Download Ordinary Differential Equations - Essay - Mathematics - Prof. Adrian Down and more Essays (high school) Mathematics in PDF only on Docsity!
Ordinary Differential Equations
Adrian Down
April 06, 2006
1 Constants in ODE problems
1.1 Motivation
Recall that the general solution of a system of n first-order differential equa- tions contains n constants. These constants must be determined from the given conditions on the system.
1.2 Initial value problems (IVPs)
1.2.1 Definition
An initial value problem (IVP) is a set of ordinary differential equations (ODEs) with accompanying initial conditions. The ODEs involved in an IVP are of the form,
x˙i = fi(t, x 1 ,... , xn) i ∈ { 1 ,... , n }
The initial conditions specify values of the dependent variables xi at par- ticular values of the independent variable t,
xi(T ) = Xi where T, Xi given
1.2.2 Existence uniqueness theorem
Theorem. If the set of functions fi and the set of their first partial deriva- tives are continuous in some neighborhood of the given initial conditions,
(t, x 1 ,... , xn) = (T, X 1 ,... , Xn)
then there is a unique solution of the IVP for some sufficiently small interval |t − T |.
Proof. The exact proof is beyond the scope of the class. The general strategy is to convert the IVP to an integral equation, since it is easier to place bounds on integrals than derivatives. The integral equations can be solved iteratively. Although we will not cover the proof, it can be done using ideas that we have covered thus far in this course.
In an IVP, the given initial conditions form a unique solution by deter- mining the particular values of the constants that appear in the general form of the solution.
1.2.3 Examples
Power law dependence Recall the ODE that we solved previously,
x˙ = −tx ⇒ x(t) = Ce−^
t 22 (1)
Suppose we are given the initial condition
x(T ) = X (2)
Substituting (2) into (1) determines the constant C,
X = Ce−^
T 2 2 ⇒ C = Xe
T 2 2
The unique solution to the ODE (1) given the initial condition (2) is thus,
x(t) = Xe
1 2 (T^2 −t^2 )
Polar spiral We also solved an ODE resulting in spiral phase space con- tours,
x(t) = Re−t^ cos (t + θ) y(t) = Re−t^ sin (t + θ) (3)
where R and θ are real constants. Suppose we are given the values of the dependent variables at t = 0. x(0) = X y(0) = Y (4)
These two initial conditions determine the constants R and θ. Substituting (3) into (4),
X = R cos θ Y = R sin θ
Variational calculation We now follow the same process as our previous variational calculation. To find the physical trial function that minimizes the surface area, take the derivative of (5) with respect to , using the product rule
S() = 2π
∫ L 2
− L 2
1 + y^2 xy +
yyxyx √ 1 + y x^2
dx (6)
Use integration by parts on the second term in the integrand of (6).
u =
yyx √ 1 + y x^2
dv = yxdx
The integration by parts will switch a derivative from one term to the other and introduce a negative. The boundary conditions will eliminate the bound- ary terms which result from the integration by parts. After integrating by parts, there will be a common factor of y in both terms of the integrand,
S() = 2π
∫ L
2 − L 2
1 + y x^2 −
yyx √ 1 + y^2 x
x
ydx + 2π
[
yyxy √ 1 + y x^2
]L 2
− L 2
As claimed above, the boundary conditions are such that the boundary term in (7) is 0. Recall that the value of y
±L 2
is fixed, regardless of how is varied. Thus,
y
L
Now, substitute = 0 into (7). By assumption, = 0 is the value of for which y(x, ) is the minimizing function. Since S(y( = 0)) is minimized, S( = 0) = 0,
0 = S(0) = 2π
∫ L 2
− L 2
1 + y^2 x −
yyx √ 1 + y x^2
x
ydx (8)
Because y could take any values as x ranges over
−L 2 , L 2
, it must be that the integrand in brackets is identically equal to 0 on this interval.
Solution The BVP, as obtained in (8), is difficult but we can obtain a simple general solution the the ODE. Using some intuition from the parabolic shape we expect for the soap film, we make a guess at the solution,
y = a cosh
x − b a
y′^ = sinh
x − b a
The derivative can be reformulated using an identity for hyperbolic trig func- tions,
cosh^2 ζ − sinh^2 ζ = 1 ⇒ cosh ζ =
1 + sinh^2 ζ
⇒
1 + (y′)^2 = cosh
x − b a
Constants We expect that the soap film should be symmetric, and hence y must be an even function. Thus we take b = 0. Substituting the boundary condition into y,
y
L
= a cosh
L
2 a
= H (9)
(9) should determine the value of the constant a given the radius of the hoops H. However, plotting H as a function of a shows some complicated behavior. The plot decreases for small a, reaches a minimum Hm, and then increases with increasing a. Hence there is no solution for H < Hm. This is a manifestation of the fact that no soap film can form if the radius of the hoops is too small relative to the separation of eth disks. For H > Hm, there are two solutions for a. One value of a corresponds to a film with moderate curvature. The other corresponds to highly-curved film. The first solution is the physical solution, as the highly-curved solution is unstable.
2 Solving linear ODEs with constant coeffi-
cients
2.1 Homogeneous case
A common example of a first order linear ODE is
y′^ + p(x)y = 0
eP^ (x)y(x) − eP^ (x^0 )y(x 0 ) =
∫ (^) x
x 0
eP^ (t)f (t)dt (11)
2.2.3 Boundary conditions
Since P (x) is any anti-derivative of p, choose P (x) such that P (x 0 ) = 0. Denoting y(x 0 ) by y 0 , (11) simplifies,
eP^ (x)y(x) − y 0 =
∫ (^) x
x 0
eP^ f
(t)dt
The final form of the solution is obtained by isolating y(x),
y(x) = y 0 e−P^ (x)^ +
∫ (^) x 1
x 0
eP^ (t)−P^ (x)f (t)dt
The first term in the solution satisfies the homogeneous equation. The second integral term is a solution to the particular equation given f (x).
