Introduction to Differential Equations National Chiao Tung Universit Chun-Jen Tsai 9/14/011
Differential Equations Definition: An equation containing the derivatives of one or more dependent variables, with respect to one or more independent variables, is said to be a differential equation (DE). Eample: 0.1t 0.1t = f ( t) e = e 0.t = 0.t = d dt ( ) d dt In this course, given a blue equation (behavior of a phenomenon), ou want to find out the red equation (the governing rule) behind it /43
Wh Differential Equations For dnamic phenomena, we want to predict their long-term behavior b observing and measuring their short-term behavior Long-term behavior of a dnamic sstem is defined b its underling rule hard to measure Short-term behavior of a dnamic sstem is described b its changing characteristics (derivatives) easier to measure 3/43
All Roads Lead to Rome If we have a phenomenon with a governing rule: Then, d d = ( ) = C Ce C R. Thus, it doesn t matter what the constant C is, = Ce is a solution of the DE d/d =. Often, a DE alone has man solutions; more information is required to resolve ambiguit, ( ) ( ) e = Ce =. 4/43
Solution is not Guaranteed Epressing a phenomenon as a differential equation does not guarantee that it has a solution. Obviousl, ( ) + = 1 has no (real-valued) solution. 5/43
Classification of DE b Tpe Ordinar differential equation (ODE): an equation contains onl ordinar derivatives of one or more dependent variables with respect to a single independent variable d d + 5 = e Partial differential equation (PDE): an equation involving the partial derivatives of one or more dependent variables of two or more independent variables u u u = t t 6/43
Classification of DE b Order The order of a differential equation is the order of the highest derivative in the equation. nd order 1st order d d 3 d + 5 4 d = e An nth-order ODE with one dependent variable can be epressed in the general form: F ( ) n (,,,,..., ) = 0 a real-valued function of n+ variables 7/43
Normal Form of ODE F() can be epressed in general in the normal form: n d ( n 1 = f (,,,,..., ) ) n d where f is a real-valued function with n+1 variables. For eample, the normal forms of first order and ndorder ODEs are: d = f (, ) d d d = f (,, ) 8/43
Classification of DE b Linearit An nth-order ODE, F, is said to be linear if F is linear in, ',., (n). That is, F can be epressed as: a where a i (), i = 0,, n depend on the independent variable onl Eample: ( )d + 4d = 0 " ' + = 0 d 3 3 d n d n d ( ) a ( ) + + a ( ) + a ( ) g ( ) d + 3 5 = d n 1 d d 1... 1 1 0 n d d n + n = e 9/43
Nonlinear ODE A differential equation with nonlinear functions of the dependent variable or its derivatives. Eamples: If is the dependent variable, (1 )' + = e d /d + sin = 0 (4) + = 0 10/43
Solution of an ODE Definition: a solution of an ODE is a function (), defined on an interval I and possessing at least n derivatives that are continuous on I, which when substituted into an nth-order ODE reduces the equation to an identit. That is, a solution () of F satisfies: F(, (), (), (),, (n) ()) = 0, I. Trivial solution of an ODE: () = 0, I 11/43
Interval of Definition A solution of an ODE includes a function () and the interval of definition, I. I is usuall referred to as the interval of definition, the interval of eistence, the interval of validit, or the domain of the solution. I can be an open interval (a, b), a closed interval [a, b], an infinite interval (a, ), and so on. 1/43
Solution Curve The graph of a solution () of an ODE is called a solution curve. Since () is a differentiable function, it is continuous on its interval of definition. There mabe a difference between the graph of () and the graph of the solution of the ODE. 1 1 1 1 = 1/, 0 = 1/, (0, ) 13/43
Eplicit and Implicit Solutions Definition: A solution in which the dependent variable is epressed solel in terms of the independent variable and constants is called an eplicit solution. Definition: A relation G(, ) = 0 is said to be an implicit solution of an ODE on an interval I provided there eists at least one function that satisfies the relation as well as the differential equation on I. 14/43
Verification of an Implicit Solution Eample: The relation + = 5 is the implicit solution of the differential equation d/d = / on the interval 5 < < 5 Verification: d d d d + + = 0 d d d = / d ( ) = ( 5) 15/43
Solving for Eplicit Solution One can solve an implicit solution for eplicit solutions. In the previous eample, 5 5 5-5 5-5 5-5 5-5 - 5 Implicit solution Eplicit solution 1 Eplicit solution + = 5 = 5, 5 < 5 5, 5 < = 5 1 < < 16/43
Families of Solutions A solution to a 1st-order DE containing an arbitrar constant represents a set G(,, c) = 0 of solutions is called a one-parameter famil of solutions. For nth-order DE, an n-parameter famil of solutions can be represented as G(,, c 1, c,, c n ) = 0. If the parameters c 1, c,, c n are resolved, then it s called a particular solution of the DE. c > 0 c = 0 = c cos is a famil of solutions of = sin c < 0 17/43
Eample: Two-Parameter Famil The functions = c 1 cos4t and = c sin4t, where c 1 and c are arbitrar constants, are both solutions of the ODE: " + 16 = 0 For = c 1 cos4t, the first two derivative w.r.t. t are ' = 4c 1 sin 4t and " = 16c 1 cos4t. Substituting " and ' into the DE gives " + 16 = 16c 1 cos4t + 16(c 1 cos4t) = 0 Similarl, for = c sin4t, we have " + 16 = 16c sin4t + 16(c sin4t) = 0 The linear combination of these two solutions is a famil of solutions. 18/43
Eample: Piecewise Solutions One can verif that = c 4 is a solution of ' 4 = 0 on the interval (, ). The following piecewise defined solution is a particular solution of the ODE: = 4, 4, < 0 0 This particular solution cannot be obtained b a single choice of c. 19/43
Piecewise Solutions (cont.) c = 1 c = 1 c = 1, < 0 c = 1, 0 0/43
Singular Solutions Definition: A singular solution is a solution that cannot be obtained b specializing an of the parameters in the famil of solutions. Eample: Both = 4 /16 and = 0 are solutions of d/d = 1/ on the interval (, ). The ODE possesses the famil of solutions = ( /4 + c). However, = 0 is not in the famil of solutions. 1/43
General Solutions Definition: If ever solution of an nth-order ODE F(,,,,, (n) ) = 0 on an interval I can be obtained from an n-parameter famil of equations G(,, c 1, c,, c n ) = 0 b appropriate choices of the parameters c i, i = 1,,, n, we then sa that the n- parameter famil of equations is the general solution of the D.E. /43
Initial Value Problem Definition: On some interval I containing 0, the problem: Solve: n d n d = f ( ( n 1),,,..., ) Subject to: ( 0 ) = 0, ( 0 ) = 1,, (n 1) ( 0 ) = n 1, where 0, 1,, n 1, are arbitraril specified real constants, is called an initial value problem (IVP). 3/43
First Order IVP A first order IVP tries to solve d/d = f(, ), subject to ( 0 ) = 0. In geometric term, we are seeking a solution so that the solution curve passes through the prescribed point ( 0, 0 ). solutions of the DE ( 0, 0 ) I 4/43
Second Order IVP A second order IVP tries to solve d /d = f(,, ), subject to ( 0 ) = 0, ( 0 ) = 1. In geometric term, we are seeking a solution so that the solution curve not onl passes through the prescribed point ( 0, 0 ), but also with a slope 1 at this point. solutions of the DE m = 1 ( 0, 0 ) I 5/43
Eample: 1st-Order IVPs It is eas to verif that = ce is a one-parameter famil of solutions of the simple first-order equation ' = on the interval (, ). (0) = 3 3 = ce 0 = c = 3e is a solution of IVP: ' =, (0) = 3. (0, 3) (1, ) 6/43
Solution b Integration If the DE can be epressed in normal form with f(, ) = g(), the solution can be obtained b integration. Since, d = d g( ) Integrating both sides, we have: = ( ) d = G( ) c g + where G() is the indefinite integral of g(). 7/43
Eample: Solution b Integration Solving the initial value problem d = + 3, (1) =. d B integrating both sides, we have = ( ) ( + 3) d = + 3 + c. 4 0 - -4-6 -8 C = C = 0 C = C = 4 C = 6 Solution that passes through the initial condition (1, ) is the curve with C = -10-6 -5-4 -3 - -1 0 1 3 4 Famil of solution curves 8/43
nd -Order Solution b Integration If we have a second-order DE of the special form: we have d d d d = g( ), = ( ) d = g( ) d = G( ) + C1, where G is an anti-derivative of g and C 1 is an arbitrar constant. Therefore, [ G( ) + C ] d = G( ) d + C, ( ) = ( ) d = 1 1 + C where C is a second arbitrar constant. 9/43
Solution Curves without Solving DE Sometimes, just b looking at the differential equation, we can learn useful information about the nature of its solutions The solution curve = () of a first order DE d/d = f (, ) on its interval of definition I must possess a tangent line at each point (, ()), and must have no breaks. The slope of the tangent line at (, ()) on a solution curve is the value of the first derivative d/d at this point. A (ver small) line segment at (, ()) that has the slope f (, ) is called lineal element of the solution curve. 30/43
Eample of Lineal Element Consider d/d = f (, ) = 0., the slope of the lineal element of the solution curve at (, 3) is f (, 3) = 1.. slope = 1. solution curve (, 3) (, 3) tangent Lineal element at (, 3) Lineal element is tangent to solution curve passes (,3) 31/43
Slope Field The collection of the lineal elements on a rectangular grid on the -plane is called a direction field or a slope field of the DE d/d = f (, ). A single solution curve on the plane will follow the flow pattern of the slope field. 4 4 - - 4 - - 4-4 - 4 The slope field of d/d = 0.. - 4-4 Solution famil: = ce 0.1 3/43
Eample: Approimating a Solution Use a slope field to approimate the IVP, d/d = sin, (0) =-3/: 1) Define the direction field around 0 ) Constraint 1: the solution must pass (0, 3/) 3) Constraint : the slope of the solution curve must be 0 when = 0 and = π 4 the solution curve can be approimated as in the figure. - - 4-4 - 4 33/43
Uniqueness of IVP Solutions (1/) Consider the IVP d/d = 1/, (0) = 0: The DE has a constant solution = 0 and a famil of solution 4 = + c. The IVP has infinite solutions: For an a 0, a = 0 a > 0 = 0, ( a ) /16, < a a (0, 0) 34/43
Uniqueness of IVP Solutions (/) Consider onl the case c 0, let c = b, b 0: = 4 ( = 4 = ( a ) 4b 4 b) /16, a = b 400 350 300 50 00 4 = + c c = 4 c = 0 c = 4 150 100 50 0-10 -8-6 -4-0 4 6 8 10 35/43
Eistence and Uniqueness Two fundamental questions of solving DE: Do solutions eist for the differential equation? Given an initial condition, is the solution unique? For eample, the differential equation ' = 1/, (0) = 0 has no solution. B integration, we have () = ln + c, but ln is not defined at 0! 36/43
Eample: Multiple-Solution IVP Each of the functions = 0 and = 4 /16 satisfies the differential equation d/d = 1/ and the initial condition (0) = 0. = 4 /16 1 = 0 (0, 0) 37/43
Eistence and Uniqueness of Solutions Theorem: Let R be a rectangular region in the plane defined b a b, c d, that contains the point ( 0, 0 ) in its interior. If f(, ) and f/ are continuous on R, then there eist some interval I 0 : 0 h < < 0 +h, h > 0, contained in a b, and a unique function () defined on I 0 that is a solution of the first-order initial-value problem: d d = f (, ), ( 0 ) = 0. d R ( 0, 0 ) c a I 0 b 38/43
Eample: Again, let s revisit the IVP: d/d = 1/, (0)=0. Since f(, ) = 1/, and f/ = /( 1/ ), the are continuous in the upper half-plane defined b > 0. Therefore, for an ( 0, 0 ), 0 > 0, there is an interval centered at 0 on which the given DE has a unique solution. However, There is no unique solution for the IVP since f/ is undefined at (0, 0). 39/43
Remark: The condition that f(, ) and f/ are continuous on R is a sufficient but not necessar condition: f(, ) continuous f/ continuous eistence of unique solution 40/43
DE as Mathematical Models Real-world situation Formulation Interpretation Mathematical model Mathematical analsis Mathematical results 41/43
Eample: Series Circuit Voltage drop across electronic components: Inductor: Resister: Capacitor: Kirchhoff s second law: L di dt ir 1 C q E(t) Voltage drop = Impressed Voltage, that is: d q dq 1 L + R + q = dt dt C E ( t) L C R 4/43
Eample: Falling Bodies Newton s second law of motion: F = ma Question: what is the position s(t) of the rock relative to the ground at time t? v 0 Acceleration of the rock: d s/dt m d s dt = mg d s dt = g rock Model: d s/dt = g, s(0) = s 0, s'(0) = v 0. s 0 s(t) Solution: s(t) = gt / + v 0 t + s 0 building ground 43/43