# Question about second order differential equation

• Apr 30th 2011, 01:55 PM
Ulysses
Question about second order differential equation
I have this problem, which says: If the graph of one solution for the equation $\displaystyle y''+P(x)y'+Q(x)y=0$ is tangent to the x axis in some point of an interval [a,b], then that solution must be identically zero. Why?

I've tried to do something with the general expression for the solution.

$\displaystyle y(x)=Ay_1(x)+By_1(x)\int \displaystyle \frac{e^{\int -P(x)dx}}{y_1^2(x)}dx \Rightarrow y'(x)=Ay_1'(x)+By_1'(x)\int \displaystyle \frac{e^{\int -P(x)dx}}{y_1^2(x)}dx +By_1(x)\displaystyle \frac{e^{\int -P(x)dx}}{y_1^2(x)}$

So I've considered y'(x)=0, because y(x) must be tangent to the x axis.
$\displaystyle 0=Ay_1'(x)+By_1'(x)\int \displaystyle \frac{e^{\int -P(x)dx}}{y_1^2(x)}dx +B \displaystyle \frac{e^{\int -P(x)dx}}{y_1(x)}\Rightarrow y_1'\left( A+B \int \displaystyle \frac{e^{\int -P(x)dx}}{y_1^2(x)}dx \right) =-B \displaystyle \frac{e^{\int -P(x)dx}}{y_1(x)}$

So, I don't know what to do next. I think this is not the right way.

I think I got it.
y(x_0)=0, it touches the x axis
y'(x_0)=0

Thats all?
• Apr 30th 2011, 04:17 PM
topsquark
Quote:

Originally Posted by Ulysses
I have this problem, which says: If the graph of one solution for the equation $\displaystyle y''+P(x)y'+Q(x)y=0$ is tangent to the x axis in some point of an interval [a,b], then that solution must be identically zero. Why?

I've tried to do something with the general expression for the solution.

$\displaystyle y(x)=Ay_1(x)+By_1(x)\int \displaystyle \frac{e^{\int -P(x)dx}}{y_1^2(x)}dx \Rightarrow y'(x)=Ay_1'(x)+By_1'(x)\int \displaystyle \frac{e^{\int -P(x)dx}}{y_1^2(x)}dx +By_1(x)\displaystyle \frac{e^{\int -P(x)dx}}{y_1^2(x)}$

So I've considered y'(x)=0, because y(x) must be tangent to the x axis.
$\displaystyle 0=Ay_1'(x)+By_1'(x)\int \displaystyle \frac{e^{\int -P(x)dx}}{y_1^2(x)}dx +B \displaystyle \frac{e^{\int -P(x)dx}}{y_1(x)}\Rightarrow y_1'\left( A+B \int \displaystyle \frac{e^{\int -P(x)dx}}{y_1^2(x)}dx \right) =-B \displaystyle \frac{e^{\int -P(x)dx}}{y_1(x)}$

So, I don't know what to do next. I think this is not the right way.

I think I got it.
y(x_0)=0, it touches the x axis
y'(x_0)=0

Thats all?

I'm going to make a comment about a theorem that I can't remember the name of and am not sure I know the details of. But it might give you an idea to work with.

Say that the point x' that y(x') = y'(x') = 0 is x' = 0 for simplicity. Then we know that
$\displaystyle y''(0) + P(0)y'(0) + Q(0)y(0) = 0 \implies y''(0) = 0$

(Providing, note, that neither P(x) nor Q(x) are singular at x = 0.) So we know that y(0) = y'(0) = y''(0) = 0. Now take the derivative of the differential equation. This gives you a third degree ODE. Again evaluate the equation at x = 0. This will give y'''(0) = 0. Rinse and repeat.

Now here's that theorem I mentioned: Any function such that $\displaystyle y^{(n)}(0) = 0$ for all non-negative integer n is going to have to be identically 0 or if not is going to be discontinuous.

Whether or not I have the theorem exactly correct I think we have to, at some point, stipulate that y(x) is continuous and that P(x) and Q(x) are $\displaystyle C^{\infty}$.

Perhaps that will generate a fruitful approach for you.

-Dan
• Apr 30th 2011, 04:37 PM
Ackbeet
Would this argument works as well?

The zero function solves the DE and the initial conditions, by inspection. Because the equation is a homogeneous linear equation, the uniqueness theorem shows it is the only solution. QED.