Special solution of inhomogeneous linear differential equations

-

The general solution of an inhomogeneous linear differential equation can be obtained as the sum of a special solution and the general solution of the corresponding homogeneous differential equation. We study the method of variation of parameters in particular.



Consider the linear differential equation of the form

\[F(D)y = q(x)\]

where \(F(t)\) is a polynomial and \(q(x)\) is a function. When \(q(x) = 0\), this is a homogeneous linear differential equation. Here, we assume \(q(x) \neq 0\).

Suppose we can factorize \(F(t)\) as \(F(t) = G(t)H(t)\). If \(y = y(x)\) is a solution of \(F(D)y = q\), then \(z = H(D)y\) is a solution of \(G(D)z = q\). Thus, the given differential equation \(F(D)y = q\) is decomposed into two parts:

  • \(G(D)z = q\) (a linear differential equation of \(z\), given \(q\)),
  • \(H(D)y = z\) (a linear differential equation of \(y\), given \(z\)),
and we can process one after the other. Thus, by factorizing the polynomial \(F(D)\), we only need to consider the case where \(F(t) = t - \alpha\) (\(\alpha \in \mathbb{C}\)). That is, \(y' - \alpha y = q.\) This ODE can be readily solved by using the method of variation of parameters. The general solution of this ODE is obtained as
\[y = e^{\alpha x}\left\{\int e^{-\alpha x}q(x)dx + C\right\}\]
where \(C\) is a constant.


Example. Let us solve \(y'' + y' - 6y = \cos x\).
Note that \(F(t) = t^2 + t - 6 = (t - 2)(t + 3)\) so the given ODE is \((D - 2)(D+3)y = \cos x\). Solving the homogeneous ODE \((D - 2)(D + 3)y = 0\), we have
  \(y = Ae^{2x} + Be^{-3x}\) where \(A, B\) are constants. Let \(z = (D+3)y\), then we have \((D-2)z = \cos x\) and its special solution is given by
\[z = e^{2x}\int e^{-2x}\cos x dx = -\frac{2}{5}\cos x + \frac{1}{5}\sin x.\]
Similarly, solving \((D+3)y = -\frac{2}{5}\cos x + \frac{1}{5}\sin x\), we have a special solution
\[y = e^{-3x}\int e^{3x}\left(-\frac{2}{5}\cos x + \frac{1}{5}\sin x\right)\,dx = -\frac{7}{50}\cos x + \frac{1}{50}\sin x.\]
Thus, the general solution is
\[y = -\frac{7}{50}\cos x + \frac{1}{50}\sin x + Ae^{2x} + Be^{-3x}\]
where \(A, B\) are constants. □

Let's consider the more general case where \(F(t) = t^2 + at + b = (t - \alpha)(t - \beta)\) with \(\alpha \neq \beta\). In particular, when \(a^2 - 4b < 0\), we have \(\beta = \bar{\alpha}\). We want to solve the following second-order ODE:
\[(D - \alpha)(D - \beta)y = q.\]
Let \(z = (D-\beta)y\), and we have
\[(D - \alpha)z = q.\tag{eq:odea}\]
We first solve the homogeneous case \((D - \alpha)z = 0\) which yields
\[z = Ae^{\alpha x}.\]
Now, regarding \(A\) as a function of \(x\), substitute it into (eq:odea), we have
\[A'(x) = e^{-\alpha x}q(x).\]
Thus, we have the special solution
\[\begin{eqnarray}z &=& e^{\alpha x}A(x)\\ &=& e^{\alpha x}\int e^{-\alpha x}q(x)\, dx. \tag{eq:odez}\end{eqnarray}\]
Next, we solve
\[(D - \beta)y = z.\tag{eq:odeb}\]
Solving the homogeneous case \((D - \beta)y = 0\), we have
\[y = Be^{\beta x}.\]
Regarding \(B\) as a function of \(x\) and substituting it into (eq:odeb), we have
\[B'(x) = e^{-\beta x}z(x).\]
Integrating this using (eq:odea), 
\[\begin{eqnarray} B(x) &=& \int e^{-\beta x}e^{\alpha x}A(x)\,dx\\ &=& \frac{e^{(\alpha - \beta)x}}{\alpha - \beta}A(x) - \int\frac{e^{(\alpha - \beta)x}}{\alpha - \beta}A'(x)\,dx ~ \text{(integration by parts)}\\ &=& \frac{e^{(\alpha - \beta)x}}{\alpha - \beta}A(x) +\int\frac{e^{(\alpha - \beta)x}}{\beta - \alpha}e^{-\alpha x}q(x)\,dx\\ &=& \frac{e^{(\alpha - \beta)x}}{\alpha - \beta}\int e^{-\alpha x}q(x)\,dx + \frac{1}{\beta - \alpha}\int e^{-\beta x}q(x)\,dx. \end{eqnarray}\]
Thus, we have the special solution
\[\begin{eqnarray}y &=& e^{\beta x}B(x)\\& = &\frac{e^{\alpha x}}{\alpha - \beta}\int e^{-\alpha x}q(x)\,dx + \frac{e^{\beta x}}{\beta - \alpha}\int e^{-\beta x}q(x)\,dx.\end{eqnarray}\]
Note the symmetry between the two terms on the right-hand side. This means that swapping \(\alpha\) and \(\beta\) does not change the result. This is expected as the differential operators are commutative: \((D-\alpha)(D - \beta) = (D -\beta)(D - \alpha)\).

Example. Let us solve
\[y'' + y = \frac{1}{\cos x}.\]
The corresponding homogeneous equation is \(y'' + y = 0\). Noting \(D^2 + 1 = (D -i)(D+i)\), its solution is
\[y = Ae^{ix} + Be^{-ix} \tag{eg:hom}\]
where \(A, B \in \mathbb{C}\) are constants. Now, we use the method of variation of parameters by regarding \(A\) and \(B\) as functions of \(x\). According to the above result, we have
\[\begin{eqnarray}A(x) &=& \int \frac{e^{-ix}}{2i\cos x}\,dx \\ &=&\frac{1}{2i}\int\left(1 - i \frac{\sin x}{\cos x}\right)\,dx\\ &=&\frac{1}{2i}(x + i\log|\cos x|) + C_1 \end{eqnarray}\]
where \(C_1\) is a constant. Similarly,
\[B(x) = -\frac{1}{2i}(x - i\log|\cos x|) + C_2\]
where \(C_2\) is a constant. Thus, the general solution is
\[y = \left[\frac{1}{2i}(x + i\log|\cos x|) + C_1\right]e^{ix} + \left[-\frac{1}{2i}(x - i\log|\cos x|) + C_2\right]e^{-ix}.\]
Using Euler's formula (\(e^{\pm i x} = \cos x \pm i \sin x\)), this can be further "simplified" (exercise!) as
\[y = (\log|\cos x| + C)\cos x + (x + D)\sin x\]
where \(C\) and \(D\) are constants. Note that this is a real-valued function if \(C\) and \(D\) are real. □

Next, let's consider \((D-\alpha)^2y = q.\) Of course, \((t - \alpha)^2 = (t-\alpha)(t-\alpha)\) so that we can apply the technique above. Let \(y_1 = (D-\alpha)y\), and we solve \((D-\alpha)y_1 = q\). As we have seen above, 
\[y_1 = e^{\alpha x}\int e^{-\alpha x}q(x)\,dx.\]
Next, we solve \((D-\alpha)y_2 = y_1\). This gives
\[y_2 = e^{\alpha x}\int e^{-\alpha x}y_1(x)\,dx.\]
We can continue the same process to solve \((D-\alpha)^my = q\) for any \(m = 3, 4, \cdots\). Let the solution of \((D-\alpha)^my = q\) be \(y_m\) for \(m = 0, 1, 2, \cdots\). We can see that
\[y_m = e^{\alpha x}\int e^{-\alpha x}y_{m-1}(x)\,dx\]
and
\[y_0(x) = q(x).\]

Example. Solve \(y'' -4y' +4y = \cos x\). It is an exercise to show that the general solution is
\[y = \frac{3}{25}\cos x -\frac{4}{25}\sin x + Ae^{2x} + Bxe^{2x}\]
where \(A\) and \(B\) are constants. □


Comments

Post a Comment

Popular posts from this blog

Birth process

-
Image
  Defining the birth process  Consider a colony of bacteria that never dies. We study the following process known as  the birth process , also known as  the Yule process . The colony starts with \(n_0\) cells at time \(t = 0\). Assume that the probability that any individual cell divides in the time interval \((t, t + \delta t)\) is proportional to \(\delta t\) for small \(\delta t\). Further assume that each cell division is independent of others. Let \(\lambda\) be the  birth rate.  The probability of a cell division for a population of \(n\) cells during \(\delta t\) is \(\lambda n \delta t\). We assume that the probability that two or more births take place in the time interval \(\delta t\) is \(o(\delta t)\). That is, it can be ignored. Consequently, the probability that no cell divides during \(\delta t\) is \(1 - \lambda n \delta t - o(\delta t)\). Note that this process is an example of the Markov chain with states \({n_0}, {n_0 + 1}, {n_0 + 2}...
Read more

Fourier series and the heat equation

-
Image
Joseph Fourier introduced the Fourier series to solve the heat equation in the 1810s. In this post, we show how the Fourier transform arises naturally in a simplified version of the heat equation. Suppose we have the unit circle \(S\) made of a metal wire. Pick an arbitrary point \(A\) on the circle. Any point \(P\) on the circle is identified by the distance \(x\) from \(A\) to \(P\) along the circle in the counter-clockwise direction (i.e., \(x\) is the angle of the section between \(A\) and \(P\) in radian). Let \(u(t,x)\) represent the temperature at position \(x\) and time \(t\). The temperature distribution at \(t = 0\) is given by \(u(0, x) = f(x)\). Assuming no radiation of heat out of the metal wire, \(u(t,x)\) for \(t > 0\) and \(0\leq x \leq 2\pi\) is determined by the following partial differential equation  (PDE) called the  heat equation : \[\gamma\frac{\partial u}{\partial t} = \kappa\frac{\partial^2 u}{\partial x^2}\] and the initial condition \[u(0,x) = f(x...
Read more

Series: Introduction

-
Image
Given a sequence \(\{a_n\}\), the expression \[\sum_{n=0}^{\infty}a_n = a_0 + a_1 + a_2 + \cdots\] is called a series (or infinite series ). This expression may or may not have value. At this point, it is purely formal. Note that the order of addition matters : We first add \(a_0\) and \(a_1\), to the result of which we add \(a_2\), to the result of which we add \(a_3\), and so on (Not something like we first add \(a_{101}\) and \(a_{58}\), then add \(a_{333051}\), and so on). We will see, however, that for a special class of series (the positive term series), the order of addition does not matter if the series converges. Example . The sum of a geometric progression \(\{ar^n\}\), that is, \(\sum_{n=0}^{\infty}ar^n\) is called a geometric series . It is understood that \(r^0 = 1\) including the case when \(r = 0\). □ Given a series \(\sum_{n=0}^{\infty}a_n\) and a number \(n\geq 0\), the sum \[\sum_{k=0}^{n}a_k = a_0 + a_1 + \cdots + a_n\] is called the \(n\)-th partial sum . We m...
Read more