Edwards-Penney Chapter 3, sections 3.1, 3.2, 3.3, 3.4 Topics, Definitions, Theorems 3.1 Second order linear differential equations ==== DEF. Linear second order differential equation A(x)y'' + B(x)y' + C(x)y=F(x) This is called the non-homogeneous equation when F(x) is nonzero. DEF. Homogeneous equation is A(x)y'' + B(x)y' + C(x)y=0 EXAMPLE. Spring-mass system mx''(t)+cx'(t)+kx(t)=0 or F(t) THEOREM [Superposition] Let y1, y2 be two solutions of a linear homogeneous second order differential equation. Let c1, c2 be any two constants. Then y=c1 y1 + c2 y2 is a solution of the differential equation. EXAMPLE. Let y1=cos(x), y2=sin(x) for y''+y=0. Both are solutions. Then y=c1 cos(x) + c2 sin(x) is also a solution of y''+y=0. THEOREM 2 [Picard Existence-Uniqueness] Let functions A(x), B(x), C(x), F(x) be continuous on an open interval containing x=a on which A(x) is never zero. Then the problem A(x)y'' + B(x)y' + C(x)y=F(x) y(0)=y0, y'(a)=y1 has a unique solution defined on the same interval, for all choices of initial values y0, y1. THEORY: The interval size is reduced for nonlinear second order equations, which in general requires additional assumptions of differentiability. SOLUTIONS CAN CROSS: Second order equation solutions graphed as x against y are curves in the plane, but they can cross. Don't expect a non-crossing theorem. EXAMPLE. Let y1=cos(x), y2=sin(x) for y''+y=0. Find the Picard theorem solution y(x) with initial conditions y(0)=3, y'(0)=-2. ANSWER: y=3 cos(x) - 2 sin(x), by linear algebra. EXAMPLE. Let y1=exp(x), y2=x exp(x) for y''-2y'+y=0. Find the Picard theorem solution y(x) with initial conditions y(0)=3, y'(0)=1. ANSWER: y=3 exp(x) - 2 x exp(x), by linear algebra. DEF. Two functions y1, y2 are independent on interval J if neither is a scalar multiple of the other on interval J. TEST: If y1/y2 and y2/y1 are both non-constant on J, then y1, y2 are independent on J. THEOREM 3 [Wronskian test for linear DE] Let y1, y2 be two solutions of a second order linear homogeneous differential equation. Then y1, y2 are independent on J if and only if their Wronskian determinant is nonzero at some point in J. THEOREM 4. [General Solution] Let A(x), B(x), C(x) be continuous on an interval J with A(x) never zero. Let y(x) be any solution on J of the homogeneous differential equation A(x)y'' + B(x)y' + C(x)y=0 Let y1, y2 be two independent solutions of this equation on J. Then there exist two constants c1, c2 such that y(x)=c1 y1(x) + c2 y2(x). EXAMPLE. Equation y''-4y=0, y1=exp(2x), y2=exp(-2x), y3=cosh(2x), y4=sinh(2x). Two bases {y1,y2} and {y3,y4} can represent a general solution. DEF. Characteristic equation found by Euler's substitution y=exp(rx) to give from DE ay''+by'+cy=0 the algebraic equation ar^2+br+c=0. THEOREM 5. [Euler's Method, real distinct roots] Let ar^+br+c=0 have real distinct roots r1, r2. Then exp(r1 x), exp(r2 x) are solution atoms for equation ay''+by'+cy=0. These two atoms are independent, therefore a general solution is a linear combination of the two solution atoms: y=c1 exp(r1 x) + c2 exp(r2 x) EXAMPLE. Solve 2y'' - 7y' + 3y = 0. EXAMPLE. Solve y'' + 2y' = 0. THEOREM 6. [Euler's Method, equal real roots] Let ar^+br+c=0 have equal real roots r1, r1. Then exp(r1 x), x exp(r1 x) are solution atoms for equation ay''+by'+cy=0. These two atoms are independent, therefore a general solution is a linear combination of the two solution atoms: y=c1 exp(r1 x) + c2 x exp(r2 x) EXAMPLE. Solve y'' + 2y' + y = 0. Then find the unique solution with initial conditions y(0)=5, y'(0)=-3. ANSWER: y = l.c. of atoms exp(-x), x exp(-x) y = c1 exp(-x) + c2 x exp(-x) Linear algebra gives c1=5, c2=-3. 3.2 General solutions of linear differential equations ==== DEF. An nth order linear differential equation is P0(x)D^n y(x) + ... + Pn(x) y(x) = F(x) Assume the coefficients are continuous on an interval J and P0(x) is never zero on J. THEOREM 1. [Superposition] Let y1, y2 be two solutions on J of the homogeneous equation D^n y(x) + p1(x)D^(n-1) y(x) + ... + pn(x) y(x) = 0 then c1 y1 + c2 y2 is also a solution. More generally, a linear combination of a finite number of solutions is also a solution. THEOREM 2. [Picard's Existence-Uniqueness] Assume coefficient functions p1 to pn and f are continous on interval J and x=a is a point in J. The differential equation D^n y(x) + p1(x)D^(n-1) y(x) + ... + pn(x) y(x) = f(x) has a unique solution for each possible initial value problem of the form y(a)=b1, y'(a)=b2, ..., D^(n-1) y(a)=bn EXAMPLE. Equation D^3 y + 3 D^2 y + 4D y + 12y=0, has a solution y=-3 exp(-3x)+3 cos(2x) -2 sin(2x), initial conditions y(0)=0, y'(0)=5, y''(0)= -39. By Picard's theorem, this is the unique solution of the DE with those initial conditions. THEOREM. The unique solution of a linear homogeneous DE with zero initial conditions is the zero solution. EXAMPLE. Equation x^2 y'' - 4xy' + 6y = 0 has two independent solution y1=x^2 and y2=x^3. Both have initial condition y(0)=y'(0)=0. Does this conflict wit the preceding theorem? ANSWER: No, Picard's theorem requires nonzero leading coefficient on J. DEF. Functions f1 to fn are independent on J provided the equation c1 f1(x)+...+cn fn(x)=0, all x in J, implies c1=...=cn=0. EXAMPLE. f1=sin(2x), f2=sin(x)cos(x), f3=exp(x) are dependent because of the trig identity sin(2x)=2sin(x)cos(x), which implies the identity (1)f1(x)+(-2)f2(x)+(0)f3(x)=0 or all x in J DEF. Wronskian of n functions. THEOREM. [Wronskian Test] If the Wronskian of functions f1 to fn is nonzero at one point of J, then f1 to fn are independent on J. EXAMPLE. y1=exp(-3x), y2=cos(2x), y3=sin(2x) are independent on J=(-inf,inf) by the Wronskian test. EXAMPLE. Show y1=x, y2=x ln(x), y3=x^2 are solutions of x^3 D^3 y - x^2 D^2 y + 2x y' - 2y = 0. Then show they are independent on J=(0,infinity). Then find the unique solution of the initial value problem y(1)=3, y'(1)=2, y''(1)=1 THEOREM 3. [Abel's Independence Theorem] Let W be the Wronskian of n solutions of a homogeneous linear differential equation with coefficients continous on J. (a) If the solutions are dependent on J, then W=0 on J. (b) If the solutions are independent on J, then W is never zero on J. (c) [Abel's Identity] W(x)=K exp(-int(p1(x),x)) THEOREM 4. [General Solution] Let y1 to yn be independent solutions of a continuous-coefficient homogeneous linear differential equation on interval J. If y is any solution of the DE on J, then there exist constants c1 to cn such that y(x) = c1 y1(x) + ... + cn yn(x) for all x in J. EXAMPLE. Find constants c1, c2, c3 for the initial value problem D^3 y + 3 D^2 y + 4D y + 12y=0, y(0)=0, y'(0)=5, y''(0)= -39. ANSWER: By linear algebra, y=-3 exp(-3x)+3 cos(2x) -2 sin(2x), THEOREM 5. [Superposition y=yh + yp] Let yp(x) be a particular solution of the continuous coefficient nonhomogeneous linear differential equation P0(x)D^n y(x) + ... + Pn(x) y(x) = F(x) on interval J, assuming P0 nonzero on J. Let y1 to yn be independent solutions on J of the homogeneous equation P0(x)D^n y(x) + ... + Pn(x) y(x) = 0 Then any solution y of the nonhomogeneous equation can be expressed for unique constants c1 to cn as y(x) = yh(x) + yp(x), yh(x) = c1 y1 + ... + cn yn(x), x in J EXAMPLE. Equation y'' + 4y = 12x, yp=3x, y1=cos(2x), y2=sin(2x). Find c1, c2 such that y=yh+yp, yh=c1 y1 + c2 y2, and y(0)=5, y'(0)=7. ANSWER: By linear algebra, c1=5, c2=2. A common error is to put yh into the equation instead of y=yh+yp (3 terms). Another error is to put y=yh+yp into y''+4y=0 (the wrong equation). 3.3 Homogeneous constant coefficient linear differential equations ==== DEF. Characteristic equation. Put y=exp(rx) into the DE (Euler's substitution) to get the characteristic equation. There is a synthetic way to do this, a fast shortcut. THEOREM 1. [Euler's Method for distinct roots] Each real root r of the characteristic equation generates one solution atom exp(rx). If there are no repeated roots and all roots are real, then there will be n independent solution atoms generated. A general solution is then a linear combination of Euler's solution atoms. EXAMPLE. D^3 y + 3 D^2 y - 10 y' = 0 with characteristic equation r^3+3r^2-10r=0 and roots r=0, 2, -5 (a) Find the Euler solution atoms. ANSWER: exp(0x), exp(2x), exp(-5x), which are independent. Then a general solution is y = c1 (1) + c2 (exp(2x)) + c3 (exp(5x)). (b) Find the unique solution for initial conditions y(0)=7, y'(0)0, y''(0)=70 . ANSWER: By linear algebra, y=2exp(-5x)+5exp(2x) THEOREM 2. [Euler's Method for repeated real roots] Let the characteristic equation have a repeated real root r of multiplicity k. Then there are k independent Euler solution atoms exp(rx), x exp(rx), ..., x^(k-1) exp(rx) corresponding to this (multiple) root r. EXAMPLE. Equation 9 D^4 y - 6 D^4 y + D^3 y = 0 has a characteristic equation with triple root r=0,0,0 and a double root r=1/3,1/3. Find a general solution. ANSWER: Euler's solution atoms are: Root r=0,0,0: exp(0x), x exp(0x), x^2 exp(0x) Root r=1/3,1/3: exp(x/3), x exp(x/3) A general solution is a linear combination of the 5 solution atoms. THEOREM 3. [Euler's Method for complex roots] (1) Let the characteristic equation have a pair of complex conjugate roots a+bi, a-bi with b>0. Then there are two independent Euler solution atoms exp(ax) cos(bx), exp(ax) sin(bx) (2) Let the characteristic equation have a repeated pair of complex roots a+bi, a-bi (with b>0) of multiplicity k. Then there are 2k independent solution atoms, obtained from the first pair above by multiplication by 1, x, ..., x^(k-1). EXAMPLE. Solve y'' + b^2 y = 0 with b>0. ANSWER: Characteristic equation r^2+b^2=0, roots bi, -bi, Euler solution atoms cos(bx), sin(bx). EXAMPLE. Solve y'' - 4y' + 5y = 0 with initial conditions y(0)=1, y'(0)=5. ANSWER: y = c1 exp(2x) cos(x) + c2 exp(2x) sin(x) By linear algebra, c1=1, c2=3. EXAMPLE. Solve for a general solution of D^4 y + 4y = 0. ANSWER: The roots are 1+i, 1-i, -1+i, -1-i. THEOREM 3 applies for each of the two complex conjugate pairs to obtain Euler solution atoms Roots 1+i, 1-i: exp(x) cos(x), exp(x) sin(x) Roots -1+i, -1-i: exp(-x) cos(x), exp(-x) sin(x) A general solution is a linear combination of the 4 independent Euler solution atoms. EXAMPLE. Solve D^4 y + 4y = 0. ANSWER: Solution y is a linear combination of the four Euler solutions e^x cos(x), e^x sin(x), e^(-x) cos(x), e^(-x) sin(x). EXAMPLE. Solve (D^2+6D + 13)^2 y = 0 ANSWER: Solution y is a linear combination of the four Euler solutions e^(-3x) cos(2x), e^(-3x) sin(2x), x e^(-3x) cos(2x), x e^(-3x) sin(2x). EXAMPLE. Solve y''' + y' -10y = 0 ANSWER: Solution y is a linear combination of the three Euler solutions e^(2x), e^(-x) cos(2x), e^(-x) sin(2x). EXAMPLE. Given the characteristic equation has roots 3, -5, 0,0,0,0, -5, 2+3i, 2-3i, 2+3i, 2-3i (total 11 roots), then find a general solution. ANSWER: Write the roots according to multiplicity 3, -5,-5, 0,0,0,0, 2+3i,2-3i, 2+3i,2-3i then apply THEOREMS 1, 2, 3 to obtain Euler solution atoms Root 3: exp(3x) Roots -5,-5: exp(-5x), x exp(-5x) Roots 0,0,0,0: exp(0x), x exp(0x), x^2 exp(0x), x^3 exp(0x) Roots 2+3i,2-3i, 2+3i,2-3i: exp(2x) cos(3x), exp(2x) sin(3x) x exp(2x) cos(3x), x exp(2x) sin(3x) There are 11 Euler solution atoms. A linear combination of the 11 atoms is a general solution. 3.4 Mechanical Vibrations ==== Spring-Mass damped oscillations, unforced mx'' + cx' + kx = 0 Spring-Mass damped oscillations, forced mx'' + cx' + kx = F(t) Simple Pendulum, unforced theta'' + (g/L) sin(theta) = 0 Derivation by conservation of mechanical energy Derivation by Newton's second law and tangential components Undamped oscillations, unforced mx'' + kx = 0 x'' + omega0^2 x = 0 x(t) = A cos(omega0 t) + B sin(omega0 t) x(t) = C cos(omega0 t - alpha) DEF. Amplitude C, frequency omega0, phase angle alpha Period T = 2 Pi/omega0 Time lag = alpha/omega0 (solve omega0 t - alpha=0 for t) EXAMPLE. m=1/2 kg, spring k=force/elongation=100/2, x(0)=1, x'(0)=-5. Verify model x'' + 100 x = 0. Solve for x(t)=A cos(10 t) + B sin(10 t) Solve for A=1, B=-1/2 by linear algebra. Evaluate omega0 = 10, T = 0.6283, frequency = 1.5915, alpha=5.8195 radians, time lag=0.5820 seconds, amplitude=C=sqrt(5)/2 Damped oscillations, unforced x'' + 2p x' + omega0^2 x = 0 omega0 = sqrt(k/m), p = c/(2m) DEF. CRITICAL DAMPING is the value of damping constant c when the square root in the quadratic formula is zero: B^2-4AC=0 This means c^2-4km is zero and the solution of the differential equation is non-oscillatory, being a combination of Euler solution atoms involving no trig functions. DEF. OVER-DAMPING is the case when the argument of the square root in the quadratic formula is positive, giving two real roots to the quadratic equation. This means c^2-4km is positive and the solution of the differential equation is non-oscillatory, being a combination of Euler solution atoms involving no trig functions. DEF. UNDER-DAMPING is the case when the argument of the square root in the quadratic formula is negative, giving two complex conjugate roots to the quadratic equation. This means c^2-4km is negative and the solution of the differential equation is oscillatory, being a combination of Euler solution atoms involving trig functions. DEF. The PSEUDO-CIRCULAR-FREQUENCY is omega1, obtained from the quadratic formula as the positive coefficient of complex unit i. This term only applies to the case when there are complex roots. DEF. The TIME-VARYING AMPLITUDE is the factor in front of the cosine factor, when the oscillating solution x(t) is presented in PHASE-AMPLITUDE FORM x(t)=C exp(-pt) cos(omega1 t - alpha) The PSEUDO-PERIOD is T = 2 Pi/omega1. It is the period of the cosine term in the phase amplitude form of the solution. EXAMPLE. m=1/2 kg, spring k=force/elongation=100/2, x(0)=1, x'(0)=-5. Attach a dashpot (dampener) with DE term cx' and assume c=force/velocity=1. Find the DE and solve it. ANSWER: x'' + 2x' + 100x = 0 ANSWER: Euler solution atoms are exp(-t)cos(sqrt(99)t), exp(-t)sin(sqrt(99)t) Calculate PSEUDO-CIRCULAR-FREQUENCY = sqrt(99) PSEUDO-PERIOD = 2 Pi/sqrt(99) = 0.6315 PSEUDO-FREQUENCY = sqrt(99)/(2 Pi) = 1.5836 x(t) = A exp(-t)cos(sqrt(99)t)+B exp(-t)sin(sqrt(99)t) A=1, B=-4/sqrt(99) x(t) = C1 exp(-t) cos(sqrt(99)t-alpha) C1 = sqrt(115/99) alpha = 5.9009 radians Find the first four equilibria t1, t2, t3, t4 [Table section 3.4] These are solutions of x(t)=0, which reduces to zeros of the cosine factor in the phase-amplitude form of the solution. === End