MHB System of differential equations

[mathmari]

Hey!

When we have a system of differential equations $$\bigwedge_{j=1}^n L_j y=f_j$$ is it possible that we reduce it into one differential equation $Ly=f$ so that the system has a solution iff the differential equation $Ly=f$ has a solution? (Wondering)

 

[Ackbach]

Hmm, that's an interesting one. While I've known about the trick for reducing a higher-order DE into a first-order DE by introducing enough new variables, I've never thought about going backwards. It might be possible; with some DE's it's definitely possible. One thing is sure: Each higher-order derivative is independent of the lower-order derivatives, and can be thought of almost as a new variable. What I'm not sure of is whether the process would yield a strange DE that doesn't bear much resemblance (in terms of its solutions) to the original.

I guess my question is, why would you want to do this? If you're trying to show existence and/or uniqueness of solutions, all the major theorems work on systems equally as well as single DE's. In fact, I would say some of the most important theorems only work on systems! In that sense, you might be better off in the system.

 

[mathmari]

I guess my question is, why would you want to do this?

I want to use this fact to eliminate a quantifier of a formula...

I thought about it again... Can we do the following? We suppose that we have a system of differential equations $$\left\{\begin{matrix}
L_1=0\\
L_2=0
\end{matrix}\right. \tag 1$$
where ther order of $L_1$ is $n$ and the order of $L_2$ is $m$ and the coefficients of the highest-order term is $1$.

Let $n<m$.

Let $d_jL_i$ be the $j$th derivative of $L_i$.

We define the differential equation $d_{m-n}L_1-L_2=0$ the order of which is smaller than the order of $L_2$.

So the system $(1)$ is equivalent to the system $$\left\{\begin{matrix}
L_1=0\\
d_{m-n}L_1-L_2=0
\end{matrix}\right.$$

So the initial system is equivalent to a system of smaller order.

We do the same procedure until we get a system of the form $$\left\{\begin{matrix}
L=0\\
\alpha
\end{matrix}\right.$$ where the order of $L$ is $0$ and $\alpha$ doesn't contain $x$, i.e., it is a relation between the parameters.

So we conclude to one differential equation.

That means that any system can be replaced by one differential equation.

Is this correct? (Wondering)

 

[Ackbach]

Hmm. Well, the problem here is that it's not clear how you're eliminating variables. A system of DE's will have, say, $m$ variables. If it's an ODE, then there's only one independent variable, call it $x$. We'll call the dependent variables $y_1, y_2, \dots, y_m$. We can assume, WLOG, that the system is first-order. Then, in general, you have the system
\begin{align*}
f_1\!\left(x,y_1,y_2,\dots,y_m,y_1',y_2',\dots,y_m'\right)&=0 \\
f_2\!\left(x,y_1,y_2,\dots,y_m,y_1',y_2',\dots,y_m'\right)&=0 \\
\vdots \\
f_p\!\left(x,y_1,y_2,\dots,y_m,y_1',y_2',\dots,y_m'\right)&=0.
\end{align*}
Your goal is to get from here to a single ODE: $f\!\left(x,y,y',y'',\dots,y^{(q)}\right)=0$. To do this, you will definitely have to eliminate variables. What is your plan to do that?

 

[mathmari]

Let $L$ be the language $\{+, \frac{d}{dx} , 0, 1\}$ and $R$ the ring $\mathbb{C}[x, e^{\lambda x} \mid \lambda \in \mathbb{C}]$.

In my notes I found the following proposition:

Let $D_1$ and $D_2$ be non-zero differential operators and $f_1$ and $f_2$ terms of $L$.
Then there id a differential operator $L$, with order less or equal to the maximum of the set $\{\text{ord}(D_1), \text{ord}(D_2)\}$, and a term $f$ of $L$, such that in the ring $R$, the following formulas are equivalent:
$$D_1(y)=f_1 \ \land \ D_2(y)=f_2 \ \ \ \ \text{ and } \ \ \ \ D(y)=f$$
Furthermore, there is an algorithm for the construction of $D$ and $f$ from $D_1$, $D_2$, $f_1$ and $f_2$. So, isn't such an algorithm the one that I described in post #3? An example of an application of this procedure is the following: $$\left\{\begin{matrix}
D_1y =y'+2y+1=0\\
D_2y =y''-y'+3x=0
\end{matrix}\right. \Leftrightarrow \left\{\begin{matrix}
D_1y =y'+2y+1=0\\
\tilde{D}_2 y =d_{1} D_1 y - D_2 y=(y''+2y)-(y''-y'+3x)=0
\end{matrix}\right. \\ \Leftrightarrow \left\{\begin{matrix}
D_1y =y'+2y+1=0\\
\tilde{D}_2 y=y'+2y-3x=0
\end{matrix}\right. \\ \Leftrightarrow \left\{\begin{matrix}
D_1y =y'+2y+1=0\\
\hat{D}_2 y=d_{0} D_1 y - \tilde{D}_2 y=(y'+2y+1)-(y'+2y-3x)=0
\end{matrix}\right. \\ \Leftrightarrow \left\{\begin{matrix}
D_1y =y'+2y+1=0\\
\hat{D}_2 y=1+3x=0
\end{matrix}\right.$$ So we end up with a system of the form $\displaystyle{\left\{\begin{matrix}
D y=0\\
\psi
\end{matrix}\right.}$ where $\psi$ doesn't contain any $y$, i.e., it is a relation between the parameters. Is this correct? (Wondering)

 

[Ackbach]

Ah, here's the nub of the issue, as I see it, right here:

$$\left\{\begin{matrix}
D_1y =y'+2y+1=0\\
D_2y =y''-y'+3x=0
\end{matrix}\right.$$

This isn't a system of differential equations, so far as I can tell - at least, it's not a normal system. It's over-determined, if you do insist on calling it a system, and the chances are, it has no solution at all. We can solve each of these differential equations separately for $y$:
\begin{align*}
y'+2y+1&=0 \qquad \to \qquad y(x)=-\frac12+Ce^{-2x} \\
y''-y'+3x&=0 \qquad \to \qquad y(x)=3x+\frac{3x^2}{2}+C_1 e^x+C_2.
\end{align*}
There is absolutely no choice of $C, C_1,$ or $C_2$ that will enable you to equate these two solutions. That is, for your example, there is no solution.

The procedure you've written down looks legal to me, but I'm not sure it's of much use. Will it work on a true system of differential equations? With multiple different dependent variables $y_1, \dots, y_m?$ That, to me, is the question. And we come back to whether or not the procedure you've outlined above is actually eliminating variables.

It did just strike me that if the system of DE's is linear with constant coefficients, then you can essentially do Gaussian elimination. But I can't remember the details as yet...