Download 0: Ordinary Differential Equations - e

0: Ordinary Differential Equations
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Some definitions:
Consider an ODE which has solution φ.
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Order of ODE: Highest derivative involved in equation.
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Linear: Equation does not involved powers of solution φ or
its derivatives
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Homogenous: Every term involves φ or its derivatives.
(Similarly for PDEs.)
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First order, linear ODEs
df
dx
= −αf
f is a function of one variable, x.
α: constant
Physics: Radioactive decay
Maths: rate of change proportional to function itself.
1 - Trial solution. Guess f (x) = Aeλx .
2 - Separate variables and integrate.
λ = −α
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Second-order, linear ODEs
a
d 2f
df
+ cf
+b
2
dx
dx
= 0
(0.1)
Again f is only dependent on one variable, x, so f (x)
Constant coefficients (a, b, c)
Homogeneous (RHS=0) - no driving force
Try exponential solution f = Aeλx
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Second derivative - need to integrate twice - two constants of
integration
f
df
dx
d 2f
dx 2
Substitute in to eqn. 0.1
Can only be true for all x if
Auxiliary eqn., quadratic, 2 roots.
So 2 solutions f1 (x)
= Aeλx
= λAeλx
= λ2 Aeλx
aλ2 + bλ + c Aeλx = 0
aλ2 + bλ + c = 0
√
−b ± b2 − 4ac
λ=
2a
= A1 eλ1 x and f2 (x) = A2 eλ2 x
with two constants to be determined from the boundary
conditions.
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Principle of Superposition:
A linear superposition (sum) of solutions to a linear equation is
also a solution to the equation.
Prove by differentiation and subsitution of
f (x) = f1 (x) + f2 (x) = A1 eλ1 x + A2 eλ2 x in to eqn. 0.1.
aλ21 + bλ1 + c A1 eλ1 x + aλ22 + bλ2 + c A2 eλ2 x = 0
Terms in parenthesis are zero. So the general solution is
f (x) = A1 eλ1 x + A2 eλ2 x
- 2 constants to be determined from boundary conditions.
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A Simple (But Important) Second Order ODE
d 2f
= α2 f
dx 2
Auxillary equation
General solution
(0.2)
λ2 = α2 → λ = ±α
f (x) = Ceαx + De−αx
Growing or decaying not oscillatory
or
f (x) = C 0 cosh αx + D 0 sinh αx
Hyperbolic functions
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Simple Harmonic Motion (SHM)
d 2f
dx 2
= −k 2 f
(0.3)
sin and cos solutions. The general solution is a linear
superposition:
f (x) = A cos(kx) + B sin(kx)
(0.4)
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Try an exponential solution: f (x) = eλx Exponentials are easier
to differentiate ! Try f = eλx
df
= λeλx
dx
d 2f
= λ2 eλx
dx 2
Compare with initial ODE and they can be identical if the
auxillary equation is λ2 = −k 2 !!!
So λ is complex! λ = ±ik where i 2 = −1
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Complex exponentials
Terms like
eikx
Easy to differentiate. Essential for quantum mechanics.
d ikx
e
dx
d 2 ikx
e
dx 2
= ikeikx
= (ik)2 eikx = −k 2 eikx
Equation of SHM with f (x) = eikx as a solution.
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Complex Solutions SHM
Showed that
f = Ceλx
where λ = ±ik
d 2f
= −k 2 f
is a solution of
dx 2
So general solution is
f (x) = Ceikx + De−ikx
(0.5)
These are wave-like/oscillatory solutions.
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Just showed the general solution to the equation of SHM is
f (x) = A cos(kx) + B sin(kx)
But complex exponentials are also a solution. How can these
be equivalent?
Look at the f (x) = eikx term at x = 0 and equate to the solution
eqn. 0.4 at x = 0.
eikx
= A cos(kx) + B sin(kx)
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At x = 0, e = 1 = A cos 0 + B sin 0 = A
Differentiate
f 0 (0) = ike0 = ik
→ ik = −Ak sin 0 + Bk cos 0 = kB
So A = 1 and B = i and eikx = cos kx + i sin kx
-Euler’s formula. Similarly for the e−ikx term or just subsitute
x = −x in to Euler’s formula, noting that cos(−x) = cos(x) and
sin(−x) = − sin(x)
Linear combinations of complex exp. equivalent to linear
combinations of sines and cosines.
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