Edwards-Penney Chapter 1, sections 1.2, 1.3, 1.4, 1.5 Topics, Definitions, Theorems, Examples 1.2 Method of Quadrature ==== DEF. General solution of y'=f(x) DEF. Particular solution of y'=f(x) ALGORITHM. To solve y'=f(x), integrate across the equation and use the Fundamental Theorem of Calculus to resolve int(y'(x),x)=y(x)+constant. Then check the answer. This is called the Method of Quadrature. See Maple documentation or a Numerical Recipes textbook. EXAMPLE. Solve y'=2x+3, y(1)=2 ANSWER: y=x^2+3x-2 THEOREM. Second order equations y''=g(x) can be solved by two applications of the method of quadrature. EXAMPLE. Solve x'' = -g, g=32 ft/sec^2 or 9.8 m/sec^2 ANSWER: x(t) = -0.5 g t^2 + v0 t + x0 EXAMPLE. Lunar lander. Model: x''(t) = 2.5, x(0)=x0, x'(0)=-450 Find x0 for a soft landing for this configuration of retro-rockets ANSWER: x0=40.5 km TABLE. Physical units of force, mass, distance, gravitational acceleration MODEL x''(t) = -g, x(0)=x0, x'(0)=v0 Solution is x(t) = -0.5 g t^2 + v0 t + x0 EXAMPLE. Solve it when x0=0, v0=96 feet per second, g=32 feet per second squared. Then find the Maximum height (ANS 144 ft) EXAMPLE. Solve it when x0=0, v0=49 meters per second, g=9.8 meters per second squared. Then find the flight time (ANS 10 seconds) APPLICATION. The swimmer's problem. EXAMPLE. Solve y'=3(1-4x^2) 1.3 Direction fields and phase diagrams ==== DEF. A DIRECTION FIELD is a graph window plus data items (x0,y0,L) where (x0,y0) is a grid point and L is a line segment going through (x0,0y). The line segment must have slope m determined by the equation m=f(x0,y0) This last equation is the ONLY CONNECTION with the differential equation, namely the RHS of the DE determines the slope of line L. ALGORITHM. Choose grid points, normally a rectangular 5x5 grid for hand work. Determine a formula for f(x,y) from the differential equation: for the example y'=x+y, define f(x,y)=x+y, the RHS. Let (x0,y0) be one of the grid points. Evaluate m=f(x0,y0) explicitly. Then (x0,y0,L) is a data item for the direction field, where L is the line through (x0,y0) with slope m. Repeat for all grid points and draw all grid points (x0,y0) and lines L on the graph window. EXAMPLE. Draw a direction field and threaded solution curves for each of the following differential equations. y'=2y, y'=y/2, y'=-y, y'=-3y MYSTERY. Why do that? What is gained by drawing such a graphic? MYSTERY. What does it mean "threads its way through the slope field, following the visible line segments as closely as possible?" EXAMPLE. Draw a direction field for y'=x-y and sketch one approximate solution curve on the graph window going through x=-4, y=4. APPLICATIONS EXAMPLE. Baseball thrown down along the gravity vector from 3000 feet. Assume air friction term 0.16(velocity). Estimate the impact speed with the ground. ANSWER: dv/dt=32-0.16v, v(infinity)=136.36 mph EXAMPLE. Logistic equation. Consider P'=0.0004P(150-P). Use a direction field to conclude that the limiting population is P(infinity)=150. EXISTENCE-UNIQUENESS of SOLUTIONS EXAMPLE. The problem y'=1/x, y(0)=0 has no solution. MYSTERY: What does this problem mean at x=0? DIRECTION FIELD: It shows there is no solution curve going through (0,0). EXAMPLE. y'=2 sqrt(y), y(0)=0 has two different solutions. MYSTERY: What does this problem mean for y<0? Can't a solution start at x=0, y=0 and become negative? DIRECTION FIELD: It shows there are two solution curves through x=0, y=0. One is y=0 and the other is y=x^2. Infinitely many solutions exist going through x=0, y=0, all non-negative. THEOREM 1. [Picard-Lindelof Existence-Uniqueness] Let f(x,y) and its partial (d/dy)f(x,y) be continuous in a rectangle R with x=a, y=b at its center. Then there is a smaller rectangle R1 with center x=a, y=b such that one solution curve y=y(x) of y'=f(x,y), y(a)=b, passes through x=a, y=b and threads across R1 from left edge to right edge. There are no other solution curves with this property. THEOREM 1a. [Peano Existence, no uniqueness] Remove the partial derivative condition from the Picard theorem. Then the conclusion is the same, except rectangle R1 may or may not have other solution curves with the same edge-to-edge threading property. Uniqueness is not guaranteed when the partial derivative condition is removed. EXAMPLE. y'=-y has continuous f(x,y)=-y, (d/dy)f(x,y)=-1, therefore the hypotheses of Theorem 1 hold and there is a unique solution curve going left edge to right edge in a smaller rectangle R1. EXAMPLE. y'=2 sqrt(|y|) has continuous f(x,y)=2 sqrt(|y|), but the partial (d/dy)f(x,y)=1/sqrt(|y|) for y>0 is discontinuous at y=0. Peano's theorem applies to produce existence but not uniqueness. Picard's theorem simply does not apply, because the partial derivative condition fails to be satisfied. EXAMPLE. y'=y^2 has continuous f(x,y)=y^2 and (d/dy)f(x,y)=2y for all values of (x,y). Picard's theorem and Peano's theorem apply. This example splits hairs: R can be any rectangle in the plane. But for initial condition x=0, y=1 there is trouble with R1: the solution is y=1/(1-x) which does not extend past x=1, so R1 cannot contain the vertical line x=1. EXAMPLE. xy'=2y has continuous f(x,y)=2y/x and (d/dy)f(x,y)=2/x provided the vertical line x=0 does not contact rectangle R. There is a unique solution on smaller rectangle R1 provided initial condition y(a)=b does not allow x=a. MYSTERY. y(x)=x^2 is a solution for x <=0 and y(x)=Cx^2 is a solution for x>0. These two curves paste continuously at x=0. Because C can be any number, then the initial condition y(a)=b looks like it has a unique solution. Not so. Why not? What restrictions must be placed on (a,b)? 1.4 Separable differential equations ==== DEF. y'=f(x,y) is called SEPARABLE provided functions F(x) and G(y) exist such that f(x,y)=F(x)G(y). DEF. The equation y'/G(y)=F(x) is called the SEPARATED FORM of the separable differentiable equation. DEF. An IMPLICIT SOLUTION is any equation in (x,y) which when differentiated implicitly (set y=y(x) and differentiate on x) reproduces the differential equation. DEF. An EXPLICIT SOLUTION is an equation of the form y=expression in x alone (or a constant and no x appears) that passes and answer check in the differential equation. Any other form of a solution is called IMPLICIT (silly, because y-2=0 is implicit while y=2 is explicit). ALGORITHM. To solve a separable equation: 1. Test y'=f(x,y) to see if it is separable. The determine formulas for F(x) and G(y). 2. Write out the separable form y'/G(y)=F(x). Solve it by the method of quadrature to find an IMPLICIT SOLUTION. 3. If possible, solve for y in terms of x to obtain an EXPLICIT SOLUTION. 4. Report any EQUILIBRIUM SOLUTIONS, which are solutions y=constant arising from the equation G(y)=0 [which was disallowed in step 2]. EXAMPLE. Solve y'=-6xy, y(0)=7. ANSWER: IMPLICIT ln|y|=-3x^2+C EXPLICIT y(x)=7 exp(-3x^2) EQUILIBRIUM none, y(0)=7 conflicts with the only equilibrium solution y=0. EXAMPLE. Solve y' = (4-2x)/(3y^2-5). ANSWER: IMPLICIT y^3-5y=4x-x^2+C EXPLICIT not found, but possible in Maple EQUILIBRIUM y=sqrt(5/3), y=-sqrt(5/3) EXAMPLE. Solve y' = (4-2x)/(3y^2-5), y(1)=3. ANSWER: IMPLICIT y^3-5y=4x-x^2+9 EQUILIBRIUM both have the wrong initial condition, so they are eliminated. EXPLICIT Maple finds two answers. This is difficult to sort out, so we make an implicit plot to find out why there are two answers, not one. See Figure 1.4.2 and sections 2.4 to 2.6 for numerical methods. DEF. SINGULAR SOLUTIONS are solutions of y'=f(x,y) that cannot be obtained from a general solution. Equilibrium solutions are examples of singular solutions. APPLICATIONS GROWTH-DECAY Malthus population growth P'=kP compound interest A'=rA radioactive decay N'=-kN drug elimination A'=-lambda A EXAMPLE. P'=kP, P(1999)=6 billion, increasing 212 million/day. Find k. MYSTERY: Why do we need two bits of information? ANSWER: P(t)=P0 exp(k(t-1999)), so to find k, P0 we need two equations in two unknowns k,P0. These equations for P in billions are obtained from the given information, which is P(1999)=6 and P'(1999) (212) 1 k = -------- = ----- 365.25 --- = 0.0129 P(1999) 10000 6 Then P(t)=P0 exp(k(t-1999)) becomes P(t) = 6 exp(0.0129(t-1999)). The middle of the 21st century is 2050. We get P(2050)=11.58. In the year 2177 the population reaches 60 billion. EXAMPLE. Charcoal at Stonehenge has 63 percent as much carbon-14 as present day samples. Find the age of the charcoal. ANSWER: 3800 years EXAMPLE. Newton's Law of Cooling and cooking a roast. A 4-lb roast at 50F is placed into a 375F oven at 5pm. After 75 minutes it's at 125F. When will it be 150F? ANSWER: After 1.75 hours at 6:45pm. EXAMPLE. Torricelli's Law v=sqrt(2gy) A hemispherical tank of radius 4 feet at the top drains at the bottom from a full tank. The hole has radius 1 inch. Find the time to drain the tank. ANSWER: 2150 seconds, which is under 36 minutes MODEL: Pi (8y-y^2) y' = -Pi (1/24)^2 sqrt(64 y) 1.5 First order linear differential equations ==== DEF. A first order linear differential equation is an equation y'=f(x,y) that can be put into the form y'+p(x)y=q(x). DEF. INTEGRATING FACTOR for y'+p(x)y=q(x) is the function W=exp(int(p(x),x)) THEOREM. The identity W(y'+p(x)y) == (Wy)' holds, called the integrating factor identity. It implies the replacement formula (Wy)'/W replaces y'+p(x)y ALGORITHM. How to solve y'+p(x)y=q(x) 1. Test y'=f(x,y) for linear. One test is: (d/dy)f(x,y) is independent of y. Or, re-arrange to y'+p(x)y=q(x). 2. Write the equation in standard form y'+p(x)y=q(x). Then replace the left side by (Wy)'/W, where W=exp(int(p(x),x)) 3. Cross-multiply to get (Wy)'=W q(x). Then apply the method of quadrature to solve for y: Wy = C + int(Wq,x) y = C/W + (1/W)int(qW,x) EXAMPLE. y'=xy+x (1) f(x,y)=xy+x, (d/dy)f(x,y)=x is independent of y (2) Rearrange y'+(-x)y=x so p=-x, q=x Then (Wy)'/W=x where W=exp(int((-x),x))=exp(-x^2/2) (3) Cross-multiply to get (Wy)'=(x)(W). Integrate Wy=C+int(x exp(-x^2/2),x) Wy=C-exp(-x^2/2) y=C exp(x^2/2) - 1 SHORTCUT. Use y=yh+yp. Guess yp=-1, then yh=constant/W=C exp(x^2/2). THEOREM [Shortcut] The equation y'+p(x)y=0 has answer y = constant/integrating factor. PROOF. Replace LHS by (Wy)'/W and cross-multiply to get (Wy)'=0. Integrate, then Wy=constant or y=C/W. THEOREM. The integrating factor W can be simplified by removing all constants from the integration. EXAMPLE. For p(x)=-x, W=exp(int(-x,x))=exp(-x^2/2+c)=exp(-x^2/2)exp(c) Then (Wy)'/W has common factor exp(c) in top and bottom, and the factor cancels. Equivalent to c=0. EXAMPLE. Solve y'-y=(11/8) exp(-x/3), y(0)=-1 ANSWER: y=exp(x)/32 - 33 exp(-x/3)/32 The terms are explained as yh=exp(x)/32 and yp= -33 exp(-x/3)/32. Then y=yh+yp. They are not found by superposition, but by the linear integrating factor method. The integrating factor is W=exp(int(-1,x)) = exp(-x) [why not -x+c?]. EXAMPLE. Solve for a general solution (x^2+1)y'+3xy=6x ANSWER: y = 2 + C (x^2+1)^(-3/2) The method is the linear integrating factor method, no shortcut possible. DEF. EQUILIBRIUM SOLUTION means a solution of the form y=C (a constant) found by substitution of y=C into the differential equation (then solve for C). EXAMPLE. The equation of the previous example has non-constant coefficients and an equilibrium solution y=2. THEOREM 1. The linear first order equation y'+p(x)y=q(x) with p, q continuous on an interval J has a unique solution y defined on J for each initial value y(x0)=y0 with x0 in J. The solution can be found by the linear integrating factor method. EXAMPLE. Solve x^2y'+xy=sin(x), y(1)=y0 ANSWER: y=(1/x)(y0 + G(x)), G(x)=int(sin(t)/t,t=1 .. x) defined for all x in J=(0,inf). MIXTURE PROBLEMS Assume the rate in is the rate out and the tank volume V is constant. MODEL: dx/dt = a - bx where a=(rate in)(concentration in) b=(rate out)/(tank volume) SOLUTION: Equilibrium solution xp=a/b, Homogeneous solution xh=C/exp(bt) x=xh+xp=C exp(-bt)+a/b EXAMPLE. Lake Erie problem, model above with rates=350, Lake Erie volume V=480 km^2. Assume c=concentration of pollutant in Lake Huron. Assume x(0)=5cV. Find the time t such that x(t)=2c=twice the pollutant concentration of Lake Huron. ANSWER: 1901 years. EXAMPLE. Brine tank of size 120 gallons with 90 lb salt dissolved in water to make 90 gallons of brine. Input brine flow 2 lb/gal at 4 gal/min, output brine flow 3 gal/min (of mixture). Find the number of pounds of salt in the tank when it is full. ANSWER: The model is dx/dt=a-rx(t)/V(t) where a=(2)(4) lb/min, r=3 gal/min, V(t)=90+t (volume increases at 1 gal/min). Solve by the integrating factor method to obtain x(t) = 180+2t-90/(1+t/90)^3 The tank fills when V(t)=120, which is for t=30 min. Then x(30) = 202 lb. ===== end