# Linearly independent functions not solutions of ODE

If I have a set of $N$ linearly independent functions $f_1,\dots,f_N$, that may NOT be the solutions of a differential equation, and I impose initial conditions $f(0)=K_0,\dots,D^{N-1}f(0)=K_{N-1}$, is it true that we will always find coefficients $a_1,\dots,a_N$ such that the function $$f(x)=a_1f_1(x)+\dots +a_Nf_N(x)$$ satisfies the initial conditions imposed? Or is only true if the linearly independent functions are solutions of a differential equation?

(Of course assuming this functions are differentiable up to order $N$)

My problem raises from thinking in $$\{sin(x),cos(x)\}$$ In this case is true the statement becasue both functions are never zero for the same value $x$. On the other side $$\{x,x^2\}$$Are also linearly independent but at $x=0$ both functions are zero so the statement is false. I don't understand why the fact that they the first set is an independent set of solutions and the second not, makes a difference.

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Presumably you meant to impose general initial conditions for $f(0),\ldots,f^{(N)}(0)$ since $a_i=0$ is always a valid solution otherwise. –  Sharkos Apr 19 '13 at 12:02
Also, I assume you only meant to impose $N$ initial conditions, in which case you should stop at $f^{(N-1)}(0)$. –  Sharkos Apr 19 '13 at 12:05
Sorry, Already fixed. –  Ambesh Apr 19 '13 at 12:16

Linear independence as functions doesn't tell you much about pointwise behaviour in general.

The set of functions $f_i(x)$ in this case are fairly irrelevant - all that matters is the set $f_i^{(k)}(0)$ which forms an $N\times N$ matrix $A_i\,^k$ if we assume we impose initial conditions $f(0)=b_0,\ldots,f^{(N-1)}(0)=b_{N-1}$. (Sorry about the weird mixture of 0-based and 1-based indices.)

Then the question is simply this:

Does there exist a vector $\mathbf a$ such that $A \mathbf a = \mathbf b$?

or in general,

Is $A$ invertible?

In general, regardless of whether the functions are linearly independent as functions, the rows of this matrix may or may not be independent. (For instance, the functions can all be piecewise defined as identically $0$ for $x\in[0,\frac{1}{2}]$ and then become linearly independent just by having different forms in $[1,2]$, say.)

The reason why a basis of solutions to a differential equations are special is because - provided they are nice and nonsingular $N$th order equations, they give a prescription for evolving any initial condition $\mathbf b$ forwards in $x$. Therefore, you can choose the rows to be independent just by choosing each $f_i$ to have initial conditions such that e.g. $A$ is the identity matrix! ($f_i^{(k)}(0) = \delta_k^i$.) This is just what we mean by choosing a basis of equations.

Note that $x,x^2$ are a basis of solutions for a differential equation of the form $$x^2f''+bxf'+cf$$ but that this is a singular equation, since you have to divide through by $x^2$.

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+1 nice explanation –  Babak S. Apr 19 '13 at 12:16
@Sharkos I don't see why is the case of differential equations is not possible $f_i^{(i)}=0$ ? Thank you –  Ambesh Apr 19 '13 at 12:20
If I understand you right, then you're thinking about this the wrong way: when you take a basis of solutions to a differential equation, it must be the case that from whatever point you choose ($x_0$) it is possible to construct any possible initial condition at that point. This is because a basis by definition spans the set of all possible solutions, and the DE has a solution with any initial conditions you set. On the other hand, you can choose a set of solutions stupidly, like $\sin x,2\sin x$. These don't have the property you want! –  Sharkos Apr 19 '13 at 12:25
@Sharkos Yes I understand. My problem is that I have a independent set of solutions of a fractional differential equation. I can't use the theorem that grants me that there is a solution to proof that the do satisfy the initial conditions. So I have N solutions, they are independent...and I'm stuck there. –  Ambesh Apr 19 '13 at 12:36
Well, I've never played with fractional differential equations, but if they're at least as singular as $x^n y^{(n)}$-type equations, you're likely out of luck, since these produce $x,x^2,\ldots,x^k, x\ln x$ type solutions. –  Sharkos Apr 19 '13 at 12:39

Let $f_1,\ldots,f_n$ be linearly independent functions and define $$D\colon f_i\mapsto \left(f_i(0),f^\prime_i(0),\ldots,f^{(n-1)}_i(0)\right).$$ For solving your initial value problem you need $Df_1,\ldots,Df_n$ to span $\mathbb{R}^n$. This is true in the case of differential equations, but not for general systems of linearly independent functions.

Since $D\colon\mathbb{R}^{[0,1]}\to\mathbb{R}^n$ its actually fairly easy to find independent functions in $\mathbb{R}^{[0,1]}$ such that their projection is no longer independent like for instance your example $x,x^2$. You just need some $g\in\mathrm{span}(f_1,\ldots,f_n)$ such that $Dg=0$, which can be done even for $g\neq 0$.

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