7.1: Initial Value Green’s Functions
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In this section we will investigate the solution of initial value problems involving nonhomogeneous differential equations using Green’s functions. Our goal is to solve the nonhomogeneous differential equation
\[a(t) y^{\prime \prime}(t)+b(t) y^{\prime}(t)+c(t) y(t)=f(t), \label{eq:1}\]
subject to the initial conditions
\[y(0)=y_{0} \quad y^{\prime}(0)=v_{0} .\nonumber \]
Since we are interested in initial value problems, we will denote the independent variable as a time variable, \(t\).
Equation \(\eqref{eq:1}\) can be written compactly as
\[L[y]=f \text {, }\nonumber \]
where \(L\) is the differential operator
\[L=a(t) \frac{d^{2}}{d t^{2}}+b(t) \frac{d}{d t}+c(t) .\nonumber \]
The solution is formally given by
\[y=L^{-1}[f] .\nonumber \]
The inverse of a differential operator is an integral operator, which we seek to write in the form
\[y(t)=\int G(t, \tau) f(\tau) d \tau .\nonumber \]
The function \(G(t, \tau)\) is referred to as the kernel of the integral operator and is called the Green’s function.
\(G(t,\tau )\) is called a Green's function.
In the last section we solved nonhomogeneous equations like Equation \(\eqref{eq:1}\) using the Method of Variation of Parameters. Letting,
\[y_{p}(t)=c_{1}(t) y_{1}(t)+c_{2}(t) y_{2}(t),\label{eq:2}\]
we found that we have to solve the system of equations
\[\begin{align} &c_{1}^{\prime}(t) y_{1}(t)+c_{2}^{\prime}(t) y_{2}(t)=0 .\nonumber \\[4pt] &c_{1}^{\prime}(t) y_{1}^{\prime}(t)+c_{2}^{\prime}(t) y_{2}^{\prime}(t)=\frac{f(t)}{q(t)} .\label{eq:3} \end{align}\]
This system is easily solved to give
\[\begin{align} &c_{1}^{\prime}(t)=-\frac{f(t) y_{2}(t)}{a(t)\left[y_{1}(t) y_{2}^{\prime}(t)-y_{1}^{\prime}(t) y_{2}(t)\right]}\nonumber \\[4pt] &c_{2}^{\prime}(t)=\frac{f(t) y_{1}(t)}{a(t)\left[y_{1}(t) y_{2}^{\prime}(t)-y_{1}^{\prime}(t) y_{2}(t)\right]} .\label{eq:4} \end{align}\]
We note that the denominator in these expressions involves the Wronskian of the solutions to the homogeneous problem, which is given by the determinant
\[W\left(y_{1}, y_{2}\right)(t)=\left|\begin{array}{ll} y_{1}(t) & y_{2}(t) \\[4pt] y_{1}^{\prime}(t) & y_{2}^{\prime}(t) \end{array}\right| .\nonumber \]
When \(y_{1}(t)\) and \(y_{2}(t)\) are linearly independent, then the Wronskian is not zero and we are guaranteed a solution to the above system.
So, after an integration, we find the parameters as
\[\begin{align} &c_{1}(t)=-\int_{t_{0}}^{t} \frac{f(\tau) y_{2}(\tau)}{a(\tau) W(\tau)} d \tau\nonumber \\[4pt] &c_{2}(t)=\int_{t_{1}}^{t} \frac{f(\tau) y_{1}(\tau)}{a(\tau) W(\tau)} d \tau\label{eq:5} \end{align}\]
where \(t_{0}\) and \(t_{1}\) are arbitrary constants to be determined from the initial conditions.
Therefore, the particular solution of Equation \(\eqref{eq:1}\) can be written as
\[y_{p}(t)=y_{2}(t) \int_{t_{1}}^{t} \frac{f(\tau) y_{1}(\tau)}{a(\tau) W(\tau)} d \tau-y_{1}(t) \int_{t_{0}}^{t} \frac{f(\tau) y_{2}(\tau)}{a(\tau) W(\tau)} d \tau .\label{eq:6}\
We begin with the particular solution (Equation \(\eqref{eq:6}\)) of the nonhomogeneous differential equation Equation \(\eqref{eq:1}\). This can be combined with the general solution of the homogeneous problem to give the general solution of the nonhomogeneous differential equation:
\[y_{p}(t)=c_{1} y_{1}(t)+c_{2} y_{2}(t)+y_{2}(t) \int_{t_{1}}^{t} \frac{f(\tau) y_{1}(\tau)}{a(\tau) W(\tau)} d \tau-y_{1}(t) \int_{t_{0}}^{t} \frac{f(\tau) y_{2}(\tau)}{a(\tau) W(\tau)} d \tau .\label{eq:7}\]
However, an appropriate choice of \(t_{0}\) and \(t_{1}\) can be found so that we need not explicitly write out the solution to the homogeneous problem, \(c_{1} y_{1}(t)+c_{2} y_{2}(t)\). However, setting up the solution in this form will allow us to use \(t_{0}\) and \(t_{1}\) to determine particular solutions which satisfies certain homogeneous conditions. In particular, we will show that Equation \(\eqref{eq:7}\) can be written in the form
\[y(t)=c_{1} y_{1}(t)+c_{2} y_{2}(t)+\int_{0}^{t} G(t, \tau) f(\tau) d \tau,\label{eq:8}\]
where the function \(G(t, \tau)\) will be identified as the Green’s function.
The goal is to develop the Green’s function technique to solve the initial value problem
\[a(t) y^{\prime \prime}(t)+b(t) y^{\prime}(t)+c(t) y(t)=f(t), \quad y(0)=y_{0}, \quad y^{\prime}(0)=v_{0} .\label{eq:9}\]
We first note that we can solve this initial value problem by solving two separate initial value problems. We assume that the solution of the homogeneous problem satisfies the original initial conditions:
\[a(t) y_{h}^{\prime \prime}(t)+b(t) y_{h}^{\prime}(t)+c(t) y_{h}(t)=0, \quad y_{h}(0)=y_{0}, \quad y_{h}^{\prime}(0)=v_{0} .\label{eq:10}\]
We then assume that the particular solution satisfies the problem
\[a(t) y_{p}^{\prime \prime}(t)+b(t) y_{p}^{\prime}(t)+c(t) y_{p}(t)=f(t), \quad y_{p}(0)=0, \quad y_{p}^{\prime}(0)=0 \text {. }\label{eq:11}\]
Since the differential equation is linear, then we know that
\[y(t)=y_{h}(t)+y_{p}(t)\nonumber \]
is a solution of the nonhomogeneous equation. Also, this solution satisfies the initial conditions:
\[\begin{aligned} &y(0)=y_{h}(0)+y_{p}(0)=y_{0}+0=y_{0}, \\[4pt] &y^{\prime}(0)=y_{h}^{\prime}(0)+y_{p}^{\prime}(0)=v_{0}+0=v_{0} . \end{aligned}\]
Therefore, we need only focus on finding a particular solution that satisfies homogeneous initial conditions. This will be done by finding values for \(t_{0}\) and \(t_{1}\) in Equation \(\eqref{eq:6}\) which satisfy the homogeneous initial conditions, \(y_{p}(0)=0\) and \(y_{p}^{\prime}(0)=0\).
First, we consider \(y_{p}(0)=0\). We have
\[y_{p}(0)=y_{2}(0) \int_{t_{1}}^{0} \frac{f(\tau) y_{1}(\tau)}{a(\tau) W(\tau)} d \tau-y_{1}(0) \int_{t_{0}}^{0} \frac{f(\tau) y_{2}(\tau)}{a(\tau) W(\tau)} d \tau .\label{eq:12}\]
Here, \(y_{1}(t)\) and \(y_{2}(t)\) are taken to be any solutions of the homogeneous differential equation. Let’s assume that \(y_{1}(0)=0\) and \(y_{2} \neq(0)=0\). Then, we have
\[y_{p}(0)=y_{2}(0) \int_{t_{1}}^{0} \frac{f(\tau) y_{1}(\tau)}{a(\tau) W(\tau)} d \tau\label{eq:13}\]
We can force \(y_{p}(0)=0\) if we set \(t_{1}=0\).
Now, we consider \(y_{p}^{\prime}(0)=0\). First we differentiate the solution and find that
\[y_{p}^{\prime}(t)=y_{2}^{\prime}(t) \int_{0}^{t} \frac{f(\tau) y_{1}(\tau)}{a(\tau) W(\tau)} d \tau-y_{1}^{\prime}(t) \int_{t_{0}}^{t} \frac{f(\tau) y_{2}(\tau)}{a(\tau) W(\tau)} d \tau,\label{eq:14}\]
since the contributions from differentiating the integrals will cancel. Evaluating this result at \(t=0\), we have
\[y_{p}^{\prime}(0)=-y_{1}^{\prime}(0) \int_{t_{0}}^{0} \frac{f(\tau) y_{2}(\tau)}{a(\tau) W(\tau)} d \tau .\label{eq:15}\]
Assuming that \(y_{1}^{\prime}(0) \neq 0\), we can set \(t_{0}=0\).
Thus, we have found that
\[\begin{align} y_{p}(x) &=y_{2}(t) \int_{0}^{t} \frac{f(\tau) y_{1}(\tau)}{a(\tau) W(\tau)} d \tau-y_{1}(t) \int_{0}^{t} \frac{f(\tau) y_{2}(\tau)}{a(\tau) W(\tau)} d \tau\nonumber \\[4pt] &=\int_{0}^{t}\left[\frac{y_{1}(\tau) y_{2}(t)-y_{1}(t) y_{2}(\tau)}{a(\tau) W(\tau)}\right] f(\tau) d \tau\label{eq:16} \end{align}\]
This result is in the correct form and we can identify the temporal, or initial value, Green’s function. So, the particular solution is given as
\[y_{p}(t)=\int_{0}^{t} G(t, \tau) f(\tau) d \tau,\label{eq:17}\]
where the initial value Green’s function is defined as
\[G(t, \tau)=\frac{y_{1}(\tau) y_{2}(t)-y_{1}(t) y_{2}(\tau)}{a(\tau) W(\tau)} .\nonumber \]
We summarize
The solution of the initial value problem,
\[a(t) y^{\prime \prime}(t)+b(t) y^{\prime}(t)+c(t) y(t)=f(t), \quad y(0)=y_{0}, \quad y^{\prime}(0)=v_{0},\nonumber \]
takes the form
\[y(t)=y_{h}(t)+\int_{0}^{t} G(t, \tau) f(\tau) d \tau,\label{eq:18}\]
where
\[G(t, \tau)=\frac{y_{1}(\tau) y_{2}(t)-y_{1}(t) y_{2}(\tau)}{a(\tau) W(\tau)}\label{eq:19}\]
is the Green’s function and \(y_{1}, y_{2}, y_{h}\) are solutions of the homogeneous equation satisfying
\[y_{1}(0)=0, y_{2}(0) \neq 0, y_{1}^{\prime}(0) \neq 0, y_{2}^{\prime}(0)=0, y_{h}(0)=y_{0}, y_{h}^{\prime}(0)=v_{0} .\nonumber \]
Solve the forced oscillator problem
\[x^{\prime \prime}+x=2 \cos t, \quad x(0)=4, \quad x^{\prime}(0)=0 .\nonumber \]
Solution
We first solve the homogeneous problem with nonhomogeneous initial conditions:
\[x_{h}^{\prime \prime}+x_{h}=0, \quad x_{h}(0)=4, \quad x_{h}^{\prime}(0)=0 .\nonumber \]
The solution is easily seen to be \(x_{h}(t)=4 \cos t\).
Next, we construct the Green’s function. We need two linearly independent solutions, \(y_{1}(x), y_{2}(x)\), to the homogeneous differential equation satisfying different homogeneous conditions, \(y_{1}(0)=0\) and \(y_{2}^{\prime}(0)=0\). The simplest solutions are \(y_{1}(t)=\sin t\) and \(y_{2}(t)=\cos t\). The Wronskian is found as
\[W(t)=y_{1}(t) y_{2}^{\prime}(t)-y_{1}^{\prime}(t) y_{2}(t)=-\sin ^{2} t-\cos ^{2} t=-1 .\nonumber \]
Since \(a(t)=1\) in this problem, we compute the Green’s function,
\[\begin{align} G(t, \tau) &=\frac{y_{1}(\tau) y_{2}(t)-y_{1}(t) y_{2}(\tau)}{a(\tau) W(\tau)}\nonumber \\[4pt] &=\sin t \cos \tau-\sin \tau \cos t\nonumber \\[4pt] &=\sin (t-\tau) .\label{eq:20} \end{align}\]
Note that the Green’s function depends on \(t-\tau\). While this is useful in some contexts, we will use the expanded form when carrying out the integration.
We can now determine the particular solution of the nonhomogeneous differential equation. We have
\[\begin{align} x_{p}(t)&=\int_{0}^{t} G(t, \tau) f(\tau) d \tau\nonumber \\[4pt] &=\int_{0}^{t}(\sin t \cos \tau-\sin \tau \cos t)(2 \cos \tau) d \tau\nonumber \\[4pt] &=2 \sin t \int_{0}^{t} \cos ^{2} \tau d \tau-2 \cos t \int_{0}^{t} \sin \tau \cos \tau d \tau\nonumber \\[4pt] &=2 \sin t\left[\frac{\tau}{2}+\frac{1}{2} \sin 2 \tau\right]_{0}^{t}-2 \cos t\left[\frac{1}{2} \sin ^{2} \tau\right]_{0}^{t}\nonumber \\[4pt] &=t \sin t .\label{eq:21}\end{align}\]
Therefore, the solution of the nonhomogeneous problem is the sum of the solution of the homogeneous problem and this particular solution: \(x(t)=4 \cos t+t \sin t\).