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2.3 Now we attack the case of arbitrary p
and q in
our differential equation
\begin{eqnarray}
y''(x) + py'(x) + qy(x) = 0.& & %&(2.24) \\
\end{eqnarray}
Remember that for p > 0 and
q > 0 this corresponds to the
differential equation Eq. (2.1 ) of the damped linear harmonic oscillator. We already know that this system performs oscillations
( ⇝\mathop{ sin}\nolimits ,\mathop{cos}\nolimits ) that can be exponentially
damped ( ⇝\mathop{ exp}\nolimits ). Therefore, we
expect something related to \mathop{sin}\nolimits ,\mathop{cos}\nolimits ,\mathop{exp}\nolimits
functions. But these are all related to each other if we recall what we have learned about complex
numbers:
\begin{eqnarray}
\mathop{exp}\nolimits (ix)& =& \mathop{cos}\nolimits (x) + i\mathop{sin}\nolimits (x),x\text{ real}%&
\\
\mathop{cos}\nolimits (x)& =& {{e}^{ix} + {e}^{−ix}\over
2} =\mathop{ Re}\nolimits [{e}^{ix}] %&
\\
\mathop{sin}\nolimits (x)& =& {{e}^{ix} − {e}^{−ix}\over
2i} =\mathop{ Im}\nolimits [{e}^{ix}]. %&(2.25) \\
\end{eqnarray}
Furthermore, for arbitrary complex z = x + iy ,
\begin{eqnarray}{
e}^{z} = {e}^{x+iy} = {e}^{x}{e}^{iy} = {e}^{x}[\mathop{cos}\nolimits (y) + i\mathop{sin}\nolimits (y)] = {e}^{x}\mathop{ cos}\nolimits (y) + i{e}^{x}\mathop{ sin}\nolimits (y).& & %&(2.26) \\
\end{eqnarray}
The function {e}^{z} with
complex z comprises the
real exponential as well as \mathop{sin}\nolimits
and \mathop{cos}\nolimits .
Let us therefore try an exponential Ansatz in Eq. (2.24 ),
\begin{eqnarray}
y(x) = {e}^{zx} ⇝ y''(x) + py'(x) + qy(x) = [{z}^{2} + pz + q]{e}^{zx} = 0.& & %&(2.27) \\
\end{eqnarray}
We recognize that y(x) = {e}^{zx} fulfills the
differential equation, if the bracket [...]
is zero:
\begin{eqnarray}
[{z}^{2} + pz + q] = 0.& & %&(2.28) \\
\end{eqnarray}
This is a quadratic equation which in general has two solutions,
\begin{eqnarray}{
z}^{2} + pz + q = 0 ⇝ {z}_{
1∕2} = −{p\over
2} ±\sqrt{{{p}^{2 } \over
4} − q}.& & %&(2.29) \\
\end{eqnarray}
Various cases arrise for different signs of the argument of the square root (the discriminant). Let us look at each of
these cases in turn.
2.3.1 In the case {{p}^{2}\over
4} − q > 0 ,
\begin{eqnarray}{
z}_{1∕2} = −{p\over
2} ±\sqrt{{{p}^{2 } \over
4} − q}& & %&(2.30) \\
\end{eqnarray}
are both real and the two solutions fulfilling Eq. (2.24 ) are
\begin{eqnarray}
{{p}^{2}\over
4} − q > 0 ⇝ {y}_{1}(x) = {y}_{1}{e}^{[−{p\over
2} +\sqrt{{{p}^{2 } \over
4} −q}]x},\quad {y}_{2}(x) = {y}_{2}{e}^{[−{p\over
2} −\sqrt{{{p}^{2 } \over
4} −q}]x}& & %&(2.31) \\
\end{eqnarray}
The general solution is the linear combination of the two,
\begin{eqnarray}
{{p}^{2}\over
4} − q > 0 ⇝ y(x) = {y}_{1}{e}^{[−{p\over
2} +\sqrt{{{p}^{2 } \over
4} −q}]x} + {y}_{2}{e}^{[−{p\over
2} −\sqrt{{{p}^{2 } \over
4} −q}]x}.& & %&(2.32) \\
\end{eqnarray}
In this case there are no oscillations at all. The ‘damping term’
py'(x) is too
strong.
2.3.2 In the case {{p}^{2}\over
4} − q < 0 ,
the two zeros become complex:
\begin{eqnarray}{
z}_{1∕2} = −{p\over
2} ±\sqrt{{{p}^{2 } \over
4} − q} = −{p\over
2} ± i\sqrt{q − {{p}^{2 } \over
4}} =: −{p\over
2} ± iΩ,& & %&(2.33) \\
\end{eqnarray}
where we define an angular frequency Ω = \sqrt{q − {p}^{2 } ∕4} .
Now, the two solutions fulfilling Eq. (2.24 ) are
\begin{eqnarray}
{{p}^{2}\over
4} − q < 0 ⇝ {y}_{1}(x) = {y}_{1}{e}^{[−{p\over
2} +iΩ]x},\quad {y}_{2}(x) = {y}_{2}{e}^{[−{p\over
2} −iΩ]x},\quad Ω := \sqrt{q − {{p}^{2 } \over
4}} .& & %&(2.34) \\
\end{eqnarray}
The general solution is the linear combination of the two,
\begin{eqnarray}
{{p}^{2}\over
4} − q < 0 ⇝ y(x) = {y}_{1}{e}^{[−{p\over
2} +iΩ]x} + {y}_{2}{e}^{[−{p\over
2} −iΩ]x},\quad Ω := \sqrt{q − {{p}^{2 } \over
4} − q}.& & %&(2.35) \\
\end{eqnarray}
We rewrite this as
\begin{eqnarray}
y(x)& =& {y}_{1}{e}^{[−{p\over
2} +iΩ]x} + {y}_{2}{e}^{[−{p\over
2} −iΩ]x} = {e}^{−px∕2}\left \{{y}_{1}{e}^{iΩx} + {y}_{2}{e}^{−iΩx}\right \} %&
\\
& =& {e}^{−px∕2}\left \{{y}_{
1}[\mathop{cos}\nolimits (Ωx) + i\mathop{sin}\nolimits (Ωx)] + {y}_{2}[\mathop{cos}\nolimits (Ωx) − i\mathop{sin}\nolimits (Ωx)]\right \}%&
\\
& =& {e}^{−px∕2}\left \{[{y}_{
1} + {y}_{2}]\mathop{cos}\nolimits (Ωx) + i[{y}_{1} − {y}_{2}]\mathop{sin}\nolimits (Ωx)\right \}. %&(2.36) \\
\end{eqnarray}
Now, this seems a bit odd since we have got a complex solution due to the term
i({y}_{1} − {y}_{2}) . However, the
constant coefficients {y}_{1}
and {y}_{2} can be complex
anyway (and still y(x)
is a solution of the differential equation). If we are only interested in real functions
y(x) , we can re–define
new constants {c}_{1} := {y}_{1} + {y}_{2}
and {c}_{2} := i[{y}_{1} − {y}_{2}]
such that the general solution becomes
\begin{eqnarray}
y''(x) + py'(x) + qy(x)& =& 0,\quad {{p}^{2}\over
4} − q < 0 ⇝ %&
\\
y(x)& =& {e}^{−px}\left \{{c}_{
1}\mathop{ cos}\nolimits (Ωx) + {c}_{2}\mathop{ sin}\nolimits (Ωx)\right \}.%&(2.37) \\
\end{eqnarray}
Still {c}_{1}
and {c}_{2}
could be complex numbers, but we can choose them real if we only want real functions
y(x) .
2.3.3 For the marginal case {{p}^{2}\over
4} − q = 0 we
have only one solution, z = −p∕2 ,
and thus only one integration constant. This is clearly insufficient! Help can be found in the expression for the
simplest marginal case,
which has as solution
We therefore substitute y(x) = (a + bx)\mathop{exp}\nolimits (−px∕2)
and find
\begin{eqnarray}
y(x)& =& (a + bx)\mathop{exp}\nolimits (−px∕2), %&
\\
y'(x)& =& b\mathop{exp}\nolimits (−px∕2) −{p\over
2}(a + bx)exp(−px∕2), %&
\\
y''(x)& =& −bp\mathop{exp}\nolimits (−px∕2) + {{p}^{2}\over
4} (a + bx)exp(−px∕2).%&(2.38) \\
\end{eqnarray}
Using the differential equation y''(x) + py'(x) + qy(x) = 0 ,
we find that
\begin{eqnarray}
y''(x) + py'(x) + qy(x) = \left ({{p}^{2}\over
4} −{{p}^{2}\over
2} + q\right )(a + bx)exp(−px∕2) − bp\mathop{exp}\nolimits (−px∕2) + bp\mathop{exp}\nolimits (−px∕2) = 0.& & %&(2.39) \\
\end{eqnarray}
Here we have used q = 4{p}^{2} .
Thus
\class{boxed}{ y(x) = (a + bx)\mathop{exp}\nolimits (−px∕2). }
2.3.4 Solutions of y''(x) + py'(x) + qy(x) = 0 :
{p}^{2} > 4q y(x) =\mathop{ exp}\nolimits (−{p\over
2}x)\left [{y}_{1}\mathop{ exp}\nolimits \left (\sqrt{{{p}^{2 } \over
4} − q}x\right ) + {y}_{2}\mathop{ exp}\nolimits \left (−\sqrt{{{p}^{2 } \over
4} − q}x\right )\right ] {p}^{2} < 4q y(x) =\mathop{ exp}\nolimits (−{p\over
2}x)\left [{c}_{1}\mathop{ cos}\nolimits \left (\sqrt{q − {{p}^{2 } \over
4}} x\right ) + {c}_{2}\mathop{ sin}\nolimits \left (\sqrt{q − {{p}^{2 } \over
4}} x\right )\right ] {p}^{2} = 4q y(x) =\mathop{ exp}\nolimits (−px∕2)(a + bx)