This section demonstrates by numerous examples how to apply the Laplace transform for solving the initial value problems for constant coefficient linear differential equations.
The Laplace transform is an alternative approach to the methods for solving initial value problems of linear
differential equations with constant coefficients. These were considered in Part IV of this tutorial. The Laplace transform is useful in dealing with discontinuous inputs (closing of a switch) and with
periodic functions (sawtooth and rectified waves). Analysis of the effect of such inputs proceeds most smoothly in the frequency domain, that is, in domain of the transform-variable, which we denote by λ.
Application of the Laplace transformation to differential equations is based on the following statements.
Theorem 1:
Suppose that f: [0, ∞) → ℝ is a continuous and its derivative f' = df/dt is piecewise continuous on any finite interval 0 ≤ t ≤ b < ∞. Suppose further that there exist constants K, γ, and M such that \( |f(t)| \le K\,e^{\gamma t} \) for t > M. Then the Laplace transform of this function, \( {\cal L}\left[ f(t) \right] (\lambda ) = f^L (\lambda ) \) exists for λ > γ, and more over,
where \( f(0^{+}) = f(0+0) = \lim_{\varepsilon \downarrow 0} f(\varepsilon ) \) is the limit value of f at zero from right.
Theorem 1 states that the Laplace transformation is a spectral representation of the derivative operator \( \texttt{D} = {\text d}/{\text d}t , \) acting in the space of functions defined on semi-line [0, ∞). This means that the Laplace transform maps the unbounded operator \( \texttt{D} \) into multiplication by a spectral parameter λ on a set of functions that vanish at the origin.
Theorem 1 can be extended for arbitrary power of the derivative operator.
Theorem 2:
Suppose that f: [0, ∞) → ℝ is continuous along with its derivatives up to the order n−1, and that its n-th derivative \( f^{(n)} \) is piecewise continuous on
any finite interval 0 ≤ t ≤ b < ∞. Suppose further that there exist constants K, γ, and M such that \( |f(t)| \le K\,e^{\gamma t} , \quad |f'(t)| \le K\,e^{\gamma t} , \quad \ldots , |f^{(n-1)} (t)| \le K\,e^{\gamma t} \) for t > M. Then the Laplace transform of this function, \( {\cal L}\left[ f(t) \right] (\lambda ) = f^L (\lambda ) \) exists for λ > γ, and moreover,
where \( \texttt{I} = \texttt{D}^0 \) is the identity operator, into multiplication by the polynomial \( L \left[ \lambda \right] = a_n \lambda^n + a_{n-1} \lambda^{n-1} + \cdots + a_1 \lambda + a_0 , \) it reduces an initial value problem into an algebraic equation. Upon solving this algebraic equation, we obtain almost immediately the Laplace transform of the unknown function---the solution of the initial value problem. There are no miracles in math, and the price you have to pay for using the beautiful operating method is hidden in the inverse Laplace transform, which is an ill-posed operation. Fortunately, the methods of determination of the inverse Laplace transformation described previously work pretty well in initial value problems for ordinary differential equations.
Schematically, this process can be illustrated starting with, for example, a second order linear
differential equation with constant coefficients, as follows.
Of course, some details need to be addressed for this to make sense. In particular, we need to
express the left side \( {\cal L} \left[ a\,y'' + b\, y' + c\,y \right] \) in terms of
\( {\cal L} \left[ y \right] = y^L . \) We illustrate the method with numerous examples. Here is an example of how you may want to utilize the Laplace transformation to solve the differential equation y'' + 4y = 0:
Upon using Theorem 2 to express \( {\cal L} \left[ y'' (t) \right] \) and \( {\cal L} \left[ y' (t) \right] \) in terms of \( {\cal L} \left[ y (t) \right] = y^L (\lambda ) , \) we find that the previous equation (1.3) becomes
Example 2:
Consider the undamped mechanical oscillator with a forcing function that is a constant f(t) = F0. Recall that it consists of a block of mass m on a table and restrained laterally by an ordinary coil spring. The displacement, denoted as x(t), of the mass (measured as positive to the right) from its equilibrium position: that is, when x = 0 the spring is neither stretched nor compressed. We are now imagining the mass to be disturbed from its equilibrium position by an initial disturbance (position s and velocity v) and applied force f(t).
Since the motion of the mass is constrained to be along a straight line, we need to consider only the forces in the x direction. Neglecting aerodynamic force and the friction force, we are left to consider the only one external (driving) force. Then according to Newton's second law and Hooke's law,
the displacement x(t) satisfies the differential equation
\[
m\, x'' + k\, x = f(t) = F_0 ,
\tag{2.1}
\]
where k is a spring constant. Of course, we need to add the initial conditions
\[
x(0) = s, \qquad x' (0) = v ,
\tag{2.2}
\]
In order to apply the Laplace transform, we multiply each term in Eq.(2.1) by the Laplace kernel \( e^{-\lambda t} \) and integrate on t from 0 to ∞. Using the standard notation for the Laplace transform, we get
where we denote by xL the Laplace transform of the unknown solution x(t) to the IVP (2.1), (2.2). The point to appreciate is that whereas in the t domain we have the unbounded linear differential operator acting on x(t), in the transform domain we now have the linear algebraic equation (2.3) on xL. By simple algebra, we find its solution almost immediately:
To determine the explicit expression for xp(t), we have two options. First, we use the explicit formula for the Laplace transform of the forcing function and get
Here Re z = \( \Re z = \Re \left( a + b{\bf j} \right) = a , \) is the real part of complex number 𝑎 + jb, and j is the unit vector in positive vertical direction on the complex plane ℂ, so j² = −1.
Since both singularities λ = ωj and λ = 0 are simple, we just differentiate the denominator and set the value of λ to be the corresponding singularity. So
P[s_] = (s + 3)*Exp[s*t]
Q[s_] = s^2 + 5*s + 6
res2 = (P[s]/D[Q[s], s]) /. s -> 2
res3 = (P[s]/D[Q[s], s]) /. s -> 3
The denominator in the second term has two real nulls \( \lambda =-2 \) and
\( \lambda =-3 \) and two complex conjugate \( \lambda = \pm 2{\bf j} . \)
We calculate residues at each pole separately.
Its solution is the sum \( y(t) = y_h (t) + y_p (t) \) of two functions, one of them
yh, is the solution of the homogeneous equation subject to the given initial conditions
With examples completed, we can make some observations about the application of the Laplace transform method. Although we considered only second order differential equations, the method can be easily extended to linear differential equations of any order.
Upon application of the Laplace transformation, the initial conditions become "build-in."
When applying the Laplace transform, we by default assume that the unknown function and all its derivatives are transformable under the Laplace method into holomorphic functions on the half-plane Reλ > γ. However, such restrictions in no way impede the solution steps in finding the Laplace transform of the solution function. If you successfully apply the inverse Laplace transform to obtain the solution, they are no longer relevant.
When solving the inhomogeneous equation
\[
L \left[ \,\texttt{D} \,\right] y (t) = f(t)
\]
for n-th order differential operator \eqref{EqIVP.3} with the aid of the Laplace transform method, we always reduce the corresponding initial value problem to an algebraic equation
If the Laplace transform fL of the input function f(t) is not available, the only way to determine the required inverse Laplace transform is to apply the convolution property:
where G(t) is the Green function. In this case, you don't need to find the Laplace transform of the input function f(t). However, determination of the convolution integral is an ill-posed problem and its practical implementation may lead to some sort of instability.
Green Functions
Definition:
Given a linear differential operator \eqref{EqIVP.3}, its Green function is
When the Green function for the linear constant coefficient differential operator \eqref{EqIVP.3} is known, then a particular solution to the inhomogeneous equation
\[
L \left[ \,\texttt{D} \,\right] y(t) = f(t)
\]
is obtained by the convolution integral \eqref{EqIVP.6}. The Green function for the linear constant coefficient operator \eqref{EqIVP.3} is the unique solution of the following initial value problem:
Theorem 3:
Suppose that f(t) is continuous function on [0, ∞) and has a Laplace transform. Then the solution of the initial value problem for the constant coefficient differential equation (𝑎 ≠ 0)
where \( \texttt{D} = {\text d}/{\text d}t \) is the derivative operator and \( \texttt{I} \) is the identity operator. Recall that the Green function for the differential operator \( L \left[ \texttt{D} \right] \) is the inverse Laplace transform of the reciprocal to the characteristic polynomial: \( \displaystyle {\cal L}^{-1} \left[ 1/L(\lambda ) \right] . \)
Since the characteristic polynomial L(λ) = (λ-1)(λ+3) has three simple real roots, we find the corresponding Green function using the inverse Laplace transform:
where \( \texttt{D} = {\text d}/{\text d}t \) is the derivative operator and \( \texttt{I} \) is the identity operator. Recall that the Green function for the differential operator \( L \left[ \texttt{D} \right] \) is the inverse Laplace transform of the reciprocal to the characteristic polynomial: \( \displaystyle {\cal L}^{-1} \left[ 1/L(\lambda ) \right] . \)
The Green function for the given differential operator is the solution of the differential equation
So the characteristic polynomial L(λ) has one double real root λ = -½ and two simple complex conjugate roots λ = 1 ±3j. Recall that j is the unit vector in positive vertical direction on the complex plane ℂ, so j² = -1. To find the explicit expression for the Green function, we need to evaluate two residues:
We plot the particular solution yp(t), given in formula (7.4), (in blue) along with the input function f(t) in black. As you can see, there is no relation of the particular solution and the input function.
Since the initial conditions are specified at point x = 1, we first make a shift in the independent variable by introducing t = x −1. This yields the IVP:
