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There is a linear homogeneous ODE (let's pick a second order one, but it can be in any order):

\begin{align*} af'' + bf' + cf = 0 \end{align*}

We know, that

\begin{align*} f(t)=e^{\lambda t} \end{align*}

is a solution, and we need to find two $\lambda$'s, so the general solution is (if $\lambda$'s are real and distinct):

\begin{align*} f(t)=c_1e^{\lambda _1 t} + c_2e^{\lambda _2 t} \end{align*}

My question is, why do the only solution is in the form of summed exponentials? What is the proof, that there is no other solution in some other form, a non-exponential one?

(I understand, that if $f_1$ and $f_2$ are solutions, then $c_1f_1+c_2f_2$ is a solution too, but I don't understand, why $f_*$ have to be exponential)

Leucippus
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geza
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  • It can have different solution. The basis is $\exp(\lambda x)$ and $x \exp(\lambda x)$ if the $\lambda1=\lambda_2=:\lambda$. – Botond Apr 25 '18 at 09:17
  • @Botond: yes, thanks, I've clarified my question – geza Apr 25 '18 at 09:19
  • I don't know how helpful is it, but you can apply laplace transform on the differential equation, do some algebra, partial fraction decomposition, and you will get terms in the form of $\frac{1}{s-\lambda}$ if all of the roots are unique. – Botond Apr 25 '18 at 09:25
  • This proof should be in virtually all (decent) calculus textbooks. Basically you let $g(t)=f(t) e^{-\lambda_1 t}$, derive a (simpler) ODE for $g$, and solve it. (You get a first-order equation for $g'$, and for that you already know that all solutions are exponentials: https://math.stackexchange.com/questions/58097/proof-that-c-expx-is-the-only-set-of-functions-for-which-fx-fx.) – Hans Lundmark Apr 25 '18 at 09:57
  • Using Picard's constructive solution we can conclude that the solution is of exponential kind. – Cesareo Apr 25 '18 at 10:06

2 Answers2

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There is a linear homogeneous ODE (let's pick a second order one, but it can be in any order):

Since you only took the second order case as an example, I'll elaborate on the more general case.

  • You can show that the solutions to an $n$-th order linear ODE form a vector space with dimension at most $n$; such a space is spanned by (at most) $n$ linearly independent functions.
  • Plugging $e^{\lambda t}$ into an $n$-th order, linear, homogeneous ODE with constant coefficients will result in an $n$-th degree polynomial which has exactly $n$ (possibly complex) solutions, if we take the multiplicities into account.
  • If the $r$ distinct roots are $\lambda_1 , \ldots , \lambda_r$ with respective multiplicities $m_1,\ldots,m_r$ (and thus we have $m_1+\ldots+m_r=n$), then you can show that for all $i$ with $1 \le i \le r$, the functions $e^{\lambda_i},te^{\lambda_i},\ldots,t^{m_i}e^{\lambda_i}$ are solutions to the ODE; there are in total $n$ such functions.
  • The $n$ functions from above are linearly independent and thus span an $n$-dimensional vector space so this contains all the solutions to the ODE; in other words: any solution will be a linear combination of these exponential functions above.

This argument is a bit indirect in the sense that it doesn't provide a direct intuition as to why the solutions have to be exponential, but it does show that there cannot be any other: all the solutions are in this vector space which is spanned by the "exponentials" (including those of the form $t^ke^{\lambda t}$ which technically aren't exponentials).

See for example here (Theorem 8.3) or here (Theorem 4.1).

StackTD
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  • Thanks for the answer! I understand your reasoning, but I've some troubles (note, I'm new to ODEs). Linearly independent function "system" differs from linearly different vectors, don't they? I mean, with linearly indep. functions, you cannot form any function-shape, but with lin.indep. vectors, you can form any vector. So, I fail to see, why a differently shaped function is not possible (why does the linear combination of the "base" solutions span the entire solution space?) – geza Apr 25 '18 at 12:09
  • Perhaps you're not very familiar with the concepts from linear algebra used in this reasoning. Vector spaces and properties such as linear (in)dependence, dimension and spanning are not limited to what you might remember as (classical, geometric) vectors. In this context, the "vectors" of the vector space mentioned are functions. And yes: any solution to the ODE can be expressed as a lineair combination of what you call the base solutions, i.e. the "exponentials". – StackTD Apr 25 '18 at 12:20
  • Yes, I understand that. The question is the "why?". As I see, basically I need the proof of your first bullet point: "You can show that the solutions to an n-th order linear ODE form a vector space with dimension at most n;". I think I understand the whole process from substituting exponential to ODE, and getting the solution. I just fail to see, why is it the only solution. For 1st order, I can solve the ODE directly, and see that exp is the only solution... – geza Apr 25 '18 at 14:26
  • ... but for higher orders, all the books/tutorials I know starts with "let's substitute exp...", and fail to properly prove that that's the only solution. Or maybe I miss something fundamental... – geza Apr 25 '18 at 14:30
  • The clue to the argument above is that once you find $n$ linearly independent solutions (which are exponentials and those of the form $t^ke^{\lambda t}$), then you know you have them all because the solution set is an $n$-dimensional vector space. Of course you then need this result about the solutions forming a vector space and you can find this in different (text)books. – StackTD Apr 25 '18 at 14:35
  • Is there a theorem (maybe a named theorem) for this: "solutions to an n-th order linear ODE form a vector space with dimension at most n"? Because it is hard to find this statement in any book. Maybe you can recommend a book which contains this? – geza Apr 25 '18 at 19:50
  • I added two references. – StackTD Apr 26 '18 at 07:46
  • Thanks! The first link contains the theorem only (no proof), but the other one seems to contain the proof too, but at a first glance, it is not an easy proof. I'll try to understand it. I'll leave the question open for a little while, maybe someone has an easier proof for this. Thanks again! – geza Apr 26 '18 at 16:23
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I don't if this answers your question. Let me impose the following IVP $$ af''+bf'+cf=0,f(0)=f_0,f'(0)=f_1. \tag{1}$$ Let $y(t)=c_1e^{\lambda _1 t} + c_2e^{\lambda _2 t}$ and then there are $c_1,c_2$ such that $y(0)=f_0,y'(0)=f_1$. Clearly $y(t)$ satisfies the following equation $$ ay''+by'+cy=0,y(0)=f_0,y'(0)=f_1 \tag{2}. $$ Let $X(t)=f(t)-y(t)$. Then $X(t)$ is the solution of the following IVP $$ aX''+bX'+cX=0,X(0)=0,X'(0)=0 \tag{3}. $$ By the existence and uniqueness theorem, (3) has the solution $X(t)\equiv0$. Namely $$ f(t)=y(t)=c_1e^{\lambda _1 t} + c_2e^{\lambda _2 t}. $$

xpaul
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  • So you basically reduce this issue to the existence and uniqueness theorem. I'm not into ODE's too much, so I don't really understand why (3) has the solution X=0. But, using the theorem, if f(0) and f'(0) is given, there must be a unique solution -> the solution space must be 2-dimensional (as we have only 2 "parameters", f(0) and f'(0)), and as the exponential solution is already 2-dimensional, there cannot be any other solutions. Am I right on this? – geza Apr 26 '18 at 17:09
  • @geza, first $X=0$ is a solution and by the uniqueness $X$ is the only solution. – xpaul Apr 26 '18 at 18:05
  • I think this answer is more readily digestible than the accepted one, but it may be benefited by expanding on the last step. – fpf3 Jun 13 '23 at 14:55