AMATH 351 Mar 15, 013 FINAL REVIEW Instructor: Jiri Najemni
ABOUT GRADES Scores I have so far will be posted on the website today sorted by the student number HW4 & Exam will be added early next wee Let me now if you see anything wrong or disagree with something by 5 th
KEEP THESE BASIC INTEGRALS AND DERIVATIVES IN MIND DERIVATIVES: d n n1 d c f x n f x f x 0 dx dx d dx exp x INTEGRALS: exp x d dx ln x sinx cos x; cosx sin x 1 x d dx ln g x n1 n x 1 x dx n 1 dx ln x n1 x expx exp xdx 0 g x g x cos xdx sin x sin xdx cos x
CLASSIFICATION OF ORDINARY DIFFERENTIAL EQUATIONS (ODEs) nth order ODE n First order ODE Second order etc y f t, y y f t, y, y Linear first order Linear second order etc y t a t y t b t y t a t y t b t y t c t Non-linear ODE is when a non-linear function of y(t), and/or its derivatives is involved; e.g. y t a t log y t b t y t c t Most ODEs (even first order) are analytically unsolvable (or unsolved) In this class, we studied solution method of linear first and second order ODEs. Also two classes of special solvable first order ODEs (exact, separable) n1 y t f t, y t, y t, y t,..., y t
Recall the direction field/integral curve intuition for the first order ODEs dy sin y subject to y0 dx dy sin x y The arrow indicates the slope (derivative) at that point 3 dx Some possible solutions Solution passing through A given point 0, 3
FIRST ORDER ODEs LINEAR WITH CONSTANT COEFFICIENTS: y t ay t b LINEAR WITH VARIABLE COEFFICIENTS: yt a t y t bt SEPARABLE: px q y y 0 EXACT: M x, y N x, y y 0, N x, y M x y y x
First order linear ODEs with constant coefficients y t ay t b Step. Step 4. Step 6. b Step 1. y t a y t a y t 1 a Step 3. dt adt b b yt yt a a b d ln y t y t a b ln y t at c b y dt a t a b y t expat c a Arbitrary real number b b Step 7. y t expcexpat expat a a
Negative sign cause the amount is decreasing y t Modeling radioactive decay y t positive constant (see later) Proportion of radioactive to-nonradioaative carbon atoms in a given fossil The number of decays per time is proportional to the current number of radioactive atoms. WHAT WE KNOW: This is a first order linear ODE with constant coefficients. y t y 0 exp t D 1 y t y the decay constant (related to half-life) proportion of radioactive carbon in the digged out fossil 0 10 = proportion of radioactive carbon in atmosphere WHAT WE DON T KNOW: t D is the unnown time since death (in years) SOLUTION: 0exp y t y t D D t D D 1 y t ln y 0
proportion of radiocarbon Radioactive decay is exponential decay 1 x 10-1 0.8 0.6 0.4 0. y t y 0 exp t half-life : 0.5y 0 y 0 exp 0.5 exp ln 0.5 ln 0.5 1 ln 0 0 0.5 1 1.5.5 3 time in years x 10 4 1 ln
First order Solving linear first ODE order with linear variable ODEs coefficients yt a t y t bt. Multiply both sides by a non-zero factor t tyt ta ty t tbt 3. t y t t a t y t t b t We are free to pic any non-zero t we want t t att t yt t y t tbt 4. The ey is to chose such that THEN d dt t y t tbt
Solving first order linear ODEs Picing t such that ta t t t 1 dln t d a t = a t ln t a tdt c dt t dt t exp c exp a t dt ; Let's pic c 0 : t exp a t dt d Plugging t exp a t dt into ty t t b t dt d exp a tdt y t tbt dt Integrating both sides: exp y t a t dt y t s b s ds c exp s b s ds a t dt c Why is the integral in the numerator written over s and not t? y 0 0exp 0 exp a 0 b a ds c dt
SEPARABLE ODEs p x It is called separable because: dy p x q y 0 dx q y dy p x dx 0 q y y Separable ODEs are very easy to solve: integral equation with variables x and y separated by equal sign 1. Separate x s and y s to different sides. Integrate the y-side with respect to y and the x-side with respect to x 3. Solve for initial condition y(t 0 ) if ased to do so AUTONOMOUS SYSTEMS are governed by separable equations: dy dt f y
A SEPARABLE EXAMPLE WITH A SUBTLETY dy Solve the equation 1 exp 1 x y dx 1 Separating the variables: dy 1 exp x dx y 1 1 Integrating the sides: dy 1 expxdx y 1 Exam will have simpler Integrals than this 0.5 0.5 dy x exp x c y 1 y 1 0.5ln y 1 0.5ln y 1 x exp x c y 1 ln x exp x c y 1 BUT NOTE THAT y x =1 and y x = 1 are also solutions, yet they are not included. This is because we divided the equation by mae it separable. WE NEED TO CHECK solutions to y 1. y 1 to
EXACT ODEs M x, y N x, y y 0 where M x, y N x, y y x HOW TO TELL IF IT S EXACT Let xy, be the solution to the exact ODE. One can show that xy, 1. M x, y x, y Mdx x xy,. N x, y x, y Ndy x 3. x, y c is implicit solution ( c is a real constant)
Solving an example exact ODE Solve xy 9x y x 1 0 subject to y 0 3 M xy 9 x ; N y x +1 x; x dy dx Step 1: Chec if the equation is exact M y Step : Recall the fact about exact ODEs M x y x y N x y x y x y,, and,, Step 3: Pic the easier one to integrate first N x x, y M x, y x xy 9x x x y 3x 3 h y Step 4: Use the other one to setup equation to solve for h y arbitrary function of y x, y h y h y N x y y x x y y y y, +1= 1
Step 5: Integrate to get Solving an example exact ODE Step 6: Plug it into the differential equation to get implicit solution 3 3 x, y c x y 3 x h y c x y 3x y y c 3 y x y x h y h y y 1 dy y y where is an arbitrary constant 1 3 0 where is an arbitrary constant Step 7: We can obtain explicit solution form 3 The implicit solution is y x 1 y 3x 0 Recall that if ay by c 0 then y In our case a 1, b x 1, c 3x y 3 4 3 x x x x 1 1 1 4 b b 4ac a quadratic formula
EULER S METHOD for approximating solutions y y 3 y f x, y The slope of this line is the same as tangent x 1 of the solution at (x 1,y 1 ) y y f x, y x x 1 0 0 0 1 0 y y f x, y x x 1 1 1 1 y y f x, y x x... 3 3 y y f x, y x x n1 n n n n1 n piecewise linear approximation of solution y 1 true solution x, y 0 0 x1 x x3 x x 4 5
SECOND ORDER LINEAR ODEs WITH CONSTANT COEFFICIENTS py t qy t ry t g t 1. Solve the corresponding homogenous system: three cases can arise -- two different real roots -- repeated real roots -- complex roots y t c y t c y t. Find one particular solution to the non-homogenous system that is Not included in the family of solutions to the homogenous system: 1 1 such that Y t py t qy t ry t g t Method of undetermined coefficients: This method is used when g(t) is a function composed of sines, cosines, exponentials, and/or polynomials (their sums and products). h. Full solution: y t y t Y t h
4 POWERFUL THEOREMS Existence & uniqueness (tae for granted) Say we have a linear second order ODE y q t y r t y s t, where q t, r t, s t are continuous on interval t. Then there is exactly one solution t to the initial value problem y t y, y t y and the solution exists for all t in the interval t. 0 0 0 0 Principle of superposition y1t yt 0, then the linear combination If and are two solutions of the homogenous ODE p t y q t y r t y c y t c y t is also a solution for any values of the constants c and c. 1 1 1
4 POWERFUL THEOREMS CONTINUED Family encompassing all solutions t0 W y1 t0 y t0 y 1 t0 y t0 y t c y t c y t If y t and y t are two solutions of p t y q t y r t y 0, 1 and if there is a point at which the Wronsian - is non-zero, then the family of solutions 1 1 with arbitrary coefficients c and c includes every solution. 1 Non-homogenous solution form The general solution of a linear second order non-homogenous ODE can be written 1 1 1 1 1 1 Y t is som i n the form y t c y t c y t Y t, where c y t c y t is the solution to the corresponding homogenous equation, and that satisfies the non-homogenous equation. e particular solution
Linear second order homogenous ODE with constant coefficients py qy ry 0 We guessed a form of solution: y exp at y a exp at ; y a exp at exp py qy ry pa qa r at We need to solve a quadratic equation for a: a We recognized three cases: q q 4 pr p 1 1 exp exp Case 1: 4 0 exp and exp q pr y t a t y t a t y t c y t c y t c a t c a t 1 1 1 1 1 1 1 1 exp exp Case : 4 0 exp and exp q pr y t a t y t ty t t a t y t c y t c y t c a t c t a t 1 1 1 1 1 Case 3: q 4 pr 0 q 4pr q 4pr q y t exp t c1cos t csin t p p p
Wronsian, linear independence, and satisfying initial conditions y1t yt Say that and are two solutions of the homogenous ODE p t y q t y r t y 0. c 1 0 0 0 0 y0 c1 y1 t0 c y t0 y c y t + c y t Given initial conditions y t = y and y t = y are satisfied? We have: + y0 y t0 y 0 y t0 y t y t y t y t 1 0 0 1 0 0 0 1 1 0 0 c y0 y 1 t0 y 0 y1 t0 y t y t y t y t 1 0 0 1 0 0 WRONSKIAN If the WRONSKIAN of the initial condition is non-zero, then c and 1 t y t y can be chosen so that to satisfy the initial conditions, and,. c 0 0 0 0 WHAT DOES NON-ZERO WRONSKIAN MEAN? If f t and g t are differentiable functions on an open interval a t b, and if Wronsian W f t g t - f t g t is non-zero for some a t b, then 0 0 0 f t and g t are linearly independent on a t b.
THE METHOD OF UNDETERMINED COEFFICIENTS py t qy t ry t g t We want to find one particular solution to the non-homogenous system that is not included in the family of solutions to the homogenous system; -- if g(t) = g 1 (t) + g (t) + + g n (t), form n sub-problems, and solve each separately. You then add the separate solutions together. -- Assume the initial form of the solution (with free parameters A, B, C, ) consistent with the form of g i (t). -- If the assumed form fits the solution of the homogenous system, multiply the form of the solution by t; if that still fits the homogenous solution, multiply by t ;
y 0 THE MASS ON A SPRING INTERPRETATION l l l s s y 0 Equilibrium y 0 yt FORCES: 1. GRAVITY: Fg mg; g 9.8m s By convention, positive force downward.. SPRING FORCE: s Negative when string stretched (h>l). Positive when string compressed (h<l). 3. DAMPING: d Opposite to velocity. By convention, positive velocity means downward. 4. EXTERNAL FORCE: F t s y t F t y t Fe t whatever g s d e my t F F t F t F t my t mg s y t y t F t ZERO my t y t y t F t e e
Duality of LRC circuit to mass on spring Q t capacitator charge l l l s s yt Kirhoff's Law: 1 V t V t V t E t L R C di t L RI t Q t E t dt C dq It dt 1 LQ t RQ t Q t E t C 1 LI t RIt I t Et C Gravity Fg Spring force Damping force External force mg s F t s y t d F t y t Fe t whatever g s d e my t F F t F t F t my t y t y t F t e
TYPES OF VIBRATIONS 1. UNDAMPED FREE VIBRATIONS (idealized scenario) g s 0 my t F F t my t y t. DAMPED FREE VIBRATIONS g d s 0 my t F F t F t my t y t y t LOOK AT OUR EXAMPLES OF THESE & UNDERSTAND THE BEHAVIOR OF SOLUTIONS 3. EXTERNALY FORCED UNDAMPED VIBRATIONS g s e my t F F t F t my t y t F t 4. EXTERNALY FORCED DAMPED VIBRATIONS e g s d e my t F F t F t F t my t y t y t F t e
IMAGINE THESE DAMPING MECHANISMS DAMPED FREE VIBRATIONS my t Fg Fs t Fd t my t yt y t 0 a m 1 ; a Case 1: 4m 0 OVER-DAMPED exp exp y t c a t c a t 1 1 Case : 4m 0 CRITICALLY DAMPED exp y t a t c c t 1 1 Case 3: 4m 0 UNDER-DAMPED y t t c1 t c t m exp sin cos where = 4 4m m m 4m
Upgrading to variable coefficients p t y t q t y t r t y t g t THE 4 THEOREMS STILL APPLY: Existence & uniqueness (tae for granted; hard to prove) Say we have a linear second order ODE y q t y r t y s t, where q t, r t, s t are continuous on interval t. Then there is exactly one solution t to the initial value problem y t y, y t y and the solution exists for all t in the interval t. 0 0 0 0 Principle of superposition (easy to prove) y1t yt 0, then the linear combination If and are two solutions of the homogenous ODE p t y q t y r t y c y t c y t is also a solution for any values of the constants c and c. 1 1 1
4 THEOREMS CONTINUED Family encompassing all solutions (relatively easy to prove) t0 W y1 t0 y t0 y 1 t0 y t0 y t c y t c y t If y t and y t are two solutions of p t y q t y r t y 0, 1 and if there is a point at which the Wronsian - is non-zero, then the family of solutions Non-homogenous solution form (easy to prove) 1 1 with arbitrary coefficients c and c includes every solution. 1 The general solution of a linear second order non-homogenous ODE can be written 1 1 1 1 1 1 Y t is som i n the form y t c y t c y t Y t, where c y t c y t is the solution to the corresponding homogenous equation, and that satisfies the non-homogenous equation. e particular solution So again, we have to solve the homogeneous equation first (Don t worry about non-homogeneous with these)
Second order linear ODE with polynomial coefficients P t y t Q t yt R t y t G t where: P t Q t R t = polynomial of t = polynomial of t polynomial of t We will see analytical solutions (written as a power series around point t ): y t a t t 0 0 Important example series - Taylor series: 0 t0 f y t t t! 0 0
Convergence of power series A power series around x, a x x converges at a given x if 0 0 0 lim a x x0 a finite number. Otherwise, the series diverges at x. 0 INTERVAL and RADIUS OF CONVERGENCE: Every power series has an interval of numbers for which the series converges There is a non-negative number called the radius of convergence such that 0 a x x converges for x x x and diverges elsewhere. 0 0 0
Ways to determine convergence radius Given a power series L= x x lim when when 0 a a 1 0 a x x 0, compute L 1, the series converges at a given x when L 1, the series diverges at a given x L 1, the test is inconclusive The radius of convergence for each of the series solutions is at least as large q t r t as the minimum of radii of convergence of the series and. p t p t 0 0 qt pt qt e of pt t p t If p t and q t are polynomials, then has a convergent power series around t if p t 0. Further, if we assume that any factors common to p t and q t have been cancelled, then the radius of convergenc 0 This can be hard to compute It is easier (& sufficient for exam) to find the bound around is preciselly the distance of to the nearest zero of (including complex zeros). t 0
EXAMPLE OF POWER SERIES SOLUTION Solve t 1 y 4ty 6y 0 using power series around t t0 0 LOWER BOUND ON CONVERGENCE RADIUS = 1 We see solution of the form y t a t t a t 0 0 0 1 1 1 y t a t y t a t We plug into our differential equation 1 0 1 t 1 1 a t 4t a t 6 a t 0 First we absorb terms into the series 1 at 1 at 4at 6at 0 1 0 WE WANT THE SAME POWERS shift indices at a t at at 0 1 0 1 1 4 6 0
EXAMPLE OF POWER SERIES SOLUTION NEXT WE WANT THE SAME STARTING INDICES pull out terms 1 1 1 a t at at 0 1 0 1 4 6... at a t at at... 1 1 4 6 0 a 6a t 4a t 6a 6 a t... 3 1 0 1... 1 a 1 a 4a 6a t 0 0 a 6a t 6a a t... 0 3 1 Add the series + algebra All polynomial coefficients must be zero... 1 a 1 a 4a 6a t 0
a 6a 0 0 6a a 0 3 1 EXAMPLE OF POWER SERIES SOLUTION 1 4 6 1 1 a 1 a 4a 6a 0 for,3,4,... 1 4 6 a 1 a 0 for,3,4,... a a for,3,4,... variables a, a a a 3a 0 1 a 3 3 1 SOLUTION: 0 1 a a 0 1 3 3 1 1 4 6 4 3 4 3 6 a a 5 3 3 3 0 1 =0 =0 THE SERIES HAPPENS TO TERMINATE 3 3 3 1 3 y t at a0 a1t at a3t a0 a1t 3a0t a1t 3 1 a 1 3t a t t 3 a 4 0 This has infinite radius of convergence, (lower bound theorem told us at least 1)
Is there always a power series solution around t=t 0 to the second order linear ODE? 0 p t y t q t y t r t y t 0 The answer is YES when t 0 is an ORDINARY POINT The answer is NO when t 0 is a SINGULAR POINT qt p t r t p t A point t is said to be an ORDINARY POINT if and are analytic around t t. If p t, q t, r t are polynomials, then this means p t 0. 0 qt r t y t yt y t p t p t 0 0 0 qt p t r t p t A point t is said to be an SINGULAR POINT if and are not analytic around t t. If p t, q t, r t are polynomials, then this means p t =0. 0 0
EULER S EQUATION is an illustrative example with a singular point t y t aty t by t where a and b are real numbers 0 NOT A WHOLE NUMBER IN GENERAL Assumed solution form: r y t t where r is to be determined Derivatives: y t rt y t r r 1 t r 1 r Plugging into our ODE: t y t aty t by t 0 r r1 r t r r 1 t atrt bt 0 r r r r r 1 t art bt 0 r t r r 1 ar b 0 r t r a r b 1 0
r t r a r b 1 0 r a r b r 1, 1 0 EULER S EQUATION 1 a a 1 4b a a DOES THIS REMIND YOU OF A CHARACTERISTIC EQUATION? Case 1: 1 4b 0 DIFFERENT REAL ROOTS Case : 1 4b 0 REPEATED REAL ROOTS a Case 3: 1 4b 0 COMPLEX ROOTS
GENERAL SOLUTIONS TO EULER S EQUATION -- employing absolute value 0 when 0 t y t aty t by t t Different real roots: when a1 4b 0 Repeated real roots: when a1 4b 0 r1 r y t c t c t r y t c t c ln t t 1 1 1 1 Complex roots: when a1 4b 0 y t = c t cos ln t c t sin ln t 1 r
Regular (wea) vs irregular (strong) singularities qt 0 where and p t p t y t q t y t r t y t 0 where p, q, r are polynomials y t u t y t v t y t u t v t REGULAR SINGULARITY: t t0 u t v t t t If appears at most to the first power in the denominator of and at most to the second power in the denominator of, then is a regular singular point. Other inds of singularities are IRREGULAR -- advanced, not in this class 0 r t p t ASSUMING COMMON FACTORS HAVE BEEN CANCELLED Proposed solution form around a regular singular point t 0 r t t a t t 0 0 0 FROBENIUS THEOREM TELLS US AT LEAST ONE OF THIS FORM EXISTS AROUND A REGULAR SINGULARITY
Example around regular singular point Solve t y t ty t t 1 y t 0; t0 0 singular Solve for t 0 0 1 0 Proposed solution form: y t t a0 a1t at... t a t a t y t a r t r1 y t a r r t r r r 0 0 r Plugging it into t y t ty t t 1 y t 0 1 1 r r 1 t a r r r t ta r t t at 0 0 0 0 r r r a r r r t a r t at at 0 0 0 0 1 1 0 ABSORBING THE TERMS INTO THE SERIES
r r r a r r r t a r t a t at 0 0 1 0 1 1 0 SHIFTING THE INDICES SO THAT POWERS ARE THE SAME a r r 1 t a rt a t... r r r 0 0 0 r r r r a r r t a r t a t at 1 1 1 1 1.. 1 0 r r r 0 0 0 1 r 1 a r r 1 t a rt a t a r r 1 a r a a t 0 SUMMING THE SERIES r PULLING OUT TERMS SO THAT STARTING INDICES ARE THE SAME r r 3r 1 a t a r r 3 a a t 0 0 1 1 r Used to solve for r r 3r1 0 1, a0 0 3 9 8 3 17 4 4 Used to get recurrence relations 1 a r r 3 a a 0 one recurrence with r r1 another recurrence with r r
r r r 3r 1 a t a r r 3 a a t 0 0 1 1 r Used to solve for r r 3r1 0 1, 3 9 8 3 17 4 4 Used to get recurrence relations 1 a r r 3 a a 0 one recurrence with r r1 another recurrence with r r a a 1 r r 3 1 r r 1 1 r 1 y1 t t a t 0 a r a 1 3 1 y t t a t 0 General solution: r 1 r y t c1 y1 t c y t c1t at ct a t 0 0
What about the case of negative t (we employ a similar tric of transforming variables lie we did with Euler eq.) t y t ty t + t 1 y t = 0 when t < 0 We can do it just via change of variables: Let t dy dz dv dz dt dv dt dv d y d dz dv d z dt dv dv dt dv Let y v z v Plugging into the ODE: Let t v WE ENDED UP WITH ANOTHER EQUATION v z v vz v + v 1 z v = 0 WE SOLVE THIS ONE & THEN PLUG IN v = -t