Ordinary Differential Equations

Transcription

1 Orinary Differential Equations Example: Harmonic Oscillator For a perfect Hooke s-law spring,force as a function of isplacement is F = kx Combine with Newton s Secon Law: F = ma with v = a = v = 2 x 2 This gives us F = m 2 x 2 an then substituting in for F gives kx = m 2 x 2 m 2 x 2 + kx = 0 This says that if we know the position x of the mass we can figure out its acceleration, an vice versa not surprising, as that s how we set up the problem. This is a linear, orinary ifferential equation that relates x position with t time. 1

2 orinary because it is an equation involving only one epenent variable x an one inepenent variable t. linear because it contains only combinations of x an erivatives of x not functions of x x 2 or functions of erivatives of x; i.e., [x t] 2. Linear First-Orer Equations The ecay rate of a sample of 210 Po is irectly proportional to the amount of 210 Po present. We write [Po] = k [Po] 2

3 One of the ways to solve this is to simply guess the answer: If I take the erivative of this function Po concentration is a function, of time, I get the same function back, times a constant I know a function that oes that! We guess an check that this works [Po] = e kt [Po] = k [Po] e kt = ke kt ke kt = ke kt Of course, I coul have guesse 2e kt, or e kt /3, or multiplie my A has a physical meaning: it exponential function by any constant, an I woul also have an answer: is just the amount of Polonium you re talking about. You can fix values of constants using starting conitions if you know that you Ae kt = kae kt have 10 µg of Po at time t = 0, kae kt = kae kt you can set It it important to provie not just a solution, but the most general solution that encompasses using constants like A, above all solutions to the problem. If guessing answers feels unsatisfactory or if you are a ba guesser we can approach this problem more systematically. Starting again with [Po] = k [Po] we rearrange to get 1 [Po] = k [Po] Now we integrate both sies with respect to t: 1 [Po] = k [Po] ln [Po] = k ln [Po] = kt + C [Po] = e kt+c = e C e kt = Ae kt or A = 10 µg. Ae 0 = 10 µg Here we ve use the funamental theorem of calculus: the integral of a erivative is the function itself. The above metho is accurate but a bit unwiel, especially for more complicate problems. 3

4 The next equation to solve is slightly more complicate: + pxy = 0 Where we wany to solve for y in terms of x yx an where px coul be any function of x. This gets solve exactly like the polonium-ecay example: = pxy = px y = px y ln y + C = px y = Ae px Now, what about the equation: + pxy = qx The following trick is often use to solve ifferential equations: If qx = 0 then we have our solution above, y = Ae px Since the function ux coul be anything, there s no loss of generality with this approach. We guess that the solutions will still look like this, an we try y = uxae px 4

5 We plug this solution back into our ifferential equation an see what we can learn about ux. This works out nicely. First, remember that the chain rule says that e fx = f efx an so again, using the funamental theorem of calculus Ae px = f px Ae px = Apxe px An now the prouct rule gives us uxae px = ux Ae px + u = Auxpxe px + A u e px Ae px Whew! Plugging this back into our original equation: A u e + pxy = qx uxae px + px uxae px = qx px Auxpxe px + Auxpxe px = qx A u px e = qx Au = qxe + px A u = qxe + px Aux = qxe + px + C an so yx = uxae px = e px qxe + px + C Totally arbitrary example; we solve: x 2 + xy = 3 To o this, we squeeze it into the form + pxy = qx 5

6 by iviing through by x 2 to get + y x = 3 x 2 With px = 1/x an qx = 3/x 2, our above equation says that yx = e px qxe + px + C px = 1 x = ln x e px = e ln x = 1 x yx = 1 x xqx + C = C x x x = C x + 3 ln x x Looking at this solution, we can figure out why C can be any constant aing in multiples of 1/x oesn t change the left-han sie of the equation, because + y x = C + C x x 2 = C x 2 + C x 2 = 0 The secon term then makes the rest of the equation work out: 3 ln x x 3x 1 x 3 ln x x 2 + y x = qx + 3 ln x x 2 = 3 x ln x x 2 = 3 x 2 3 x 2 3 ln x x ln x x 2 = 3 x 2 3 x 2 = 3 x 2 Your book provies the nice example of a two-step reaction with first-orer kinetics: A k1 B k2 C 6

7 which gives the ifferential equations A B C = k 1 A = k 1 A k 2 B = k 2 B If, at time t = 0, we have A=A 0, B=0, an C=0, we can solve this exactly. The first equation we ve alrea seen with our polonium example; it has the solution At = Ce k1t which, for our initial conitions, becomes At = A 0 e k1t The secon equation is now trickier: B = k 1 A k 2 B B + k 2B = k 1 A 0 e k1t This gives us pt = k 2 an qt = k 1 A 0 e k1t, an so yt = e pt qte + pt + β Bt = e k2t k 1 A 0 e k1t e k2t + β = e k2t k 1 A 0 e k2 k1t + β = e k2t k1 A 0 e k2 k1t + β = k 1 A 0 e k1t + βe k2t Since B=0 when t = 0, we get 0 = k 1 A 0 + β β = k 1A 0 Bt = k 1 A 0 e k1t k 1A 0 e k2t = k 1 A 0 e k 1t e k2t 7

8 We coul go ahea an solve the ifferential equation for C, but it s easier to note that an so At + Bt + Ct = A0 + B0 + C0 = A 0 Ct = A 0 A 0 e k1t k 1A 0 e k 1t e k2t 8