Differential Equations

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Transcription:

October 19, 2017

The photo of me on the title page was taken in Taipei on August 1. I put it up on Instagram (realkenribet) on Tuesday.

Midterm #2 The second midterm will be one week from today, 8:10 9:30AM in 155 Dwinelle. It is cumulative in principle, but it will stress the second third of the course. It covers everything through today s course meeting. Questions? Post them on piazza that seems best.

Breakfast, anyone? New breakfast: Wednesday, October 25 at 9AM. Please send me email to sign up. (This continues the tradition of breakfasts the day before our exams.) I d be down for November 1 or November 3 as well. Time to be determined. Neither is scheduled for now I await your requests.

Lunch and Dinner Pop-in lunch tomorrow (Friday) at noon. Meet in the Great Hall of the Faculty Club. Foothill DC dinner, Friday, Nov. 3, 6:30PM.

Slides for this course Last Thursday, our class was visited by Richard Freishtat, the director of the UC Berkeley Center for Teaching and Learning. I asked him to come to give me feedback on my teaching. I was pleased to learn that he thinks I m doing a good job. Hey, I was like blushing.

Slides for this course Last Thursday, our class was visited by Richard Freishtat, the director of the UC Berkeley Center for Teaching and Learning. I asked him to come to give me feedback on my teaching. I was pleased to learn that he thinks I m doing a good job. Hey, I was like blushing.

Slides for this course Last Thursday, our class was visited by Richard Freishtat, the director of the UC Berkeley Center for Teaching and Learning. I asked him to come to give me feedback on my teaching. I was pleased to learn that he thinks I m doing a good job. Hey, I was like blushing.

But he did write this in his notes: never reads/repeats slide text, but truly complements and expands on it. There is a unique value in coming to Ribet s class, to get his explanation that cannot be found anywhere else. The takeaway: when you need to miss an occasional class, you won t be able to pick up on everything that took place during the class meeting just by looking at slides. On bcourses, don t forget that there s a course capture for the class and that the document camera pages are available as scans.

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Today We will study differential equations, equations involving a function and its derivative(s). The aim is to determine the function. For example: Find f (x) given that f (x) = cos x and f (0) = 1. Here, we recognize an antiderivative problem and see that f (x) = sin x + C. Setting x = 0, we find that C = 1 and emerge with the formula f (x) = 1 + sin x.

Important principle If F = 0, then F is constant.

Why is this true? Heuristic reason: if a function has derivative identically equal to 0, it s not changing. It s a constant. Sophisticated reason: this follows from the mean value theorem. Namely, if a and b are two different numbers, then F(b) F(a) = F (c)(b a) for some c between a and b. Since F (c) = 0, F(b) = F(a). To say that all values of a function are the same is to say the function is constant.

Important principle If two functions have the same derivative, their difference is a constant. Why? Apply previous principle to the difference between the two functions. The derivative of the difference is 0.

Today Two distinct goals: To recognize a differential equation from a real-life scenario. (The mathematician s concept of real life may not be yours.) A key word is modeling. To solve a differential equation symbolically. I will begin with modeling.

Exponential growth and the logistic equation A scenario is presented in which the derivative of a function of time is a constant times the function: dn dt = RN, where R is a constant. For example, N might be the number of bacteria in some colony.... This is a DE (differential equation) and the general solution is N(t) = Ce Rt, where C is a constant. If we set t = 0, we see that C = N(0). When the initial value N(0) is furnished, and R is known, then N(t) is completely determined.

Analysis We can first check that N(t) = Ce Rt satisfies the equation. We just take the derivative of both sides of N(t) = Ce Rt and get N (t) = Ce Rt R = RCe Rt = RN(t). We used the chain rule to differentiate e Rt. Conversely, how do we know that every solution to the equation may be written Ce Rt for some constant?

Reformulation If N is a solution, we want to know that N/e Rt = Ne Rt is a constant. To see this, we apply the first important principle. To show something is a constant, it s enough to check that the derivative is 0. This is not hard (see doc camera).

Growth and decay When R is positive, N is growing exponentially. When R is negative, N is decaying exponentially. When R = 0, N is constant (boring).

Growth and decay When R is positive, N is growing exponentially. When R is negative, N is decaying exponentially. When R = 0, N is constant (boring).

Logistic stuff In biology (a.k.a. real life), things don t continue to grow forever. There are brakes. Here is the book s notation: If dn = RN where R is a constant, dt we have exponential growth (or decay). Replace the constant R by a product r(1 N ), where K is K some limiting value of N (the most N could possibly be). The quantities r and K are constants; N continues to be a function of t.

Logistic stuff dn dt = r(1 N )N, where r and K are constants K When N is small, the brake (1 N ) is near 1, so we re dealing K essentially with dn = rn. dt When N is very close to K, (1 N ) is close to 0, so the growth K of N slows down. What we expect is that N grows exponentially until it starts getting close to K and then its growth gets slower and slower.

A logistic function Let N(t) = on HW. et 1 + e t = 1. You ve seen this kind of function 1 + e t 1 0.8 0.6 0.4 y = e t /(1 + e t ) 0.2-4 -2 2 4

1 0.8 0.6 y = e t /(1 + e t ) 0.4 0.2-4 -2 2 4 For t, the function approaches 0 = 0. For t +, the 1 1 function approaches = 1. At t = 0, the value of the 1 + 0 function is 1. You can see this behavior pretty well on the plot, 2 which was made for t between 5 and +5.

Looking ahead The domain of the logistic function N is the set of all real numbers. The function is defined everywhere! Meanwhile, the values of N are the real numbers between 0 and 1 (excluding both of them). Since probabilities are numbers between 0 and 1, the logistic function is important in probability theory. Just sayin.

Looking ahead The domain of the logistic function N is the set of all real numbers. The function is defined everywhere! Meanwhile, the values of N are the real numbers between 0 and 1 (excluding both of them). Since probabilities are numbers between 0 and 1, the logistic function is important in probability theory. Just sayin.

Logistic function, logistic equation Check this out: dn = N(1 N). dt In other words, the logistic function N satisfies the logistic equation (with r = K = 1). Note: sometimes mathematicians write N(t) for the function, and sometimes just N. It s the same thing.

Here s the verification: N(t) = et 1 + e t, 1 N(t) = (1 + et ) e t 1 + e t = 1 1 + e t, N (t) = (1 + et )e t e t e t (1 + e t ) 2 = e t (1 + e t ) 2. The right-hand function in the third equation is the product of the two functions above it. In other words, N (t) = N(t)(1 N(t)).

Separable differential equations A separable DE has the initial form dy dt = some messy function of y and t. What makes it separable is that we can torment it algebraically until it has the form function of y dy = function of t dt, say We can then write f (t) dt = g(y) dy. f (t) dt = g(y) dy and solve the equation.

Separable differential equations For example, consider the DE dy dx = 3xy, ( 6.2, problem 23). (We have x instead of t.) This is separable: 1 dy = 3x dx. y Integrating gives 1 y dy = 3x dx, ln y = 3 2 x 2 + C y = (some constant) e 3 2 x 2.

Check the answer If y = Ke 3x 2 /2, then dy dx = Ke3x 2 /2 3x = 3xy. One might ask: why not have ln( y ) instead of ln y when we integrate 1 dy? If y is negative, then we re indeed dealing with y ln( y); we d get and we could write instead y = (some constant) e 3 2 x 2 y = (some constant) e 3 2 x 2 by changing the sign of the constant.

Check the answer If y = Ke 3x 2 /2, then dy dx = Ke3x 2 /2 3x = 3xy. One might ask: why not have ln( y ) instead of ln y when we integrate 1 dy? If y is negative, then we re indeed dealing with y ln( y); we d get and we could write instead y = (some constant) e 3 2 x 2 y = (some constant) e 3 2 x 2 by changing the sign of the constant.