Homogeneous, linear ODEs of the form $$\mathrm{f}^{(n)}(x)+a_{n-1}\mathrm{f}^{(n-1)}(x)+\cdots+a_1\mathrm{f}'(x)+a_0\mathrm{f}(x)=0$$ where each $a_i \in \mathbb{R}$ are known to have "solution spaces" of dimension $n$.

For example, the first order ODE $$\mathrm{f}'(x)+\mathrm{f}(x)=0$$ has the solution space $\mathrm{f}(x)=k\,\mathrm{e}^{-x}$ which is spanned by $\mathrm{e}^{-x}$. The second order ODE $$\mathrm{f}''(x)+\mathrm{f}(x)=0$$ has the solution space $\mathrm{f}(x)=a\sin x + b\cos x$ which is spanned by $\sin x$ and $\cos x$.

What is the name of this result, and can someone please recommend a good on-line reference?


This is the uniqueness theorem for linear differential equations. It is usually proved by variation of parameters using Wronskians, as is done below The proof below easily generalizes to higher order. An analogous proof works also for difference equations (recurrences).

Theorem $\ $ If $\rm\:f,g,h\: $ are solutions on an interval I of

$$\rm y'' =\ p\ y' + q\ y,\ \ \ \ p,q\ \ continuous\ on\ I $$

and the Wronskian $\rm\ \ W = g\:h'-g'h \ne 0\:$ for all $\rm\:x\in I$

then $\,\exists\,$ constants $\rm\: c,d\:$ such that $\rm\: f = c\: g + d\: h\:$ on $\rm\,I.$

Proof $\ $ The equations $[0],[1]$ below have unique solution $\rm\:(c,d)\:$ via det $\rm = W \ne 0\:$ on $\rm\,I.$

$\rm[0]\qquad f\ =\ c\: g \: + d\: h $

$\rm[1]\qquad f' =\ c\: g' + d\: h'$

Now $\rm\:q\:[0] + p\:[1]\ $ yields, $ $ on $ $ LHS: $\rm\,\ q\:f+p\:f'\: =\ f'',\ $ similar on RHS below

$\rm[2]\qquad f'' =\ c\: g'' + d\: h''\ $ via RHS: $\rm\ \, q\:g+p\:g'\: =\ g'',\,\ \ q\:h+p\:h'\: =\ h''$

$\rm[3]\qquad 0\ =\ c'\:g \:+ d'\:h\:\ \ $ via $\ \ [0]'-[1]$

$\rm[4]\qquad 0\ =\ c'\:g' + d'\:h'\ \ $ via $\ \ [1]'-[2]$

$[3],[4]\:$ have solution $\rm\:(c',d') = (0,0),\:$ which is unique by $\rm\ det = W = g\:h'-g'\:h \ne 0\:$ on $\rm\,I.\:$ Therefore $\rm\:c,d\:$ are constants. $\ \ $ QED


L. E. Pursell. A simple uniqueness theory for ordinary linear homogeneous differential equations, Amer. Math. Monthly, 74, 1967, 47-50

Variation of Parameters:
https://planetmath.org/variationofparameters http://ltcconline.net/greenl/courses/204/appsHigherOrder/variationHigher.htm

Marius van der Put. Symbolic analysis of differential equations.

This post: 2003-11-12, There are no other solutions... how to prove it?

The above post is excerpted from my sci.math post on Apr 27 2004 in the thread "number of indep. soloutions to diffyqs?" (sic)


The solution space $\mathscr X$ of $$ \mathrm{f}^{(n)}(x)+a_{n-1}\mathrm{f}^{(n-1)}(x)+\cdots+a_1\mathrm{f}'(x)+a_0\mathrm{f}(x)=0, $$ is an $n-$dimensional subspace of $C^\infty(\mathbb R)$.

I can only think of rather advanced textbooks for reference. The one I prefer is the textbook of Hartmann.

The best way to show that $\mathscr X$ is $n-$dimensional is the following:

a. For every $\xi_1,\ldots,\xi_n$ the initial value problem $$ \mathrm{f}^{(n)}(x)+a_{n-1}\mathrm{f}^{(n-1)}(x)+\cdots+a_1\mathrm{f}'(x)+a_0\mathrm{f}(x)=0, \quad f^{(0)}=\xi_1,\ldots,f^{(n-1)}(0)=\xi_{n}, $$ possesses a unique, and global solution $\varphi=\varphi(t;\xi_1,\ldots,\xi_n)$, which is linear with respect to $\xi_1,\ldots,\xi_n$, and hence $$ \varphi(t;\xi_1,\ldots,\xi_n)=\sum_{k=1}^n\xi_k\varphi(t;\delta_{1k},\ldots,\delta_{1n}) =\sum_{k=1}\xi_k\varphi_k(t). $$ So, these $\varphi_1,\ldots,\varphi_n$ span $\mathscr X$. They are linearly dependent because if $$ c_1\varphi_1(t)+\cdots+c_n\varphi_n(t)=0, \quad \text{for all $t$}, $$ then $$ \varphi(t;c_1,\ldots,c_n)=c_1\varphi_1(t)+\cdots+c_n\varphi_n(t)=0, \quad \text{for all $t$}, $$ and thus for every $k=1,\ldots,n$ we would have $\varphi^{(k-1)}(0;c_1,\ldots,c_n)=0$. But $$ \varphi^{(k-1)}(0;c_1,\ldots,c_n)=c_k. $$

  • $\begingroup$ Thank you for your answer. I wish that I could give it more that one up-vote. First question: how can we be sure that the solutions must all be smooth functions? (I am inclined to believe the statement because all of my "counter examples" require non-linear operations.) $\endgroup$ Jan 21 '14 at 18:49

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.