ENGI 940 Lecture Notes - ODEs Page.0. Ordinary Differential Equations An equation involving a function of one independent variable and the derivative(s) of that function is an ordinary differential equation (ODE). The highest order derivative present determines the order of the ODE and the power to which that highest order derivative appears is the degree of the ODE. A general n th order ODE is Example.00. ( n) ( y ) F xyy,,, y,, = 0 d y d x dy + x = dx xy is a Example.00. dy x dx = xy is a In this course we will usually consider first degree ODEs of first or second order only. The topics in this chapter are treated briefly, because it is assumed that graduate students will have seen this material during their undergraduate years. Sections in this Chapter:.0 First Order ODEs - Separation of Variables.0 Exact First Order ODEs.03 Integrating Factor.04 First Order Linear ODEs [ + Integration by Parts].05 Bernoulli ODEs.06 Second Order Homogeneous Linear ODEs.07 Variation of Parameters.08 Method of Undetermined Coefficients.09 Laplace Transforms.0 Series Solutions of ODEs. The Gamma Function. Bessel and Legendre ODEs
ENGI 940.0 Separation of Variables Page.0.0 First Order ODEs - Separation of Variables Example.0. A particle falls under gravity from rest through a viscous medium such that the drag force is proportional to the square of the speed. Find the speed v(t) at any time t > 0 and find the terminal speed v.
ENGI 940.0 Separation of Variables Page.03 Example.0. (continued)
ENGI 940.0 Separation of Variables Page.04 Example.0. (continued) General solution: Initial condition: ( ) v t = k pt ( + Ae ) Ae pt Complete solution: Terminal speed v : The terminal speed can also be found directly from the ODE. At terminal speed, the acceleration is zero, so that the ODE simplifies to Graph of speed against time: [For a 90 kg person in air, b kg m k 30 ms 00 km/h. v(t) is approximately linear at first, but air resistance builds quickly. One accelerates to within 0 km/h of terminal velocity very fast, in just a few seconds.]
ENGI 940.0 Exact ODEs Page.05.0 Exact First Order ODEs If x and y are related implicitly by the equation u(x, y) = c (constant), then the chain rule for differentiation leads to the ODE u u du = dx + dy = 0 x y Therefore, for the functions M(x, y) and N(x, y) in the first order ODE M dx + N dy = 0, if a potential function u(x, y) exists such that u u M = and N = x y, then u(x, y) = c is the general solution to the ODE and the ODE is said to be exact. Note that, for nearly all functions of interest, Clairault s theorem results in the identity u u y x x y This leads to a simple test to determine whether or not an ODE is exact: M y N M dx + N dy = x 0 is exact A separable first order ODE is also exact (after suitable rearrangement). However, the converse is false. One counter-example will suffice.
ENGI 940.0 Exact ODEs Page.06 Example.0. The ODE x x y e x dx + e dy = is exact, ( ) 0 Example.0. Is the ODE y dx + x dy = 0 exact?
ENGI 940.0 Exact ODEs Page.07 Example.0.3 Is the ODE n+ n+ n+ n A x y dx + x y dy = 0 ( ) (where n is any constant and A is any non-zero constant) exact? Find the general solution. Note that the exact ODE in example.0.3 is just the non-exact ODE of example.0. n+ n multiplied by the factor I( x, y) = Ax y. The ODEs are therefore equivalent and n+ share the same general solution. The function ( ) factor for the ODE of example.0.. n I x, y = Ax y is an integrating Also note that the integrating factor is not unique. In this case, any two distinct values of n generate two distinct integrating factors that both convert the non-exact ODE into an exact form. However, we need to guard against introducing a spurious singular solution y 0.
ENGI 940.03 Integrating Factor Page.08.03 Integrating Factor Occasionally it is possible to transform a non-exact first order ODE into exact form, using an integrating factor I (x, y). Suppose that P dx + Q dy = 0 is not exact, but that IP dx + IQ dy = 0 is exact. Then, using the product rule,
ENGI 940.03 Integrating Factor Page.09 Example.03. (Example.0. again) Find the general solution of the ODE y dx + x dy = 0
ENGI 940.03 Integrating Factor Page.0 Example.03. Find the general solution of the ODE xy dx + (x + 3y) dy = 0
ENGI 940.04 Linear ODEs Page..04 First Order Linear ODEs [ + Integration by Parts] A special case of a first order ODE is the linear ODE: [or, in some cases, dy P x y R x dx + = dx dy ( ) ( ) ( y) x = S( y) + Q ] Rearranging the first ODE into standard form, (P(x) y R(x)) dx + dy = 0 dy + is dx hx ( ) hx ( ) y ( x) = e e R ( x) dx C +, where h( x) = P ( x) dx Therefore the general solution of P( x) y = R( x)
ENGI 940.04 Linear ODEs Page. Example.04. Solve the ordinary differential equation dy dx + y = x
ENGI 940.04 Linear ODEs Page.3 Example.04. (continued)
ENGI 940.04 Linear ODEs Page.4 Examples of Integration by Parts The method of integration by parts will be required in the next example of a first order linear ODE (Example.04.4). There are three main cases for integration by parts: Example.04. Integrate x 3 e x with respect to x. Example.04.3 Integrate ln x with respect to x.
ENGI 940.04 Linear ODEs Page.5 Example.04.4 An electrical circuit that contains a resistor, R = 8 Ω (ohm), an inductor, L = 0.0 millihenry, and an applied emf, E(t) = cos (5t), is governed by the differential equation Determine the current at any time t > 0, if initially there is a current of ampere in the circuit. First note that the inductance L = 0 5 H is very small. The ODE is therefore not very different from 0 + R i = de/dt which has the immediate solution i = (/R) de/dt = (/8) ( 0 sin 5t) We therefore anticipate that i = (5/4) sin 5t will be a good approximation to the exact solution. Substituting all values (R = 8, L = 0 5, E = cos 5t E' = 0 sin 5t) into the ODE yields which is a linear first order ODE. P(t) = 400 000 and R(t) = 500 000 sin 5t h = P dt = 400000 t integrating factor = e h 400 000t = e Integration by parts of the general case e ax sin bx dx :
ENGI 940.04 Linear ODEs Page.6 Example.04.4 (continued) ax ax e sin bx dx = e a sin bx b cosbx + C ( ) a + b Set a = 400 000, b = 5 and x = t: h 400000t e R dt = 500 000 e 400 000sin 5t 5cos 5t 400 000 + 5 The general solution is h h ( ) = ( + ) i t e e R dt C ( ) t ( ) ( 400000sin 5 5cos5 ) 400000 500 000 i t = Ae t t 400 000 + 5 But i(0) = A = (400 000 + 5 500 000) / (400 000 + 5) Therefore the complete solution is [exactly] ( ) i t = 400000t ( ) 59997 500 05 e 500 000 400 000sin 5t 5cos 5t 6000000005 To an excellent approximation, this complete solution is 400000t 5 ( ) sin 5 i t e t 4 After only a few microseconds, the transient term is negligible. The complete solution is then, to an excellent approximation, as before.
ENGI 940.05 Bernoulli ODEs Page.7.05 Bernoulli ODEs The first order linear ODE is a special case of the Bernoulli ODE dy dx + P( x) y = n R( x) y If n = 0 then the ODE is linear. If n = then the ODE is separable. n y For any other value of n, the change of variables u = n will convert the Bernoulli ODE for y into a linear ODE for u. du du dy n ndy dy ndu = = y = y dx dy dx n dx dx dx The ODE transforms to n du n du n y + P x y = R x y + P x y dx dx = R x We therefore obtain the linear ODE for u: du (( n) P( x) ) u dx + = R( x) whose solution is n ( ) ( ) ( ) ( ) ( ( ) ) ( ) ( ) ( ) y hx ( ) hx ( ) = u ( x) = e e R x dx + C, where h x = n P x dx n together with the singular solution y 0 in the cases where n > 0.
ENGI 940.05 Bernoulli ODEs Page.8 Example.05. Find the general solution of the logistic population model dy ay by dx = where a, b are positive constants. The Bernoulli equation is dy dx + a y = b y with P = a, R = b, n =. ( ) ( )
ENGI 940.06 nd Order Linear ODEs Page.9.06 Second Order Homogeneous Linear ODEs The general second order linear ordinary differential equation with constant real coefficients may be written in the form d y dy + p + qy = r ( x) dx dx If, in addition, the right-side function r(x) is identically zero, then the ODE is said to be homogeneous. Otherwise it is inhomogeneous. The most general possible solution y C to the homogeneous ODE y + py + qy = 0 is called the complementary function. A solution y P to the inhomogeneous ODE y + py + qy = r( x) is called the particular solution. The linearity of the ODE leads to the following two properties: Any linear combination of two solutions to the homogeneous ODE is another solution to the homogeneous ODE; and The sum of any solution to the homogeneous ODE and a particular solution is another solution to the inhomogeneous ODE. It can be shown that the following is a valid method for obtaining the complementary function: From the ODE y py qy r( x) + + = form the auxiliary equation (or characteristic equation ) λ + pλ + q = 0 If the roots λ, λ of this quadratic equation are distinct, then a basis for the entire set of λx λx y, y = e, e. possible complementary functions is { } { } If the roots are not real (and therefore form a complex conjugate pair a ± bj ), then the ax ax e cos bx, e sin bx. basis can be expressed instead as the equivalent real set { } If the roots are equal (and therefore real), then a basis for the entire set of possible λx λx y, y = e, xe. complementary functions is { } { } The complementary function, in the form that captures all possibilities, is then yc = Ay + By where A and B are arbitrary constants.
ENGI 940.06 nd Order Linear ODEs Page.0 Example.06. A simple unforced mass-spring system (with damping coefficient per unit mass = 6 s and restoring coefficient per unit mass = 9 s ) is released from rest at an extension m beyond its equilibrium position (s = 0). Find the position s(t) at all subsequent times t. The simple mass-spring system may be modelled by a second order linear ODE. ds The term represents the acceleration of the mass, due to the net force. dt The ds term represents the friction (damping) term. dt The s term represents the restoring force. The model is d s ds + 6 + 9s = 0 dt dt
ENGI 940.06 nd Order Linear ODEs Page. Example.06. (continued) This is an example of critical damping. Real distinct roots for λ correspond to over-damping. Complex conjugate roots for λ correspond to under-damping (damped oscillations). Illustrated here are a critically damped case ( ) ( ) 3 Example.06.), an over-damped case 4 ( ) 3 ( 4 t s t e e t ) t case s( t) e ( cos 6t sin 6t) 3 s t t = + 3t e (the solution to = and an under-damped = +, all of which share the same initial conditions s(0) = and s ( 0) = 0.
ENGI 940.07 Variation of Parameters Page..07 Variation of Parameters A particular solution y P to the inhomogeneous ODE y py qy r( x) constructed from the set of basis functions {, } varying the parameters: + + = may be y y for the complementary function by Try yp( x) u( x) y( x) v( x) y( x) (i) y is a solution of y py qy r( x) (ii) = +, where the functions u(x) and v(x) are such that P + + = and one free constraint is imposed, to ease the search for u(x) and v(x). Substituting yp = uy + vy into the ODE,
ENGI 940.07 Variation of Parameters Page.3 [space to continue the derivation of the method of variation of parameters]
ENGI 940.07 Variation of Parameters Page.4 Example.07. A mass spring system is at rest until the instant t = 3, when a sudden hammer blow, of impulse 0 Ns, sets the system into motion. No further external force is applied to the system, which has a mass of kg, a restoring force coefficient of 6 kg s and a friction coefficient of kg s. The response x(t) at any time t > 0 is governed by the differential equation d x dx + + 6 x = 0 δ ( t 3) dt dt (where δ (t a) is the Dirac delta function), together with the initial conditions x(0) = x' (0) = 0. Find the complete solution to this initial value problem.
ENGI 940.07 Variation of Parameters Page.5 Example.07. (continued) This complete solution is continuous at t = 3. It is not differentiable at t = 3, because of the infinite discontinuity of the Dirac delta function inside r(t) at t = 3. Note: δ ( t a) = lim g( ta ;, ε) H( t a) ε 0 [Total area = ] 0 = ( t < a) ( t a)
ENGI 940.07 Variation of Parameters Page.6 Example.07. Find the general solution of the ODE y" + y' 3y = x + e x.
ENGI 940.07 Variation of Parameters Page.7 Example.07. (continued)
ENGI 940.08 Undetermined Coefficients Page.8.08 Method of Undetermined Coefficients When trying to find the particular solution of the inhomogeneous ODE d y dy + p + qy = r ( x) dx dx an alternative method to variation of parameters is available only when r( x ) is one of the following special types: n kx k e, cos kx, sin kx, ak x and any linear combinations of these types and any k= products of these types. When it is available, this method is often faster than the method of variation of parameters. The method involves the substitution of a form for P r x, with coefficients yet to be determined, into the ODE. kx kx If r( x) = ce, then try yp = de, with the coefficient d to be determined. If r ( x) = a cos kx or bsin kx, then try yp = c cos kx + d sin kx, with the coefficients c and d to be determined. If r( x ) is an n th order polynomial function of x, then set y P equal to an n th order polynomial function of x, with all (n + ) coefficients to be determined. y that resembles ( ) However, if r( x ) contains a constant multiple of either part of the complementary function ( or ) y. y y, then that part must be multiplied by x in the trial function for P
ENGI 940.08 Undetermined Coefficients Page.9 Example.08. (Example.07. again) Find the general solution of the ODE y" + y' 3y = x + e x. A.E.: λ + λ 3 = 0 (λ + 3) (λ ) = 0 λ = 3, C.F.: C 3x y = Ae + Be x Particular Solution by Undetermined Coefficients:
ENGI 940.08 Undetermined Coefficients Page.30 Example.08. Find the general solution of the ODE d y dy + 4 dx dx + 4y = e x
ENGI 940.09 Laplace Transforms Page.3.09 Laplace Transforms Laplace transforms can convert some initial value problems into algebra problems. It is assumed here that students have met Laplace transforms before. Only the key results are displayed here, before they are employed to solve some initial value problems. The Laplace transform of a function f (t) is the integral where the integral exists. st F( s) = L { f ( t) } = e f ( t) dt 0 Some standard transforms and properties are: Linearity: L a f t + bg t = a L f t + b L g t a, b= constants { ( ) ( ) } ( ) { } { ( ) } ( ) Polynomial functions: n n! L{ t } L n+ n s s = = t n ( n )! First Shift Theorem: at { ( ) } ( ) ( ) { } ( ) L f t = F s L e f t = F s a at t L e and L s a n = = n at e! n ( s a) ( ) Trigonometric Functions: at at ω e sinωt { e sinωt} L = L = ( s a) + ω ( s a) + ω ω at s a s a at { e cosωt} L = L e cosωt = ( s a) + ω ( s a) + ω Derivatives: L f t = s L f t f { ( ) } ( ) ( ) { } ( 0) { f t } = s { f ( t) } s f ( 0) f ( 0) L L
ENGI 940.09 Laplace Transforms Page.3 Integration: d ds Second shift theorem: t L G( s) = { G( s) } dτ s L 0 L{ f ( t) } = L t f ( t) ( ) { } { ( ) } ( ) ( ) as { F s } = t { F( s) } L L { } ( ) ( ) L F s = f t L e F s = H t a f t a where H( t a) 0 = ( t < a) ( t a) is the Heaviside (unit step) function. Dirac delta function where ( ) δ ( ) as { δ ( t a) } = e ( ) ( if < < ) 0 ( a< c or a > d) d f a c a d f t t a dt =. c For a periodic function f (t) with fundamental period p, Convolution: L L f t ps e f t dt e 0 st { ( ) } = ( ) { F( s) G( s) } = { F( s) } { G( s) } L L L where (f * g)(t) denotes the convolution of f (t) and g(t) and is defined by ( )( ) = ( ) ( ) t f g t f τ g t τ dτ The identity function for convolution is the Dirac delta function: δ t a f t = f t a H t a δ t f t = f t 0 ( ) ( ) ( ) ( ) ( ) ( ) ( ) p
ENGI 940.09 Laplace Transforms Page.33 Here is a summary of inverse Laplace transforms. F (s) f (t) F (s) f (t) st e f ( t) dt f (t) s 0 n s s s a ( s a) n (n ø) (n ø) π t e at n t e ( n )! e as δ (t a) e as s + ω ( s a) + ω ( s a) b ( s a) ( s a) + ω ( s a) ( s a) b t n ( n )! at H (t a) sin ω t ω e at sin ω t ω e at sinh bt b e at cos ω t e at cosh bt cos s ( s + ω ) ( s + ω ) { s n F (s) s n f (0) s n f N(0) s n 3 f O(0)... s f (n ) (0) f (n ) (0) } ω t ω ω t sin ω t ω s 3 ( s + ω ) s ( s + ω ) s ω ( s + ω ) as tanh s s b as as tanh s F( s) s df ds b as ( e ) sin ω t ω t cos ω t 3 ω t sin ω t ω t cos ω t Square wave, period a, amplitude Triangular wave, period a, amplitude a Sawtooth wave, period a, amplitude b t 0 n d f n dt f ( τ ) dτ t f (t)
ENGI 940.09 Laplace Transforms Page.34 Example.09. (Example.08. again) Find the general solution of the ODE d y dy + 4 dx dx + 4y = e x The initial conditions are unknown, so let a = y(0) and b = y' (0). Taking the Laplace transform of the initial value problem,
ENGI 940.09 Laplace Transforms Page.35 Example.09. (Example.07. again) Find the complete solution to the initial value problem d x dx + + 6 x = 0 δ t 3 dt dt (where δ (t a) is the Dirac delta function), together with the initial conditions x(0) = x' (0) = 0. ( ) Let X(s) = L {x(t)} be the Laplace transform of the solution x(t). Taking the Laplace transform of the initial value problem,
ENGI 940.0 Series Solutions Page.36.0 Series Solutions of ODEs If the functions p(x), q(x) and r(x) in the ODE d y dy + p ( x) + q( x) y = r( x) dx dx are all analytic in some interval xo h < x < xo + h (and therefore possess Taylor series expansions around x o with radii of convergence of at least h), then a series solution to the ODE around x with a radius of convergence of at least h exists: o n = 0 ( n) ( ) n y x y( x) = an( x xo ), an = n! o Example.0. Find a series solution as far as the term in x 3, to the initial value problem d y dy x x + e y = 4 ; y ( 0) =, y ( 0) = 4 dx dx None of our previous methods apply to this problem. The functions x, e x and 4 are all analytic everywhere. The solution of this ODE, expressed as a power series, is y ( 0) y ( 0) 3 y( x) = y( 0) + y ( 0) x + x + x +! 3! But y(0) = and y'(0) = 4. From the ODE,
ENGI 940.0 Series Solutions Page.37 Example.0. Find the general solution (as a power series about x = 0) to the ordinary differential equation d y + xy= 0 dx n = ax n = 0 n. n n n n= n= Let the general solution be y( x). Then n ( ) = and ( ) = ( ) y x nax y x nn ax Substitute into the ODE:
ENGI 940. Gamma Function Page.38. The Gamma Function The gamma function Γ ( x) is a special function that will be needed in the solution of Bessel s ODE. Γ ( x) is a generalisation of the factorial function n! from positive integers to most real numbers. For any positive integer n, n! = n n n 3 (with 0! defined to be ) ( ) ( ) When x is a positive integer n, Γ ( n) = ( n ) We know that n! = n ( n! ) The gamma function has a similar recurrence relationship: Γ ( x+ ) = x Γ ( x) Γ ( x + ) This allows Γ ( x) to be defined for non-integer negative x, using Γ ( x) = For example, it can be shown that Γ ( ) = π Γ ( + ) Γ ( ) = = π 3 ( ) Γ ( x) is infinite when x is a negative integer or zero. It is well defined for all other real numbers x.! ( ) Γ 4 π Γ = = +, etc. 3 3 x In this graph of y ( x) = Γ, values of the factorial function (at positive integer values of x) are highlighted. There are several ways to define the gamma function, such as and x t ( x) t e dt ( x 0) Γ = > ( x) Γ = 0 lim n x nn! x x x n ( + ) ( + )
ENGI 940. Gamma Function Page.39 A related special function is the beta function: ( ) ( ) ( ) ( ) ( m n) / n π = = θ θ θ = 0 0 Γ + m m n Γ m Γ n B m, n t t dt sin cos d Among the many results involving the gamma function are: For the closed region V in the first octant, bounded by the coordinate planes and the α β γ x y z surface + + =, with all constants positive, a b c V p q r ( α) ( β ) ( γ ) p q ( r α β γ ) p q r p q r abc Γ Γ Γ I = x y z dx dy dz = αβγ Γ + + + For the closed area A in the first quadrant, bounded by the coordinate axes and the curve α β x y + =, with all constants positive, a b A p q ( α ) Γ( β ) p q ( α β ) p q p q a b Γ I = x y dx dy = αβ Γ + + Example.. Establish the formula for the area enclosed by an ellipse. The Cartesian equation of a standard ellipse is x y + =. a b
ENGI 940. Bessel and Legendre ODEs Page.40. Bessel and Legendre ODEs Frobenius Series Solution of an ODE If the ODE P( x) y + Q( x) y + R( x) y = F( x) ( ) ( ) is such that P( x o ) = 0, but ( x xo) ( x xo) ( ) ( ) Q x R x F x, and are all P x P x P x analytic at x o, then x = x o is a regular singular point of the ODE. A Frobenius series solution of the ODE about x= x exists: ( ) = ( ) y x c x x n = 0 n o o n+ r for some real number(s) r and for some set of values { c n }. ( ) ( ) Example.. Find a solution of Bessel s ordinary differential equation of order ν, (ν > 0), ( ν ) x y + xy + x y = 0
ENGI 940. Bessel and Legendre ODEs Page.4 Example.. (continued)
ENGI 940. Bessel and Legendre ODEs Page.4 Example.. (continued) One Frobenius solution of Bessel s equation of order ν is therefore where J ( x) k ( ) k+ ν 0 k 0 k! Γ ( ν + k+ ) k = 0 ( ) ( ν ) ( ν ) ( ) y x = c Γ + x = c Γ + Jν x ν is the Bessel function of the first kind of order ν. It turns out that the Frobenius series found by setting r = ν generates a second linearly independent solution J ν (x) of the Bessel equation only if ν is not an integer.
ENGI 940. Bessel and Legendre ODEs Page.43 The Bessel ODE in standard form, x y + xy + x ν y = 0 has the general solution ( ) y( x) = AJν ( x) + BYν ( x) unless ν is not an integer, in which case Yν ( x) can be replaced by J ν ( x) Yν ( x) is the Bessel function of the second kind. ν When ν is an integer, J ( x) ( ) J ( x) ν = ν. Graphs of Bessel functions of the first kind, for ν = 0,, :. The series expression for the Bessel function of the first kind is k k+ ν ( ) x Jν ( x) = k! Γ k+ ν + k = 0 ( ) This function has a simpler form when ν is an odd half-integer. For example, J/ ( x) = sin x, J / ( x) = cos x πx πx
ENGI 940. Bessel and Legendre ODEs Page.44 The Bessel function of the second kind is Jν x cos νπ J ν x Yν ( x) = sin νπ ( x) Yν is unbounded as 0 ( ) ( ) ( ) ( ) x : lim Y ( x) = x 0+ Bessel functions of the second kind (all of which have a singularity at x = 0): ν Bessel functions arise frequently in situations where cylindrical or spherical polar coordinates are used. A generalised Bessel ODE is d y dy c x + ( a) x + ( bcx + ( a cν )) y dx dx = 0 whose general solution is y x = a x A Jν c bx + BYν c bx For a generalised Bessel ODE with a 0 as 0 ( ) ( ) ( ) ( ), whenever the solution must remain bounded a c y x = Ax Jν bx. x, the general solution simplifies to ( ) ( )
ENGI 940. Bessel and Legendre ODEs Page.45 Example.. Find a Maclaurin series solution to Legendre s ODE ( ) dy x x dx dx + p( p ) y = 0 in the case when p is a non-negative integer.
ENGI 940. Bessel and Legendre ODEs Page.46 Example.. (continued)
ENGI 940. Bessel and Legendre ODEs Page.47 If we set a = 0 when p is even, then the series solution terminates as a p th order polynomial (and therefore converges for all x). If we set a 0 = 0 when p is odd, then the series solution terminates as a p th order polynomial (and therefore converges for all x). With suitable choices of a 0 and we have the set of Legendre polynomials: P0 ( x ) =, P ( x) 3 P =, a, so that ( ) n = x, P ( ) ( x = 3x ), ( ) = ( 3 5 3 ), P ( x 4 4 ) = 8 ( 35 x 30 x + 3), P ( x) ( 5 3 ) 5 8 63 x 70 x 5 x ( ) = 6 4 ( 3 35 + 05 5), etc. P x x x P x x x x 6 6 Each ( ) n P x is a solution of Legendre s ODE with p = n. Rodrigues formula generates all of the Legendre polynomials: d Pn x n n x n! dx n ( ) ( ) = ( ) n = +, Among the properties of Legendre polynomials is their orthogonality on [, ]: + m ( ) ( ) P x P x dx n = 0 n + ( m n) ( m= n)
ENGI 940. ODEs Page.48 [Space for any additional notes] END OF CHAPTER