A solution of differential equation and their derivates.

Question

I been solving some problems of the existence and uniqueness theorem, and i'm trying to justify this: Consider the equation $5x^2y^{(5)} − (6 \sin x)y''' + 2xy'' + \pi x^3y' + (3x − 5)y = 0$

Suppose that $Y(x)$ is a solution of this equation such that $Y(1) = 0, Y' (1) = 0, Y'' (1) = 0, Y''' (1) = 0, Y^{(4)}(1) = 0$, and $Y^{(5)}(1) = 0$. Why must $Y(x)$ be equal to 0 for all values of $x$?

i think that the reason is, because the existence and uniqueness theorem says that if $y(t_0)=y_0$ implies $y'(t_0)=y_0'$ for a case of $n=2$ in general $\vec{y}^{(n)}=y_0^{(n)}$, then if any $Y^{(k)} \neq 0$ contradicts the hypotheses of the theorem, am i correct? thanks.

Answer 1

Write the equation as $$ y^{(5)} =\frac{6 \sin x}{5\,x^2}\,y''' - \frac{2}{5\,x}\,y'' - \frac{\pi\,x}{5}\,y' - \frac{3\,x − 5}{5\,x^2}\,y. $$ The right hand side is continuous and Lipschitz in the variables $y,y',y'',y''',y''''$ on a neighborhood of $(1,0,0,0,0,0)$. By the existence and uniqueness theorem there is a unique solution defined on a neighborhood of $x=1$ such that $y(1)=y'(1)=y''(1)=y'''(1)=y''''(1)=0$, and $y=0$ is clearly a solution satisfying the initial conditions.

If you do not have seen the theorem for $n$-th order equations, transform the equation into a first order system: \begin{align} y'&=y_1\\ y_1'&=y_2\\ y_2'&=y_3\\ y_3'&=y_4\\ y_4'&=\frac{6 \sin x}{5\,x^2}\,y_3 - \frac{2}{5\,x}\,y_2 - \frac{\pi\,x}{5}\,y_1 - \frac{3\,x − 5}{5\,x^2}\,y\\ \end{align} with initial conditions $y(1)=y_1(1)=y_2(1)=y_3(1)=y_4(1)=0$.

Answer 2

An easy way to understand the behaviour of a solution of an ODE is studying the derivatives that are involved in it.

What I mean is, colloquially, that $y'$ gives you the behaviour (in fact it shows you the change of $y$ with respect to $x$) of $y$, $y''$ shows the same with $y'$, etc.

Then, if you have that in your initial condition the derivatives of $y$ are $0$, this means that:

• $y^{(5)}(1) = 0 \implies y^{(4)}(1)\;$ "doesn't change".
• ...
• ...
• $y'(1) = 0 \implies y(1)\;$ "doesn't change".

That implies that your initial condition is, in fact, a fixed point (and then, your entire solution).

I think that the phase portrait theory is quite good to see these illustrations.

Maybe seeing an easier problem of first order makes me explain this better:

Consider the following Cauchy Problem, with an autonomous ODE (it doesn't depend on $t$), \begin{equation} x' = f(x)\\ x(t_0) = x_0 \end{equation} such that $f(x_0) = 0$. Assume too that $f$ is locally Lipschitz (this implies that we are assuming the existence and uniqueness of the solution).

$f$ gives you the velocity of the solution in each point of the phase portrait. As you are trying to find the solution that contains the point $x_0$ at time $t_0$ and the velocity in this point is exactly $f(x_0) = 0$, the solution won't move from that point. That means it will be a fixed point and your solution will be $x(t) = x_0$.

Excuse me, I've just seen that I used different notations in your example and in the one that I added. I hope it doesn't confuse... By the way, I used in my example $t$ as the independent variable because thinking as if it was the time in the evolution of your dynamical system maybe helps you.