# Fun Little Differential Equation Exercise

• December 1st 2011, 09:50 AM
Ackbeet
Fun Little Differential Equation Exercise
Saw this problem in Zill, 6th Edition (it's Problem 4.3.62, in case you're curious), and had enough fun solving it that I thought I'd post it here for your pleasure. It's not an overly challenging problem, but it is more subtle than it appears at first.

What conditions should be imposed on the constant coefficients $a, b,$ and $c$ in order to guarantee that all solutions of the second-order differential equation $ay''+by'+cy=0$ are bounded on the interval $[0,\infty)?$ (Assume the coefficients are all real.)
• December 2nd 2011, 07:30 AM
chisigma
Re: Fun Little Differential Equation Exercise
Quote:

Originally Posted by Ackbeet
Saw this problem in Zill, 6th Edition (it's Problem 4.3.62, in case you're curious), and had enough fun solving it that I thought I'd post it here for your pleasure. It's not an overly challenging problem, but it is more subtle than it appears at first.

What conditions should be imposed on the constant coefficients $a, b,$ and $c$ in order to guarantee that all solutions of the second-order differential equation $ay''+by'+cy=0$ are bounded on the interval $[0,\infty)?$ (Assume the coefficients are all real.)

The condition is that both the solutions of the algebraic equation $a\ t^{2}+ b\ t + c$ do have real part not greater than 0. The same is for a non-homogeneous linear DE with constant coefficient of any order. For an homogeneous requation like $a\ y^{''} + b\ y^{'} + c= h(x)$, where h(*) is a bounded function in $[0,\infty)$ the condition is that both the solutions of $a\ t^{2}+ b\ t + c$ do have real part strictly less than 0...

Kind regards

$\chi$ $\sigma$
• December 2nd 2011, 07:51 AM
Ackbeet
Re: Fun Little Differential Equation Exercise
Quote:

Originally Posted by chisigma
The condition is that both the solutions of the algebraic equation $a\ t^{2}+ b\ t + c$ do have real part not greater than 0. The same is for a non-homogeneous linear DE with constant coefficient of any order. For an homogeneous requation like $a\ y^{''} + b\ y^{'} + c= h(x)$, where h(*) is a bounded function in $[0,\infty)$ the condition is that both the solutions of $a\ t^{2}+ b\ t + c$ do have real part strictly less than 0...

Kind regards

$\chi$ $\sigma$

All very true. However, the question is asking for conditions on a, b, and c; not on the characteristic equation. So I think the question is asking you to go further.
• December 2nd 2011, 08:30 AM
wnvl
Re: Fun Little Differential Equation Exercise
$Re\left [\frac{-b\pm \sqrt{b^{2}-4ac}}{2a} \right ]<0$

assume a>0

if $b^{2}-4ac < 0$
then b>0

if $b^{2}-4ac > 0$
then $\sqrt{b^{2}-4ac}>|b|$
• December 2nd 2011, 09:06 AM
Ackbeet
Re: Fun Little Differential Equation Exercise
Quote:

Originally Posted by wnvl
$Re\left [\frac{-b\pm \sqrt{b^{2}-4ac}}{2a} \right ]<0$

What about $=0$? Can that happen and the solution still be bounded?

Quote:

assume a>0

if $b^{2}-4ac < 0$
then b>0

if $b^{2}-4ac > 0$
then $\sqrt{b^{2}-4ac}>|b|$
What if $a\ge 0?$ What if $b^{2}-4ac=0?$ Can you then generalize a condition that works for every subcase?
• December 2nd 2011, 11:15 AM
wnvl
Re: Fun Little Differential Equation Exercise
Quote:

Originally Posted by Ackbeet
What about $=0$? Can that happen and the solution still be bounded?

if $Re\left [\frac{-b\pm \sqrt{b^{2}-4ac}}{2a} \right ]=0$ then a bounded solution is only possible if $b^{2}-4ac < 0 \right ]$.

Quote:

Originally Posted by Ackbeet
What if $a\ge 0?$ What if $b^{2}-4ac=0?$ Can you then generalize a condition that works for every subcase?

If a = 0, then we need -c/b < 0 for bounded solution.
If i $b^{2}-4ac = 0 \right ]$ then we need b>0.
• December 2nd 2011, 06:43 PM
chisigma
Re: Fun Little Differential Equation Exercise
Quote:

Originally Posted by chisigma
The condition is that both the solutions of the algebraic equation $a\ t^{2}+ b\ t + c$ do have real part not greater than 0. The same is for a non-homogeneous linear DE with constant coefficient of any order. For an homogeneous requation like $a\ y^{''} + b\ y^{'} + c= h(x)$, where h(*) is a bounded function in $[0,\infty)$ the condition is that both the solutions of $a\ t^{2}+ b\ t + c$ do have real part strictly less than 0...

Kind regards

$\chi$ $\sigma$

A polynomial the roots of which have all non-positive real part is called Hurwitz Polynomial. Neccesary but not sufficient condition for a polynomial to be Hurvitz is that all coefficients have the same sign. For neccesary and suffiicient conditions see...

Hurwitz polynomial

Routh Hurwitz stability criterion

Kind regards

$\chi$ $\sigma$
• December 2nd 2011, 07:35 PM
chisigma
Re: Fun Little Differential Equation Exercise
Quote:

Originally Posted by chisigma
A polynomial the roots of which have all non-positive real part is called Hurwitz Polynomial. Neccesary but not sufficient condition for a polynomial to be Hurvitz is that all coefficients have the same sign. For neccesary and suffiicient conditions see...

Hurwitz polynomial

Routh Hurwitz stability criterion

In order to complete the exam of the question we observe that with the expression 'all the coefficients are of the same sign' we intend that all the coefficients are different from 0. For a second degree polynomial $a\ s^{2} + b\ s + c$ where a, b and c are all different from 0, the necessary condition is also sufficient because we can have two cases...

a) $b^{2}- a c >0$ and the solutions are both real negative...

b) $b^{2}- a c \le 0$ and the solutions have both negative real part...

At this point we have to investigate three more cases...

c) $b=0,\ c \ne 0$, where if a and c have the same sign, then the two solution have real part equal to 0 and the polynomial is Hurwitz...

d) $c=0,\ b \ne 0$, where if a and b have the same sign, then the two solution are real and both $\le 0$ so that the polynomial is Hurwitz...

e) $b=c=0$ where the two solution are both 0 and the polynomial is Hurwitz...

In the last case it has to be specified that the polynomial $p(s)= a\ s^{2}$, on the basis of definition, is Hurvitz but the DE $a\ y^{''}=0$ has unbounded solutions...

Kind regards

$\chi$ $\sigma$