# Why do we need n initial conditions to specify the solution of an nth order linear DE?

Particular cases show that this is the case. E.g. free fall $\frac{d^2 x}{d t^2} = -g$, and the simple harmonic oscillator $\frac{d^2 x}{d t^2} + \omega^2x = 0$.

I can see why this is so physically, as in the above cases, but not in the general sense.

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Am I missing something here? Surely in solving $\frac{\mathrm d^2x}{\mathrm dt^2}=-g$ you would need 2 initial conditions, equal to the order. Intuitively, if you want to know the position of a free falling object at a certain time what do you need to know except for the acceleration which is given by the equation? The position at a given $(x,t)$ and the velocity at a given $(x,t)$, right? As far as I know this isn't any different for the simple harmonic oscillator or other linear ODEs. –  user50407 Dec 31 '12 at 18:43
@Ron, I want to make sure you know about the "accept an answer" feature of this site. There is a checkmark to the left of each answer to your question, which you can use to mark the answer that was most helpful to you. It may be too early to do so for the present question, but you should accept some answers to your older questions. For details, see Why should we accept answers? –  user53153 Jan 1 '13 at 2:02

About the title: Consider an experiment, that at time, say $t=0$ (Starting) a body in $10$ feeet from an origin and is moving with a velocity of $20$ ft/sec. And assume the motion of the body is given by 2-parameter familiy for example: $x=16t^2+c_1t+c_2$ wherein $x$ is the distence of the particle from an origin at time $t$. We have, then : $v=32t+c_1$. Here we have to choose the constant $c_1,c_2$ and so when $t=0$, $x=10$ and$v=20$ and regard two equations above we find $c_1=10$ and $c_2=20$. Hence the particular solution is $x=16t^2+20t+10$. The conditions which enabled us to find a certain solution is called initial conditions. Accodring to defenition; if we have $c_1,c_2,...c_n$ constand in our $n-$ parameter family then we need $n$ conditions. Remmember the constant in indefinte integral and think what would happened if we have to integral twice?

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Seen as I have written lots in comments already I would like to make an answer. I saw Hagen's edited answer pop up while writing, but I'm not sure I understand it.

Your idea is wrong. A solution of a linear differential equation of order $n$ (with constant coefficients) has $n$ degrees of freedom.

I don't know if you have had linear algebra, but I will try to give an explanation of how it is used in the theory of linear differential equations (and hope that this is correct, if not any corrections are appreciated.)

First of all you need to know that certain sets of functions, with a natural addition and scalar multiplication are vector spaces, i.e. they satisfy the axioms of a vector space. This means we will be able to apply theorems established in linear algebra to solve problems regarding these sets of functions. In particular the set we will consider is $\mathsf C^\infty$, the set of all functions from $\mathbb R\to \mathbb C$ with derivatives of all orders.

It turns out that $\mathsf{D}:\mathsf C^\infty\to\mathsf C^\infty,f(x)\mapsto f'(x)$ is a linear operator and so is any sum, scalar multiple or composition of $\mathsf D$. This means we can apply theorems we know about linear transformations to $\mathsf D$. It also is the case that when you have a differential operator $p(\mathsf D)$, basically a polynomial in $\mathsf D$, it can be factored just like the characteristic polynomial of the differential equation. This can be used to solve the homogeneous equation, example: $f''(x)+3f'(x)+2f(x)=0$, can be written as $\mathsf D^2(f)+3\mathsf D(f)+2\mathrm{id_{\mathsf C^\infty}}(f)=0$ and just as the characteristic polynomial is factored, this is $$(\mathsf D+2)(\mathsf D +1)(f)=\\(\mathsf D+2)(f'(x)+f(x))\\f''(x)+f'(x)+2f'(x)+2f(x)=\\f''(x)+3f'(x)+2f(x)=0$$ which we see is the same as the equation we started from. This can be used in a similar way as factoring polynomials to say that $f''(x)+2f'(x)+f(x)=0\iff f\in\mathrm{ker}(\mathsf D+2)+\mathrm{ker}(\mathsf D+1)$ (you might expect it to lie in the union of the kernels, but this is not the case.) This works analogously for higher orders and by the similarity between the factorization of the differential operator and the characteristic polynomial we have that any $f$ which satisfies the homogeneous equation lies in the sum of these kernels, each with dimension $1$, so we end up with a solution space with the same dimension as the order of the equation. This is a bit of a hand waving argument and here I have assumed there aren't any repeated roots, but it works in general.

Now I will explain the way I see it for the case of an inhomogeneous equation (I hope it is right). You probably know the idea of a line defined by a support vector and a direction vector. I will try to show the same sort of idea applies here. Any function in the kernel which we saw in the previous paragraph can be added to a solution of a differential equation of the form $f''(x)+af'(x)+bf(x)=g(x)$. In this way we can get the set of solutions if we know one solution $f_p$ to the equation above and we know the solution space of the homogeneous equation $\mathsf K_p$ (by the method above), that is $f_p+\mathsf K_H$. So if you think of it geometrically, all solutions to $f''(x)+af'(x)+bf(x)=g(x)$ will lie in a "plane" which is parallel to the "plane" we get from the homogeneous equation and which is supported by a (not unique) support vector which you can find for example by educated guess. It follows (think of the plane or line, where the support vector adds no degree of freedom) that there are just as many degrees of freedom in the solution to the inhomogeneous equation as in the solution to the homogeneous equation, which we have seen to be the order of the equation.

I have written quite a lot and hope it is at least in some way helpful and correct. Else, just post a comment.

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Simply put, we are looking for the kernel of a linear form on a $(n+1)$-dimensional space. This kernel then has dimension $n$, i.e. an element of it can be specified by $n$ coordinates.

To be (slightly) more explicit: Given an equation of the form $a_0y+a_1y'+\ldots +a_ny^{(n)}=0$, we can consider the linear map $$D\colon (f_0,f_1,\ldots, f_n)\mapsto \left(f_1-\frac d{dx}f_0,f_2-\frac d{dx}f_1,\ldots, f_n-\frac d{dx}f_{n-1}, a_0f_0+\ldots a_nf_n\right)$$ which is an endomorphism of the space $V$ of smooth functions. A tuple $(f_0,\ldots, f_n)$ is in the kernel iff $f_i=y^{(i)}$ for some solution $y$ of the DE. Now if the $f_i$ were just elements of our base field (e.g. $\mathbb R$), then clearly $D$ would be defined on an $(n+1)$ dimensional space and hence its kernel is (if it is not the whole space for trivial reasons) of dimension $n$. Now two strategies are possible: Either one constructs a field whose elements are in fact functions. Or one does not consider the space of all smooth functions, but merely a finite dimensional space $V$ that is big enough to contain all solutions of the DE. At any rate, since the kernel of $D$ is $n$-dimensional, it should not be surprising that a "typical" linear map $V\to \mathbb R^n$ (i.e. one that is not "badly chosen"), such as $$\tag1(f_0,f_1,\ldots, f_n)\mapsto (f_0(1),f_0(2),f_1(42), f_0(0)+f_1(0)+f_2(0), \ldots)$$ can induce an isomorphism of $\ker D$ with $\mathbb R^n$. In this fashion, e.g. $(1)$ would correspond to initial conditions $y(1)=?$, $y(2)=?$, $y'(42)=?$, $y(0)+y'(0)+y''(0)=?$ and so on.

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Are you sure this is right? Solving a linear homogeneous DE is finding the kernel of the corresponding differential operator. This kernel can be proven to have the same dimension as the order of the diff. operator. What you are saying (if I am understanding it right) seems like it can't be right, usually you can restrict the differential operator to a subspace that contains the vectors of the kernel but is of lesser dimension than the space the differential operator was defined on. I think your argument then implies that the two kernels (despite being the same) have different dimensions. –  user50407 Jan 1 '13 at 16:43
LOL. Apparently I wrote this result-oriented: The original formulation of the question was about $n-1$ initial conditions :) –  Hagen von Eitzen Jan 1 '13 at 16:59
Haha yes, I thought it was something like that. OP was wrong in his question. –  user50407 Jan 1 '13 at 17:00
Sorry guys -__- –  Ron Jan 10 '13 at 15:22
No problem, errors and mistypings happen everyday and by the best and belong to the lesser sins, no big need to excuse for them. But did you notice Pavel M's comment to your question? –  Hagen von Eitzen Jan 10 '13 at 21:25