advertisement

Prof. Dobrushkin APMA 0330: Solution to Homework 1 September 23, 2015 In each problem, draw a direction field for the given differential equation. Based on the direction field, determine the behavior of y as t 7→ ∞. If this behavior depends on the initial value of y at t = 0, describe this dependency. 1 Problem 1.1 (20 pts): y ′ = y(2 − y) Consider the initial condition y(0) = y0 . • For y0 > 2, the slope of the solution is initially negative, so the solution is decreasing (in magnitude) close to 2, then it converges to y = 2. • For y0 = 2, equilibrium, slope is zero, then solution remains constant y = 2. • For 0 < y0 < 2 , the slope of the solution is negative, so the solution is decreasing (in magnitude) approaching 2, therefore, it converges to y = 2. • For y0 = 0, equilibrium, slope is zero, then solution is y = 0. • For y0 < 0 , the slope of the solution is negative, then the solution diverges to y = −∞. Below is matlab script. % Problem 1a from hw1 % Direction field for y’=y(2-y) 5 clear a l l close a l l [t, y] = meshgrid(0:0.2:7,-2:0.2:5); 10 dy = y.*(2 - y); dt = ones( s i z e (dy)); dyu = dy./sqrt(dt.ˆ2 + dy.ˆ2); dtu = dt./sqrt(dt.ˆ2 + dy.ˆ2); quiver(t,y,dtu,dyu,1); y ′ = y(y − 1)2 let the initial condition be y(0) = y0 . • For y0 > 1, the slope of the solution is positive, then the streamline diverges to y = +∞. • For y0 = 1, equilibrium, slope is zero, the solution remains constant y = 1. • For 0 < y0 < 1 , the slope of the solution is positive and the solution curve increases (in magnitude) approaching 1, so the solution converges to y = 1. • For y0 = 0, equilibrium, slope is zero, the solution is y = 0. • For y0 < 0 , the slope of the solution is negative, then the solution curve diverges to y = −∞. % Problem 1b from hw1 % Direction field for y’=y(y-1)^2 5 clear a l l close a l l Page 1 of 8 Prof. Dobrushkin APMA 0330: Solution to Homework 1 September 23, 2015 [t, y] = meshgrid(0:0.2:7,-2:0.2:5); 10 dy = y.*(y - 1).ˆ2; dt = ones( s i z e (dy)); dyu = dy./sqrt(dt.ˆ2 + dy.ˆ2); dtu = dt./sqrt(dt.ˆ2 + dy.ˆ2); quiver(t,y,dtu,dyu,1); 2 Problem 1.2 (10 pts): A spherical raindrop evaporates at a rate proportional to its surface area. Write a differential equation for the volume of the raindrop as a function of time. 4 Solution: The surface area of a sphere is A = 4π r2 and its volume is V = π r3 . Expressing volume 3 through the area, we get 3/2 4 4 A . V = π r3 = π 3 3 4π Therefore, the required differential equation becomes dV = −k V 2/3 dt for some constant k. 3 Problem 1.3 (10 pts): Newton’s law of cooling states that the temperature u(t) of an object changes at a rate proportional to the difference between the temperature of the object itself and the temperature of its surroundings (the ambient air temperature in most cases): u̇(t) = −k(u − T ), where T is the ambient temperature and k is a positive constant. Suppose that the initial temperature of the object is u(0) = u0 , find its temperature at any time t. Solution We compute: du u−T ln(u − T ) − ln(u0 − T ) u(t) = −kdt = −k(t − t0 ) = T + (u0 − T )e−k(t−t0 ) Or, since we have t0 = 0: u(t) = T + (u0 − T )e−kt 4 Problem 1.4 (20 pts): A certain drug is being administrated intravenously to a hospital patient. Fluid containing 6 mg/cm3 of the drug enters the patient’s bloodstream at a rate of 120 cm3 /h. The drug is absorbed by body tissues or otherwise leaves the bloodstream at a rate proportional to the amount present, with a rate of 0.5 (h)−1 . (a) Assuming that the drug is always uniformly distributed throughout the bloodstream, write a differential equation for the amount of the drug that is present in the bloodstream at any time. Page 2 of 8 Prof. Dobrushkin APMA 0330: Solution to Homework 1 September 23, 2015 (b) How much of the drug is present in the bloodstream after a long time? Solution: (a) Let m(t) mg be the total amount of drug in bloodstream at time t. Consider a short time interval [t, t + ∆t]. The increment of the amount of the drug that is present in the bloodstream can be calculated in two ways: m(t + ∆t) − m(t) and 6 × 120∆t − 0.5m∆t Therefore, we have m(t + ∆t) − m(t) = 6 × 120∆t − 0.5m∆t. Dividing both sides by ∆t and taking the limit: ∆t → 0, we get m′ (t) = 720 − 0.5m(t) (b) First, we plot the direction field using Mathematica: StreamPlot[{1, 720 - 0.5*m}, {t, 10, 20}, {m, 1430, 1450}] From the direction field, we know that after a long time, the drug present in bloodstream should be stable, and there should be m′ (t) = 0. Then we have 720 − 0.5m(t) = 0 =⇒ m(t) = 1440. 1450 1445 1440 1435 1430 10 12 14 16 18 20 Figure 1: Direction field for Problem 1.4, plotted with Mathematica. Page 3 of 8 Prof. Dobrushkin 5 APMA 0330: Solution to Homework 1 September 23, 2015 Problem 1.5 (20 pts): Consider the falling object of mass 5 kg that experiences the drag force, which is assumed to be proportional to the square of the velocity: v̇ = 492 − v 2 /245. (a) Determine an equilibrium solution. (b) Plot a slope field for the given differential equation using one of your lovely software package. Provide the codes of your plot or state what resources did you use. Based on the direction field, determine the behavior of v(t) as t → ∞. (c) Find the limiting velocity v∞ = limt→∞ v(t) if initially v(0) = 0, and determine the time that must elapse for the object to reach 98% of its limiting velocity. (d) Find the time it takes the object to fall 300 m. Solution: (a) The equilibrium is achieved when 492 − v 2 = 0, 245 i.e., v = ±49. However, note that v = −49 is nonphysical: we designate positive velocity as downwards velocity, so v = −49 would be falling upwards. Thus, the only equilibrium in which we are interested is v = 49. (b) We have that v → 49 as t → ∞ for all physical trajectories. Note that with this model, if the object were moving upwards fast enough, it would continue to accelerate upwards forever! Now we plot using Mathematica: VectorPlot[{1, (49ˆ2 - yˆ2)/245}, {x, -100, 100}, {y, -100, 100}, VectorPoints -> 20, VectorStyle -> Arrowheads[0.028]] Then we do the same job using matlab: 5 [t, v] = meshgrid(0:.5:10,40:1:60); dv = (49ˆ2-v.ˆ2)/245; dt = ones( s i z e (dy)); dvu = dv./sqrt(dt. 2 + dv. 2 ); dtu = dt./sqrt(dt. 2 + dv. 2 ); quiver(t,v,dtu,dvu,1); Page 4 of 8 Prof. Dobrushkin APMA 0330: Solution to Homework 1 September 23, 2015 100 50 0 -50 -100 -100 -50 0 50 100 Figure 2: Direction field for Problem 1.5, plotted with Mathematica. (c) If v(0) = 0, then v∞ is indeed 49. To find the elapsed time we first solve the differential equation dv 492 − v 2 dv 1 1 + 98 49 + v 49 − v Z 1 1 dv + 49 + v 49 − v 1 [ln(49 + v) − ln(49 − v)] 98 49 + v ln 49 − v 49 + v 49 − v 49 + v = = = = = dt 245 dt 245 Z 2 dt 5 t 245 2t 5 2 = e5t = (49 − v)e 5 t v(1 + e 5 t ) = 49(1 − e 5 t ) v = 2 = 2 2 1 − e2t/5 1 + e2t/5 t 49 tanh 5 49 Page 5 of 8 Prof. Dobrushkin APMA 0330: Solution to Homework 1 September 23, 2015 V Figure 3: The electric circuit of Problem 1.6. We seek t such that v(t) = 0.98 · v∞ = 0.98 · 49, i.e., t 5 0.98 · 49 = 49 tanh t = ≈ 5 tanh−1 (0.98) 11.5 (d) We can arbitrarily define the starting point as x0 = x(0) = 0. then the position of the object is the same as the distance traveled, and is given by Z t Z t s t 49 tanh v(s)ds = x(t) = . ds = 245 ln cosh 5 5 0 0 We seek t such that x(t) = 300, then 300 = 245 ln cosh 6 t 5 =⇒ t = 5 cosh−1 e300/245 ≈ 9.48. Problem 1.6 (20 pts): Consider an electrical circuit containing a capacitor, resistor, and battery. The charge Q(t) on the capacitor satisfies the equation dQ Q R + = V, dt C where R is the (constant) resistance, C is the (constant) capacitance, and V is the constant voltage supplied by the battery. (a) If Q(0) = 0, find Q(t) at any time, and sketch the graph of Q versus t. (b) Find the limiting value QL that Q(t) approaches after a long time. (c) Suppose that Q(t1 ) = QL and that at time t = t1 the battery is removed and the circuit is closed again. Find Q(t) for t > t1 and sketch its graph. Solution: (a) The corresponding initial value problem is: dQ Q R dt + C = V Q(0) = 0 Page 6 of 8 Prof. Dobrushkin We subtract Q C APMA 0330: Solution to Homework 1 September 23, 2015 from both sides: dQ Q =− +V dt C The equation is clearly separable. Gathering the Q terms on the left-hand side and the t terms on the right-hand side, we get dQ = dt R Q −C + V R Now integrate both sides: Z dQ = R Q −C + V Z dt The right-hand side is a simple integral, the left-hand side is a little trickier but still doable: Q −RC ln − + V = t + k1 C In the above question, k is the constant of integration. Moving the −RC to the right-hand side: Q t + k1 ln − + V = − . C RC Exponentiate both sides: − t+k1 k1 t t Q + V = e− RC = e− RC e− RC = k2 e− RC . C k1 Here, k2 = e− RC is a new constant, for convenience. Now, solving for Q: t Q = −C k2 e− RC − V Now we apply the initial condition Q(0) = 0 to solve for the constant k2 : 0 0 = −C k2 e− RC − V = −C (k2 − V ) =⇒ C = 0 or k2 − V = 0. Presumably our capacitance C is nonzero, we thus have that k2 = V . So the solution to our initial value problem is: t Q(t) = −C V e− RC − V Cleaning it up a bit: t Q(t) = CV 1 − e− RC t≥0 25 20 15 10 5 0 0 2 4 6 8 10 12 14 16 18 20 Page 7 of 8 Prof. Dobrushkin APMA 0330: Solution to Homework 1 September 23, 2015 In order to sketch a graph of Q versus t, we need to pick some actual values for C, V , and R. The specific values don’t matter too much here; we just want to get an idea of what the graph looks at. Arbitrarily choosing C = 2.5, V = 8, R = 1, we obtain: The above graph was generated using the following matlab code: t = linspace(0,20,10000); C = 2.5; V = 8; R = 1; Q = C*V*(1-exp(-t/(R*C))); plot(t,Q) axis([0, 20, 0, 25]) 5 (b) t t = lim CV − CV e− RC CV 1 − e− RC t→∞ t→∞ t − RC = lim (CV ) − lim CV e = CV − 0 = CV QL = lim t→∞ t→∞ So, since the exponential term goes to zero as t goes to infinity, we find that QL = CV . (c) We now have a different initial value problem. Since the battery has been removed, the voltage V = 0. So our initial value problem is: dQ Q R dt + C = 0 Q(t1 ) = QL = CV Our differential equation is again separable, and is relatively straightforward: R dQ Q dQ Q dQ Q + = 0 =⇒ R =− =⇒ =− dt C dt C dt RC Z Z 1 1 1 t dQ =− dt =⇒ dQ = − dt =⇒ ln (Q) = − + k1 =⇒ Q RC Q RC RC t t t =⇒ Q = e− RC +k = e− RC ek1 = k2 e− RC We now apply the initial condition to solve for the constant of integration k2 : t1 t1 QL = CV = k2 e− RC =⇒ k2 = CV e RC Cleaning things up t1 t Q(t) = CV e RC e− RC = CV e t1 −t RC = CV e −(t−t1 ) RC So the solution to our initial value problem is C = 2.5, V = 8, R = 1, t1 = 10: Q(t) = CV e −(t−t1 ) RC , t ≥ t1 To plot, we again arbitrarily pick values. Using C = 2.5, V = 8, R = 1, t1 = 10: The above graph was generated using the following matlab code: 5 t = linspace(0,20,10000); C = 2.5; V = 8; R = 1; t1 = 10; Q = C*V*exp(-(t-t1)/(R*C)); plot(t,Q) axis([0, 20, 0, 25]) Page 8 of 8