← IA Differential Equations – Full Text
These are Zixuan’s notes for Part IA – Differential Equations at the University of Cambridge in 2025. The notes are not endorsed by the lecturers or the University, and all errors are my own.
The latest version of this document is available at academic.micfong.space. Please direct any comments to my CRSid email or use the contact details listed on the site.
This document is typeset using Typst. Almost all figures are created using Inkscape and Mathematica.
Differential equations appear in almost all branches of science and applied mathematics.
For example, we have the following differential equation,
which relates the rate of change of position , the dependent variable, with time , the independent variable.
The main purpose of this course is to solve such equations.
1 Differentiation
1.1 Introduction
The derivative of a function w.r.t. its argument is the function
Remark. For the derivative to exist, we require the left-handed and right-handed limit to exist and be equal. We write
for the limits. e.g. is not differentiable at , despite being differentiable at everywhere else.
Informally, if , then can be made arbitrarily close to by making sufficiently close to . Nonetheless, we do not require . [See IA Analysis I for a more formal discussion.]
Notation. We write
We can higher derivatives for sufficiently smooth functions. For example,
To refer to the th derivative, we write
1.2 Big O and Little O
If we wish to compare the behavior of functions close to a limiting point , we can use order parameters. There are two kinds of them: Big O and Little O.
-
Big O — “can be bounded by”
Definition 1.2 (Big O)-
if is finite, then is as if and such that with , we have
We often write It follows that is bounded as .
e.g. if , then and .
e.g. as since .
-
if , then is as if and such that ,
e.g. as since ,
-
-
Little O — “much smaller than”
Definition 1.3 (Little O)is as if such that ,
If in vicinity of (regardless of the behavior at ), equivalently
We often write .
e.g. as , since .
e.g. as .
Remark.
-
is a stronger statement than . Big O shows that a function is bounded by a given multiple, whereas Little O shows that it is bounded by any multiple.
So, , but not the converse.
e.g. but as .
-
Multiplicative constants do not matter for Big O. i.e. If , and for any non-zero constant .
Order parameters are useful to classify the remainder terms before taking limits. So we can write
Therefore as .
Hence
as . This result extends to Theorem 1.10 (Taylor’s Theorem).
1.3 Rules for Differentiation
[Differentiating a function of a function.] Given , then
Given , then
[Special case of the product rule.] Given , then
Consider By repeated applying the product rule, we have
c.f. Pascal’s triangle, we can generalize this into Leibniz’s Rule.
[Generalization of the product rule.] Given , we have
1.4 Taylor Series
For a function which is infinitely differentiable at , the Taylor series about is
For a function which is infinitely differentiable at , the Taylor polynomial of degree is
Note that are the partial sums of .
For a function which is differentiable times at ,
where as .
Remark. Note that is a stronger statement than .
e.g. is but not as .
With , Theorem 1.10 (Taylor’s theorem) gives
This is to say, that provides a local approximation to in the vicinity of with error or .
Consider, about , the function . Then for ,
where .
Then the fractional error
Therefore, for a given target accurarcy at , this can be used to specify how large must be.
1.5 L’Hôpital’s Rule
L’Hôpital’s rule allows us to deal with limits of indeterminate forms.
Let and be differentiable at with continuous first derivatives there, and
then if ,
provided that the limit on the RHS exists.
Remark. L’Hôpital’s rule can be generalized. For example, if , then
This can be generalized even further to produce a limit on RHS results.
Let
Then
Therefore, we can find the limit
2 Integration
We have seen the definition of integration in terms of area under a curve and inverse of differentiation. We shall review what these really mean.
2.1 Integrals as Riemann Sums
This section talks about the area under a curve idea.
The integral of a suitably well-defined function is the limit of a sum
where and .
Following Definition 2.1, a function is Riemann integrable if the generalized Riemann sum does not depend on exactly how we choose the rectangles in the limit that all .
[This includes the cases where is non-uniform, or we cannot evaluate on the LHS, etc.]
How are we sure this limit evaluates to the area under a curve idea? We shall consider one rectangle at first.
We will need to borrow this following idea from IA Analysis I for now:
For a continuous function , the area under curve from to is
for some where .
Hence, if is differentiable, then by Taylor’s theorem
Hence,
Therefore, the total area under curve from to is
2.2 Fundamental Theorem of Calculus (FTC)
This section talks about the inverse of differentiation idea.
Let for some Riemann integrable function Then
Proof. By definition,
Remark. is a solution to the differential equation
with .
Let for some Riemann integrable function . Then we have
Notation. Indefinite integrals are written in the form
The unspecified lower limit gives an integration constant.
2.3 Integration Techniques
2.3.1 Integration by Substitution
If the integrand contains a function of a function, it might help to substitute for the inner function.
Consider . Then let
and
Another important class of substitutions are trigonometric substitutions. These make use of the following identities:
We can use the following table as a substitution reference.
| Integrand contains | Substitution |
Consider
We should try with , so would be unique for .
2.3.2 Integration by Parts
Recall the product rule
we can derive the integration by parts technique
Let
Then we have
Let . We can let and , and then
This integration by parts method also works for inverse trignometric functions and inverse hyperbolic functions.
3 Partial Differentiation
3.1 Functions of Several Variables
We shall now generalize to functions of more than one independent variables, which are also called multivariate functions.
- Height of terrain,
- Temperature in a room,
- Pressure of gas as a function of volume and temperature,
A way of representing these functions is using a contour plot.
The slope of a point on the plot depends on the direction. Firstly, let us consider what happens along the coodinate directions.
3.2 Partial Derivatives
Partial derivatives ar ederivatives of multivariate functions with respect to one variable, while holding the other fixed.
Given a function of several variables, e.g. , the partial derivative of with respect to at fixed is
Roughly speaking, this is the slope of when moving in the positive direction.
can be defined similarly.
Consider . Then
We can similarly calculate higher derivatives.
Also, we can do mixed derivatives.
Notation. The notation is getting somewhat cumbersome. We usually omit and use of to indicate that all other variables are kept fixed. i.e.
Alternatively, there is a even more compact notation:
Note that in Example 3.3, it seems that
This is indeed the case under certain conditions.
If is a multivariate function with continuous 2nd derivatives, then
if the conditions of Schwarz’s theorem is satisfied.
3.3 Multivariate Chain Rule
Give path and a function , consider along the path.
Differential of a function is
Proof. Consider the change in under
Then
Consider the two brackets separately. Then, as ,
Note that
Hence,
Taking the limit gives the required results.
Thus, for the path , we have
If instead we parameterize path by coordinate , so we are left with , then
We can also reach the integral form of chain rule:
For a function , the change in between two endpoints is
For ,
so the final result does not depend on a particular path for given endpoints.
3.4 Applications of Multivariate Chain Rule
3.4.1 Change of Variables
It is often useful to write a differential equation in a different coordinate system.
For example, we can change Cartesian coordinates to polar coordinates with
Think of as . From the chain rule,
3.4.2 Implicit Differentiation
For , we have
Consider , which represents a surface in 3D space. In this case, it implicitly defines
but we may not be able to find the solutions explicitly.
However, we can still evaluate derivatives like .
In the equation
we cannot find explicitly. However, we can take the derivative with respect to holding fixed only, we get
If , then
Proof. In general, if , then
Therefore, we can’t vary independently and stay on the surface.
Thus (be aware that is not fixed),
We can similarly find and .
Therefore, we can conclude
The reciprocal rule applies if the same variables are held fixed:
Therefore,
Important. For the change in variables , we have
but
3.5 Differentiation of an Integral w.r.t. a Parameter
For a family of functions where is a parameter, define
Then
Proof. By definition,
Now let us consider the parts of the expression separately.
Hence
Consider
Then
Suppose that we want to evaluate
Then, let
with . Hence
Setting , we get .
4 First-Order Linear Differential Equations
is a first-order linear ordinary differential equation.
4.1 The Exponential Function
In order to introduce first-order linear ordinary differential equations, we first need to explore the exponential function.
The exponential function is defined the infinite series
This can also be written as
Differentiaing the exponential function, we have
This allows us to define the exponential function in another way. We can define to be the solution of
with the initial condition .
The exponential function satisfies this key property:
Following this property, it suggests that we should write as , where
The inverse function is defined as the function where
It follows that
Hence
An eigenfunction of an operator is a function that is unchanged up to multiplicative scaling by the eigenvalue, under action of the operator.
In case of the differential operator, an eigenfunction satisfies
Hence, the eigenfunctions of the differential operator are of the form
4.2 First Order Linear Ordinary Differential Equations
Here are some properties of first-order linear ordinary differential equations.
- Any th order linear ODE has independent solutions.
- For any linear homogenous ODE, any constant multiple of a solution is also a solution.
4.2.1 Homogeneous Linear ODEs with Constant Coefficients
Consider the equation
We should try . Then
Which leads to (given that )
So .
This is the general solution, as it contains an arbitrary constant .
To specify unique solutions, it requires us to apply suitable initial conditions. [ conditions are needed for a th order ODE.]
For example, in Example 4.12, if we have , then .
4.2.2 Discrete Equations
It is sometimes useful to consider functions evaluated at discrete points.
Consider again with .
We can approximate this equation by discrete form at with and . With , we have
[This is called the Forward Euler scheme, which is not a great approximation numerically.]
Substituting this approximation into the original equation gives
Hence we have
Now take [this represents steps from to ] as . Then
which, thankfully, agrees with the continuous case.
4.2.3 Series Solutions
This is a powerful way to solve ODEs. Essentially, we are looking for soluions in the form of a power series,
where we will determine by substituting into the ODE.
We shall get back to the example
Then, we have
Thus,
Also, by multiplying our original series by ,
Therefore, substituting into the ODE gives
Hence we can derive the recurrence relation
Therefore
and
4.3 Forced (non-homogeneous) ODEs
In a forced ODE, there are terms not involving the dependent variable or its derivatives. In this case, in general, is no longer a solution. To solve these equations, we should do the followings.
-
Find any solution of the forced equation, called a particular integral (PI) .
-
Write the general solution as , and find the complementary function (CF) that satisfies the corresponding homogeneous equation.
-
Combine and to get the general solution.
This method is general for linear ODEs.
4.3.1 Constant Forcing
Consider
A particular integral is , since substituting it gives
Then the complementary function is the solution of the homogeneous equation , which we have already solved as .
Hence, the general solution is
4.3.2 Eigenfunction Forcing
The forcing term may also be an eigenfunction of the differential operator.
Consider the decay between three isotopes , with decay constants for and respectively.
Thus we have
and also
We shall try the particular integral of the form . Substituting it gives
Hence is the solution of the homogeneous equation . Thus
Thus, the general solution is
Now, if we were given the initial conditions , then
Hence,
4.4 Non-constant Coefficients
The general form of such equations is
We can get the standard form by dividing both sides by (assuming ):
To solve these equations, we use an integrating factor (IF) . Multiplying our standard form by gives
We want the left-hand side to be , so we require by the product rule. This is a separable equation, so
Therefore,
which is unique up to an irrelevant constant factor. Hence, the original equation becomes
Consider the equation
The integrating factor is
Therefore,
Let us get back to Example 4.15.
We have
We can identify and . Thus, the integrating factor is
Hence,
Let us consider two cases.
-
If , then
This is exactly the solution we arrived at in Example 4.15, using the PI and CF method.
-
If , then
Thus,
Note that the particular integral is now proportional to , which is different from the previous case. This is called the resonance case.
5 Nonlinear First-Order ODEs
Recall that a non-linear ODE is one in which the dependent variable [usually ] and its derivatives appear with exponents other than 1 or are multiplied together.
The general form of a first-order nonlinear ODE is
[The term in could be nonlinear, but it is not considered here.]
We have two special cases of nonlinear first-order ODEs that we can solve: separable equations and exact equations.
5.1 Separable Equations
A first-order ODE is called separable if it can be written in the form
These equations can be solved by integrating both sides.
Consider the equation
Rearranging the equation gives
5.2 Exact equations
The ODE Equation 1 is called exact if is an exact differential, i.e. there exists a function such that
In particular, if Equation 1 is exact, then and being a constant is a solution.
If an ODE is exact, then using the multivariate chain rule, we have
So, we can solve exact equations by finding a function such that and .
Since partial derivatives commute, we have
This is a necessary condition but not sufficient for exactness:
A domain is path connected if every pair of points in can be connected by a path in .
Consider the equation
Rewriting gives
Hence we have
Hence, the equation is exact in any simply connected domain.
Thus
and similarly,
Hence , and we have the implicit solution
5.3 Solution Curves and Isoclines
The general idea is that nonlinear ODEs are often impossible to solve in simple closed forms. We can still analyse the behaviour of solutions using graphical methods.
5.3.1 Solution Curves
Consider
Then, each initial condition (e.g. ) generates a distinct solution curve (trajectory).
We can still sketch these solution curves without actually solving the ODE.
We can solve the following equation to illustrate the ideas:
This is a separable equation:
Hence, we have a family of solution curves parameterised by .
If , then . Now, we can sketch the solution curves for various .
Note that, some general behavior follows directly from the ODE:
-
for all if . Hence we have two constant solution curves at and .
-
at for any . Hence, all solution curves have a horizontal tangent at .
The equation gives us the gradient of the solution through .
Note that the solution curves can’t cross if is single valued.
Consider
Then, for ,
- for and
- for .
For
we have
for some constant .
5.4 Fixed (Equilibrium) Points and Stability
These points often reveal important features of ODEs.
Consider
The fixed points are . They have very different character as seen from the sketchs above. The solution curves as , while those near diverge away from it.
A fixed point is
- stable if whenever deviates slightly from , converges to as .
- unstable if whenever deviates slightly from , diverges from as .
5.4.1 Perturbation Analysis and Stability
Let be a fixed point of [i.e. for all ]. Consider a small perturbation about the fixed point:
We want to analyze how behaves as .
Then
Linearize for small epsilon:
Not that this is a linear ODE in .
Note that if , then we need higher order terms to determine stability.
Consider the equation
We are aware that the fixed points are . We have
Near :
Hence, as , and the fixed point at is stable.
Near :
Hence, as , and the fixed point at is unstable.
5.4.2 Autonomous Systems and Phase Portraits
Autonomous systems are ODEs in which the independent variable (e.g. ) does not appear explicitly in the equation. e.g.
First of all, these autonomous systems are separable. We can write
Hence, if is a solution, then so is for any constant .
Although this equation is separable, we may not be able to solve it in closed form. Consider, near a fixed point , we have
Hence, if
- , then the fixed point is stable.
- , then the fixed point is unstable.
Consider a chemical reaction
The number of molecules of at time are respectively.
The initial numbers are . Hence, we have the conservation laws
since one of and is consumed to produce one of and . Assume that the rate of reaction is proportional to the product of the numbers of and molecules (e.g. we considering dilute gases):
Therefore, we have an example of an autonomous non-linear first-order ODE.
Note that the fixed points are and (corresponding to the complete consumption of either or ).
Now assume . Then, is unphysical. We shall now carry out perturbation analysis to determine the stability of the fixed points.
For , is a stable fixed point, while is an unstable fixed point.
We can sketch a 1D phase portrait to visualise the behaviour of solutions.
Consider a population of size . We have
-
birth rate: with ,
-
death rate: , where
- models isolated deaths
- models deaths due to overcrowding.
Thus we have the ODE
To make things simpler, let and . Then,
This is called a differential logistic equation. It is an example of an autonomous system. The fixed points are and . We can carry out perturbation analysis to determine their stability.
For (), is an unstable fixed point, while is a stable fixed point.
We can sketch a 1D phase portrait to visualise the behaviour of solutions.
This equation can be solved exactly and sketched.
5.5 Fixed Points in Discrete Equations
A first-order discrete equation is a recurrence relation of the form
Note that the RHS is independent of , which is like an autonomous system in continuous ODEs.
We can also analyse the stability of fixed points in discrete equations using perturbation analysis. Let be a fixed point, and write , where .
Then,
Therefore, is
-
stable if ,
-
unstable if .
Consider the discrete equation
This is called a discrete logistic equation, or the logistic map. We can compare this with the continuous logistic equation in Example 5.19.
[This is useful to model population dynamics when we consider births at discrete time intervals only.]
We are only interested in . We can sketch the graph of against .
From the graph, if , then stay within .
The fixed points satisfy . Hence, the fixed points are
However, be aware that the second fixed point only makes sense for (in order to be non-negative).
We can carry out perturbation analysis to determine their stability.
Therefore,
-
For the fixed point at :
- stable if ,
- unstable if .
-
For the fixed point at :
- stable if ,
- unstable if .
We can illustrate the behaviour of solutions using cobweb diagrams.
-
Consider .
This shows that is a stable fixed point: solutions for any initial condition in will converge to .
6 Higher Order Linear ODEs
We shall focus on 2nd order linear ODEs, but many methods are also applicable to higher order linear ODEs.
Different to the 1st order case, closed form solutions to 2nd order linear ODEs don’t always exist.
6.1 2nd Order ODEs with Constant Coefficients
The general form of a 2nd order linear ODE with constant coefficients is
where are constants and is a given function, and is the differential operator
which is linear.
A differential operator is linear if for any functions and constants ,
which is called the principle of superposition.
We can exploit the linearity of to solve 2nd order linear ODEs.
-
Find the complementary functions that satisfy the homogeneous equation:
-
Find a particular integral that satisfies the non-homogenous equation:
-
A solution of the full equation is then given by
since by linearity, .
A 2nd order ODE has two linearly independent complementary functions, so the general solution to the full equation is
A set of functions is linearly dependent if there exist constants , not all zero, such that
One can compare the two definitions above with the definition of vectors.
6.2 Complementary Functions
Recall that
so is also an eigenfunction of , where
The complementary functions satisfy , i.e. are eigenfunctions of with eigenvalue .
Therefore, with satisfying the characteristic equation of :
There are two roots, , to the characteristic equation, which can be real or complex.
-
Case 1:
We have two linearly independent complementary functions:
Hence, the most general complementary function is
In the language of linear algebra, the space of complementary functions is a 2-dimensional vector space spanned by the basis .
Remark. The roots may be complex, which will lead to oscillations. -
Case 2: (degenerate case)
We only have one linearly independent complementary function . See the following examples to find the second complementary function.
Consider the ODE
The characteristic equation is
which has roots .
Therefore, the general complementary function is
where are constants.
Consider the ODE
The characteristic equation is
which has roots .
Therefore, the general complementary function is
Note that using Euler’s formula, we can rewrite this as
where are constants (, ).
Consider the ODE
The characteristic equation is
which has a degenerate root .
We only have one complementary function .
To find the second complementary function, we can “detune” the equation slightly to remove the degeneracy, by considering a slightly modified equation:
where .
Then the characteristic equation is
which has roots .
Hence, the general complementary function is
Consider the initial conditions , to both the original and detuned equations. We have
Therefore,
We have
Therefore, we can write the general complementary function as
General rule. If is a degenerate complementary function of linear ODE with constant coefficients, then
is a second linearly independent complementary function.
6.3 Homogeneous 2nd Order ODEs with Non-Constant Coefficients
The general form of this kind of ODE is
6.3.1 Second Complementary Function – Reduction of Order
We wish to find a 2nd solution given one solution .
We will try from the form
where is an unknown function to be determined. We have
Substituting into the ODE, we get
Therefore,
Let , then
This is a separable 1st order ODE for :
Integrating both sides, we get
Hence, we can integrate to find , and then find .
We shall see an example with constant coefficients first.
Consider the ODE
where we have , .
One solution is . Therefore,
Hence, , so .
Therefore, the second solution is
Hence is a second linearly independent solution.
6.3.2 Phase Space
For an th order linear ODE
is determined by , and higher derivatives are determined by differentiating the ODE.
Hence, we can construct Taylor series about if are specified.
We say that the state of the system at any is fully specified by an -dimensional solution vector.
Remark.
-
At any , defines a point in -dimensional phase space.
-
As varies, traces out a trajectory in phase space.
Consider the 2nd order linear ODE
We know that the solutions are
Hence the solution vectors are
In this case we have a 2D phase space.
Note that and are linearly independent vectors, so we can use them as a basis for the phase space.
6.3.3 Wronskian and Linear Dependence
Recall that are linearly dependent if there exist constants , not all zero, such that
Hence, we can differentiate this equation times to get
so being linearly dependent implies that is linearly dependent.
Given solution vectors of an th order linear ODE, the fundamental matrix is the matrix whose columns are the solution vectors:
The Wronskian of the functions is defined as the determinant of the fundamental matrix:
From the above, if are linearly dependent, then for all .
It follows that if for some , then are linearly independent.
Consider the ODE
The Wronskian of the two solutions , is
Hence the two solutions are linearly independent.
6.3.4 Abel’s Theorem
Given any 2 solutions of
if and are continuous on an interval , then either the Wronskian for all or for all .
Sketch Proof. We have
This is a separable ODE for :
Hence, if , then for all ; otherwise, for all .
The geometric interpretation is that the solution vectors are either always collinear or never collinear in the phase space.
We can find without knowing the solutions explicitly.
Consider
Then, Abel’s identity gives
Application. Abel’s identity can be usde to find a second solution given a known solution . Consider that we have
Dividing both sides by , we get
This is the same as we had using reduction of order.
Abel’s theorem can be generalised.
6.4 Linear Equidimensional ODEs
These ODEs are related to ODEs with constant coefficients.
A linear 2nd order ODE is equidimensional if it is of the form
where are constants.
Proof. We have
Therefore,
6.4.1 Solving by Eigenfunctions
so is an eigenfunction of the operator with eigenvalue .
It suggests that we should look for complementary functions of the form . Substituting into the homogeneous equation, we get
Let the solutions be . Then the complementary functions are
if .
6.4.2 Solving by Substitution
Substitute , then
Substituting into the ODE, we get
Now we have a linear ODE with constant coefficients in the variable , which we can solve using the methods discussed earlier. The characteristic equation is
This is the same characteristic equation as before when we tried the eigenfunction method. The complementary functions are therefore, if ,
We can now deal with the degenerate case similarly as before.
6.5 Inhomogeneous (Forced) 2nd Order ODEs
We will discuss the methods to find particular integrals.
6.5.1 Constant Coefficient ODEs
We have the form
Now, use the following ansatz for depending on the form of :
| Try of the form | |
| or | |
| Polynomial of degree |
We can determine the constants by substituting into the ODE. Since the ODE is linear, we can superpose terms.
Consider the ODE
We shall try a particular integral of the form
Substituting into the ODE, we get
Equating coefficients, we have
Therefore, , , .
The complmentary function, as discussed earlier, is
Hence, the general solution is
where are constants.
If the forcing term involves a term that is in a complementary function, a resonance case occurs. In this case, we carry out detuning.
Consider the ODE
which represents a simple harmonic oscillator driven at its natural frequency . We say that this oscillator is driven resonantly. We have the comcomplementary functions
Since is already in the complementary function, we consider detuning by looking at the slightly modified equation
with .
We try a particular integral of the form
We can see that must be zero since there is no term on the RHS. Substituting into the ODE, we get
Note that the limit does not exist since diverges. We can add in a complementary function to regularise the limit:
We know that this satisfies the detuned equation. Now, taking the limit , we have
Therefore, a particular integral for the resonant case is
The general rule is that if the forcing term is a linear combination of linearly independent complementary functions, the particular integral is of the form
Remark. If the homogeneous equation is degenerate, we may need to multiply by higher powers of to find a particular integral, in the form
for a 2nd order degenerate case.
6.5.2 Equidimensional ODEs
Consider the equidimensional ODE
We have seen that the complementary functions are of the form
assuming .
If , we try a particular integral of the form
for and .
For the resonance cases or , then the particular integral is of the form
which follows from Section 6.4.2.
Remark. If the homogeneous equation is degenerate, we may need to multiply by higher powers of to find a particular integral, in the form
for a 2nd order degenerate case.
6.6 Variation of Parameters
This is a systematic method to find particular integrals given two linearly independent complementary functions.
Consider
with linearly independent complementary functions and .
We will use solution vectors and as a basis in phase space at any to write the solution vector for the particular integral.
We have
The components are
If we differentiate again, we get
Subsituting into the ODE, we have
Since are complementary functions, we have
Therefore,
Note that the second component of must be consistent with the derivative of the first component. Therefore, we have the additional constraint
We now have two equations for and :
We can solve for and :
We can therefore invert the fundamental matrix to get
Hence, we can integrate to find and :
The particular integral is therefore
Note that changing the lower limits of the integrals only adds multiples of complementary functions to .
Consider
The complementary functions are, as we have seen before
The Wronskian is
Note that the forcing term is resonant.
6.7 Forced ODEs, Transients and Damping
In the diagram above, by Newton’s 2nd Law we have
Rearranging, we get
where are positive constants.
For and , we have simple harmonic motion at angular frequency
For convenience, we will add in a dimensionless time coordinate . Then we have . Therefore,
where
6.7.1 Free (Unforced) Response
The behavior is described by one dimensionless parameter . We have
The characteristic equation is
The roots are
Depending on the value of , we have three cases:
-
Light damping (underdamping) when .
In this case, the roots are complex, so we can write
So the general solution is
where are constants.
This is a oscillation at , which tends to as , with an exponentially decaying amplitude.
The period is
-
Critical damping when .
We have a degenerate root .
The general solution is
where are constants.
-
Heavy damping when .
The roots are real and negative. WLOG take . Then
The general solution is
where are constants. [ indicates that the exponentials decay to zero as .] Note that , so the first term dominates the long-term behavior if present.
Unforced response decays eventually in all cases.
6.7.2 Forced Response
Initially, the behavior is determined by , which is called the transient response.
Over time, decays, and the behavior is dominated by the , called the steady-state response.
Consider the ODE
[We can relate this back with and .]
Assume light damping with . Then the complementary functions are
where .
For the particular integral, we try
Substituting into the ODE, we get
Equating coefficients, we have
Eliminating , we have
Therefore,
Hence,
6.8 Impulses and Point Forces
Conside a system that experiences a sudden force between time and .
e.g. striking a mass on a spring, or a car going over a curb.
The equation can be in the form
which is a forced, damped oscillator.
It is mathamatically convienient to consider the limit of a sudden impluse, as .
We can integrate the ODE from to , and take the limit .
So we see that
and the velocity is discontinuous.
As only impluse matters for subsequent motion.
6.8.1 Dirac Delta Function
We shall formalise the idea of an impulsive force.
Consider a family of functions such that
and
So the impulsive force we considered earlier is:
One example of such a family is
[See Example Sheet 1, Q14 for normalisation.]
Note that the family of such functions is not unique, but for any family, yields the Dirac delta function.
The Dirac delta function is defined by
It is technically not a function, but a distribution. It only makes sense under an integral.
The Dirac delta function has the following properties:
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-
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(Sampling property.) For all functions that are continuous at ,
More generally, for a function that is continuous at ,
6.8.2 Delta Function Forcing
Consider the ODE
assuming that and are continuous.
For and , we have the homogeneous equation
But there is a discontinuity in at :
Therefore,
which is referred to as a jump condition.
The general rule for higher order ODEs is that the highest-order term in ODE addresses the delta-function forcing.
Consider the ODE
with at and . We wish to find for .
Note that we will need to solve the two regions and separately, and match them at using the jump condition.
For , we have , which solves to .
For , we have , which solves to .
We shall now join the two solutions at .
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is continuous, so
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The jump condition gives
Therefore, the solution is
6.8.3 Heaviside Step Function
The Heaviside step function is defined as
By the fundamental theorem of calculus, we have
The ramp function is defined as
Note that functions get smoother as we integrate them.
6.8.4 Forcing with
Consider the ODE
assuming that and are continuous at .
So, we have
Therefore, and is continuous at . We have around . So, evaluating both sides around ,
The jump conditions for Heaviside function are
A typical situation is that for , and we are asked to find for . This leads to two arbitrary constants, and we can determine them using the two jump conditions.
Or alternatively, we can solve the ODE for and by matching at using the jump conditions.
7 Higher Order Discrete Equations
Consider a linear, discrete, 2nd order equation with constant coefficients of the form
where , , are constants.
This may arise when discretising a 2nd order ODE:
which we can correspond to
We can solve this using similar methods to solving 2nd order ODEs, with general solution
where is the complementary function, and is a particular integral.
7.1 Complementary Functions
The complementary function must satisfy
We can try , since is an eigenfunction. This gives the characteristic equation
The general complementary function is
where , are the roots of the characteristic equation, and , are arbitrary constants.
7.2 Particular integrals
We can find particular integrals based on the form of :
Consider the sequence with conditions
[The sequence starts with ]
We can rewrite this as
The characteristic equation is
with roots
Note that the roots are golden ratios:
Thus the complementary function is
To find and , we use the initial conditions:
Solving this system gives
Therefore the closed form solution is
Interestingly, an integer sequence can be expressed in terms of irrational numbers.
Since , as . Therefore,
8 Series Solutions
When we cannot find simple closed form solutions to ODEs, series solutions may be useful.
Consider
The feasibility of finding series solution around depends on nature of , , and around .
8.1 Classification of Singular Points
The point is an ordinary point of the ODE
if and are both analytic at .
[For the purpose of this course, a function is analytic at if it have a convergent Taylor series about .]
Otherwise, is a singular point.
If is a singular point, but
are analytic, then is a regular singular point; otherwise, it is an irregular singular point.
Compare this to an equidimensional equation
where is a regular singular point. We can state that regular singular points are no more singular than in equidimensional equations.
Consider the ODE
So we have
and
Therefore, and are singular points.
Now, consider . We have
which is analytic at with value . Also,
which is analytic at with value .
Hence, is a regular singular point. Similarly, is also a regular singular point.
Consider the ODE
So we have
It may appear that is an ordinary point. However, the 2nd derivative of is not defined at . Therfore, is a singular point. Now, consider
It does not have a Taylor series about again. Hence, is an irregular singular point.
8.2 Method of Frobenius
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If is an ordinary point of the ODE
then there are two linearly independent solutions of the form
which converge for some neighborhood of . i.e. there is a Taylor series solution about .
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If is a regular singular point of the ODE, then there is at least one solution of the form
where can be real or complex, and [so that is unique]. The series converges for some neighborhood of .
This is called a Frobenius series.
Note that there is no guarantee of two linearly independent solutions in this case.
The series solution method may fail completely at irregular singular points.
Suppose we want to find a series solution about to the ODE
Since is an ordinary point, we expect two linearly independent series solutions.
We shall try
So we have
and
For convenience, we shall multiply the ODE by to get
Substituting the series into the ODE gives
Equating coefficients of for gives
Hence we have a recurrence relation for . Therefore, and are arbitrary constants in the general solution.
Consider the odd terms. Note that . Hence, all odd terms are zero. Therefore, one solution is
Consider the even terms. We have
Therefore
Hence, the other solution is
Note that
Therefore, we can write the even solution as
Hence, we can write the general solution as
which is a closed form solution as well.
Note the behavior near , near the regular singular points.
Consider the ODE
We know that is a regular singular point. For convenience, multiply the ODE by to get
We shall try a Frobenius series solution of the form
Since the ODE now looks like an equidimensional equation, we expect to be able to extract a sum when we substitute it in. So we have
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We shall look at the lowest power of to determine . In this case, it is with .
Since , we have or .
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The next lowerest power is with .
Since , we must have in both cases.
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For more generality, consider with .
Now we should consider the two cases for separately.
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For , we have
Since , all odd terms are zero. Now, for the even terms, we have
Therefore, the even terms give one (Taylor series) solution
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For , we have
Therefore, for the even terms,
Again, all odd terms are zero. Therefore, the even terms give another solution (relabelling constant to ):
Note that this is not a Taylor series but a Frobenius series solution.
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Therefore, there are two linearly independent solutions at this regular singular point.
8.3 Second Solutions
Note that we are guaranteed to get one series solution about a regular singular point, but whether we get a second linearly independent such solution depends on the roots of the indicial equation, and .
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If is not an integer, then we get two linearly independent series solutions, in the form
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If is a non-zero integer, then we get one series solution involving the larger root, say .
The second solution has the form:
where may or may not be zero. is a constant determined in terms of and , so that we have two arbitrary constants as required.
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If , our solutions is similar to case (2), but the logarithmic term is always present.
Consider the ODE
We know that is a regular singular point. We shall try a Frobenius series solution of the form
Substituting into the ODE gives
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The lowest power is with .
Therefore, and . Since is a non-zero integer, we are in case (2).
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For , we get
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For , we have
Therefore,
which is a Taylor series solution.
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For , we have
Note that when , we have , which is a contradiction. Therefore, we cannot find a series solution for . Hence consider the form
We can determine and by direct substitution into the ODE, or we can use the reduction of order method discussed earlier.
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9 Multivariate Functions: Applications
In this section we will discuss
- directional derivatives
- extrema
- coupled systems of 1st order ODEs
- partial differential equations
9.1 Directional derivatives
Consider and a vector displacement .
The infinestimal change in along is given by
where we have defined the gradient operator
If we write where is a unit vector in the direction of , then we have
The directional derivative of in the direction of the unit vector is defined as
where is the angle between and .
This is the rate of change of in the direction of .
Remark. We can define the gradient vector geometrically in the other way round, as the vector such that
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The direction of is the direction of maximum increase of .
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The magnitude of is the maximum rate of change of , i.e.
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If is parallel to contours of , then
Therefore, is perpendicular to the contours of .
9.2 Stationary Points
There is always at least one direction where is zero at a given point, namely the direction parallel to the contours of at that point.
So stationary points are the points where
9.2.1 Types of Stationary Points
Note that contours cross at (and only at) saddle points.
9.3 Classification of Stationary Points
We shall consider how change in vicinity of a stationary point.
9.3.1 Taylor Series for Multivariate Functions
Consider how varies along the line
Along the line, is a function of , and we can use the usual Taylor series for single variable functions:
We have
and also
The Hessian matrix of is defined as
where , , etc.
This is a symmetric matrix since .
Thus, the multivariate Taylor series expansion of about the point is
We can also write this in coordinate-independent form as
9.3.2 Nature of Stationary Points and the Hessian
Suppose is a stationary point with
Around :
where .
A real symmetric matrix is positive definite if
It is negative definite if
Otherwise, it is indefinite.
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If is positive definite at , then for all near , so is a local minimum.
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If is negative definite at , then for all near , so is a local maximum.
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If is indefinite at , then it may be a maximum, minimum or saddle point.
9.3.2.1 Definiteness and Eigenvalues
If is a real symmetric matrix, then we can diagonalise it by an orthogonal transformation (by results in IA Vectors and Matrices). Using coordinates along the principal axes (eigenvectors), in dimensions:
Hence,
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is positive definite iff all eigenvalues (minimum),
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is negative definite iff all eigenvalues (maximum),
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If all eigenvalues are non-zero, but are of mixed signs, then this corresponds to a saddle point.
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If any of the eigenvalues are zero, then we need higher order terms in the Taylor series to classify the stationary point.
Consider .
This function has a (global) minimum at since for all . We have
At the stationary point , the Hessian matrix is
This has eigenvalues and . Therefore, the Hessian is positive semi-definite, and we need to consider higher order terms to classify the stationary point.
9.3.2.2 Definiteness and Signature
An alternative method to determine definiteness without having to compute eigenvalues is to use signatures.
For a function , the signature if given by the signs of
We shall call these determinants
Let be a real symmetric matrix of size . Then
9.3.3 Contours Near Stationary Points
Suppose has a stationary point at . Using coordinates aligned with the principal axes of the Hessian matrix at , we have
Assume that the eigenvalues are non-zero. Then, consider
then around we have
On contours near , since is constant, we have
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At a maximum or minimum, and have the same sign, and the contours are ellipses.
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At a saddle point, and have opposite signs, and the contours are hyperbolae.
Consider the stationary points of .
We have
The stationary points are found by solving and simultaenously:
Thus, the stationary points are at and .
We have
Hence, the Hessian matrix is
At the stationary point , we have
The leading principal minors are
Thus, the signature is , so it is indefinite. See that , [so that eigenvalues are all non-zero,] and hence is a saddle point.
At the stationary point , we have
The leading principal minors are
The signature is , so it is positive definite, and hence is a local minimum.
Near the saddle points, the contours satisfy
Here are some plots of the function and its contours:
9.4 Systems of Linear ODEs
Consider and with
where , , , are constants. We can write this in vector form as
where
There are two ways to solve this system:
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Convert to a single higher order ODE for one variable.
We have
Now we have a linear 2nd order ODE with constant coefficients.
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Solve directly with matrix methods. [This may be more convenient.]
Remark. Under some cases, we write higher order ODE as a set of 1st order ODEs, esentially reversing the process above.
Example 9.10Consider the equation
We can let , and . We then have
Hence,
9.4.1 Matrix methods
Consider
where is a constant matrix.
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Write where is the complementary function satisfying , and is a particular integral.
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Look for of the form where is a constant vector. Then,
Since holds for all , taking we have
Hence is an eigenvalue of , and is the corresponding eigenvector.
For a system of equations, we have such complementary functions if eigenvalues are distinct.
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Find a that satisfies by trying an appropriate form.
Consider
Write , and for consider .
Then,
We have
The corresponding eigenvectors are
Hence,
Try . Then,
Note that an inverse exists since ( is not an eigenvalue of ). We have ƒ
Thus, the general solution is
9.4.2 Non-Degenerate Phase portraits
For first-order ODEs, the phase space is an -dimensional space with coordinates given by
For autonomous systems, there is one trajectory through each point in phase space, except at fixed points.
Consider the homogeneous equation
There is a fixed point at . For , the general solution for (non-degenerate case) is
where are constants.
For and , we have the following cases:
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and are real and of opposite signs. WLOG suppose . In this case, can be chosen to be real. The fixed point is a saddle node.
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and are real and have the same sign. WLOG suppose .
- If both are positive, then the fixed point is an unstable node.
- If both are negative, then the fixed point is a stable node.
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and are complex conjugates, then and . Then,
where .
- If , then we have a unstable spiral.
- If , then we have a stable spiral.
- If , then we have a centre, with closed elliptical trajectories.
In order to determine the direction of motion along the trajectories, we can evaluate at some points on the trajectory.
For example, if at , then motion is upwards at that point, so the direction of motion is counter-clockwise.
9.5 Non-Linear Dynamical Systems
We aim to use techniques for linear systems to investigate the nature of equilibrium points.
Consider an autonomous system of 2 non-linear first-order ODEs:
where and are general non-linear functions of and . [They do not depend explicitly on .]
An equilibrium (fixed) point of the system is a point at which and , i.e.
We need to solve simultaneously to determine the fixed points.
Stabilities of the fixed points can be deduced from perturbation analysis:
where and are small perturbations. We have
We can write this in matrix form as
This is a linear system of homogenous ODEs, and hence the eigenvalues of determine the stability of the fixed point.
Consider a population of prey and predators with the equations
where are positive constants.
Consider a specific case with
The fixed points satisfy
There are three fixed points: , and .
We have
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At , we have
The eigenvalues are and , with eigenvectors and .
Thus, it is a saddle point.
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At , we have
The eigenvalues are and , with eigenvectors and .
Thus, it is a saddle point.
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At , we have
The eigenvalues are found by solving
Since , the fixed point is a stable spiral.
At , we have , so the motion is counter-clockwise.
Now, we can sketch the overall phase portrait.
9.6 Partial Differential Equations
Partial differential equations (PDEs) involve several independent variables. We will illustrate some ideas with wave equations.
9.6.1 First-Order Wave Equation
Consider , where is the spatial coordinate and is time, with
where is a constant with dimensions of velocity.
We can solve this by the method of characteristics. We consider how vary along a path , so that we consider . We have
If we choose such that , then where is a constant, and we have
Paths where are called characteristics of .
Since is constant along characteristics, the general solution of is
where is an arbitrary function.
This expression translates the -dependence of at to the left by at time .
The solutions are left-moving wave solutions.
We have
with .
The general solution is
Using the initial condition, we have
Therefore, the specific solution is
Consider
with .
The characteristics are of the form .
Along these characteristics, we have
So this gives
where is an arbitrary function.
Using the initial condition at , we have
Thus, the specific solution is
9.6.2 Second-Order Wave Equation
A lot of physical systems allow waves to propagate in both directions. This is modelled by second-order wave equations.
Consider
where is a constant with dimensions of velocity.
Since the differential operator can be factorised as
we can write the wave equation as
These two operators commute, so both and are solutions, where and are arbitrary functions.
This suggests that the general solution is
where and are arbitrary functions.
Remark. We can show that this is indeed the most general solution. Let and .
so, we have
So the wave equation becomes
Therefore,
for arbitrary functions and .
Consider
with and . The general solution is
Using the initial conditions, we have
where is a constant. Solving these two equations gives
Thus, the specific solution is
