# MAT244H1

MAT244H1a.doc Introduction INTRODUCTION TO DIFFERENTIAL EQUATIONS A differential equation is an equation involving some hypothetical function and its derivatives. Example y ′′ + 2 y ′ = x is an differential equation. As such, the differential equation is a description of some function (exists or not). A solution to a differential equation is a function that satisfies the differential equation. Example y = x 3 + 5x • • • • is a solution to y ′′ = x + y − x 3 . Some differential equations are famous/important: y′ = y y = e x , . ax y ′ = ay y = e , . y ′′ + y = 0 y = cos x , . y ′′ + ay = 0 y = cos a x , . Recall that a differential equation describes a phenomenon in terms of changes. For example, if P = mv , then dP =F Fdt = dm ⋅ v or dt . Example A pool contains V liters of water which contains M kg of salt. Pure water enters the pool at a constant rate of v liters per minute, and after mixing, exits at the same rate. Write a differential equation that describes the density of salt in the pool at an arbitrary time t. M (t ) ρ (t ) = ρ ( t) V . • Let be the density at time t. Then • To model change in ρ 2 − ρ1 v V ≈ − ρ1 t 2 − t1 V ρ (t ) , let ρ1 = ρ (t1 ) and ρ 2 = ρ (t 2 ) . Then (ρ 2 − ρ1 )V ≈ − ρ 1v(t 2 − t1 ) , so ρ (t + ∆t ) − ρ (t ) v V ≈ − ρ (t ) or t + ∆t − t dρ v v = − ρ (t ) ρ ′ = −ρ V or V . • Now, as t → 0 , dt V . • Without solving this equation, we can predict facts about this system. As t → ∞ , ρ → 0 . dρ v = − dt ρ V • To solve this differential equation, write . Integrating both sides, we get ln ρ (t ) = − v t +C V ρ (t ) = e v − t +C V = eC e v − t V = Ae Page 1 of 32 v − t V . MAT244H1a.doc ρ (0 ) = ρ to • If we add ρ ′ = −ρ v V , then we have an IVP (initial value problem). ISSUES ABOUT THE USE OF DIFFERENTIAL EQUATIONS 1) How to translate a real problem to a differential equation. Keep your eyes open! 2) Some patterns of nature are ill-defined. Use different points of view and different mental differential equation models to reformulate them. 3) Differential equations have infinitely many solutions. Which one is yours? The initial value are extremely important. 4) There may be no analytic solution found. • Is there a solution? • Is this solution unique? • If a numeric answer is required, i.e. the value of the solution at one particular point, then use numerical approximations. It does not give any feelings for the pattern, nor does it give elbow room. • Use theoretical analysis if you need the behavior of the solution. This does not give any values. • To know the behavior locally/in a neighborhood, solve in series. 5) The data does not fit you solution. You need to repeat (as in a feedback/controlled system). NOTATIONS WITH REGARD TO THE INPUT/OUTPUT SYSTEMS Example xy ′′ + 2 y ′ − sin x ⋅ y = tan x can be written as L[ y ] = tan x . Solve it, and the answer is the output. • L[ y ] is the “black box system”. • tan x is the ‘input”. For theoretical purposes, mathematicians use these equivalents: y ′′′ = y ′′′ = f ( x, y , y ′) or F (x, y, y ′, y ′′, y ′′′) = 0 . LINEAR VS. NON-LINEAR DIFFERENTIAL EQUATIONS 1 1 y′ + e x y = is linear. x 1 + tan x • tan x ⋅ y ′′ + • y ′′′ + y ′ + y 2 = 0 is non-linear. First Order Differential Equations LINEAR EQUATIONS Page 2 of 32 2 y′ tan x sin x + y− is the same as x x x MAT244H1a.doc First order linear equations have the form y ′ + p(t ) y = g (t ) . Derivation • Suppose I can find µ (t ) so that µ (t ) p(t ) = µ ′(t ) . • Multiply both sides of y ′ + p(t ) y = g (t ) by µ (t ) : µ (t ) y ′ + µ (t ) p(t ) y = µ (t )g (t ) which is µ (t ) y ′ + µ ′(t ) y = µ (t )g (t ) , i.e. (µ (t ) y )′ = µ (t )g (t ) . • Integrate both sides: µ (t ) y = µ (t )g (t )dt + C , so y (t ) = • But what is µ (t ) ? Since µ (t ) p(t ) = µ ′ ⇔ 1 µ (t )g (t )dt + C . µ (t ) µ ′(t ) = p(t ) , therefore ln µ (t ) = µ (t ) p(t )dt . So µ (t ) = e p (t )dt . General Solution To solve y ′ + p(t ) y = g (t ) , 1) Let µ (t ) = e p (t )dt (no constant needed). 1 2) The solution is y (t ) = µ (t )g (t )dt + C . µ (t ) [ ] Example 2 Solve y ′ + 2ty = 2te −t . • Here, p(t ) = 2t , g (t ) = 2te −t . 2 • Let µ (t ) = e 2tdt 2 = et . • The solution is y (t ) = 1 t2 [e e • Note that as t → ∞ , y → 0 . t2 2 ] 2te −t dt + C = 1 e t2 [ 2tdt + C ]= e1 (t t2 2 ) 2 2 + C = t 2 e −t + Ce −t . Importance of Analysis One needs to have an understand of the solution before (even after) solving it, with respect to: 1) Behavior of the solution as t → ∞ . 2) The nature and behavior of the solution within a family (depends on y0). Variation of Parameter Recall that the family of solutions of a first order linear differential equation y ′ + p (t ) y = g (t ) , µ (t ) = e p (t )dt , 1 1 C y (t ) = µ (t )g (t )dt + C = µ (t )g (t )dt + . Notice that the family of solutions is generated by µ (t ) µ (t ) µ (t ) µ (t ) . This leads to the technique of variation of parameter. [ ] Recall a differential equation L[ y ] = g (x ) . If g (x ) = 0 , then we have a zero-input system, or a homogeneous differential equation L[ y ] = 0 which describes the solutions to a great extent. Page 3 of 32 MAT244H1a.doc 1 1 For example, consider y ′ + y = 3 cos 2t . First solve the corresponding homogeneous equation y ′ + y = 0 t t 1 dt t 1 (0 + C ) = C . Then the general solution to µ (t ) t 1 1 1 1 y ′ + y = 3 cos 2t looks like y (t ) = A(t ) . Then y ′ = A′(t ) − A(t ) 2 , so t t t t 1 1 1 1 A′(t ) − A(t ) 2 + A(t ) = 3 cos 2t A′(t ) = 3t cos 2t A(t ) = (3t cos 2t )dt . Thus, t t t t 1 (3t cos 2t )dt + C . y (t ) = t to find y 0 (t ) . Since µ (t ) = e [ = t , y 0 (t ) = ] ASYMPTOTIC BEHAVIOR OF SOLUTIONS Recall from calculus that f (x ) and g ( x ) are asymptotes of one another if lim ( f (x ) − g (x )) = 0 . x →∞ Example 2t − 5 and 2t − 5 + ce −t and 2t − 5 + c are asymptotic to each other. t SEPARATION OF VARIABLES Idea A differential (not necessarily linear) may appear as g ( y )dy = dy f (x ) = . Then, g ( y )dy = f (x )dx , and the solution is dx g ( y ) f (x )dx . Note Other ways a separable differential equation can appear as: M ( y ) = dy dy N (x ) or f (x )g ( y ) = . dx dx Example dy 1 = x y . So dy = xdx dx y y − 1 2 dy 1 = xdx 2y 2 = x2 +C. 2 Note The solution about is an implicit solution. When write a solution explicitly, be careful! Pay attention to the domain and range. Page 4 of 32 MAT244H1a.doc Example dy = x3 y dx y 2 1 3 dy 2 y3 3 3 x2 1 y = − 2 2 2 x2 1 − >0 3 3 − 2 = 3 3 3 x2 y = + C . Now, suppose y (2) = 1 , then = 2 + C 2 2 2 xdx x2 1 = − 3 3 x 2 −1 > 0 x2 >1 x2 1 y= − 3 3 3 2 C=− 1 . So 2 . We need x > 1, x < −1 . IMPLICIT VS. EXPLICIT SOLUTIONS For separable (not linear in general), we have implicit solutions (not necessarily functions). So as the solutions to non-linear equations are implicit, the analysis of the solution is very difficult, and we need to know if such solutions have explicit forms or not (and where). We need to know an interval on which explicit solutions exists. Example Consider 2 y dy = 1 . The solution is y 2 = x + c . dx • Now, if y (1) = 1 , then c = 0 and the solution is y 2 = x . An explicit solution y = x exists on the interval (0, ∞) . This solution can’t be extended to y < 0 because it doesn’t pass the vertical line test. dy is undefined. This indicates the possibility of a problem with defining an dx explicit solution to the differential equation. • At the point (0,0) , Theorem: Implicit Function Theorem If F (x, y ) = 0 and (a, b ) is such that F (a, b ) = 0 and if F y (a, b ) ≠ 0 , then we have an explicit function y = f (x ) on an interval containing (a, b ) . Conclusion: On any interval as long as dy is defined, we will have an explicit solution. dx Example dy x + 3 y Solve = using separation of variables. dx x− y • Let v = y x y = vx dy dv = v+ x . dx dx Page 5 of 32 MAT244H1a.doc dy x + 3 y = = • So dx x− y 1− v (1 + v ) 2 dv = y x becomes v + x dv = 1 + 3v y dx 1 − v 1− x 1+ 3 dv 1 + 3v (1 + v )2 . So we have = −v = dx 1 − v 1− v 1 dx . x • Let w = 1 + v , dv = dw . So 2 − − ln w = ln x + C w − x 2−w 2 dw = 1 dx x 2 2 w w 2 − − ln 1 + v = ln x + C − 1+ v x+ y 2x − ln = ln x + C x+ y x − 1 dw = ln x + C w dw − 2 1+ y x − ln 1 + 2x − ln x + y + ln x = ln x + C x+ y y = ln x + C x − 2x − ln x + y = C . x+ y • Our techniques (integration) limits us to 1 + v = w ≠ 0 , but are we excluding some solutions? Now x + 3 y −2 x y x+ y dy 1+ v = 1+ = = 0 means y = − x . Notice that if y = − x , = −1 and = = −1 , so x− y 2x x x dx we can add y = − x to the family of solutions. ISSUES ON MODELING Example: Money Growth • If we say that annual rate of interest is 5%, we mean that on $100, we get $5 in one year. So P(1) − P(0) = 0.05P(0 ) . • Equivalently, dP(t ) = 0.05P (t ) . So P(t ) = Ae 0.05t dt P(1) = P (0)e 0.05 = P(0) 1 + 0.05 + P(1) − P(0) = 0.05P(0) + (0.05)2 2! + P(t ) = P(0)e 0.05t . So = P (0) + 0.05P(0) + (0.05)2 P(0) + 2! , so (0.05)2 P(0) + > 0.05P(0) . 2! • If we say the annual interest rate r is compounded semi-annually, we mean P r r r r 1 r 1 r 1 − P(0) = P (0) ⇔ P = P(0) 1 + . So, P(1) = P 1+ = P(0) 1 + 1+ = P(0) 1 + 2 2 2 2 2 2 2 2 2 • In general, when we have an annual interest rate r compounding n times a year, • Compounding continuously means n → ∞ . So P(1) = lim P(0 ) 1 + n →∞ • If we contribute continuously to the account, then r n n r P(1) = P (0) 1 + n Page 6 of 32 . n . = P(0)e r . Similarly, P(1) = P(0)e rt . dP(t ) = rP (t ) + k where k is the constant contribution. dt EXISTENCE OF A UNIQUE SOLUTION 2 MAT244H1a.doc Example • Recall that a differential equation can be built starting from the solution like y= x y=− x . They both can be “expressed” by y 2 = x and 2 yy ′ = 1 . • Notice that the differential equation 2 yy ′ = 1 at (0,0) is a confused initial value problem. Note Recall that given a first order differential equation y ′ + p(t ) y = g (t ) , the solution is [ ] 1 µ (t )g (t )dt + C where µ (t ) = e µ (t ) functions, then we have a good solution. y= p (t )dt . So as long as p(t ) and g (t ) are integrable (continuous) Theorem If p(t ) and g (t ) are continuous on I = [α , β ] , then for any value y0 (that is already given), there is a unique solution y = φ (t ) that on I it satisfies y ′ + p(t ) y = g (t ) , y 0 = y (t 0 ) where t 0 ∈ I . Theorem Let y ′ = f ( y, t ) . If there is an “open window” I × J on which f and ∂f are continuous, then the initial value ∂y problem y ′ = f ( y, t ) , y 0 = y (t 0 ) has an unique solution (for any t 0 ∈ I and y 0 ∈ J ⇔ (t 0 , y 0 ) ∈ I × J ). Note Notice that y ′ = 1 is not continuous at any neighborhood of (0,0) , so the theorem doesn’t apply. 2y Example ( ) 3 For which initial values does y ′ = t 2 + y 2 2 = f ( y, t ) have a unique solution? • f is continuous everywhere. 1 ∂f 3 2 • = t + y 2 2 2 y is also continuous everywhere. ∂y 2 ( ) ( ) • So y ′ = t 2 + y 2 3 2 has a unique solution for all initial values. Example For which initial values does y ′ = cot y = f ( y, t ) have a unique solution? 1+ y • f is discontinuous at y = −1 and y = kπ . • ∂f csc 2 y (1 + y ) − cot y = is also discontinuous at y = −1 and y = kπ . ∂y (1 + y )2 Page 7 of 32 MAT244H1a.doc BERNOULLI Bernoulli solved y ′ + p(t ) y = y n g (t ) as follows: v = y 1− n , and v ′ = y′ y n + p(t ) dv = (1 − n ) y 1− n −1 y ′ = (1 − n ) y − n y ′ dt y yn = g (t ) y − n y ′ + y 1− n p(t ) = g (t ) . Let 1 v ′ = y − n y ′ . So now we get 1− n 1 v ′ + p(t )v = g (t ) . 1− n AUTONOMOUS DIFFERENTIAL EQUATIONS Autonomous differential equations look like y ′ = f ( y ) . Examples 1) y ′ = 0 ; y = c . 2) y ′ = k ; y = kx + c . 3) y ′ = ry exponential growth. 4) y ′ = y (1 − y ) is know as logistic growth. Example: Logistic Growth Model Consider the spread of a disease. If y (t ) is the number of infected population, then dy ≈ y (k − y ) . So dt dy y y = ay(k − y ) = aky 1 − = ry 1 − , k is “environmental carrying capacity” or “saturation level” and r is dt k k the “intrinsic growth rate”. To solve it, dy y 1− ln y y 1− k = rt + C y k = rdt y y 1− k = Ae rt 1 + y 1 1− y k dy = rdt y = Ae rt − Ae rt y k ln y − ln 1 − y + Ae rt y = rt + C k y = Ae rt k y 1+ A rt e = Ae rt k y0 Ae rt A kA k = = − rt =A . Now, y (0 ) = y 0 means = A . So y0 A A ke + A k − rt 1 + e rt e − rt + e +1 1− k k A k ky 0 . As t → ∞ , y → k . y (t ) = = y 0 + (k + y 0 )e − rt y (t ) = Example Page 8 of 32 MAT244H1a.doc An important model: dy k = ry (M − ln y ) = ry (ln k − ln y ) = ry ln . dt y Example: Logistic Growth With Threshold Some times we need enough of y0 to start the epidemic. Let points at y = 0, k , T . Now we get dy = −ry (k − y )(T − y ) = f ( y ), r > 0 , with critical dt dy y y = ry 1 − 1− . dt k T CRITICAL POINTS OR EQUILIBRIUMS OF AN AUTONOMOUS DIFFERENTIAL EQUATION Example y ′ = y (1 − y ) = 0 gives constant solutions y (t ) = 0 and y (t ) = 1 . These are the critical points or the equilibriums. y semi-stable equilibrium y (t ) = 1 unstable equilibrium t y (t ) = 0 Example y ′ = y (1 − y ) = 0 gives constant solutions y (t ) = 0 and y (t ) = 1 . y asymptotically stable equilibrium y (t ) = 1 y (t ) = 0 t Example: Schaefer Model Page 9 of 32 MAT244H1a.doc Let y (t ) be the fish population at time t, which follows the logistic growth model. Then where E the rate of harvest which is proportional to the fish population. Rewriting equilibriums are at y (t ) = 0 and r 1 − is at y = y (t ) −E =0 k dy y = ry 1 − − Ey , dt k dy y = y r 1− − E , the dt k k y (t ) = (r − E ) . Now, if r > E , the stable equilibrium r k (r − E ) . But if r < E , the stable equilibrium is at y = 0 . r PARAMETRIC DIFFERENTIAL EQUATIONS Example Consider y ′ = ay − y 2 = y (a − y ) . The equilibriums are at y = 0 and y = a . • When a > 0 , we have a stable equilibrium around y = a . • When a = 0 , y ′ = y 2 and we have a semi-stable solution around y = 0 . • When a < 0 , we have a unstable solution around y = a . Here, a = 0 is called a bifurcation point. EXACT DIFFERENTIAL EQUATIONS Suppose ψ ( x, y ) = c . It implicitly defines a function y ( x ) . Now dy dy ∂ψ ∂ψ dy d ψ ( x, y ) = 0 = + := M ( x, y ) + N (x, y ) . So M (x, y ) + N (x, y ) = 0 or dx ∂x ∂y dx dx dx M (x, y )dx + N (x, y )dy = 0 . This type of differential equations are called exact equations. Existence of Solution If we are given M (x, y )dx + N (x, y )dy = 0 , how would we know there is such a ψ (x, y ) corresponding to it? Indeed the condition Example Consider y ∂M 1 ∂N 1 + 6 x dx + (ln x − 2)dy = 0 = M (x, y ) + N (x, y ) . Since = and = , the equation is x ∂y x ∂x x exact. Notice that Now, ∂M ∂N = = is necessary and sufficient for the existence of such ψ (x, y ) . ∂y ∂x y ∂ψ = M (x, y ) = + 6 x . Integrating with respect to x, we get ψ (x, y ) = y ln x + 3x 2 + f ( y ) . ∂x x ∂ψ = N ( x, y ) = ln x + f ′( y ) = ln x − 2 , so f ′( y ) = −2 ∂y ψ (x, y ) = y ln x + 3x 2 − 2 y + c 0 . The solution is ψ (x, y ) = C Page 10 of 32 f ( y ) = −2 y + c 0 . Hence y ln x + 3 x 2 − 2 y = c . MAT244H1a.doc INTEGRATING FACTOR • What if M y ≠ N x ? Does that mean ψ (x, y ) doesn’t exists? Yes, but we may be able to change the differential equation M (x, y )dx + N ( x, y )dy = 0 to a new “better” one. • If we multiply M (x, y )dx + N ( x, y )dy = 0 by µ (x, y ) , we may get (µM ) y = (µN ) x . • How do we know if we should look for such µ? Finding µ is very difficult, unless µ is somehow a µ (x ) or µ ( y ) only. Example • Suppose (µM )dx + (µN )dy = 0 . We check if it is exact, and if it is, that fact should lead us to the answer. • To be exact, we need (µM ) y = (µN ) x µ y M − µ x N = µ (N x − M y ) . µ y M + µM y = µ x N + µN x ( • Now if µ ( x, y ) = µ (x ) , then µ y = 0 . Then − µ x N = µ N x − M y (N x − M y ) µx =− . µ N ) Example This tells us a criterion for finding integrating factors of different type: µ ( xy ) . If there exists a µ ( xy ) . Nx −M y xM − yN = R( xy ) , then Example Consider 3x + Nx −M y xM − yN y dy 6 x2 + +3 = 0 . It is obvious M y ≠ N x , but y y x dx 2 = y 6 x −3 2 + 2 y x y 2 2 = y 6 x −3 2 + 2 y x y 2 y y x x − x2 −3 2x 2 + 6 − 3 y x y x makes this differential equation exact. 3x 2 + 6 = 1 xy 2x 2 − 3 2x 2 + 6 y2 x +6 x y 2 y x −3 y x = 1 . So there is µ ( xy ) that xy Note We were looking for an integrating factor µ ( xy ) so that µM + µNy ′ = 0 is exact, i.e. (µM ) y = (µN ) x . This means µ y M + µM y = µ x N + µN x . But since we are requiring that µ ( xy ) is a function of xy, then by letting w = xy , µ ( xy ) becomes µ (w ) , and µ x = becomes µ ′xM + µM y = µ ′yN + µN x dµ ∂w = µ ′y and µ y = µ ′x . Therefore the condition of exactness dw ∂x µ′ Nx − M y µ ′(xM − yN ) = µ N x − M y = = R(w) = R(xy ) . µ xM − yN ( ) Now, we can easily solve for µ. NUMERICAL APPROXIMATION SOLUTIONS TO DIFFERENTIAL EQUATIONS Page 11 of 32 MAT244H1a.doc If g (t ) is a solution to the differential equation y ′ = f (t , y ) and we need to know y1 = g (t1 ) and an approximate value is good enough, then we can use the tangent line instead of g (t ) . y ′ = y1 − y 0 = y ′(t1 − t 0 ) y1 = y ′(t1 − t 0 ) + y 0 y1 = y 0 + f (t 0 , y 0 )(t1 − t 0 ) . y1 − y 0 t1 − t 0 Euler suggested to evaluate y1 = y 0 + f (t 0 , y 0 )(t1 − t 0 ) , y 2 = y1 + f (t1 , y1 )(t 2 − t1 ) , y 3 = y 2 + f (t 2 , y 2 )(t 3 − t 2 ) . So, to calculate g (T ) , we can take [t 0 , T ] instead of using y1 = y 0 + f (t 0 , y 0 )(t1 − t 0 ) and subdivide it into t 0 , t1 , , t n −1 , T . This way, the answer is a lot closer. Convergence of Euler’s Method If we let h → 0 (step size), then the approximate answer equals the actual answer. So for first order differential equations, it is a good idea to let h → 0 to get a better solution. Example Prove that Euler’s method converges for y ′ = y, y (0 ) = 1 . We know the solution is y (t ) = e t . • Let h = t . n • Now, y1 = y 0 + y ′(0,1)h = 1 + h , y 2 = 1 + h + y ′(t 2 ,1 + h )h = 1 + h + (1 + h )h = (1 + h )2 , so y n = (1 + h )n . • If n → ∞ , then (1 + h )n = 1 + t n n = e t = y (t ) . EXISTENCE AND UNIQUENESS OF A SOLUTION TO AN INITIAL VALUE PROBLEM Technique An initial value problem y ′ = f (t , y ), y (0 ) = y 0 can be transformed into w ′ = g (t , w), w(0 ) = 0 . Example Consider y ′ = t 2 + y 2 , y (1) = 2 . Let t = s + 1 , then y (t ) = z (s ) , so the problem becomes z ′ = (s − 1)2 + z 2 , z (0) = 2 . Now let w = z − 2 ⇔ z = w + 2 , then z ′ = w ′ , so w ′ = (s + 1)2 + (w + 2 )2 . Now, w(0 ) = z (0 ) − 2 = 2 − 2 = 0 . Theorem If y ′ = f ( y, t ) and if f and ∂f are continuous on the rectangle I × J , then for some h > 0 there is an unique ∂y solution of the form y = φ (t ), ∀t 0 − h < t < t 0 + h . Note Page 12 of 32 MAT244H1a.doc To prove this theorem, we first showed it is alright to assume y 0 = 0 and t 0 = 0 , so we assume that our IVP is of the form y ′ = f ( y, t ), y (0) = 0 . The method of the proof is Picard’s Method Note Observe that y = φ (t ) is a solution to y ′ = f ( y, t ), y (0) = 0 . So it also satisfies y ′ = φ ′(t ) = f (φ (t ), t ) and t t 0 0 y = φ (t ) = φ ′(s )ds = f (φ (s ), s )ds . So the IVP is equivalent to an integral equation. 0 Notice that φ (0) = φ ′(s )ds = 0 , i.e. y (0) = 0 . 0 Picard’s Method 1) Find a function φ 0 (t ) that satisfies φ 0 (0) = 0 (ex: φ 0 (t ) = 0 ). 2) Define φ n (t ) such that φ k +1 = t 0 f (φ k (s ), t )ds, k = 0,1, . So we create a sequence of {φ n (t ), n = 0,1, } . Now: 1) Is this an infinite sequence? • If at some point we stop producing new functions when n = k , then we already have the solution φ k (t ) of the integral equation. • If the sequence is infinite, then does it converge? To converge, we need g n (t ) = φ n +1 (t ) − φ n (t ) to converge, i.e. lim g n (t ) = 0, ∀t ∈ [t 0 − h, t 0 + h] , or lim φ n (t ) = φ (t ), ∀t ∈ [t 0 − h, t 0 + h] . n →∞ • n →∞ } is defined, and φ (t ) = lim φ n (t ) exists, then what properties does φ (t ) have? 2) Suppose {φ n (t ), n = 0,1, n→ ∞ φ (0) = lim φ n (0) = lim 0 = 0 . n →∞ t • 0 f (φ (s ), s )ds = n →∞ t 0 f lim φ n (s ), s ds = So φ (t ) is a solution. n →∞ t lim f (φ n (s ), s )ds = lim 0n →∞ t n →∞ 0 f (φ n (s ), s )ds = lim φ n −1 (t ) = φ (t ) . n →∞ Steps of the Proof of the Existence and Uniqueness Theorem We constructed a sequence of functions {φ n (t ), n = 0,1, } with the hopes that the sequence converges to the solution φ (t ) . We saw that lim φ n (t ) satisfies the criteria for the solution. n →∞ 1) When does this sequence converge? If f (t , y1 ) − f (t , y 2 ) ≤ K y1 − y 2 ⇔ 2) ∂f exits and is continuous in D, then there is a number K such that ∂y f (t , y1 ) − f (t , y 2 ) y1 − y 2 ≤K. f (t , φ n (t )) − f (t , φ n −1 (t )) ≤ K φ n (t ) − φ n −1 (t ) . 3) Because f (t , y ) is continuous on D, then for some h, for all t 0 − h < t < t 0 + h , f (t , y ) ≤ M . Then φ1 (t ) = t 0 f (s, φ 0 (s ))ds ≤ M t , and because φ 0 = 0 , φ1 (t ) − φ 0 (t ) ≤ M t . Also, Page 13 of 32 MAT244H1a.doc t φ 2 (t ) − φ1 (t ) ≤ 0 t t 2 0 0 2 (MKs )ds = MK t f (s, φ1 (s )) − f (s, φ 0 (s ))ds ≤ K φ1 (s ) − φ 0 (s )ds ≤ MK 2 t n +1 . (n + 1)! 4) Now, notice that φ n (t ) − φ1 (t ) = (φ n (t ) − φ n −1 (t )) + (φ n −1 (t ) − φ n − 2 (t )) + . So, φ n (t ) − φ n −1 (t ) ≤ inequality, φ n (t ) − φ1 (t ) ≤ φ n (t ) − φ n −1 (t ) + φ n (t ) ≤ φ n (t ) + φ n (t ) − φ n −1 (t ) + + φ 2 (t ) − φ1 (t ) . Finally, + φ 2 (t ) − φ1 (t ) ≤ M t + MK t2 + 2 + (φ 2 (t ) − φ1 (t )) , so by the triangle = (Kh )2 + M Kh + K 2! . As n → ∞ , it M M , so all the φ n are bounded by e Kh . K K 5) We need to prove the uniqueness of this solution. Suppose there are two solutions, φ (t ) and ψ (t ) . Then converges to e Kh φ (t ) −ψ (t ) = U (t ) = t 0 t f (s, φ (s )) − f (s,ψ (s ))ds ≤ 0 0 t f (s, φ (s )) − f (s,ψ (s )) ds ≤ K φ (s ) −ψ (s ) ds . Now let 0 t φ (s ) −ψ (s ) ds , where U (t ) ≥ 0 . Notice U ′(t ) = φ (t ) −ψ (t ) ≤ K φ (s ) −ψ (s ) ds = KU (t ) . So e − kt U ′ − e − kt KU ≤ 0 U ′ − KU ≤ 0 But t U (t ) ≥ 0 U (t ) = 0 U (t ) ≤ 0 (e − kt U ) ≤ 0 , therefore, (e ′ t 0 − ks 0 ) ′ U (s ) ds = e − kt U (t ) ≤ 0 U (t ) ≤ 0 . φ (t ) = ψ (t ) . First Order Difference Equations Examples of famous discrete process are stochastic processes y n +1 = f (n, y n , y n −1 , , y 0 , b) . However, we can only study simple ones y n +1 = f (n, y n ) . Compare it with y ′ = f (t , y ) . Notice that y n +1 − y n (n + 1) − n = f (n, y n ) − y n = g (n, y n ) ≈ y n′ = g (n, y n ) . (n + 1) − n A solution to a difference equation is a sequence of numbers y = y1 , Notice that the sequence defines a function with domain {0,1,2, } . So, , y n that satisfies y n +1 = f (n, y n ) . y 0 = y (0) , y1 = y (1) , etc. Notice that if the increments become small, the difference equation becomes a differential equation. Example Solve y n +1 = • y1 = n+3 y n with respect to y0. n +1 0+3 3 1+ 3 y0 = y0 , y2 = y1 = 0 +1 1 1+1 • So, y n = 4×3 y0 , y3 = 2 ×1 (n + 2)(n + 1) (4)(3) y = (n + 2)(n + 1) y . 0 0 n(n − 1) (2 )(1) 2 Page 14 of 32 2+3 y2 = 2 +1 5× 4×3 y0 . 3 × 2 ×1 MAT244H1a.doc Definitions 1) We say the annual rate is r compounding monthly to mean the period is one monthly and the rate on that r . period is 12 2) Effective annual rate equivalent to rate r compounding monthly is the rate re (compounding annually) such r that the effect of re over a year is the same as the effect of for 12 periods. That is, if Y1 = (1 + re ) y 0 and 12 y12 = 1 + r 12 12 y 0 , then Y1 = y12 . Note To calculate the effective annual rate, 1 + re = 1 + r 12 12 . AUTONOMOUS DIFFERENCE EQUATIONS The simplest of difference equations is autonomous linear y n +1 = f ( y n ) where f ( y n ) = ρy n + b . Solution To solve an autonomous linear difference equation, note that: • y1 = f ( y 0 ) = ρy 0 + b . • • y 2 = f ( y1 ) = ρy1 + b = ρ (ρy 0 + b ) + b = ρ 2 y 0 + ρb + b . ( ) y 3 = f ( y 2 ) = ρ ρ 2 y 0 + ρb + b + b = ρ 3 y 0 + ρ 2 b + ρb + b . So y n = ρ n y 0 + ρ n −1b + ( ) + ρb + b = ρ n y 0 + b ρ n −1 + + 1 = ρ n y0 + b ρ n −1 . ρ −1 Note Notice that if ρ < 1 , then as n → ∞ , y n = ρ n y 0 + b y n = ρ n y0 + b ρ n −1 1 →b . However, if ρ > 1, ρ > 0 , as n → ∞ , ρ −1 1− ρ ρ n −1 1 →b ; but if ρ > 1, ρ < 0 , we are confused (the limit doesn’t exist)! ρ −1 1− ρ Example Consider a 20 year mortgage at rate 10% (compounded semi-annually) with monthly payment of b = 1000 . What will the maximum loan be under the these restrictions? • Model: y n +1 = (1 + i ) y n − 1000 . At n = 12 × 20 = 240 , y 240 = 0 . r • What is the periodic rate i? 1 + 2 2 = (1 + i )12 1 (1.05) 6 Page 15 of 32 = 1 + i = 1.00816 . MAT244H1a.doc (1 − i )n − 1 . y = 0 (1 − i )n y − b (1 − i )n − 1 = 0 0 (1 − i ) − 1 240 (1 − i ) − 1 240 (1.00816) − 1 = 105121.37 . y 0 = 1000 0.00816 × (1.00816)240 • So y n = (1 − i )n y 0 − b • So the maximum loan is $105121.37. LOGISTIC GROWTH EQUATION Logistic growth equations have the form u n +1 = ρu n (1 − u n ) . Equilibrium Equilibrium solutions are obtained when u n +1 = u n , i.e. u n = ρu n (1 − u n ) ρu n − ρu n 2 − u n = 0 u n (ρ − ρu n − 1) = 0 u n = ρu n − ρu n 2 un = 0 ρ − 1 . So different values of ρ present different ρ equilibriums: if ρ = 1 , then the two solutions are the same; if ρ ≈ 1 , then the two equilibriums are very close. un = Stability of the Equilibriums 1) Suppose a system starts near the 0 equilibrium, and suppose for simplicity that it is linear, i.e. u n +1 = ρu n . If u n ≈ 0 (very near the equilibrium 0), then we may assume any logistic growth would be u n +1 = ρu n − ρu n 2 . Then u n 2 is closer to 0, so we can assume u n +1 ≈ ρu n . We know as n → ∞ , u n → ρ n u 0 , so u n → 0 if ρ < 1 . Therefore the logistic growth difference equation u n +1 = ρu n (1 − u n ) will converge to 0 (the equilibrium) if we start near 0 and ρ < 1 . More difficult analysis guarantees this situation is true for nonlinear ones. 2) The behavior near the ρ −1 equilibrium. Let vn be an amount of perturbation that defines a new solution near ρ ρ −1 ρ −1 , that is u n = + v n , v n ≈ 0 . This is a solution and needs to be reformulated to a process. Notice that ρ ρ ρ −1 ρ −1 ρ −1 u n +1 = + v n +1 , u n = + v n , and u n +1 = ρu n (1 − u n ) , so v n +1 = u n +1 − ρ ρ ρ ρ −1 v n +1 = ρu n (1 − u n ) − v n +1 = (2 − ρ )v n − ρv n 2 . Notice that since v n ≈ 0 , v n +1 ≈ (2 − ρ )v n , so this new ρ process converges if 2 − ρ < 1 ⇔ 1 < ρ < 3 . Second Order Differential Equations A second order differential equation is of the form y ′′ = f (t , y, y ′) . Page 16 of 32 MAT244H1a.doc Note Because of y , y ′ in f, we have two degrees of freedom: at t = t 0 , we have y 0 = y (t 0 ) . y 0′ = y ′(t 0 ) Example Consider y ′′ + y = 0 . Solutions to this are y = sin x , y = cos x , y = 0 ,…there are infinitely many. Specifying y 0 = y (0) = 0 , we have y = sin x , y = 2 sin x , y = − sin x ,…there are still infinite number of solutions. To specify one solution, we need to fix y ′ also. If y 0 = y (0) = 0 1 1 , then the solution is y = − sin x . 3 y 0′ = y ′(0) = − 3 Note Sometimes, we can point to a solution by specifying two points on the plane. y (0) = 0 ′ ′ , then y (t ) = 3 sin t is the solution. This is a boundary value problem. • If y + y = 0, π y =3 2 • If y ′′ + y = 0, y (0) = y 0 , then this is a initial value problem. y ′(0) = y 0′ Example y (0 ) = 0 y (0 ) = 0 may have no solutions, but y ′′ + y = 0, have infinitely many solutions. So y (π ) = 2 y (π ) = 0 BVP are not guaranteed to give a unique solution, but IVP will. y ′′ + y = 0, SECOND ORDER LINEAR DIFFERENTIAL EQUATIONS Definitions 1) A second order linear differential equation occurs when y ′′ = f (t , y, y ′) = g (t ) − p(t ) y ′ − q(t ) y or y ′′ + p(t ) y ′ + q(t ) y = g (t ) . 2) If p(t ) and q(t ) are constants, when we have a constant coefficient second order linear differential equation. Note Notice that if y (t ) = e rt is a solution to ay ′′ + by ′ + cy = 0 , then ar 2 e rt + bre rt + ce rt = 0 ar 2 + br + c = 0 . So if r is a solution to ar 2 + br + c = 0 (characteristic equation of the differential equation), then surely y (t ) = e rt is a solution to ay ′′ + by ′ + cy = 0 . If the solution to ar 2 + br + c = 0 is imaginary, then e it = cos t + i sin t . Page 17 of 32 MAT244H1a.doc Note Notice that because the input is 0, any linear combination of y1 and y 2 is also a solution, i.e. y ′(t ) = c1 r1e r1t + c 2 r2 e r2t y (t ) = c1 y1 (t ) + c 2 y 2 (t ) is the general solution to the differential equation. Since ( ) ( ) ( ) y ′′(t ) = c1 r1 2 e r1t + c 2 r2 2 e r2t , so ay ′′ + by ′ + cy = a c1 r1 2 e r1t + c 2 r2 2 e r2t + b c1 r1e r1t + c 2 r2 e r2t + c c1 e r1t + c 2 e r2t = ( ) ( ) c1 ar1 2 e r1t + br1e r1t + cre r1t + c 2 ar2 2 e r2t + br2 e r2t + ce r2t = y1 (t ) + y 2 (t ) = 0. Example The second order differential equation y ′′ − 2 y ′ − 2 y = 0 has the characteristic equation r 2 − 2r − 2 = 0 with r1 = 1 + 3 and general solution y (t ) = c1 e r1t + c 2 e r2t (the solution space, since all possible vectors in r2 = 1 − 3 a vector space can be written as a linear combination of basis vectors). roots y (0) = c1 + c 2 = 1 ( ) ( ) y ′(t ) = 1 + 3 c1 + 1 + 3 c 2 = 2 1 c1 = ( 1 = 1− 3 −1− 3 = 3 +1 2 3 ) 1 . The augmented matrix is 1− 3 − 2 ) 3t + c 2 e (1− y ′(t ) = 1 + 3 c1 e (1+ and c1 = 1+ 3 1− 3 ) 3t 1 1 1+ 3 1− 3 1 2 1− 3 1 y (t ) = c1 e (1+ y (0 ) = 1 , we solve y ′(0 ) = 2 To solve the IVP y ′′ − 2 y ′ − 2 y = 0, 1 1 1+ 3 2 1 1 = ) 3t ( ) + 1 + 3 c 2 e (1− ) 3t , so by Cramer’s Rule, 2 2 −1− 3 1 − 3 −1 − 3 = 3 −1 2 3 . 1+ 3 1− 3 Example The solution to y ′′ − 2 y ′ − 2 y = 0 is y (t ) = c1 e (1+ 3 )t + c 2 e (1− 3 )t . We can analyze the behavior of the family, ex: where are the solutions positive, 0, negative; where are the max/min points; the behavior as t → ±∞ . To answer these questions, it is better to factor. To find y (t ) = 0 , let y (t ) = e (1+ As t → ∞ , y (t ) = e (1+ ) 3t ) 3t c1 + c 2 e − 2 c1 + c 2 e − 2 3t 3t → c1e (1+ =0 ) 3t t=− 1 2 3 ln c2 . c1 . As t → −∞ , y (t ) = e (1− EQUILIBRIUM SOLUTIONS Equilibrium solutions are the constant solutions, i.e. y = c . Example The equilibrium solution to y ′′ − 2 y ′ − 2 y = 3 is y e = − 3 . 2 Page 18 of 32 ) 3t c1 e 2 3t + c 2 → c 2 e (1− ). 3t MAT244H1a.doc We can transform this non-homogenous differential equation to a homogenous one. Let Y = y − ye Y ′ = y′ Y ′′ − 2Y ′ − 2Y = 0 . This transformation is shift. Y ′′ = y ′′ Example Consider y ′′ + y ′ = e −t . To transform it, let v = f (t ) + c1 dy = f (t ) + c1 , then y = dt v = y′ . Then v ′ + v = e −t . Now solve v ′ = y ′′ f (t )dt + c1t + c 2 . Example Consider 2 y 2 y ′′ + 2 y ( y ′)2 = 1 (t is not involved). Let v = y ′ = So we have 2 y 2 v dy d dy dv dv dy dv , then y ′′ = = = =v . dt dt dt dy dt dy dt [ ] dv d 2 2 + 2 yv 2 = 1 . Now y is independent, i.e. y v = 1. dy dy FUNDAMENTAL SOLUTIONS OF LINEAR HOMOGENEOUS EQUATIONS Solutions to a second order differential equation are real valued functions which are at least twice differentiable. The collection of such functions is denoted by C 2 . We know from linear algebra that C 2 and addition of functions ( f + g )(x ) = f (x ) + g ( x ) and scalar multiplication (αf )(x ) = αf (x ) form a vector space. Linear Transformations • The derivative D( f ) = f ′ , D 2 ( f ) = D(D( f )) = f ′′ . • p(t )D( f ) = p(t ) f ′ , p(t )D( f + g ) = p(t )( f + g )′ = p(t ) f ′ + p(t )g ′ = p(t )D ( f ) + p(t )D (g ) . • q(t ) : R 2 → R 2 , q(t ) = q(t )1 ⇔ q(t ) f = q(t )1 f . Note Any linear differential equation can be captured as a linear transformation. So y ′′ + p(t ) y ′ + q(t ) y = 0 can be ( ) written as L[ y ] = D 2 + p(t )D + q(t ) ( y ) = D 2 ( y ) + p(t )D ( y ) + q(t ) y = 0 . Note Notice that if y1 and y 2 are solutions to L[ y ] , then so is y = c1 y1 + c 2 y 2 . Proof: L[ y1 ] = 0 L[ y 2 ] = 0 L[c1 y1 + c 2 y 2 ] = c1 L[c1 y1 ] + c 2 L[ y 2 ] = 0 + 0 = 0 . Page 19 of 32 MAT244H1a.doc Theorem: Existence and Uniqueness Theorem y (t 0 ) = y 0 The initial value problem L[ y ] = g (t ), has a unique solution on any interval I on which t 0 ∈ I y ′(t 0 ) = y 0′ and p(t ) , q(t ) , g (t ) are all continuous. Example Determine the largest interval I on which the IVP (x − 2 ) y ′′ + y ′ + ((x − 2 ) tan x ) y = 0, y (3) = 1 has a unique y ′(3) = 2 solution. • At x = 2 , y ′ = 0 , so we have no solution at x = 2 . • Since x ≠ 2 , the IVP is now y ′′ + 1 π y ′ + (tan x ) y = 0 . Note that tan x is discontinuous at and 2 x−2 3π . 2 • So 2, 3π 2 is the largest interval on which we have a unique solution. LINEAR INDEPENDENCE AND THE WRONSKIAN Recall that in the case of a first order differential equation, whenever we found a solution, we added a constant to it in order to included all possible solutions (the general/family of solutions). In the case of second order differential equations, we find two solutions y1 and y 2 and claim that c1 y1 + c 2 y 2 is the general solution. How do we decide if two functions are linearly independent on an interval I? Use Wronskian of the two functions. Wronskian If the Wronskian W ( y1 , y 2 ) = y1 y1′ y2 is non zero for some t ∈ I , then y1 and y 2 are linearly independent. y 2′ Theorem: Abel’s Theorem If y1 and y 2 are solutions to L[ y ] = y ′′ + p(t ) y ′ + q(t ) y = 0 where p(t ) and q(t ) are continuous on an open interval I, then the Wronskian (W ( y1 , y 2 ))(t ) = y1 y1′ y2 − p (t )dt (t ) = ce where c is a constant that depends on y ′2 y1 and y 2 but not on t. COMPLEX ROOTS OF THE CHARACTERISTIC EQUATION Recall: To find solutions to ay ′′ + by ′ + cy = 0 we decided to find the roots r1 and r2 of ar 2 + br + c = 0 , and y1 (t ) = e r1t , y 2 (t ) = e r2t . Page 20 of 32 MAT244H1a.doc What if r1 = r2 (repeated root), or there are no real roots (complex roots)? If ∆ = b 2 − 4ac < 0 , then r = As usual we say − b ± ∆ − b ± (− ∆ )(− 1) − b ± = = 2a 2a (− 1) (− ∆ ) 2a =− b i (− ∆ ) ± = λ ± iµ . 2a 2a y1 (t ) = e = e (λ +iµ )t = e λt e iµt = e λt [cos(µt ) + i sin (µt )] are the solutions. y (t ) = e r2t = e (λ −iµ )t = e λt e −iµt = e λt [cos(µt ) − i sin (µt )] r1t 2 Real-Valued Solutions But we are looking for a real solution. Indeed, y1 (t ) + y 2 (t ) = 2e λt cos(µt ) − iy1 (t ) + iy 2 (t ) = 2e λt sin (µt ) are solutions too. So the general solution is y (t ) = c1 e λt cos(µt ) + c 2 e λt sin (µt ) = e λt [c1 cos(µt ) + c 2 sin (µt )] . If λ = 0 i.e. − b =0 2a b = 0 , then the solution becomes y (t ) = c1 cos(µt ) + c 2 sin (µt ) = c 2 c c1 sin α cos(µt ) + sin (µt ) . By letting 1 = tan α = , we get c2 cos α c2 c sin α cos(µt ) + sin (µt ) = 2 [(sin α )(cos(µt )) + (cos α )(sin (µt ))] = A sin (µt + α ) . cos α cos α y (t ) = c 2 Note Given y ′′ + p(t ) y ′ + q(t ) y = 0 we can change the variable t to x so y (t ) d2y dt 2 = d dy dx dy d dx d dy ⋅ + ⋅ = dt dx dt dx dt dt dt dx dy dx + dt dx then y ′′ + p(t ) y ′ + q(t ) y = 0 becomes y ′′ + by ′ + c(t ) y = 0 . Example ( ) Consider ty ′′ + t 2 − 1 y ′ + t 3 y = 0 , or y ′′ + x(t ) = (y ′′t 2 1 q(t ) 2 dt = tdt = ) + y′ + d 2x dt 2 y (x ) . Now, dy dy dx = , dt dx dt 1 . It turns out that if we let x(t ) = q(t ) 2 dt , t 2 −1 y ′ + t 2 y = 0 . So q(t ) = t 2 . Now let t t2 dx d2x t 2 −1 ′′ + , and so =t, = 1 . Now, y y ′ + t 2 y = 0 becomes 2 dt t dt 2 t 2 −1 ( y ′t ) + t 2 y = 0 t y ′′t 2 + y ′ + t 2 y ′ − y ′ + t 2 y = 0 y ′′ + y ′ + y = 0 . REPEATED ROOTS b − t −b± ∆ b Recall that r = . If ∆ = b 2 − 4ac = 0 , then r = − . So only one solution exists: y1 = e 2 a . But if 2a 2a we let y (t ) = v(t ) y1 , then maybe we can come up with some other solutions as well. Page 21 of 32 MAT244H1a.doc If y (t ) is to satisfy ay ′′ + by ′ + cy = 0 , then: y (t ) = v(t ) y1 y ′(t ) = v ′(t ) y1 + v(t ) y1′ • • y ′′(t ) = v ′′(t ) y1 + v ′(t ) y ′ + v ′(t ) y ′ + v(t ) y1′′ = v ′′(t ) y1 + 2v ′(t ) y ′ + v(t ) y1′′ ′ ′ and ay + by ′ + cy = 0 becomes a(v ′′(t ) y1 + 2v ′(t ) y ′ + v(t ) y1′′ ) + b(v ′(t ) y1 + v(t ) y1′ ) + c(v(t ) y1 ) = 0 [av ′′(t )y1 + 2av ′(t )y ′ + bv ′(t )y1 ] + v(t )[ay1′′ + by1′ + cy1 ] = 0 av ′′(t ) y1 + 2av ′(t )y ′ + bv ′(t )y1 = 0 • av ′′(t )e − b t 2a − bv ′(t )e − b t 2a b − t e 2a y 2 (t ) = (c1t + c 2 ) + bv ′(t )e − b t 2a =0 v ′(t ) = c1 v ′′(t ) = 0 v(t ) = c1t + c 2 . So another solution is . So the general solution is y (t ) = C1e − b t 2a + C 2 (c1t + c 2 )e − b t 2a = (C1 + C 2 c1t + c 2 )e − b t 2a = (at + b )e − b t 2a . NON-HOMOGENOUS DIFFERENTIAL EQUATIONS A non-homogenous differential equation has the form L[ y ] = y ′′ + p(t ) y ′ + q(t ) y = g (t ) . L[ y ] = 0 decides the nature of the system. General Solution Suppose y1 and y 2 are two solutions to L[ y ] = g (t ) , i.e. L[ y1 ] = g (t ) = L[ y 2 ] . Then L[ y1 ] − L[ y 2 ] = g (t ) = g (t ) = 0 , so y1 − y 2 is a solution to L[ y ] = 0 . So if y (t ) = c1 y1 (t ) + c 2 y 2 (t ) is the general solution to L[ y ] = 0 , then y1 − y 2 = c1 y1 + c 2 y 2 for some c1 and c 2 . Therefore, y1 = y 2 + c1 y1 + c 2 y 2 , which means if one solution to L[ y ] = g (t ) is known ( y p a particular solution) then the general solution to L[ y ] = g (t ) is y (t ) = y p (t ) + c1 y1 (t ) + c 2 y 2 (t ) . So to find the generals solution y (t ) to L[ y ] = g (t ) , we need to first guess y p (t ) then solve L[ y ] = 0 . How to Guess a Particular Solution? Notice that a (linear) differential equation does not completely disfigure the input g (t ) (at least the “reasonable” inputs will be recognizable). “Reasonable” inputs: exponential e rt , polynomial p(t ) , trigonometric functions sin (αt ) , cos(αt ) , tan (αt ) = sin (αt ) , roots cos(αt ) n x. Example Consider L[ y ] = y ′′ + 2 y ′ + 5 y = 3 sin (2t ) . Guess a y p (t ) . y p (t ) = A sin (2t ) + B cos(2t ) Let y ′p (t ) = 2 A cos(2t ) − 2 B sin (2t ) y ′p′ (t ) = −4 A sin (2t ) − 4 B cos(2t ) = −4 y p (t ) [ ] , then L y p = 3 sin (2t ) − 4 y p + 2[2 A cos(2t ) − 2 B sin (2t )] + 5 y p = 3 sin (2t ) Page 22 of 32 y p + 4 A cos(2t ) − 4 B sin (2t ) = 3 sin (2t ) MAT244H1a.doc y p + 4 A cos(2t ) − 4 B sin (2t ) = 3 sin (2t ) A sin (2t ) + B cos(2t ) + 4 A cos(2t ) − 4 B sin (2t ) = 3 sin (2t ) ( A − 4 B ) sin (2t ) + (4 A + B ) cos(2t ) = 3 sin (2t ) . Since sin (2t ) and cos(2t ) are linearly independent, 3 17 . 12 B=− 17 1 So guess y p (t ) = [3 sin (2t ) − 12 cos(2t )] . 17 A − 4B = 3 4A + B = 0 A= More Complicated Inputs What if L[ y ] = g (t ) , but g (t ) = g 1 (t ) + g 2 (t ) + g 3 (t ) ? It can be reduced to finding particular solutions to L[ y ] = g 1 (t ) L[ y ] = g 2 (t ) because L[ y1 + y 2 + y 3 ] = L[ y1 ] + L[ y 2 ] + L[ y 3 ] = g1 (t ) + g 2 (t ) + g 3 (t ) = g (t ) L[ y ] = g 3 (t ) y p (t ) = y1 (t ) + y 2 (t ) + y 3 (t ) . Example Consider y ′′ + y ′ = e −t . Let y p (t ) = Ae −t y ′p (t ) = − Ae −t Ae −t − Ae −t = e −t 0 = e −t ! This problem happens y ′p′ (t ) = Ae −t [ ] because y p is already a solution to L y p = 0 . Try replacing y p with ty p . Repeated Roots [ ] [y ′p + y ′p + ty ′p′ ]+ p(t )[y p + ty ′p ]+ q(t )t[y p ] = g (t ) (tp y )″ + p(t )(ty p )′ + q(t )(ty p ) = g (t ) t [ty ′p′ + p(t ) y ′p + q(t ) y p ]+ 2 y ′p + p(t ) y p = g (t ) If L[ y ] = y ′′ + p(t ) y ′ + q(t ) y = g (t ) , then L ty p = g (t ) 2 y ′p + p(t ) y p = g (t ) . Now solve for y p (t ) . METHOD OF UNDETERMINED COEFFICIENTS Example Solve the IVP y ′′ + 4 y = t 2 + 3e t , 1) Solve y ′′ + 4 y = 0 . r 2 + 4 = 0 y (0) = 0 . y ′(0) = 2 r = 0 ± 2i , so the fundamental solution is y c (t ) = c1 cos(2t ) + c 2 sin (2t ) . Page 23 of 32 MAT244H1a.doc y p (t ) = At 2 + Bt + C + De t 2) Look for a particular solution y p (t ) . Since g (t ) = t 2 + 3e t , let y ′p′ + 4 y p = t 2 + 3e t [2 A + De ]+ 4[At t 2 ] y ′p (t ) = 2 At + B + De t y ′p′ (t ) = 2 A + De t + Bt + C + De t = t 2 + 3e t 4A = 1 4B = 0 2 A + 4C = 0 5D = 3 4 At 2 + 4 Bt + (2 A + 4C ) + 5 De t = t 2 + 3e t A= (t ) = c1 cos(2t ) + c 2 sin (2t ) + 1 t 2 − 1 + 3 e t 4 y ′(t ) y (0) = 0 y ′(0) = 2 8 5 1 4 B=0 C = − 18 D= 3) The general solution is y (t ) = y c (t ) + y p (t ) = c1 cos(2t ) + c 2 sin (2t ) + 4) Now, y . So . 3 5 1 2 1 3 t t − + e . 4 8 5 = −2c1 sin (2t ) + 2c 2 cos(2t ) + 1 3 t + e t , so 2 5 − 19 19 7 1 1 3 40 . So the solution to the IVP is y (t ) = − cos(2t ) + sin (2t ) + t 2 − + e t . 7 40 10 4 8 5 c2 = 10 c1 = Example Consider y ′′ + 4 y = 3 sin 2t . We already know the fundamental solutions y c (t ) = c1 cos(2t ) + c 2 sin (2t ) . So when we “guess” the particular solution, use y p (t ) = At sin (2t ) + Bt cos(2t ) . Then the general solution is y (t ) = − 3 t cos(2t ) + c1 cos(2t ) + c 2 sin (2t ) . 4 VARIATION OF PARAMETERS This technique works for general linear differential equations with arbitrary input g (t ) . Idea If y1 (t ) , y 2 (t ) are fundamental solutions to L[ y ] = y ′′ + p(t ) y ′ + q(t ) y = 0 , then the general solution to L[ y ] = g (t ) “can” be y (t ) = u1 (t ) y1 (t ) + u 2 (t ) y 2 (t ) . We want more restrictions on u1 and u 2 . If u1′ y1 + u 2′ y 2 = 0 , then (u1′ y1 + u 2′ y 2 )′ = 0 u1′′ y1 + u1′ y1′ + u 2′′ y 2 + u 2′ y ′2 = 0 . So u1′ y1 + u 2′ y 2 = 0 . Now, we can solve for u1 (t ) and u 2 (t ) . u1′ y1′ + u 2′ y ′2 = g (t ) Example Page 24 of 32 MAT244H1a.doc Consider x 2 y ′′ − 3xy ′ + 4 y = x 2 ln x , and y1 = x 2 y 2 = x 2 ln x fundamental solutions. Rewrite as u1′ x 2 + u 2′ x 2 ln x = 0 3 4 y ′′ − y ′ + 2 y = ln x . Let x u1′ (2 x ) + u 2′ (2 x ln x + x ) = ln x x u1′ x 2 + u 2′ x 2 ln x = 0 u1′ + u 2′ ln x + 1 ln x . = 2 2x Observation 0 ′ ′ g (t ) u y + u2 y2 = 0 To solve 1 1 , use Cramer’s rule. So u1′ = u1′ y1′ + u ′2 y ′2 = g (t ) y1 y1′ u 2′ = y1 y1′ g (t ) 0 y1 y1′ y2 y ′2 = y2 y 2′ = y2 y ′2 − y 2 g (t ) and W ( y1 , y 2 ) y1 g (t ) . Thus, the general solution is W ( y1 , y 2 ) y (t ) = u1 (t ) y1 (t ) + u 2 (t ) y 2 (t ) = y1 (t ) t t0 − y1 (s )g (s ) ds + c1 + y 2 (t ) W ( y1 , y 2 )(s ) t t0 y 2 (s )g (s ) ds + c 2 . W ( y1 , y 2 )(s ) REDUCTION OF ORDER If only y1 (t ) is given as a solution to L[ y ] = y ′′ + p(t ) y ′ + q(t ) y = 0 , we claim that y (t ) = v(t ) y1 (t ) is a solution to L[ y ] = g (t ) . So v ′′y1 + 2v ′y1′ + p(t )v ′y1 = g (t ) or v ′′y1 + (2 y1′ + p(t ) y1 )v ′ = g (t ) , which we can solve! SPRING VIBRATIONS Hooke’s Law “The more the spring stretches, the more force.” Fs = −kL . Equilibrium mg = −kL ⇔ mg + kL = 0 . Example Given a spring, a 100g mass is hanged on it. The spring stretches by 2cm. What is k? mg (0.1)(− 9.8) = 49 . mg = −kL k = − =− L 0.02 Page 25 of 32 MAT244H1a.doc General Solution On the way down, the mass passes through u (t ) at time t. So m(u ′′(t ) + g ) + k (L + u (t )) = 0 mu ′′(t ) + mg + kL + ku (t ) = 0 mu ′′(t ) + ku(t ) = 0 . Therefore, at any point u (with respect to any reference point), we have mu ′′ + ku = 0 as the differential equation form of Hooke’s Law. λ =0 k Now, u ′′ + u = 0 , so . The solution is u (t ) = c1 cos(ω 0 t ) + c 2 sin (ω 0 t ) = k m µ= = ω0 m (A is the amplitude, δ is the phase). The period is T = = A cos(ω 0 t − δ ) 2π m = 2π . ω0 k Fluid Resistance The fluid that the mass is moving in causes resistance. Fr ~ v , so Fr = λu ′ . So now, the differential equation becomes mu ′′ + λu ′ + ku = 0 . So r = − λ 2m ± λ 2 − 4mk 2m =− ωq = λ 2m 2m λ ±i ± λ2 4m 2 − k . m k . m If λ = 0 , we have undamped motion with frequency ω 0 = If λ 2 < 4mk , then r = − λ − k λ2 2 m cos − . So u ( t ) = Ae m 4m 2 k λ2 − t − δ , where m 4m 2 k λ2 λ2 2 − = ω − < ω 0 is the quasi-frequency. 0 m 4m 2 4m 2 Change in Mass k λ2 k λ2 λ2 k − > . As m increases, ω increases. As m decreases, increases faster than , , q 2 2 2 m 4m m 4m m 4m and there will eventually be no oscillation. ωq = Pendulum The motion of the pendulum is governed by θ ′′ + ω0 = T= g θ = 0. L g is the number of oscillation per second. L 2π L = 2π is the period. ω0 g ELECTRIC CIRCUITS Components Page 26 of 32 θ L MAT244H1a.doc dQ dQ 1 V= ⇔ VR = R , R is resistance in ohms. R dt dt 1 2) Capacitor: V L = Q , C is capacity in Faraday. C 1) Resistor: I = d 2Q dI = L 2 , L is in Henry. dt dt 3) Coil: V L = L LC Circuit VC + V L = 0 , V L = LQ ′′ , VC = So LQ ′′ + 1 Q = 0 with ω 0 = C 1 Q. C 1 . LC LCR Circuit 1 Q =0. C VC + V R + V L = 0 . So LQ ′′ + RQ ′ + FORCED VIBRATION Vibration with input is mu ′′ + γu ′ + ku = F (t ) . Cases of Interest 1) F (t ) = F0 cos(ωt ) where ω ≠ ω 0 . Then u (t ) = Ae − rt cos(ω 0 t − δ ) + R cos(ωt − δ ) , where R= F0 ( m 2 ω0 2 − ω ) 2 2 + γ 2ω 2 F0 then R ≈ = m 2ω 4 + γ 2ω 2 . Since ω 0 2 = F0 ω m 2ω 2 + γ 2 k , so if ω ≈ 0 then R ≈ m F0 2 m ω0 4 = F0 mω 0 2 = F0 ; if ω >> ω 0 k which approaches 0 as ω approaches infinity. 2) F (t ) = F0 cos(ωt ) where ω ≈ ω 0 . Then u (t ) = Ae − rt cos(ω 0 t − δ ) + R cos(ωt − δ ) , where R= 2 ( F0 2 m ω0 − ω ) 2 2 2 +γ ω 2 . Since ω ≈ ω 0 , so R ≈ F0 γ 2ω 2 = F0 γω ; if γ ≈ 0 then R is large (momentary resonance). 3) F (t ) = F0 cos(ωt ) where γ = 0 and ω = ω 0 . Then u (t ) = c1 cos(ω 0 t ) + c 2 sin (ω 0 t ) + u (0) = 0 , then we have resonance. u ′(0) = 0 Page 27 of 32 F0 t sin (ω 0 t ) . If 2mω 0 MAT244H1a.doc 4) F (t ) = F0 cos(ωt ) where γ = 0 and ω ≠ ω 0 . Then u (t ) = c1 cos(ω 0 t ) + c 2 sin (ω 0 t ) + u (0) = 0 c1 = − u ′(0) = 0 c2 = 0 u (t ) = ( F0 m ω0 2 − ω ) ( F0 ( F0 m ω0 2 − ω 2 ) cos(ωt ) . If ) m ω 0 2 − ω 2 , then [cos(ωt ) − cos(ω 0 t )] = F0 ( m ω0 2 − ω ) ω0 2 − ω 2 2 sin 2 t sin ω0 2 + ω 2 2 t . We let the actual frequency to be ω 0 2 + ω 2 , and then the amplitude keeps changing with time (with frequency ω0 2 − ω 2 < ω0 2 + ω 2 . Series Solution of Second Order Linear Equations Definition: Regular Point A point x 0 is said to be a regular point of the differential equation P( x ) y ′′ + Q(x ) y ′ + R( x ) y = 0 if P( x 0 ) ≠ 0 . That is, if Q( x ) R(x ) and are continuous at x 0 , then the IVP has a solution defined near y (0 ) = x 0 . P(x ) P(x ) Example Consider the IVP (1 − x )2 y ′′ − 2 xy ′ + 12 y = 0, ∞ y= ∞ a n ( x − 0 )n = n =0 a n x n . Then − 2 xy ′ = −2 x n =0 (1 − x )y ′′ = (1 − x ) 2 x(0) = 0 . Assuming y exists near x 0 = 0 , we have x ′(0) = 1 2 ∞ ∞ n =0 ∞ n(n − 1)a n x n − 2 = n=0 ∞ na n x n −1 = − 2na n x n and n =1 n(n − 1)a n x n − 2 − n =2 ∞ n(n − 1)a n x n . n=2 So (1 − x )2 y ′′ − 2 xy ′ + 12 y = 0 becomes ∞ n(n − 1)a n x n − 2 + n=2 ∞ − n(n − 1)a n x n + n= 2 ∞ have (n + 1)(n + 2 )a n + 2 x n + n =0 ∞ − 2na n x n + n =1 ∞ 12a n x n = 0 . Making the powers look alike, we n=0 − n(n − 1)a n x n + n =2 ∞ ∞ − 2na n x n + n =1 ∞ 12a n x n = 0 . n=0 So the DE becomes ∞ [2a2 +12a0 ] + [6a3 x +10a1 x] + [(n +1)(n + 2)an+2 − n(n −1)an − 2nan +12an ]x n = 0 n=2 Page 28 of 32 MAT244H1a.doc a 2 = −6a 0 6a 0 + a 2 = 0 5a1 + 3a 3 = 0 (n + 1)(n + 2 )a n + 2 − n(n − 1)a n − 2na n + 12a n that =0 a2 = 0 y (0) = a 0 = 0 , so (n + 3)(n − 3) a a n+2 = y ′(0) = a1 = 1 (n + 1)(n + 2) n a3 = − 5 3 a 5 = a 3+ 2 = 0 a 3 = 0 a n+2 = Now, if (n + 3)(n − 3) a (n + 1)(n + 2) n a 2 n +1 = 0, ∀n = 2,3, a1 = 0 y (0) = 1 , then a3 = 0 y ′(0) = 0 5 . Notice a 3 = − a1 3 n(n − 1) + 2n − 12 (n + 3)(n − 3) a a n+2 = a = (n + 1)(n + 2) n (n + 1)(n + 2) n a 2 n = 0, ∀n = 0,1, and . So the solution is y1 ( x ) = x − 5 3 x . 3 a0 = 1 a 2 n +1 = 0, ∀n = 0,1, and a 2 = −6 . So y 2 ( x ) = 1 − 6 x 2 + 3x 4 + . a4 = 3 Therefore, the general solution is y ( x ) = c1 y1 + c 2 y 2 . Notice that if y (0) = b1 , then y ( x ) = b2 y1 + b1 y 2 . y ′(0) = b2 Definition: Regular Point A regular point is a point where all the derivatives of the solution function are defined. In other words, the Taylor series exists. RESTRICTIONS ON POWER SERIES SOLUTION Example: Radius of Convergence Consider x 2 − 2 x − 3 y ′′ + xy ′ + 4 y = 0 at various points: x 0 = 4 , x 0 = −4 , and x 0 = 0 . Find a lower bound for the radius of convergence of the given points. • The issues arise when the coefficient of y ′′ becomes 0. x 2 − 2 x − 3 = ( x − 3)( x + 1) , so at x = 3 and x = −1 , we have issues. • The solution shall make sense on the intervals (− ∞,−1) , (− 1,3) , (3, ∞ ) . ( ) • At x 0 = 4 , the radius of convergence is at least 1 ( ρ ≥ 1 ). At x 0 = −4 , the radius of convergence is at least 3 ( ρ ≥ 3 ). At x 0 = 0 , the radius of convergence is at least 1 ( ρ ≥ 1 ). Another technique for solving differential equations (by series) is by only calculating a few terms of the solution. Example Page 29 of 32 MAT244H1a.doc Evaluate four terms of the solution to (cos x ) y ′′ + xy ′ − 2 y = 0 . 1− Rewriting, we get ( (2a x2 + 2 2 + 6a 3 x + 12a 4 x 2 + 20a 5 x 5 + − 2 a 0 + a1 x + a 2 x 2 + a 3 x 3 + [(2a + 6a x + 12a x − [2a + 2a x + 2a x 2 3 0 4 1 2 2 2 5 + 20a 5 x + 3 + 2a 3 x + )− (a ]= 0 2x 2 )= 0 + 3a 3 x 3 + )+ x(a )]+ [a x + 2a 1 1 2x 2 + 2a 2 x + 3a 3 x 2 + 4a 4 x 3 + + 3a 3 x 3 + 4a 4 x 4 + ) ] − a0 + a2 = 0 (2a 2 − 2a 0 ) + (6a 3 − a1 )x + (12a 4 − a 2 )x 2 + (20a 5 − 2a 3 )x 3 + − a1 + 6a 3 = 0 =0 − a 2 + 12a 4 = 0 . Solve in terms of − a 3 + 10a 5 = 0 a 0 and a1 . System of First Order Linear Equations INTRODUCTION x1′ = F1 (t , x1 , x 2 , x 3 ) A system of differential equation is like x ′2 = F2 (t , x1 , x 2 , x 3 ) . x 3′ = F3 (t , x1 , x 2 , x 3 ) Example Consider the system x1′ = 2 x1 − x 2 + cos t x′ 2 − 1 x1 cos t . Such a system is written as 1 = + or x 2′ = 3x1 x 2′ 3 0 x2 0 X ′ = AX + g (t ) , which is a first order differential equation. Example We can use a system approach to solve any differential equation of any order. Consider u ′′′ + 2u ′ − u = 1 . Let x1 = u x1′ = x 2 x1′ 0 1 0 x1 0 x 2 = x1′ = u ′ , then x ′2 = x1 . So we have x 2′ = 0 0 1 x 2 + 0 or x 3 = x 2′ = u ′′ x 3′ = u ′′′ = u − 2u ′ + 1 = x1 − x 2 + 1 0 X ′ = AX + 0 . 1 Example Page 30 of 32 x 3′ 1 − 2 0 x3 1 MAT244H1a.doc Conversely, we one can translate a system to an ordinary differential equation. Consider u = x1 . We have u ′ = x1′ = 2 x 2 x 2 (0) = 4 u ′′ = 2 x ′2 = −4 x1 = −4u x1 (0) = 3 x1′ = 2 x 2 where x ′2 = −2 x1 u (0) = 3 . u ′(0) = 2 x 2 (0) = 8 u ′′ + 4u = 0 where APPLICATIONS IN MODELING Example: A Double Tank Mixing Problem ρ g/min x 2 (0) = 300 2 L/min x1 (0) = 500 V1 = 20L dx1 x (t ) x (t ) = 2ρ + 3 2 − 5 1 dt V2 V1 dx 2 x (t ) x (t ) = 5 1 −5 2 dt V1 V2 − 5 x1′ = 520 x ′2 20 3 50 5 − 50 x1 x2 + 3 L/min V2 = 50L 5 L/min x1′ = 2 ρ + − , so we have 2ρ 0 ⇔ X′= 2 L/min 5 3 x1 + x2 V1 V2 5 5 x 2′ = x1 − x2 V1 V2 − 14 1 4 3 50 − 101 X+ Example: Double Mass-Spring Problem F1(t) k1 m1 The variables are x1 and x 2 . So m2 d 2 x1 dt 2 d 2 x2 dt 2 2ρ 0 k2 where where X (0) = F2(t) m1 m2 x1 x2 = F1 (t ) − k1 x1 − k 2 (x 2 − x1 ) . = F2 (t ) − k 2 (x 2 − x1 ) EIGENVALUES AND EIGENVECTORS Recall Page 31 of 32 x1 (0) = 500 x 2 (0) = 300 500 300 . . So MAT244H1a.doc Two solutions y1 and y 2 to a differential equation are independent on an interval I if W ( y1 , y 2 ) = y1 y1′ y2 ≠ 0 on I (where I contains the initial values). y 2′ x1(1) x1(2 ) x1(3) For systems X ′ = AX we say solution X (1) (t ) = x 2(1) , X (2 ) (t ) = x 2(2 ) , X (3) (t ) = x 2(3) x 3(1) x 3(2 ) x 3(3) ( ) on I if W X (1) , X (2 ) , X (3) = X (1) X (2 ) are independent X (3) ≠ 0 . Recall If r is an eigenvalue of A and ξ is the corresponding eigenvectors, i.e. Aξ = rξ , then in particular X ′ = AX = rX Now, if x1 x2 x1′ = rx1 x 2′ = rx 2 x1 (t ) = ae rt x 2 (t ) = be rt or X = e rt a . b is an eigenvector then x 2 = αx1 , then be rt = αae rt . Notice that we may have two independent eigenvectors for A. In this case, two solutions in two independent directions can be found so that if ξ (1) and ξ (2 ) are independent eigenvectors of A corresponding to eigenvalues r1 and r2 , then we have two independent solutions X (t ) = c1 X (1) (t ) + c 2 X (2 ) (t ) . Page 32 of 32 X (1) = e r1t ξ (1) X (2 ) = e r2t ξ (2 ) and the general solution is

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