Modelling 1

Question 1

What is the SEIR Model and its differential equations?

Answer 1

S(t) + E(t) + I(t) + R(t) = N

dS/dt (t) = -betaI(t)S(t)/N
dE/dt (t) = betaI(t)S(t)/N - epsilonE(t)
dI/dt (t) = epsilonE(t) - gammaI(t)
dR/dt (t) = gammaI(t)

Question 2

How are Differential Equations formed from a Reaction Diagram?

Answer 2

A + B (k1)⇋(k2) C

For the reactants A, B, C in the above example (1.1), we obtain
d/dt a(t) = k2c(t) − k1a(t)b(t),
d/dt b(t) = k2c(t) − k1a(t)b(t),
d/dt c(t) = k1a(t)b(t) − k2c(t).

Or more generally, nAa + nBb (k)→ nCc + nDd gives the rate equations:
d/dt a(t) = −nAka(t)^nA * b(t)^nB ,
d/dt b(t) = −nBka(t)^nA * b(t)^nB ,
d/dt c(t) = nCka(t)^nA * b(t)^nB ,
d/dt d(t) = nDka(t)^nA * b(t)^nB .

Question 3

What are the three types of problem?

Answer 3

1. Forward Problem - Given all the information and use this to quantitively predict an outcome. E.g. the maximum speed of a car.
2. Inverse Problem - When some of the information is not directly known, but may be able to be ‘reverse engineered’. E.g. using CT scans to locate a tumour.
3. Control Problem - Trying to make a solution to best fit a goal. E.g. the best paper airplane.

Question 4

What is the RLC differential equation?

Answer 4

Ld^2/dt”2 q(t) + R d/dt q(t) + 1/C q(t) = 0

Question 5

What is the model for Population Growth?

Answer 5

d/dt p(t) = (kp(t))(1 - p(t)/pm)
where pm is the maximum population density.

Question 6

What are the Fundamental Dimensions?

Answer 6

T - time - seconds, days
L - length - nanometers, feet, lightyears
M - mass - micrograms, how about elephants?
A - quantity - head count, mol
Θ - temperature - kelvin, degrees fahrenheit
Q - charge - coulomb

Question 7

How do we denote the dimension of a variable?

Answer 7

The dimension of a variable v is denoted by [v].
The dimension of a product is equal to the product of its parts’ dimensions.

Question 8

What is the process of non-dimensionalisation?

Answer 8

1. Write the dimensions of all variables in terms of fundamental dimensions.
2. Express all fundamental dimensions required for the dependent variables as a product of suitable independent variables. The choice made for each fundamental dimension is called a scale.
3. Consider any dependent variable dn and let us write Dn = [dn] for its dimension. This dimension is made up of fundamental dimensions. We have expressed these in terms of the independent variables by defining suitable scales, hence Dn = [pn(i1, . . . , iM)] for some product pn of some independent variables. This product is then called the scale for the variable dn and usually denoted with a line of the letter in the following, ¬dn = pn(i1, . . . , iM). Actually, it is then the scale for all variables with the same dimension. Let now ˜dn = dn/¬dn. This variable has no dimension because
   [˜dn] = [dn/dn] = [dn]/[¬dn] = 1.
4. Write down the nondimensional problem and reduce it.
5. Add the dimensions again by transforming the reduced equation back.

Question 9

What is a Differential Equation?

Answer 9

For a differential equation of the form
F(t, x(t), x′(t), x′′(t), . . . , x^(k−1)(t), x^(k)(t)) = 0
we define:
1. The order of a differential equation is the order of its highest derivative.
2. The differential equation is autonomous if F does not explicitly depend on the independent variable, so F : R^k+1 → R, and the differential equation reads
F(x(t), x′(t), . . . , x^(k)(t)) = 0.
i.e. No t constant.
3. The differential equation is linear if F can be written in the form
F(t, x(t), x′(t), . . . , x^(k)(t)) = s(t) + Σi=0,k (ai(t)x^(i)(t))
with some functions s, ai : (α, β) → R, i = 0, . . . , k.
4. Let s be the function obtained if x(t) = x′(t) = x^(k)(t) = 0,
s(t) = F(t, 0, . . . , 0), t ∈ (α, β).
The equation is called homogeneous if s = 0. I.e. if all derivative terms are 0 and the constant is 0, s is homogeneous.
Otherwise, it is called inhomogeneous.

Question 10

When is a solution to a DE explicit or implicit?

Answer 10

A function x : (α, β) → R is called a solution to the differential equation if it satisfies the equation.
* The solution is called an explicit solution, if the dependent variable is given in terms of the independent variable as a combination of algebraic expressions or elementary functions.
* An implicit form of the solution is an equation that relates the dependent and independent variables and involves no derivatives.

Question 11

What is the FTC?

Answer 11

Suppose g : [a, b] → R is continuous and let
G(x) := ∫a,x (g(˜x)dx), x ˜ ∈ [a, b].
Then G is an anti-derivative of g, i.e., it satisfies
d/dx G(x) = g(x), x ∈ (a, b).
Moreover, if G˜ is any other anti-derivative of g then
∫a,b (g(˜x)dx˜) = G˜(b) − G˜(a),
and the two anti-derivatives of g differ by a constant only, so for some c ∈ R
G˜(x) = G(x) + c, x ∈ [a, b].

Question 12

What is the difference between a general solution and particular solution?

Answer 12

The general solution to the trivial differential equation is given by
x(t) = F(t) + c, t ∈ (α, β)
where F is an anti-derivative of f and c ∈ R is any number.
For a specific number c, x(t) = F(t) + c is a particular solution.

Ie. the general solution is a family of solutions, particular solutions involves a specific c value.

Question 13

What is a stable stationary point?

Answer 13

A stationary point x∗ of an autonomous differential equation of the form
d/dt x(t) = f(x(t)), is called stable if nearby solutions remain close as the independent variable grows. In mathematical terms, the distance dist(x(t), x∗) remains bounded as t → ∞.

We have that:
-if f(x∗) = 0, f′(x∗) > 0 : x∗ is unstable,
-if f(x∗) = 0, f′(x∗) < 0 : x∗ is stable.
with f being of the form defined.

Question 14

How to solve first order homogeneous DEs?

Answer 14

We have the equation d/dt x(t) = r(t)x(t), if we first consider r(t) = r to be constant, we have a function whose derivative is proportional to itself, the exponential.
And so all solutions will be of the form x(t) = ce^rt, with initial conditions determining c.
Now for r(t) depending on t, by trying x(t) = ce^R(t), we find that R must be the anti-derivate of r, find R by integrating r from the initial condition to t. And then using x0 to find c.

Question 15

What is an integrating factor?

Answer 15

The function I(t) = e^−R(t), where R(t) = ∫r(t˜)dt˜ is an anti-derivative of r, is called an integrating factor for the differential equation d/dt x(t) = r(t)x(t) + s(t).

Question 16

How to solve inhomogeneous DEs with integrating factor?

Answer 16

We have the equation d/dt x(t) = r(t)x(t) + s(t).
Multiplying through by the integrating factor gives
I(t) d/dt x(t) − I(t)r(t)x(t) = I(t)s(t)
This can then be written as d/dt [I(t)x(t)] = I(t)s(t).

And so the solution x(t) can be given by
x(t) = e^(R(t)x0) + ∫t,t0 [e^(R(t)−R(t˜)) * s(t˜)] dt˜ with R(t) = ∫t,t0 r(t˜) dt.

Question 17

When are two functions linearly independent?

Answer 17

Two functions x1, x2 on an interval [α, β] are linearly independent if the only solution to l1x1(t) + l2x2(t) = 0 for any l1, l2 ∈ R is l1 = l2

Assume that x1 and x2 are two solutions of a second order homogeneous differential equation. Then so is
x(t) = l1x1(t) + l2x2(t), t ∈ [α, β],
for all l1, l2 ∈ R.

Question 18

How do linearly independent functions span the solution set of second order homogeneous differential equations?

Answer 18

Any two linearly independent solutions x1, x2 span the solution space of second order homogeneous differential equations.
i.e., all solutions are of the form x(t) = l1x1(t) + l2x2(t), t ∈ [α, β], for some l1, l2

Question 19

What is a complementary function?

Answer 19

Given any two linearly independent solutions x1, x2 of the linear, homogeneous, second order differential equation, the general solution (or set of solutions)
l1x1(t) + l2x2(t) = 0 for any l1, l2 ∈ R is called complementary function.

Question 20

How to solve second order homogeneous differential equations?

Answer 20

We have an equation of the form a(t)d^2/dt^2 x(t) + b(t)d/dt x(t) + c(t)x(t) = 0 and initial conditions x(t0) = x0, d/dt x(t0) = v0.
We have to find a complementary function and then account for the two initial conditions.
To do this we must find two linearly independent solutions of the DE.
Then the solution to the initial value problem with these initial conditions is given by
x(t) = l1x1(t) + l2x2(t), t ∈ [α, β],
where l1, l2 are such that
x0 = l1x1(t0) + l2x2(t0),
v0 = l1d/dt x1(t0) + l2d/dt x2(t0).

From this we can derive specific solutions to the DE that make it easier to find such l1 and l2.
Let xd(t) denote the solution with initial conditions
xd(t0) = 1, x’d(t0) = 0,
and xn(t) the solution with initial condition
xn(t0) = 0, x’n(t0) = 1.

We now have the solution given by x(t) = x0xd(t) + v0xn(t).
I.e. here we have substituted x1 and x2 for xd and xn, with them having easier derivatives to work with, making the solution nicer, allowing for l1 = x0 and l2 = v0.

Question 21

How to solve second order inhomogeneous differential equations?

Answer 21

Same method to get complementary function from the homogeneous case. Now we have a particular integral xp such that xp(t0) = 0, d/dt xp(t0) = 0. (The particular integral is a solution to the DE, whereby its second derivative is equal to the constant)
Then the function x(t) = xp(t) + x0xd(t) + v0xn(t) is a solution to the DE, with x0, xd, v0, xn defined as before.

Question 22

What is an Auxiliary Equation?

Answer 22

The equation
aλ^2 + bλ + c = 0
is called the auxiliary equation or characteristic equation of the differential equation
ad^2/dt^2 x(t) + bd/dt x(t) + cx(t) = s(t). (A second order DE with constant coefficients)

If λ is a solution to the auxiliary equation then e^λt is a solution to the DE.

Question 23

What is the case where the auxiliary equation has two real roots?

Answer 23

We have b^2 − 4ac > 0 then λ1 /= λ2 are two real roots of the auxiliary equation. The functions
x1(t) = e^λ1t, x2(t) = e^λ2t then are two linearly independent solutions.

We then need to find xd and xn, assuming that to = 0.
We have
xd(t) = l1dx1(t) + l2dx2(t) = l1de^λ1t + l2de^λ2t (by simply writing xd as the sum of two linearly independent solutions)
and so using the initial conditions for xd we get,
1 = xd(0) = l1dx1(0) + l2dx2(0) = l1d + l2d,
0 = xd’(0) = l1dx1’(0) + l2dx2’(0) = l1dλ1 + l2dλ2
From here we can solve for l1 and l2 (in terms of λ), and so getting a solution for xd.

We repeat the process for xn, using its initial conditions.
And finally write the solution in the form
x(t) = x0xd(t) + v0xn(t)

Question 24

What is the case where the auxiliary equation has no real roots?

Answer 24

We have b^2 − 4ac < 0 and λ1,2 = p ± iq
We can have x˜1(t) = e^(p+iq)t, x˜2(t) = e^(p−iq)t be two linearly independent complex-valued solutions.
And so get real-valued solutions we use the fact that
e^±iqt = cos(qt) ± isin(qt).
This gives x1(t) = e^pt * cos(qt), x2(t) = e^pt * sin(qt) and
x(t) = l1x1 + l2x2

Now solve for xd and xn similarly.
*Note it may be easier to solve for x(t) directly, substituting for the initial conditions straight away, rather thana solving for xn and xd.

Question 25

What is the case where the auxiliary equation has one real root?

Answer 25

In the case b^2 − 4ac = 0 the auxiliary equation has only one root λ := λ1,2 = −b/2a.
Then ˜x1(t) = e^λt is only one solution. A convenient second, linearly independent solution is given by x˜2(t) = te^λt.
Taking x1(t) = e^λ(t−t0), x2(t) = (t − t0)e^λ(t−t0).
And then write x(t) as the sum of these two solutions.

Question 26

How to solve homogeneous, linear, second order differential equation with constant coefficients?

Answer 26

-Solve auxiliary equation to find the roots
-Then apply the roots to two linearly independent solutions
-Write the solution as a sum of these, accounting for initial conditions.

Question 27

How to find a particular integral?

Answer 27

There is no general method to find the particular integral other than ‘guesswork’.
Try to use a general solution of the same form as the constant term.
E.g. if s(t) = t^2, try using xp(t) = at^2 + bt + c, solving for a, b and c.

Question 28

How to solve inhomogeneous, linear, second order differential equation with constant coefficients?

Answer 28

First solve homogeneous case to find complementary function via solving auxiliary equation. Then find a particular integral.

Question 29

What are the different types of damping?

Answer 29

-Underdamped - no real roots
-Overdamped - two real roots
-Critically damped - one real root

Roots are from auxiliary equation.

Question 30

What is a system of differential equations?

Answer 30

When there are multiple differential equations accounting for multiple independent variables.

Question 31

What are direction field and phase portraits?

Answer 31

1. A direction field is the attachment of an arrow proportional to (f1(x1, x2), f2(x1, x2)) at each point (x1, x2) ∈ R^2.
2. A phase portrait (also called phase diagram) of (4.5) is the set of curves in the x1-x2-plane that are parametrised by solutions to the system. These curves traced out by solutions are so-called trajectories.

See notes for examples.

Question 32

How do we check if a stationary solution is linearly stable?

Answer 32

Let (¯x1, x¯2) denote a stationary solution (or point). In order to get some more insight into whether (¯x1, x¯2) is stable we study solutions of the form
(x1(t), x2(t)) = (¯x1 + ξ1(t), x¯2 + ξ2(t))
where (ξ1(t), ξ2(t)) is a ’small’ deviation from the stationary point.

We then work out the partial derivatives of the system so that we can write,
d/dt ξ1(t) = ∂/∂x1 f1(¯x1, x¯2)ξ1(t) + ∂/∂x2 f1(¯x1, x¯2)ξ2(t),
d/dt ξ2(t) = ∂/∂x1 f2(¯x1, x¯2)ξ1(t) + ∂/∂x2 f2(¯x1, x¯2)ξ2(t),
With the partial derivatives of f1 and f2 being worked out, with f1 being the RHS of DE1 and f2 being the RHS of DE2.

This then gives values, in terms of ξ1 and ξ2, for their derivatives.

Question 33

How to solve homogeneous linear systems with constant coefficients?

Answer 33

Let us look into linear systems of the form
d/dt x1(t) = a1,1x1(t) + a1,2x2(t),
d/dt x2(t) = a2,1x1(t) + a2,2x2(t),
These can be formulated in the matrix-vector form
d/dt x(t) = Ax(t), (*)
A = (a1,1 a1,2)∈ R^2×2
(a2,1 a2,2)
x(t) = (x1(t))
(x2(t))

By trying an exponential as a solution with
x(t) = e^λtv and
v = (v1)
(v2)

We can substitute this into the system () to get λv = Av.
So x(t) = e^λtv is a solution if v is an eigenvector of A with corresponding eigenvalue λ.

And so by finding the 2 eigenvectors and eigenvalues, we get two linearly independent solution and so the general solution is given by
x(t) = l1e^λ1tv1 + l2e^λ2tv2

Question 34

What are eigenvalues and eigenvectors?

Answer 34

Eigenvalues are found by working out the roots to a matrix’s characteristic polynomial. That is solving for λ in:
0 = det(λI − A)
= det (λ − a1,1 −a1,2)
(−a2,1 λ − a2,2)

Corresponding eigenvectors then are the solutions vi ∈ C^2, vi /= (0, 0), of
(λi*I − A)vi = 0, i = 1, 2.

That is, for a given eigenvalue λ, its eigenvector v is the solution to (λI − A)v = 0

Question 35

What does it mean for a DE to be well posed?

Answer 35

well-posedness = existence + uniqueness + stability.