I have a question about ordinary differential equations. I have an initial condition that results in a set of equations. I only have one unknown function ($f(x)$) and only one variable $x$. I should find a solution that simultaneously satisfies all equations. For example, the general case is in the following form:
$ 1) a1 f(x)+b2 f'(x)+c1 f''(x)=0\\ 2) a2 f(x)+b3 f'(x)+c3 f''(x)=0\\ 3) a3 f(x)+b3 f'(x)+c3 f''(x)=0\\ 4) a4 f(x)+b4 f'(x)+c4 f''(x)=0\\ $
Now, what is the proper method for solving these equations? And in more general case if we have the following form:
$ 1) a1 f(x)+b2 f'(x)+c1 f''(x)+d1 f(x) f'(x)+e1 f'(x) f''(x)+h1 f(x) f''(x)=0\\ 2) a2 f(x)+b3 f'(x)+c3 f''(x)+d2 f(x) f'(x)+e2 f'(x) f''(x)+h2 f(x) f''(x)=0\\ 3) a3 f(x)+b3 f'(x)+c3 f''(x)+d3 f(x) f'(x)+e3 f'(x) f''(x)+h3 f(x) f''(x)=0\\ 4) a4 f(x)+b4 f'(x)+c4 f''(x)+d4 f(x) f'(x)+e4 f'(x) f''(x)+h4 f(x) f''(x)=0\\ $
Can we solve that? And How can we understand if these equations have a solution? Thanks


1 Answer 1


These equations can be written as matrix form: $$\begin{bmatrix} a_1&b_1&c_1\\ a_2&b_2&c_2\\ a_3&b_3&c_3\\ a_4&b_4&c_4\\ \end{bmatrix} \begin{bmatrix} f(x)\\f'(x)\\f''(x) \end{bmatrix}= \begin{bmatrix} 0\\0\\0\\0 \end{bmatrix}$$.

To solve the differential equation, we need numerical values for $f(x)$, $f'(x)$, $f''(x)$ for some value of $x$. These numerical values can be obtained via row operations(see https://en.wikipedia.org/wiki/Elementary_matrix) on 4x4 matrix: $$ \begin{bmatrix} a_1&b_1&c_1&|&0\\ a_2&b_2&c_2&|&0\\ a_3&b_3&c_3&|&0\\ a_4&b_4&c_4&|&0\\ \end{bmatrix}\rightarrow\begin{bmatrix}1&0&0&a\\0&1&0&b\\0&0&1&c\\0&0&0&d\end{bmatrix}$$,

once we have $[a,b,c]$, we can do $f(x)=a, f'(x)=b, f''(x)=c$, and then the formula for the sequence is: $$ g(z) = a{z\choose{0}}+b{z\choose{1}}+c{z\choose{2}}, z \in N$$. To get closed form, you just need to simplify the expression involving binomial co-efficient to get a sequence of numbers for each $x$.

Then the solution of the differential equation is(for each x): $$ f(x) = g(0), f(x+dx)=g(1), f(x+dx+dx)=g(2)$$.

I don't know how to continue from here. The solution only have point-wise samples, but we need to have smooth function f(x). (also it seems to want large amount of input data, since $a_1..c_4$ variables need to be available for each x)

EDIT: I've figured out how to go further from this. Basically you need to move from newton-graham technique to the taylor series of the form: $$ f(x+\epsilon) = e^{\epsilon \frac{d}{dx}} f(x) $$, and use the $$ e^z = \sum_{n=0}^\infty \frac{z^n}{n!} $$, then the taylor series will be $$ f(x+\epsilon) = \sum_{n=0}^\infty \frac{\epsilon^n \frac{d^n}{dx^n}}{n!} f(x)$$, this taylor series being an operator where we can ignore the terms with $n\geq 3$, the operator uses the $f(x), f'(x), f''(x)$ that we calculated earlier.

  • $\begingroup$ Thank you very much. It solved my problem. $\endgroup$ Feb 17 at 11:41

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