ENGG2013 Unit 24 Linear DE and Applications Apr, 2011.

Outline Method of separating variable Method of integrating factor System of linear and first-order differential equations – Graphical method using phase plane kshum2

Nomenclatures “First-order”: only the first derivative is involved. “Autonomous”: the independent variable does not appear in the DE “Linear”: – “Homogeneous” – “Non-homogeneous” c(t) not identically zero kshum3

Separable DE “Separable”: A first-order DE is called separable if it can be written in the following form Examples – x’ = cos(t) – x’ = x+1 – x’ = t 2 sin(x) – t  x’ = x 2 –1 – All linear homogeneous DE kshum4

SEPARABLE DE AND METHOD OF SEPARATING VARIABLES kshum5

How to solve separable DE Write x’= f(x) g(t) as. Separate variable x and t (move all “x” to the LHS and all “t” to the RHS) Integrate both sides kshum6

Example kshum7 Solve (1) Write the DE as (2) Separate the variables (3) Integrate both sides General solution to x’=t/x

Solution curves The solutions are hyperbolae kshum8 Some constant Sample solutions

Example: Newton’s law of cooling Suppose that the room temperature is T r = 24 degree Celsius. The temperature of a can of coffee is 15 o C at T=0 and rises to 16 o C after one minute. – T(0) = 15, T(1) = 16. Find the temperature after 10 minutes kshum9 Proportionality constant

LINEAR NON-HOMOGENEOUS DE METHOD OF INTEGRATING FACTOR kshum10

Example: RC in series Physical laws – Voltage drop across resistor = V R (t) = R I(t) – Voltage drop across inductor = C V C (t) = Q(t) kshum11 Charge From Kirchoff voltage law V C (t) + V R (t) = sin(  t) Linear non-homogeneous

Linear DE in standard form Linear equation has the following form By dividing both sides by p(t), we can write the differential equation in standard form kshum12

Product rule of differentiation Idea: Given a DE in standard form Multiply both sides by some function u(t) so that the product rule can be applied. kshum13

Illustrations 1.Solve the initial value problem 2.Find the general solution to kshum14

Example: Mixing problem In-flow of water: 10 L per minute Out-flow of water: 10 L per minute In-flowing water contains Caesium with concentration 5  Bq/L Describe the concentration of Ce in the water tank as a function of time. kshum15 Water tank 1000 L Initial Caesium concentration = 1  Bq/L

Henri Becquerel French physicist Dec 1852 ~ Aug 1908 Nobel prize laureate of Physics in 1903 (together with Marie Curie and Pierre Curie) for the discovery of radioactivity. Bq is the SI unit for radioactivity – Defined as the number of nucleus decays per second. kshum16

Back to the RC example Write it in standard form Multiply by an unknown function u(t) kshum17

Integrating factor Is there any function u(t) such that u’(t) = u(t)/RC ?  Choose u(t) = exp(t/RC) kshum18

Now we can integrate kshum19 Use a standard fact from calculus

Solution to RC in series General solution If it is known that Q(0) = 0, then kshum20 approaches zero as t   Steady-state solution

Sample solution curves Take R=C = 1,  =10 for example. kshum21 Steady state Transient state Different solutions correspond to different initial values.

SYSTEM OF DIFFERENTIAL EQUATIONS kshum22

Interaction between components If we have two or more objects, each and they interact with each other, we need a system of differential equations. Metronomes synchronization – Double pendulum – kshum23

General form of a system of linear differential equation System variables: x 1 (t), x 2 (t), …, x n (t). A system of DE kshum24 Some functions

System of linear constant-coeff. differential equations System variables: x 1 (t), x 2 (t), x 3 (t). Constant-coefficient linear DE – a ij are constants, – g 1 (t), g 2 (t) and g 3 (t) are some function of t. Matrix form: kshum25

Application 1: Mixing C 1 (t) and C 2 (t) are concentrations of a substance, e.g. salt, in tank 1 and 2. Given – Initial concentrations C 1 (0) = a, C 1 (0) = b. – In-low to tank 1 = f 1 m 3 /s, with concentration c. – Flow from tank 1 to tank 2 = f 12 m 3 /s – Flow from tank 2 to tank 1 = f 21 m 3 /s – Out-flow from tank 2 = f 2 m 3 /s Objective: Find C 1 (t) and C 2 (t). kshum26 Water tank 1 Volume = V 1 m 3 Concentration = C 1 (t) Water tank 2 Volume = V 2 m 3 Concentration = C 2 (t) f1f1 f 12 f 21 f2f2

Modeling Consider a short time interval [t, t+  t]  C 1 = C 1 (t+  t)–C 1 (t) = cf 1  t + f 21 C 2  t – f 12 C 1  t  C 2 = C 2 (t+  t)–C 2 (t) = f 12 C 1  t – f 21 C 2  t – f 2 C 2  t Take  t  0, we have C 1 ’ = – f 12 C 1 + f 21 C 2 + cf 1 C 2 ’ = f 12 C 1 – (f 21 + f 2 ) C 2 kshum27

Graphical method For autonomous system, we can plot the phase plane (aka phase portrait) to understand the system qualitatively. Select a grid of points, and draw an arrow for each point. The direction of each arrow is kshum28

Phase Plane Suppose – f 1 = 5 – f 2 = 5 – f 12 = 6 – f 21 = 1 – c = 2 – Initial concentrations are zero kshum29 Converges to (2,2) C 1 ’ = – 6C 1 + C C 2 ’ = 6C 1 – 6 C 2

Convergence (C 1,C 2 )=(2,2) is a critical point. – C 1 ’ and C 2 ’ are both zero when C 1 = C 2 =2. The analyze the stability of critical point, we usually make a change of coordinates and move the critical point to the origin. Let x 1 = C 1 –2, x 2 = C 2 –2. kshum30 C 1 ’ = – 6C 1 + C C 2 ’ = 6C 1 – 6 C 2 x 1 ’ = – 6x 1 + x 2 x 2 ’ = 6x 1 – 6 x 2

Phase plane of a system with stable node kshum31 All arrows points towards the origin

Sample solution curves kshum32 The origin is a stable node

Theoretical explanation for convergence The eigenvalues of the coefficient matrix are negative. Indeed, they are equal to – and – The corresponding eigenvectors are [ ] and [– ] kshum33

Eigen-direction If we start on any point in the direction of the eigenvectors, the system converges to the critical point in a straight line. This is another geometric interpretation of the eigenvectors. kshum34

Application 2: RLC mesh circuit Suppose that the initial charge at the capacity is Q 0. Describe the currents in the two loops after the switch is closed. kshum35 i 1 (t) i 2 (t) Physical Laws Resistor: V=R i Inductor: V=L i’ Capacitor: V=Q/C KVL, KCL Homework exercise

An expanding system Both eigenvalues are positive. kshum36

Phase Plane of a system with unstable node kshum37 The origin is an unstable node. The red arrows indicate the eigenvectors

A system with saddle point One eigenvalue is positive, and another eigenvalue is negative kshum38

Phase Plane of a system with saddle node kshum39 The origin is a saddle point. The thick red arrows indicate the eigenvectors

Conclusion The convergence and stability of a system of linear equations is intimately related to the signs of eigenvalues. kshum40