7
$\begingroup$

Prove that the real vector space of $C^\infty$ functions $f$ such that $$f''(x) + f(x) = 0$$ is two-dimensional.

I feel like there must be an obvious answer to this question, but I don't know what it is. In particular, I can see that $\sin(x)$ and $\cos(x)$ are linearly independent, but how do I know that they span the space?

If we restrict the problem to analytic functions, then two-dimensionality is easy: once you know the values of $f(0)$ and $f'(0)$, the rest of the Maclaurin series follows from the identity $f^{(n+2)}(0) = -f^{(n)}(0)$ for all $n$. But how do we know there are no non-analytic solutions?

Before you ask: this isn't quite a homework question. I'm going to present this example as an algebra TA, and I'd like some confidence that my answer is actually rue.

$\endgroup$
5
  • $\begingroup$ The set of solutions to $f''(x)+f(x)=0$ is $f(x)=c_1\cos(x)+c_2\sin(x)$. So any solution is a linear combination of $\cos(x)$ and $\sin(x)$, so these functions span the space and are linearly independent. $\endgroup$ – Dave Aug 28 '17 at 16:34
  • 6
    $\begingroup$ @Dave He is asking how we prove there are no non-analytic solutions, not what the solutions are. $\endgroup$ – Isaac Browne Aug 28 '17 at 16:36
  • $\begingroup$ @IsaacBrowne Sorry but I don't get it: does not the fact that the space of solution is two-dimensional prove the result? Or are you saying Dave has not proven that the linear combination is all the space of solutions? $\endgroup$ – anderstood Aug 28 '17 at 17:41
  • $\begingroup$ @anderstood I am saying that Dave has not shown the proof that the linear combinations are $all$ the space of solutions. Anschel's question asks how we know there are no non-analytic solutions, and that would be the way to answer it. $\endgroup$ – Isaac Browne Aug 28 '17 at 17:56
  • $\begingroup$ @anderstood Dave hasn't proven that this is the set of all solutions. That's precisely what I'm trying to prove. $\endgroup$ – Anschel Schaffer-Cohen Aug 28 '17 at 18:38
7
$\begingroup$

I shall prove a general statement. Let $p(t)\in\mathbb{C}[t]$ be a monic polynomial of degree $n\in\mathbb{Z}_{>0}$ and suppose that $V:=\mathcal{C}^\infty(\Omega,\mathbb{C})$ is the $\mathbb{C}$-vector space of complex-valued functions on $\Omega$, where $\Omega$ is a nonempty connected open set of $\mathbb{R}$. Write $D:V\to V$ for the differentiation map: $$\big(D(f)\big)(x):=f'(x)=\left.\left(\frac{\text{d}}{\text{d}u}f(u)\right)\right|_{u=x}$$ for all $x\in \Omega$. Then, $\ker\big(p(D)\big)$ is an $n$-dimensional $\mathbb{C}$-vector subspace of $V$. (Even more generally, if $\Omega$ is an open set of $\mathbb{R}$ with $c$ connected components, then $\ker\big(p(D)\big)$ is $cn$-dimensional over $\mathbb{C}$. Also, $V$ can be replaced by the space of holomorphic functions on $\Omega$, if $\Omega$ is taken to be an open subset of $\mathbb{C}$.)

First, write $p(t)=\prod_{i=1}^k\,\left(t-z_i\right)^{m_i}$, where $z_1,z_2,\ldots,z_k\in\mathbb{C}$ are pairwise distinct, and $m_1,m_2,\ldots,m_k\in\mathbb{Z}_{>0}$. Define $$p_i(t):=\left(t-z_i\right)^{m_i}\text{ and }q_i(t):=\frac{p(t)}{p_i(t)}\,.$$ Then, the fact that $$p(t)=p_i(t)\,q_i(t)\text{ and }p_i(t)\,f_i(t)+q_i(t)\,g_i(t)=1$$ for some $f_i(t),g_i(t)\in\mathbb{C}[t]$ implies that $$\ker\big(p(D)\big)=\ker\big(p_i(D)\big)\oplus \ker\big(q_i(D)\big)\,.$$ By induction, we see that $$\ker\big(p(D)\big)=\bigoplus_{i=1}^k\,\ker\big(p_i(D)\big)\,.$$

Thus, it boils down to studying $\ker\big(p(D)\big)$, where $p(t)=(t-z)^m$ for some $z\in\mathbb{C}$ and $m\in\mathbb{Z}_{>0}$. However, consider the map $M_z:V\to V$ given by $$\big(M_z(f)\big)(x):=\exp(zx)\,f(x)$$ for all $x\in \Omega$. As $$p(D)=M_z\,D^m\,M_{-z}=M_z\,D^m\,\left(M_z\right)^{-1}\,,$$ $p(D)$ and $D^m$ are conjugate linear maps. Therefore, $$\ker\big(p(D)\big)=M_z\big(\ker\left(D^m\right)\big)\,.$$ Since $\ker\left(D^m\right)$ is $m$-dimensional and $M_z$ is a vector-space automorphism, $$\dim\Big(\ker\big(p(D)\big)\Big)=\dim\Big(\ker\big(D^m\big)\Big)=m\,.$$ In fact, $\ker\big(p(D)\big)$ consists of elements of the form $M_z(f)$, where $f:\Omega\to\mathbb{C}$ is a polynomial function of degree less than $m$.


Alternatively, let $U:=\ker\big(p(D)\big)$. Then, show that $p(t)$ is the minimal polynomial of $D|_U:U\to U$. From my post here, $U$ decomposes as $$U=\bigoplus_{i=1}^m\,\ker\left(\big(D-z_i\,\text{id}_U\big)^{m_i}\right)\,,$$ if $p(t)=\prod_{i=1}^k\,\left(t-z_i\right)^{m_i}$.


Let $\mathcal{V}$ denote the $\mathbb{C}$-vector space of $\mathbb{C}$-valued sequences $\left(X_N\right)_{N\in\mathbb{Z}_{\geq 0}}$. Fix $a_0,a_1,\ldots,a_{n-1}\in\mathbb{C}$. The same technique can be employed to show that the solutions $\left(X_N\right)_{N\in\mathbb{Z}_{\geq 0}}\in\mathcal{V}$ of a recurrence relation $$X_{N+n}+a_{n-1}\,X_{N+n-1}+\ldots+a_1\,X_{N+1}+a_0\,X_N=0$$ for all $N\in\mathbb{Z}_{\geq 0}$ form an $n$-dimensional $\mathbb{C}$-vector subspace of $\mathcal{V}$.


P.S.: If you want to consider $V_\mathbb{R}:=\mathcal{C}^\infty(\Omega,\mathbb{R})$ instead, then use the fact that $V=\mathbb{C}\underset{\mathbb{R}}{\otimes} V_{\mathbb{R}}$. That is, if $p(t)\in\mathbb{R}[t]$, then the kernel $K_\mathbb{R}$ of $\big.p(D)\big|_{V_\mathbb{R}}:V_\mathbb{R}\to V_\mathbb{R}$ satisfies $$\mathbb{C}\underset{\mathbb{R}}{\otimes} K_\mathbb{R}=\ker\big(p(D)\big)\,.$$

$\endgroup$
3
$\begingroup$

Read about the Wronskian.

For three supposedly linearly independent $C^{\infty}$ solutions $f$ $g$, and $h$, the Wronskian determinant should be nonzero. But since the last row will be $-1$ times the first row, the Wronskian will be $0$. This contradicts the possibility for three independent solutions.

$\endgroup$
3
$\begingroup$

HINT: Show that the function $$ \mathbb{R} \ni t \mapsto \left (\matrix {\cos t & - \sin t \\ \sin t & \cos t } \right )\cdot \left ( \matrix{ f(t) \\ f'(t) } \right ) \in \mathbb{R}^2$$ is constant

$\bf{Added:}$

Let's show that if $n$ solutions of the linear equation $y'(t) = A(t) y(t)$ have an invertible Wronskian $W(t)$, and $f$ is another solution then $f(t)$ is a linear combination of these solutions. Recall that the columns of $W$ are these $n$ solutions, so we have $W'(t) = A(t) W(t)$.

Indeed, consider the function $$t\mapsto W(t)^{-1} f(t)$$ Its derivative equals $$(W(t)^{-1})' f(t) + W(t)^{-1} f'(t)$$ Now we have $$(W(t)^{-1})'= - W(t)^{-1} W(t)' W(t)^{-1} $$ We get $$(W(t)^{-1})' f(t) + W(t)^{-1} f'(t) = \\=- W(t)^{-1} A(t) W(t) W(t)^{-1} f(t) + W(t)^{-1} A(t) f(t)= \\ =- W(t)^{-1} A(t) f(t) + W(t)^{-1} A(t) f(t) = 0$$

Hence the function $$ W(t)^{-1} f(t) $$ is a constant $c$ (in $\mathbb{R}^n)$ and so $f(t) = W(t) \cdot c$ is a linear combination of the columns of $W(t)$.

$\endgroup$
2
  • $\begingroup$ the derivative is $0$ $\endgroup$ – orangeskid Aug 28 '17 at 17:06
  • $\begingroup$ In general, if we have a linear differential equation $y'(t) = A(t) \cdot y(t)$ and $n$ solutions such that their Wronskian is invertible then any other solution $f$ is a linear combination of them. Just check that the function $W(t)^{-1} \cdot f(t)$ is constant. Works like magic. $\endgroup$ – orangeskid Aug 28 '17 at 18:00
0
$\begingroup$

Here is an approach that essentially uses reduction of order.

Define $g(x) = f(x)/\cos(x)$, so that $f(x) = g(x) \cos(x)$. Plugging this in to the differential equation yields $$ 0 = f''(x) + f(x) = [\cos(x) g''(x) - 2\sin(x)g'(x) - \cos(x) g(x)] + \cos(x)g(x)\\ = \cos(x) g''(x) - 2 \sin(x) g'(x). $$ In other words, $h(x) = g'(x)$ solves the differential equation $$ \cos(x)h'(x) - 2 \sin(x)h(x) = 0 \implies h'(x) = 2 \tan(x) h(x). $$ This is a separable first order differential equation, which we can routinely solve to get $$ h(x) = C_1 \sec^2(x). $$ With that, we have $$ g'(x) = C_1 \sec^2(x) \implies g(x) = C_1 \tan(x) + C_2, $$ which leads to our solution $$ f(x) = g(x) \cos(x) = C_1 \sin(x) + C_2 \cos(x), $$ which was what we wanted.


There are some weaknesses to this approach. Division by $\cos(x)$ means that this is technically only an argument that a solution exists over intervals that exclude the zeros of $\cos(x)$. However, the time-invariant nature of this differential equation means that this existence extends to arbitrary intervals "by symmetry."

Now, we can establish that the solution space is two dimensional by arguing that separable first order differential equations have a one-dimensional solution spaces.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.