Differential Equations – Analytical Techniques for 1d

● Special cases of linear 1d equations● Integrals as solutions● Separable equations● General 1d linear differential equations –

Method of integrating factor

Reminder: Malthusian Growth

● Remember that the solution to

is: ● In 1d linear differential equations can be slightly

more general, i.e. of the form

How to solve those?

dP(t )dt

=rP (t) , P(0)=P0

P(t )=P0 er t

dx (t)dt

=a+bx (t) , x (0)=x0

Newton's Law of Cooling

● After a murder what a forensic scientist first does is take the temperature of the victim, a little later the temperature is taken again and these two data points usually allow to extrapolate the time of death

● Newton's law of cooling: rate of change in temperature of a cooling body is proportional to the difference in the body's temperature T and the surrounding environmental temperature T

e,

i.e.ddt

T=−k (T−T e)

Problem

● The victim is found at 8.30am and the temperature is 30 degrees. An hour later the temperature of the victim is found to be 28 degrees

● The temperature of the room in which the victim was found is constant at 22 degrees

● Determine the approximate time of death!● Need a solution of

ddt

T=−k (T−T e)

Solution (1)

● Can substitute z(t)=T-Te and then

● Hence: transforms into

● We already now this solution:● Re-susbtituting

ddt

T=−k (T−T e)

ddt

z=ddt

(T−T e)=ddt

T

ddt

z=−k z

z (t )=z0 e−kt

T (t)=z (t )+T e=T e+(T 0−T e)e−kt

Solution (2)

● To solve our forensic problem (ignoring units):

● Two equations for two unknowns (t, k). E.g. divide (2) by (1)

● Insert this into (1):

● The death occurred around 6.19am

30=22+(37−22)e−kt

28=22+(37−22)e−k (t+1 )

(1)

(2)

e−k=6 /8

8 /15=(6 /8)t

t=log(8 /15)/ log(6/ 8)≈2.19

Solution (3)

● Our substitution trick only works if Te=const.

● What if the temperature of the environment would be changing?

● To see what to do in the more general case, let's study the behaviour of a falling cat ... ● Surprisingly, cats sometimes fare better if falling

from a greater height than when falling from lower height

● Crucial question:Will it land on its feet?

● There are more detailed studies on this important topic, but we develop some simpler theory in this lecture ...

Falling Cats ...

● Cats usually need around 0.3 seconds to turn around. What is the minimum height a cat must fall from to land on its feet?

● Some physics: sum of forces acting on cat are only gravity

● Hence: ma=−mg

mass of catacceleration of cat g=979cm/sec2

Falling Cats (2)

● We need to solve the following differential equation:

● This is a second order linear equation● Let's assume:

– h(0)=h0 (height from which cat drops)– v(0)=h'(0)=0 (initial velocity zero)

● Need to find h0 such that h(0.3s)>0● (*) can be written as a system of two equations

h ' '=−g (*)

v '=−g h '=vand

Falling Cats (3)

● Let's investigate● More generally, this is of the form

(where the r.h.s. only depends on t and NOT on x)● Need a function x(t) which if differentiated yields f(t)● This is the integral:

(Since by the fundamental theorem of diff-int calc.

)● In our case:

d /dt v=−g

d /dt x=f (t )

x (t)=x0+∫t 0

tf (t ')dt '

d /dt (x0+ x (t))=d /dt∫t 0

tf (t ' )dt '=f (t)

v (t)=v 0−∫t 0

tgdt '=v0−g( t−t 0)

Falling Cats (4)

● Since v(0)=0 and t0=0 -> v(t)=-gt ● Also need to solve the second equation

● Again, r.h.s. independent of h and hence

● “Critical point” is h(0.3s)=0

h '=v

d /dt h=−t g

h( t)=h0−∫t0

tt ' gdt '

h( t)=h0−g/2 t 2

h0=g /2 (0.3s)2=44.05cm

“cats are fine if they jumpfrom heights larger than44.05cm”

Important to Remember

● If a 1d ODE is of the form

● with a r.h.s. independent of the unknown function x, it can be solved by integration:

d /dt x=f (t )

x (t)=x0+∫t 0

tf (t ' )dt '

Back to Malthus

● The Malthusian growth model

has a simple exponential solution:

● The model can be used in some form to predict human population growth, but a constant growth rate r is not realistic

● Add changes in r to model changing societal conditions?

d /dt P=r P (t) , P(t 0)=P0

P(t )=P0 exp(r (t−t 0))

d /dt P=r (t)P( t) ,P (t 0)=P0

Malthus (1)

● We have:

and let's say

(Technology improves and allows faster growth of population ...)

● This is still a linear (but non-autonomous) differential equation, but cannot be solved with the techniques discussed so far

● Q: Can we model the US census data for population from 1790 to 1990 with some choice of a and b?

d /dt P=r (t)P( t) ,P (t 0)=P0

r ( t)=a+bt

Separable Differential Equations

● Above Diff. Eq. has the form

● Rewrite this as:

(I.e. l.h.s. depends only on x and r.h.s. depends only on t)

● Solution:

dxdt

=f (x , t)=N (x )M ( t)

dxN (x )

=M (t)dt

∫dx

N (x )=∫M ( t)dt

Separable Equations

● Example: dxdt

=2 t x2

dx

x2 =2 t dt

∫dx

x2 =∫2 t dt

−1 / x=t 2+cIntegration constant

x (t)=−1

t 2+c

Back to Malthusian Growth

● Best fit of log P to US census data gives

a=-7 10-5 and b=0.03476

dPdt

=(a+bt)P , P(t 0)=P0

∫dPP

=∫(a+bt)dt

ln P=at+b /2 t2+c

P(t )=P0 exp(a(t−t 0)+b ' /2( t−t 0)2)

US Census Data and Malthusian Growth

General Linear Equations

● Our general formulation for a 1d diff equation is

● Which is linear if f is linear in y. Most general form of f is then:

● Conventionally, linear 1d equations are written as:

● Possible to give an algorithm to solve those● However: only for linear equations!

dydx

=f (x , y)

f (x , y)=g( x)−p (x ) y

dydx

+ p (x) y=g( x)

Integrating Factors ...

● Idea is to write l.h.s. as a total derivative. To achieve this multiply equation by a factor µ(x) (and µ(x) is later chosen suitably ...)

● Need:● Then Eq. can be written as

(and we can just integrate it ...)

dydx

+ p (x) y=g( x)

μ(x)dydx

+μ(x ) p (x) y=g( x)μ(x )

dμ( x)/dx=μ( x) p (x )ddx

(μ(x ) y)=g(x )μ(x )

(*)

(**)

Integrating Factors (2)

● Strategy: First solve (*) to find µ(x) and then integrate (**)

● (*) is:

this is a separable diff. equation and we employ the strategy developed earlier in the lecture:

● Now (**) becomes:

dμ( x)/dx=μ( x) p (x )

dμ(x )

μ=p(x )dx

μ(x)=exp (∫xp (s )ds)

y (x )=1/μ∫x0

xg (z)μ(z)dz

Integrating Factors (3)

● And putting all together we obtain a solution formula:

● Let's see how this is done in an example

● We first identify p(x)=-2/x● Then:

y (x )=exp(−∫xp(s)ds)(∫x0

xg(z )exp(−∫

zp(s)ds )dz+C )

d ydx

−2y / x=0

μ(x)=exp (∫xp (s )ds)=exp(∫

x−2/s ds)

=exp (−2 ln(x ))=x−2

(***)

Integrating Factor (4)

● Next: multiply both sides of (***) by µ(x)

x−2 d y

dx−2y / x

−3=0

x2 y '−2xy

x4 =0

ddx ( y

x2 )=0

y (x )=C x2

So ...

● This becomes all very complicated already for 1d linear differential equations!

● In the general case analytical solutions not available

● What can we do?● Numerical integration● Solutions available for certain classes of equations

(e.g. systems of linear differential equations)● Equilibrium analysis (and qualitative analysis

techniques for non-linear systems)

Pollution in a Lake

● Consider the scenario of a new pesticide that is applied upstream from a lake of volume V

● River receives a constant amount of this pesticide into its water and flows into the lake at a constant rate f (i.e. River has a constant concentration p of this pesticide)

● Lake is well-mixed and has constant volume, i.e. another river flows out of it at rate f

● Assuming the lake is initially clean, what is the concentration of the pesticide over time?