DIFFERENTIAL EQUATIONS AND FAMILIES OF CURVES

Transcription

1 DIFFERENTIAL EQUATIONS AND FAMILIES OF CURVES General solution of a first-order ODE y = f(x,y) contains an arbitrary constant: y = (x,c) one curve in x,y plane for each value of c general solution can be thought of as one-parameter family of curves Example: y = x/y. separable equation y dy = x dx y / = x /+c i.e., x +y = constant : family of circles centered at origin y x Fig.1

2 Orthogonal trajectories Given the family of curves representing solutions of ODE y = f(x,y), orthogonal trajectories are given by a second family of curves which are solutions of y = 1/f(x,y). Then each curve in either family is perpendicular to every curve in the other family. Example: Find the orthogonal trajectories to the family of circles y = x/y. Solve y = y/x. dy dx y = lny = lnx+constant x i.e., y = cx : family of straight lines through the origin

3 a) Find the family of curves corresponding to solutions of the ODE y = (y x )/(xy). b) Find the orthogonal trajectories to the above family of curves. homogeneous equation y = f(y/x) with f(y/x) = (y/x x/y)/ solvable by y v = y/x and separation of variables x +y = cx : family of circles tangent to y axis at 0 y x Fig. orthogonal trajectories found by solving y = xy/(y x ) x +y = ky : family of circles tangent to x axis at 0

4 EXPLOITING FIRST-ORDER METHODS TO TREAT EQUATIONS OF HIGHER ORDER IN SPECIAL CASES y not present in nd-order equation F(x,y,y,y ) = 0 setting y = q yields 1st-order equation for q(x). x not present in nd-order equation F(x,y,y,y ) = 0 setting y = q, y = dq/dx = q(dq/dy) yields G(y,q,dq/dy) = 0. Example: homogeneous, flexible chain hanging under its own weight y s T θ ρ = linear mass density T x Using Newton s law, the shape y(x) of the chain obeys the nd order nonlinear differential equation y = a 1 + (y ), a ρ g / T Setting y = q q = a 1 + q

5 Separation of variables 1 dq = a 1+q dx Using q = dy/dx = 0 at x = 0 ln(q + 1+q ) = ax Solving for q q = dy/dx = (e ax e ax )/ Thus y(x) = 1 a e ax +e ax +constant = 1 a coshax+constant This curve is called a catenary. Historical note. The problem of the catenary was the subject of a challenge posed by Jakob Bernoulli in 1690, in response to which the problem was solved the following year indipendently by Johann Bernoulli, Leibniz and Huygens.

6 Summary No general method of solution for 1st-order ODEs beyond linear case; rather, a variety of techniques that work on a case-by-case basis. Main guiding criteria: methods to bring equation to separated-variables form methods to bring equation to exact differential form transformations that linearize the equation 1st-order ODEs correspond to families of curves in x, y plane geometric interpretation of solutions Equations of higher order may be reduceable to first-order problems in special cases e.g. when y or x variables are missing from nd order equations

7 SECOND-ORDER LINEAR ODEs f +p(x)f +q(x)f = h(x) Generalities Structure of general solution Equations with constant coefficients

8 These equations are incompatible in most cases. Second order linear equations General form : d f dx df + p( x) + q( x) f = h( x). dx Integrating factor? Suppose I( x ) such that d If dx = Ih di d I = Ip and = Iq.. dx dx

9 Structure of the general solution (GS) f +p(x)f +q(x)f = h(x) The general solution f is the sum of a particular solution f 0 (the particular integral, PI) and the general solution f 1 of the associated homogeneous equation (the complementary function, CF): f = f 0 +f 1, i.e., GS = PI + CF. The complementary function CF is given by linear combination of two linearly independent ( see next) solutions u 1 and u : CF = c 1 u 1 (x)+c u (x) c 1 and c arbitrary constants

10 Two functions u 1 (x) and u (x) are linearly independent if the relation αu 1 (x)+βu (x) = 0 implies α = β = 0. Let αu 1 (x)+βu (x) = 0. Differentiating αu 1(x)+βu (x) = 0. If W(u 1,u ) = u 1 u u 1 u = u 1u u u 1 0 then α = β = 0, and u 1 and u are linearly independent. If W(u 1,u ) = u 1 u u u 1 = 0 then u = constant u 1 u 1 and u not linearly independent W(u 1,u ) = wronskian determinant of functions u 1 and u

11 Example: The functions u 1 (x) = sinx and u (x) = cosx are linearly independent. Let αsinx+βcosx = 0. Differentiating αcosx βsinx = 0. So α = β sinx cosx β ( sin ) x cosx +cosx = 0 β 1 cosx = 0 β = 0. Thus α = β = 0. Alternatively: W(u 1,u ) = u 1 u u u 1 = sin x cos x = 1 linear independence

12 Homogeneous equation: f +p(x)f +q(x)f = 0 Any solution u can be written as a linear combination of linearly independent solutions u 1 and u. Since u, u 1 and u all solve the homogeneous eq., with nonzero coefficients of the second-derivative, first-derivative and no-derivative terms, we must have det = 0 u u 1 u u u 1 u u u 1 u = 0 W(u,u 1,u ) = 0 for solutions u, u 1, u αu+βu 1 +γu = 0 for α, β and γ not all zero. Solving for u expresses the solution u as a linear combination of u 1 and u. CF = c 1 u 1 (x)+c u (x) solutions span whole set of linear combinations of two independent u 1, u

13 Example: The general solution of the nd-order linear ODE y +y = 0 is Asinx+Bcosx. (simple harmonic oscillator) To show this, it is sufficient to show that i) sinx and cosx solve the equation (e.g. by direct computation) and ii) sin x and cos x are linearly independent (see previous Example). Then general theorem CF = c 1 u 1 (x)+c u (x) yields the result.

14 Useful reference for this part of the course, with worked problems and examples, is Schaum s Outline Series Differential Equations R. Bronson and G. Costa McGraw-Hill (Third Edition, 006) See chapters 8 to 14.

15 Next we will specialize to equations with constant coefficients: p(x) = const., q(x) = const. general methods of solution available arise in many physical applications Second order linear equation with constant coefficients d f df Lf = a a 1 a0f h( x) dx + dx + =.