
A Ricatti ODE is a differential equation of the form
x′(t) = p(t) + q(t)x(t) + r(t)x^2 (t).
We can solve this ODE if we have a particular solution x 1 (t), i.e. a function that satisfies the ODE above. If this particular solution is found, we can consider the translation u(t) = x(t) − x 1 (t). Substituting this relation into the above ODE tells us that
u′^ − (q + 2rx 1 )u − ru^2 = 0,
which is a Bernoulli ODE with n = 2. Therefore, to solve a Ricatti ODE one considers the following steps:
- Find a particular solution x 1 (t).
- Solve the Bernoulli ODE u′(t) = (q(t) + 2r(t)x 1 (t))u(t) + r(t)u^2 (t).
- The general solution will be x(t) = x 1 (t) + u(t). We will study some other ODEs which can only be solved after finding a particular solution.
Example 1.5. Consider the ODE
x′^ = 2t − x t
x^2 t^3
We need to find a particular solution. Since the coefficients involves only powers of t, we guess a solution of the form x(t) = ctα.
Substituting this into the ODE tells us that α = 2 and c = 1, 2, so we pick the particular solution x 1 (t) = t^2. The associated Bernoulli ODE is
u′^ =
t
u +
t^3
u^2.
To solve u(t) one considers the change of variables z = u^1 −^2 = u−^1 , transforming this ODE into
z′^ = −
t
z −
t^3
2 Lecture 2 – Matrices
This is the first of three lectures giving a crash course in linear algebra. After these three lectures we will introduce additional concepts in linear algebra that we need as we go along.
2.1 Row Echelon Form
The primary goal of introducing matrices in our course is to solve a system of linear equations
a 11 x 1 + · · · + a 1 nxn = b 1 , a 21 x 1 + · · · + a 2 nxn = b 2 , .. .
am 1 x 1 + · · · + amnxn = bm.
We can arrange it in terms of matrices in two ways. The first way is to write the coefficients as a m × (n + 1) array of numbers
[
A|
b
]
, or more
explicitly (^)
a 11 · · · a 1 n b 1 a 21 · · · a 2 n b 2 .. .
am 1 · · · amn bm
We can then try to solve this equation via row operations, which we explain in a bit. The second way is to write it in terms of matrix multiplication AX =
b , or more explicitly
a 11 · · · a 1 n .. .
am 1 · · · amn
x 1 .. . xn
b 1 .. . bm
To make sense of the left-hand side we define matrix multiplication between an m×n matrix and an n × 1 matrix in the obvious sense, such that the rows corresponds to our system of linear equations. (We define a matrix to simply be an array of numbers; we will give it a deeper meaning next time.) The general formula for matrix multiplication will be given later, and again in the next lecture. Let us now concentrate on the first way of writing coefficients. Clearly, subtracting, scaling, and adding rows with one another corresponds to solving equations, just like how you would in high school. Recall we denoted[ A be the m × n array of aij ’s corresponding to the array of numbers
A|
b
]
Definition 2.1. A matrix A is in row echelon form (REF) if
- rows with all zeros are below any row with nonzero entries,
- the nonzero leading coefficient, or pivot of a row is strictly to the right of the nonzero leading coefficient of the row above it. In addition, if every nonzero leading coefficient of A equals 1, and every column containing a 1 has zero in every other entry, then A is in reduced row echelon form (RREF).
Every matrix can be reduced to REF after performing a sequence of row operations. For example, the left matrix below is in REF but not the right matrix.
The REF for both matrices above are the same, and after performing more row operations their RREF is (^)
There exists a systematic way of computing REF and RREF using something called Gaussian elimination, but this is quite obvious and will not be outlined here. The method is really what you think it is.
Definition 2.2. The rank of A is the number of leading coefficients after performing REF.
Let us return to our system of equations
[
A|
b
]
. Such a systems always have zero, one, or
infinitely many solutions, and the way to determine this is by performing row operations until A is in REF or RREF. The system has:
- no solutions if there are inconsistencies, i.e. if the last nonzero row after performing RREF is
[ 0 · · · 0 γ ], γ 6 = 0,
corresponding to 0 = 1;
- one solution if the system is consistent and there are no free variables, i.e. rank A equals the number of columns of A;
- infinitely many solutions if the system is consistent and there are free variables, i.e. rank A is less than the number of columns of A.
Example 2.3. Let us consider the system of equations
After REF, we obtain (^)
There are no solutions for the first system, and for the second system
x 3 = t, x 2 = 5 − 3 t, x 1 = − 9 − 5 t.
2.2 Determinants and Invertible Matrices
We now restrict to the case where A is an n × n matrix. The situation is the same as before: we want to solve a system of linear equations. The previous subsection tells us how to determine if this systems has solutions. We now give a criteria to determine if the system has a unique solution. To do this, we use the second way of thinking: writing the system of equations as AX =
b.
Definition 2.4. The determinant of an n×n matrix A = (aij ) can be recursively defined as follow.
- If n = 1, then det A = a 11.
- If n = 2, then det A = a 11 a 22 − a 12 a 21.
- If n > 2, then det A can be computed in one of two ways.
Definition 2.9. Let A be an n × n matrix. If there exists another n × n matrix B such that AB = BA = I, then A is invertible, and B is called the inverse matrix of A. We denote B by A−^1.
Theorem 2.10. Let A = (aij ) be an n × n invertible matrix. Define
Aij = (−1)i+j^ det Mij ,
where Mij is the (n − 1) × (n − 1) matrix obtained by removing the ith^ row and jth^ column of A. Then
A−^1 =
det A
A 11 A 21 · · · An 1 A 12 A 22 · · · An 2 .. .
A 1 n A 2 n · · · Ann
(Note the arrangements of Aij in A.) In particular, the inverse of A is unique.
Proof. Computation.
Example 2.11. Let us consider the system AX =
b given explicitly by
x 1 x 2 x 3
One computes det A = −7, so the system has a unique solution X = A−^1
b. By the theorem above
A−^1 = −
so (^)
x 1 x 2 x 3
2.3 More Determinant Facts
Proposition 2.12. Let A and B be n × n matrices.
- det(cA) = cn^ det A.
- det At^ = det A, where At^ is the transpose of A obtained by writing the rows as columns.
- det(AB) = det(A) det(B).
- If A is invertible, then det(A−^1 ) = (det A)−^1.
- det A remains the same after adding or subtracting rows (or columns) with each other.
Proof. Computation.
Proposition 2.13. The following are equivalent for an n × n matrix A. (a) A is invertible. (b) det A 6 = 0. (c) rank A = n.
Proof. Statements (b) and (c) are equivalent by the last assertion in the above proposition. State- ments (a) and (b) are equivalent by the third assertion and Theorem 2.10.
Example 2.14. It is an exercise on row reduction to see that the determinant of the Vandermonde matrix (^)
1 a 1 a^21 · · · an 1 −^1 1 a 2 a^22 · · · an 2 −^1 .. .
1 an a^2 n · · · an n−^1
is (^) ∏
1 ≤i<j≤n
(aj − ai).
We have now come to the most important theorem of this lecture.
Theorem 2.15. Consider a system of linear equations AX =
b.
- If det A 6 = 0, then there is a unique solution for X given by X = A−^1
b.
- If det A = 0, then there are either zero or infinitely many solutions for
b. There is no solution if, after performing RREF on
[
A|
b
]
, the last nonzero row is of the form
[ 0 · · · 0 γ ], γ 6 = 0;
otherwise, there are infinitely many solutions.
Proof. This is a consolidation of everything said in this lecture.
To end this lecture we present another method to find the solution of AX =
b if A is invertible.
Theorem 2.16 (Cramer’s Rule). Consider a system of linear equations AX =
b where A is an n × n invertible matrix. Then
xi =
det Ai det A
where Ai is the matrix A with the ith^ column replaced by
b.
Proof. Computation.
Example 2.17. We now solve
x 1 x 2 x 3
using Cramer’s Rule. One has det A = −7, and
det A 1 = det
and similarly det A 2 = −3 and det A 3 = −5. Hence the computation agrees with Example 2.11.
Given any linear map, we can construct a matrix associated to it.
Definition 3.8. Let ϕ : F m^ −→ F n^ be a linear map. The standard matrix for ϕ is the n × m matrix A such that ϕ(vi) = Avi for all i.
Computing the standard matrix for ϕ is relatively simple. Let e 1 ,... , em be the standard basis for F m. For each ei, write ϕ(ei) = a 1 ie 1 + · · · + anien.
Then the standard matrix for F n^ is the matrix
a 11 · · · a 1 m .. .
an 1 · · · anm
Example 3.9. The standard matrix associated to the linear map in Example 3.5 is [ 1 2
]
It is important to write down the matrix with respect to different bases as well.
Definition 3.10. Let v 1 ,... , vm be a basis for F m, and let w 1 ,... , wn be a basis for F n. The matrix for ϕ with respect to these bases is the n × m matrix A = (aij ) such that
ϕ(vj ) = a 1 j w 1 + · · · + anj wn
for all j.
Again, after fixing bases, computing the matrix for ϕ is relatively simple: write
ϕ(vj ) = a 1 j w 1 + · · · + anj wn
for all j. Then the matrix we want is
a 11 · · · a 1 m .. .
an 1 · · · anm
Example 3.11. Let us consider the basis (1, 1), (1, 0) for F 2 and 2 for F. Then, using the linear map in Example 3.5,
ϕ(1, 1) = 3 =
ϕ(1, 0) = 1 =
so the matrix of ϕ with respect to these bases is [ 3 / 2 1 / 2
]
One last thing to mention about matrix representations is the matrix for map composition.
Definition 3.12. Let A = (aij ) and B = (bi′j′^ ) be two n × n matrices. Then the product AB is the n × n matrix with (i, j)th^ entry
ai 1 b 1 j + ai 2 b 2 j + · · · + ainbnj.
Remark. Consider
A =
[
]
, B =
[
]
Then
AB =
[
]
, BA =
[
]
so matrices do not commute in general. In fact, matrices do not every satisfy the property AB = AC implies B = C. An example is
C =
[
]
, AC =
[
]
Proposition 3.13. Fix F n, F m, F l, their bases, and linear maps ϕ : F n^ −→ F m^ and ψ : F m^ −→ F l. If A and B corresponds to the matrices of the respective maps, then the matrix of their com- position ψ ◦ ϕ : F n^ −→ F l^ is BA.
Proof. Computation.
3.3 Change of Basis
Earlier we mentioned that there are many different choice of basis for F n. Suppose v 1 ,... , vn and v 1 ′,... , v′ n are two such bases. Can we construct an n × n matrix A such that Avi = v′ i for all i? Equivalently, we write to write down the matrix of the linear map f : F n^ −→ F n^ defined by
f (a 1 v 1 + · · · + anvn) = a 1 v′ 1 + · · · + anv′ n,
where we fixed the basis v 1 ,... , vn on both sides. How do we construct this matrix, called the change of basis matrix? We use the same method as before.
- For each i, write down v′ i as a linear combinations of the v 1 ,... , vn, i.e.
v′ i = a 1 iv 1 + · · · + anivn.
- The matrix we want is then (^)
a 11 · · · a 1 n .. .
an 1 · · · ann
Example 3.14. To compute the change of basis matrix from (1, 2), (3, 1) to e 1 , e 2 , we observe that
e 1 = −
e 2 =
Hence the matrix we want is (^) [ − 1 / 5 3 / 5 2 / 5 − 1 / 5
]
Clearly every change of basis matrix is invertible, for we can easily construct its inverse linear map. We now have the following corollary.
Theorem 3.21 (Rank-Nullity). Let ϕ : F m^ −→ F n^ be a linear map. Then
dim ker ϕ + dim im ϕ = m.
In the language of matrices, if A is an n × m matrix, then
null A + rank A = m.
Proof. This is easy to see by doing REF on A. One can also prove it using abstract linear algebra, which we will not do here.
Proposition 3.22. Let ϕ : F m^ −→ F n^ be a linear map. If ϕ is bijective, then m = n. Furthermore, the following are equivalent. (a) ϕ is bijective. (b) ker ϕ = { 0 }. (c) ϕ is injective. (d) ϕ is surjective.
Proof. If ϕ is bijective, then ker ϕ = { 0 } necessarily. By the Rank-Nullity Theorem dim im ϕ = m. But by bijectivity im ϕ = F n, so m = n necessarily. Now we show the equivalences. (a) ⇒ (b) is clear. For (b) ⇒ (c), if ker ϕ = { 0 }, then ϕ(v) = ϕ(w) implies ϕ(v − w) = 0, so v − w ∈ ker ϕ and v − w = 0, implying injectivity. For (c) ⇒ (d), if ϕ is injective, then ker ϕ = { 0 }, so dim im ϕ = n. But this implies im ϕ is a subset of F n^ of the same dimension, so im ϕ = F n, implying surjectivity. For (d) ⇒ (a), the Rank-Nullity Theorem tells us that dim ker ϕ = 0, so ker ϕ = { 0 } necessarily, implying injectivity, and together with surjectivity gives bijectivity.
The proposition above gives the following corollary. This corollary tells us that, although matrices do not commute in general, an invertible matrix and its inverse do.
Corollary 3.23. Let A and B be two n × n matrices. If AB = I then BA = I.
Proof. Since A is an invertible matrix, ker A = { 0 }. Consider the matrix I − BA. Note that A(I − BA) = 0. If I − BA were nonzero, then there exists a vector v such that (I − BA)v 6 = 0. This would imply A(I − BA)v 6 = 0 since ker A is trivial, a contradiction.
3.5 An Overview of Abstract Vector Spaces
In general, one can give a notion of an abstract vector space V over F and show that every such V is isomorphic to F n^ for some n. We will not dwell on the precise definition here, but we will just say the following. A vector space over F is a set V , together with addition and scalar multiplication satisfying some obvious axioms. A subspace of V is a subset W such that cw 1 + w 2 ∈ W for every w 1 , w 2 ∈ W and c ∈ F. In this course our vector space V will almost always be one of the following:
- a set of nice functions Rn^ −→ R, or
- the set of polynomials Pn of degree at most n with coefficients in R. The subspace W will almost always be the functions of V satisfying some linear ODE. (One can check that this is indeed a subspace.) If U and V are two vector spaces, a linear map ϕ : U −→ V is still defined to be a function satisfying ϕ(cu 1 + u 2 ) = cϕ(u 1 ) + ϕ(u 2 ) for any constant c ∈ F and any vector u 1 , u 2 ∈ U. In our course, a linear map is a map defined almost exclusively by a linear ODE. We use the same method as before to construct the matrix associated to a linear map.
Example 3.24. Let S be the space of real polynomials of the form ax^3 + bx, and consider the linear map ϕ : S −→ R^2 defined by
ϕ(ax^3 + bx) = (a, a + b).
Pick bases x, x^3 and (1, 0), (0, 1). Then
ϕ(x) = (0, 1) = 0 · (1, 0) + 1 · (0, 1), ϕ(x^3 ) = (1, 1) = 1 · (1, 0) + 1 · (0, 1).
Thus the matrix of ϕ is (^) [ 0 1 1 1
]
Example 3.25. Consider the linear map ϕ : P 2 −→ R defined by ϕ(p) = p(1), or in other words
ϕ(ax^2 + bx + c) = a + b + c.
Since the dimensions of P 2 and R are 3 and 1 respectively, the matrix of ϕ should be 1 × 3. With respect to the bases 1, x, x^2 and 1, it is (^) [ 1 1 1
]
Example 3.26. Let M 2 (R) be the space of 2 × 2 real matrices. The map ϕ : M 2 (R) −→ R defined by ϕ(A) = det(A) is not a linear map, for det(A + B) 6 = det(A) + det(B) in general.
Example 3.27. Consider the linear map ϕ : P 2 −→ P 2 defined by ϕ(p) = 2p′′^ + (x − 1)p, or in other words ϕ(ax^2 + bx + c) = 4a − b + (b − 2 a)x + 2ax^2.
With respect to the basis 1, x, x^2 ,
ϕ(1) = 0, ϕ(x) = −1 + x, ϕ(x^2 ) = 4 − 2 x + 2x^2.
Thus the matrix of ϕ is (^)
Example 3.28. Consider the vector space V of real polynomials in two variables x and y of degree at most two. This vector space is six dimensional, with basis 1, x, y, x^2 , xy, y^2. Let us consider the linear map ϕ : V −→ V by
T (p(x, y)) = y ∂p ∂x
A computation tells us that the matrix of ϕ with respect to this basis is
