Thoughts for Jan. 30 – Feb. 5.

Michael Stone, February 5, 2012, , (src)

As I mentioned last week, I’ve recently become interested in trying to fill in some gaps in my mathematical education, beginning with the theory of differential equations. To that end, I began by recalling a (very mildly careful) definition of the word “equation”. Now, let’s try out my definitions on Coddington’s exercise §1.6.1(c).

To begin, let

• $\mathbb{N}$ be the non-negative integers,
• $\mathbb{C}$ be the complex numbers,
• $I = [p, q] \subset \mathbb{R}$ be a closed real interval,
• $(A, B)$ be the space of all functions from $A$ to $B$,
• $C^1\!\!(A, B)$ be the space of continuously differentiable functions from $A$ to $B$,
• $L(A, B)$ be the space of linear transformations from $A$ to $B$,
• $D : C^1\!\!(A, B) \to (A, L(A,B))$ be the total derivative operator.

Exercise §1.6.1(c) asks us to find all the solutions of the equation

$y' - 2y = x^2 + x$

This means that we’re looking for a parameterization of the subset $K \subset C^1\!\!(I, \mathbb{C})$ satisfying the identity, for all $k \in K$ and $x \in I$:

$(Dk)(x) - 2k(x) - x^2 - x = 0$

We are also given, via Theorem §1.6.2 that for all $a \in \mathbb{C}$ and $b \in C^0\!\!(I, \mathbb{C})$ continuous on $I$, the sets of solutions $V$ of differential equations of the form:

$y' + ay = b(x)$

all have, for some $x_0 \in I$ and for all $c \in \mathbb{C}$, the form

$V = \left\{\phi : \phi(x) = e^{-ax} \int_{x_0}^{x} e^{at} b(t) \mathrm{d}t + ce^{-ax}\right\}$

Unifying the signature of Theorem §1.6.2 with the equation of Exercise §1.6.1(c), we find that $a = -2$ and $b(x) = x^2 + x$ and that

$K = \left\{\lambda x. e^{2x} \int_{x_0}^{x} e^{-2t}(t^2 + t) \mathrm{d}t + ce^{2x}\right\}_{c \in \mathbb{C}}$

Our remaining task is to simplify $K$ by simplifying

$\int e^{-2t}(t^2 + t) \mathrm{d}t$

To that end, we start off knowing that, for all $a, b, c \in \mathbb{C}$ and for all $n \in N$, $\int e^{c t} = \frac{1}{c} e^{c t}, \int t^n = \frac{1}{n+1} t^{n+1}, \mbox{ and } \int \left(af + bg\right) = a \int f + b \int g$ Next, via the product rule for differentiation, we know that $D(e^{ct} t^n) = D(e^{ct}) t^n + e^{ct} D(t^n) = c e^{ct} t^{n} + n e^{ct} t^{n-1}$ Next, while we don’t yet know how to compute $\int e^{-2t}(t^2 + t)$, we might hope to find an appropriate antiderivative by solving $e^{-2t}(t^2 + t) = D\left(aF + bG + cH\right)$ for some $a, b, c \in \mathbb{C}$ and for $F = e^{-2t} t^2$, $G = e^{-2t} t^1$, and $H = e^{-2t} t^0$.

Fortunately indeed, this question reduces to solving the following system of three linear equations in the Euclidean vector space generated by the basis $\left\{x^a e^{-2bx}\right\}_{a, b \in \mathbb{N}, a < 3}$ with coefficients in $\mathbb{C}$:

\begin{align*} \left[F + G\right] &= \left[\begin{array}{c}DF \\ DG \\ DH\end{array}\right]^T \left[\begin{array}{c}a \\ b \\ c\end{array}\right] \\ \\ \left[F + G\right] &= \left[\begin{array}{c}-2F + 2G \\ -2G + H \\ -2H \end{array}\right]^T \left[\begin{array}{c}a \\ b \\ c\end{array}\right] \\ \\ \left[\begin{array}{c}F \\ G \\ H\end{array}\right]^T \left[\begin{array}{c}1 \\ 1 \\ 0\end{array}\right] &= \left[\begin{array}{c}F \\ G \\ H\end{array}\right]^T \left[\begin{array}{ccc}-2 & 0 & 0 \\ 2 & -2 & 0 \\ 0 & 1 & -2 \end{array}\right] \left[\begin{array}{c}a \\ b \\ c\end{array}\right] \\ \\ \left[\begin{array}{c}1 \\ 1 \\ 0\end{array}\right] &= \left[\begin{array}{ccc}-2 & 0 & 0 \\ 2 & -2 & 0 \\ 0 & 1 & -2 \end{array}\right] \left[\begin{array}{c}a \\ b \\ c\end{array}\right] \end{align*}

From here, a quick trip through Octave yields:

octave:1> A = [-2, 0, 0; 2, -2, 0; 0, 1, -2]
A =

-2   0   0
2  -2   0
0   1  -2

octave:2> inv(A)
ans =

-0.50000   0.00000   0.00000
-0.50000  -0.50000   0.00000
-0.25000  -0.25000  -0.50000

octave:3> inv(A) * [1; 1; 0]
ans =

-0.50000
-1.00000
-0.50000

Thus we have $\left[\begin{array}{c}a \\ b \\ c\end{array}\right] = \frac{-1}{2} \left[\begin{array}{c} 1 \\ 2 \\ 1 \end{array}\right]$ and $\int e^{-2t}(t^2 + t) \mathrm{d}t = \int D\left(aF + bG + cH\right) = aF + bG + cH$ and, finally, $K = \left\{\lambda x. \frac{-1}{2} (x^2 + 2x + 1) + ce^{2x}\right\}_{c \in \mathbb{C}}$