This problem does not fit into any place; but, I am asked this question often enough that I may as well post it *someplace*.
We will discuss the linear second-order ordinary differential-equation, with constant coefficients, in terms of a series LRC circuit. The inductor,
of inductance L, the resistor, of resistance R, and the condenser, of capacity C, are in series. We will write the equation in terms of the charge
q, as a function of time t.
When the driving-function is zero, the equation is said to be homogeneous. The differential equation is
L q" + R q' + q / C = 0.
For brevity of notation, let lower-case Greek omega wo be
wo = 1 / sqrt(L C).
Then, we may solve for C, to obtain
1 / C = wo^2 L.
Also, for brevity of notation, let lower-case Greek lambda m be
m = R / (2 L).
Granted that this glyph does not resemble the lambda; but, we have to make-do with the available glyphs. While, ordinarily, m is strictly positive, a negative (including zero) value may be achieved by the use of an active element. A superconductor also has a zero value for m. However, we will write all the equations for positive m, only. Then, we may solve for R, to obtain
R = 2 m L.
Substitute these two into the differential equation, to obtain
L q" + 2 m L q' + wo^2 L q = 0.
Since L is not zero, we may divide through by L, to obtain
q" + 2 m q' + wo^2 q = 0.
Considered as an algebraic quadratic equation in q, by the formula, its roots are
- m +- sqrt(m^2 - wo^2).
Let w be
w = sqrt(wo^2 - m^2).
Substitute, to obtain the roots as
- m +- i w.
There are four cases to consider.
This is the easiest of the homogeneous situations; because we have only the trigonometric sine and cosine functions of time. Then, the differential equation becomes
q" + wo^2 q = 0.
Considered as an algebraic quadratic equation in q, its roots are
+- i wo.
The complementary solution is
q = A cos(wo t) + B sin(wo t),
where A and B are the constants of integration. They are determined by the initial conditions. The two derivatives of q are
q' = wo B cos(wo t) - wo A sin(wo t)
q" = - wo^2 A cos(wo t) - wo^2 B sin(wo t).
As a check, substitute into the differential equation, to obtain
0 ?= ((- wo^2 + wo^2) A) cos(wo t) + ((- wo^2 + wo^2) B) sin(wo t) = 0.
At the time t = 0, the initial conditions are
q(0) = A
q'(0) = wo B.
From which we obtain
A = q(0)
B = q'(0) / wo.
Substitute these into the expression for the charge q, to obtain
q = q(0) cos(wo t) + (q'(0) / wo) sin(wo t).
This situation is only slightly more complicated, than the foregoing one; because, we have an exponential function of time multiplied by a linear function of time. Then, the differential equation becomes
q" + 2 wo q' + wo^2 q = 0.
Considered as an algebraic quadratic equation in q, it is a perfect square. Hence, each of its two roots is
- wo
The complementary solution is
q = exp(- wo t) (A + B t),
where A and B are the constants of integration. They are determined by the initial conditions. Please observe that the constant B is a coefficient of time. The two derivatives of q are
q' = exp(- wo t) ((- wo A + B) - wo B t)
q" = exp(- wo t) ((wo^2 A - 2 wo B) + wo^2 B t).
As a check, substitute into the differential equation, to obtain
0 ?= exp(- wo t) (((wo^2 - 2 wo^2 + wo^2) A + (- 2 wo + 2 wo) B) + (wo^2 - 2 wo^2 + wo^2) B t)) = 0.
At the time t = 0, the initial conditions are
q(0) = A
q'(0) = - wo A + B.
From which we obtain
A = q(0)
B = wo q(0) + q'(0).
Substitute these into the expression for the charge q, to obtain
q = exp(- wo t) (q(0) + (wo q(0) + q'(0)) t).
We observe again that we have a linear function of time multiplying the exponential function of time. Usually, wo would be positive; but, the solution is valid for negative wo, as well.
Actually, either solution is valid in both domains. We provide the two distinct solutions only to obviate the necessity of dealing with complex values explicitly.
This is the most common situation. The under-damped condition is oscillatory. It also is the most complicated situation; because, we have a product of an exponential function of time times the trigonometric sine and cosine functions of time. The natural omega w is real. The complementary solution is
q = exp(- m t) (A cos(w t) + B sin(w t)),
where A and B are the constants of integration. They are determined by the initial conditions. The two derivatives of q are
q' = exp(- m t) ((- m A + w B) cos(w t) + (- w A - m B) sin(w t))
q" = exp(- m t) (((- w^2 + m^2) A - 2 w m B) cos(w t) + (2 w m A + (- w^2 + m^2) B)) sin(w t)).
As a check, substitute into the differential equation, to obtain
0 ?= exp(- m t) (((- w^2 + m^2) A - 2 w m B) + 2 m (- m A + w B) + wo^2 (A)) cos(w t) +
+ ((2 w m A + (- w^2 + m^2) B) + 2 m (- w A - m B) + wo^2 (B)) sin(w t)) =
= exp(- m t) (((- w^2 + m^2 - 2 m^2 + wo^2) A + (- 2 w m + 2 w m) B) cos(w t) + ((2 w m - 2 w m) A + (- w^2 + m^2 - 2 m^2 + wo^2) B) sin(w t)) =
0,
where we have made use of the definition of w. At the time t = 0, the initial conditions are
q(0) = A
q'(0) = - m A + w B.
From which we obtain
A = q(0)
B = (m q(0) + q'(0)) / w.
Substitute these into the expression for the charge q, to obtain
q = exp(- m t) (q(0) cos(w t) + ((m q(0) + q'(0)) / w) sin(w t).
Usually, m would be positive; but, the solution is valid for negative m, as well.
The natural omega w is purely imaginary; hence, (i w) is real. The complementary solution is
q = A exp((- m + i w) t) + B exp((- m - i w) t),
where A and B are the constants of integration. They are determined by the initial conditions. The two derivatives of q are
q' = (- m + i w) A exp((- m + i w) t) + (- m - i w) B exp((- m - i w) t)
q" = (- m + i w)^2 A exp((- m + i w) t) + (- m - i w)^2 B exp((- m - i w) t).
As a check, substitute into the differential equation, to obtain
0 ?= ((- m + i w)^2 + (- 2 m^2 + 2 i m w) + (wo^2)) A exp((- m + i w) t) + ((- m - i w)^2 + (- 2 m^2 - 2 i m w) + (wo^2)) B exp((- m - i w) t) = 0,
where we have made use of the definition of w. At the time t = 0, the initial conditions are
q(0) = A + B
q'(0) = (- m + i w) A + (- m - i w) B.
From which we obtain
A = ((m + i w) q(0) + q'(0)) / (2 i w)
B = ((m - i w) q(0) + q'(0)) / (- 2 i w).
Substitute these into the expression for the charge q, to obtain
q = (((m + i w) q(0) + q'(0)) / (2 i w)) exp((- m + i w) t) + (((m - i w) q(0) + q'(0)) / (- 2 i w)) exp((- m - i w) t),
Usually, m would be positive; but, the solution is valid for negative m, as well.
Now, the right-hand side of the differential equation -- called the driving function -- is not zero. It consists of a sum of terms in the sine
and cosine of constant multiples of time t.
There are three cases to consider.
Nearly this situation is called resonance. It requires other techniques, which will not be considered here.
For pedagogical reasons, please first read the following case. This situation is the easiest among the inhomogeneous ones; because we handle the
whole problem at once. The differential equation is
q" + wo^2 q = (u cos(wo t) + v sin(wo t)) / L.
The particular solution is that of the following case multiplied by the time, namely
q = A t cos(wo t) + B t sin(wo t),
where A and B are constants, which we will have to evaluate. Please note that these A and B are *new* constants -- they are distinct from those which we employed in the homogeneous case, as constants of integration. Please observe that time multiplies each of the cosine and sine functions of time. This function is known as the particular solution. Its two derivatives are
q' = - A t wo sin(wo t) + B t wo cos(wo t) + A cos(wo t) + B sin(wo t)
q" = - A t wo^2 cos(wo t) - B t wo^2 sin(wo t) + 2 wo B cos(wo t) - 2 wo A sin(wo t).
When we substitute into the differential equation, the first two terms cancel. Thus, we obtain
2 wo B cos(wo t) - 2 wo A sin(wo t) = (u cos(wo t) + v sin(wo t)) / L.
From which we obtain
wo A = - v / (2 L)
wo B = u / (2 L).
Substitute into the particular solution, to obtain
q = (- v t cos(wo t) + u t sin(wo t) / (2 wo L).
Now, add the complementary solution, to obtain
q = ((A - v t) cos(wo t) + (B + u t) sin(wo t) / (2 wo L).
Its two derivatives are
q' = ((wo B + wo u t - v) cos(wo t) + (- wo A + wo v t + u) sin(wo t)) / (2 wo L)
q" = ((- wo^2 A + wo^2 v t + 2 u wo) cos(wo t) + (- wo^2 B - wo^2 u t + 2 v wo) sin(wo t)) / (2 wo L).
It obviously checks, when substituted into the differential equation. At time t = 0, we have
q(0) = A / (2 wo L)
q'(0) = (wo B - v) / (2 wo L).
From which we obtain
A = (2 wo L) q(0)
wo B = v + (2 wo L) q'(0).
Substitute into q, q', and q", to obtain
q = (((2 wo L) q(0) - v t) cos(wo t) + ((v + (2 wo L) q'(0)) / wo + u t) sin(wo t)) / (2 wo L) =
= (((2 wo L) q(0) - v t) cos(wo t) + ((v / wo + 2 L) q'(0) + u t) sin(wo t)) / (2 wo L).
q' = (((v + (2 wo L) q'(0)) + wo u t - v) cos(wo t) + (- wo ((2 wo L) q(0)) + wo v t + u) sin(wo t)) / (2 wo L) =
= ((v + (2 wo L) q'(0)) + wo u t - v) cos(wo t) + (- 2 wo^2 L q(0) + wo v t + u) sin(wo t)) / (2 wo L)
q" = ((- wo^2 ((2 wo L) q(0)) + wo^2 v t + 2 u wo) cos(wo t) + (- wo (v + (2 wo L) q'(0)) - wo^2 u t + 2 v wo) sin(wo t)) / (2 wo L)
= ((- 2 wo^3 L q(0) + wo^2 v t + 2 u wo) cos(wo t) + (- wo (v + (2 wo L) q'(0)) - wo^2 u t + 2 v wo) sin(wo t)) / (2 wo L).
We observe again that we have a linear function of time multiplying the each of the trigonometric cosine and sine functions of time. Usually, wo would be positive; but, the solution is valid for negative wo, as well.
Let us consider a typical such function
q = A cos(p t) + B sin(p t),
where A and B are constants, which we will have to evaluate, and the constant p is strictly positive. Please note that these A and B are *new* constants -- they are distinct from those which we employed in the homogeneous case, as constants of integration. This function is known as the particular solution. Its two derivatives are
q' = p B cos(p t) - p A sin(p t)
q" = - p^2 A cos(p t) - p^2 B sin(p t).
Substitution into the differential equation yields
((- p^2 + wo^2) A + 2 m p B) cos(p t) + ((- 2 m p) A + (- p^2 + wo^2) B) sin(p t) =
= (u cos(p t) + v sin(p t)) / L,
where u and v are arbitrary given constants. (Remember, we had divided the original differential equation by L.) For this to be an identity in the trigonometric functions, we have the set of two simultaneous linear algebraic equations in two unknowns
(- p^2 + wo^2) A + 2 m p B = u / L
- 2 m p A + (- p^2 + wo^2) B = v / L.
To solve for A, multiply the first equation by (- p^2 + wo^2) and the second by (- 2 m p) and add, to obtain
((- p^2 + wo^2)^2 + (2 m p)^2) A = ((- p^2 + wo^2) u - 2 m p v) / L.
For brevity of notation, let the impedance Z be given by
|Z|^2 = L^2 ((- p^2 + wo^2)^2 + (2 m p)^2) / p^2.
Then, we have
|Z|^2 p^2 A = ((- p^2 + wo^2) u - 2 m p v) L.
To solve for B, multiply the first equation by (2 m p) and the second by (- p^2 + wo^2) and add, to obtain
((- p^2) + wo^2)^2 + (2 m p)^2) B = (2 m p u + (- p^2 + wo^2) v) / L.
Again, we employ the same Z, to obtain
|Z|^2 p^2 B = (2 m p u + (- p^2 + wo^2) v) L.
Thus, we have
- p A = L (2 m p v - (- p^2 + wo^2) u) / (p |Z|^2)
p B = L (2 m p u + (- p^2 + wo^2) v) / (p |Z|^2).
Let us evaluate these in terms of the L, R, and C. We have
|Z|^2 = (L p - wo^2 L / p)^2 + (2 m L)^2 = (L p - 1 / (C p))^2 + R^2
- p A = (R v - (L p - 1 / (L C)) u) / Z^2
p B = (R u - (L p - 1 / (L C)) v) / Z^2.
The reactance X is
X = L p - 1 / (C p).
Then, we have the impedance Z and its magnitude
Z = R + i X = R + i (L p - 1 / (C p))
|Z|^2 = R^2 + X^2 = R^2 + (L p - 1 / (C p))^2.
The reciprocal of the impedance Z is called the admittance Y and is
Y = 1 / Z = (complex conjugate of Z) / |Z|^2 = (R - i X) / (R^2 + X^2).
When p is such that the reactance X is zero, we say that we have resonance. The coefficients become
- p A = (R v - X u) / |Z|^2
p B = (R u - X v) / |Z|^2.
Substitute into q, q', and q", to obtain
q = (- (R v - X u) cos(p t) + (R u - X v) sin(p t)) / (p |Z|^2).
q' = ((R u - X v) cos(p t) + (R v - X u) sin(p t)) / |Z|^2
q" = p ((R v - X u) cos(p t) - (R u - X v) sin(p t)) / |Z|^2.
At time t = 0, we have
q(0) = A
q'(0) = p B.
Since, by now, we already know the values of A and B, by substitution, we obtain
q(0) = (R v - X u) / (- p |Z|^2)
q'(0) = (R u - X v) / |Z|^2.
The direct current component for the driving-function u is is given by
q = C u.
Each of its two derivatives is identically zero:
q' = 0
q" = 0.
Theorem: Linear additivity. For any integer n in the semi-open interval [0, oo), if for any integer i in the closed interval [0, n], qi = qi(t) is a solution of the differential equation
L q" + R q' + q / C = gi(t);
then q = sum over i in [0, n] of qi is a solution of the differential equation
L q" + R q' + q / C = g(t),
where g = sum over i in [0, n] of gi.
The proof is obvious by Mathematical Induction.
Perform the following steps, in the indicated sequence:
There are four interesting situations.
We are given
First, the particular solution: The reactance X is
The impedance Z becomes
Its square of its magnitude is
The particular charge q becomes
Its two derivatives are
At time t=0, they become
Second, the complementary solution: We have to subtract these from the given initial conditions, to obtain those for use in the homogeneous differential equation. Since the given initial conditions each was zero, the result of the subtraction is just the negative of the foregoing, namely
Now, we have to compute the natural omega wo and the damping constant lambda. They are
Since 0 < m < wo, we are under-damped. Then, the complementary charge q is given as
Its two derivatives are
The solution of the differential equation is the sum of these two, namely
At time t=0, the q and q' become
OK. It checks.
The potential drops across each of the three components are:
Their sum, which should equal the driving function, is
OK. It checks. Grammercy.
We are given
This problem is the easiest one. First, let us compute the natural omega wo
The solution for the charge q is
Its two derivatives are
The initial values at t=0 are
OK. It checks.
The potential drops across each of the three components are:
Their sum, which should equal the driving function, is
OK. It checks. As I said, this is the easiest of the four problems.
We are given
First, the particular solution: The reactance X is
Thus, we are at resonance. However, since the resistance is non-zero, being at resonance has no specific consequence, except for making some of the calculations a little easier. The impedance Z becomes
Its square of its magnitude is
The particular charge q becomes
Its two derivatives are
At time t=0, they become
Second, the complementary solution: We have to subtract these from the given initial conditions, to obtain those for use in the homogeneous differential equation. Since the given initial conditions each was zero, the result of the subtraction is just the negative of the foregoing, namely
Now, we have to compute the natural omega wo and the damping constant lambda. They are
Since m > wo > 0, we are over-damped. Then, the complementary charge q is given as
Its two derivatives are
The solution of the differential equation is the sum of these two, namely
At time t=0, the q and q' become
OK. It checks.
The potential drops across each of the three components are:
Their sum, which should equal the driving function, is
OK. It checks. Grammercy.
We are given
First, the particular solution: The reactance X is
Thus, we are at resonance. However, since the resistance is non-zero, being at resonance has no specific consequence, except for making some of the calculations a little easier. The impedance Z becomes
Its square of its magnitude is
The particular charge q becomes
Its two derivatives are
At time t=0, they become
Second, the complementary solution: We have to subtract these from the given initial conditions, to obtain those for use in the homogeneous differential equation. Since the given initial conditions each was zero, the result of the subtraction is just the negative of the foregoing, namely
Now, we have to compute the natural omega wo and the damping constant lambda. They are
Since m = wo > 0, we are critically-damped. Then, the complementary charge q is given as
Its two derivatives are
The solution of the differential equation is the sum of these two, namely
At time t=0, the q and q' become
OK. It checks.
The potential drops across each of the three components are:
Their sum, which should equal the driving function, is
OK. It checks. Grammercy.
I did attempt to be careful. I believe that the foregoing is correct. However, I am notoriously error-prone. Hence, beware! :-) If you should discover any errors or if you have comments or suggestions, please contact me. Grammercy.
Copyright (c) 2001 by R.I. 'Scibor-Marchocki. Webmaster@rism.com Last modified Thursday 19-th July 2001.