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where A and B are arbitrary constants. Both of these are homogeneous linear differential equations with constant coefficients. An nth-order linear differential is of the form Linear Differential Equations n 00 0 Math 130 Linear anf (t) + ··· + a2f (t) + a1f (t) + a0f(t) = b. D Joyce, Fall 2013 that is, some linear combination of the It’s important to know some of the applications up through the nth is equal to b. In gen- of , and one of those applications is eral, the coefficients an, . . . , a1, a0, and b can be any to homogeneous linear differential equations with functions of t, but when they’re all scalars (either constant coefficients. real or complex numbers), then it’s a linear differ- ential equation with constant coefficients. When b What is a differential equation? It’s an equa- is 0, then it’s homogeneous. tion where the unknown is a and the equa- For the rest of this discussion, we’ll only consider tion is a statement about how the derivatives of the at homogeneous linear differential equations with function are related. constant coefficients. The two example equations 0 Perhaps the most important differential equation are such; the first being f −kf = 0, and the second 00 is the exponential differential equation. That’s the being f + f = 0. one that says a quantity is proportional to its rate of change. If we let t be the independent , f(t) What do they have to do with linear alge- the quantity that depends on t, and k the constant bra? Their solutions form vector spaces, and the of proportionality, then the differential equation is dimension of the is the order n of the equation. 0 f (t) = kf(t). The first example was the first-order equation f 0 − kf = 0. Its solutions were Aekt where A was Its general solution is an arbitrary constant. The set of these solutions f(t) = Aekt is a vector space because the 0 function is one so- lution, they’re closed under addition, and they’re where A is an arbitrary constant. This differential closed under multiplication by constants. equation has applications as the exponential model The second example was the second-order equa- of population growth, , compound tion f 00 + f = 0. Its solutions, A cos t + B sin t form interest, Newton’s law of cooling, and lot’s more. a two-dimensional vector space. Another important differential equation is the In general the set of solutions will always be a one that the and cosine function satisfies. For vector space. The zero function is always a solution; both of those functions, their is that’s because we’re only considering homogeneous their negation. They satisfy the differential equa- differential equations. If f and g both satisfy the tion equation, then so does f +g. Finally, if c is a scalar f 00(t) = −f(t). and f satisfies the equation, so does cf. That’s called a second order differential equation since the second derivative is involved. The general So, how do you solve these equations, and solution to this differential equation is why does the dimension of the solution space equal the order of the equation? Let’s take f(t) = A cos t + B sin t an example to illustrate it. It should be general

1 enough to show the method, but not two compli- You can use Euler’s identity to convert between the cated. Consider the fourth-order equation two forms. 2 0000 000 00 0 We still have the other factor (D − 2) to deal f − 4f + 5f − 4f − 4f = 0. with. The double root at 2 complicates the solu- The trick is to treat derivatives as differential tions. I’ll leave out the derivation of the solution 2 operators—things which operate on functions. We to the differential equation (D − 2) t = 0, and just d give its general solution, f(t) = Cte2t + Det2. You could denote this as as is common in dt can check that it works. Analogous solutions can , but a simple uppercase D makes it eas- be found when there are multiple roots rather than ier to see what’s going on. Then the differential just double roots. equation looks like In summary, this example differential equation, f 0000 − 4f 000 + 5f 00 − 4f 0 − 4f = 0, has the general D4f − 4D3f + 5D2f − 4Df − 4f = 0, solution and that can be rewritten as f(t) = Aeit + Be−it + Cte2t + Det2. (D4 − 4D3 + 5D2 − 4D − 4)f = 0. We haven’t shown that there aren’t any other solu- tions, but we’ll skip that part since we’re just doing Now, I chose this example to factor nicely. The a survey on the topic, not a whole course. differential operator D4−4D3+5D2−4D−4 factors as (D2 + 1)(D2 − 4D + 4), which further factors as Dynamical systems. You might ask, are these (D2 + 1)(D − 2)2. That first quadratic polynomial, really important? Yes. Linear dynamical systems D2 + 1, is irreducible over R but factors over C, are modelled using these equations. so to make things easier, we’ll work over C. That In such a system you have several things that gives us the factorization into linear factors change over time. Suppost they’re the quantities (D − i)(D + i)(D − 2)2. x, y, and z. Some of these things affect others pos- itively, some negatively. If x affects y positively, The Fundamental Theorem of Algebra assures us then y0 will have a term kx where k is positive, but that we can always factor an nth degree polynomial if negatively, then k is negative. A linear dynamical into linear factors over C, so what happened in this system is a system of linear equations like example will also happen in the general case. The  x0 = + 2y − z Fundamental Theorem of Algebra connnects the or-  y0 = 2x der of the differential equation to the dimension of  z0 = y − 3z the solution space. Now we can break down the differential equations For this example, x and y affect each other posi- into linear pieces. We can solve (D − i)f = 0 since tively in a positive feedback loop, and y positively it’s just the exponential differential equation. It has affects z, but z negatively affects x and itself. solutions f(t) = Aeit. Likewise (D+i)f = 0 has the Although it’s a system of three linear differential solutions f(t) = Be−it. That gives us solutions for equations, it’s equivalent to one third order differ- (D2 + 1)f = 0, namely Aeit + Be−it. Note that we ential equation. already saw a different description of the solutions Dynamical systems like this are used in , for this equation, f 00 − f = 0, and that was as , , , and almost any sub- A cos t + B sin t. The between the two ject that calls itself a science. forms of the solutions is Euler’s identity Math 130 Home Page at eit = cos t + i sin t. http://math.clarku.edu/~djoyce/ma130/

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