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I am given the following ODE : $y'' - 2ay'+by = 0$ for some real constants $a,b$, along with the initial conditions $y(0) = y(1) = 0$. Furthermore I know that some function $y$ is a solution of this ODE.

How do I show that $y(n) = 0$ for all natural numbers $n$?

The idea would be to solve the equation : it is a second order linear differential equation. Hence, (skipping rigour) we solve $r^2 - 2ar + b = 0$ to get solutions $r_1,r_2$ which may or may not be equal (and may be complex).

The solution is now given by :

1 : $y(t) = c_1 e^{r_1 t} + c_2 e^{r_2 t}$ if $r_1 \neq r_2$ (note that if they are complex then by Euler's formula we have a linear combination of trigonometric functions coming in)

2 : $y(t) = c_1 e^{r_1t} + c_2te^{r_1}t$ if $r_1 = r_2$.

In case 1 , substituting $y(0) = 0$ gives $c_1 + c_2 = 0$ and combining with $y(1)=0$ gives $e^{r_1} = e^{r_2}$ and $c_1 = -c_2$, so $\color{red}{e^{r_1 t} = e^{r_2 t} \mbox{ for all } t}$ hence the solution is identically zero.

In case 2, substituting $y(0) = 0$ gives $c_1 = 0$ and then $y(1) = 0$ gives $c_2 = 0$ so the solution is identically zero.

There is definitely something wrong here. I'd like people to point it out, since I think I've blindly used the formula given to me here. This function is supposed to be non-zero at least at the integers, and the fact that no more is provided hints that there are non-trivial solutions.

EDIT : I found the error thanks to the great people below. You may find the error colored in red above.

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In 1., for a nontrivial solution with $f(0)=f(1)=0$ has $c_1+c_2=0$ and $c_1e^{r_1}+c_1e^{r_2}=0$. Thus $c_2=-c_1\ne0$ and $c_1(1-e^{r_2-r_1})=0$. This means that $r_2-r_1=2k\pi i$. Then we easily get $f(n)=0$.

There aren't any solutions in case 2.

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  • $\begingroup$ If $1 - e^{r_1 - r_2} = 0$ then $e^{r_1} = e^{r_2}$, but from the comment in the answer below the next step is incorrect. Thank you for pointing out the error. $\endgroup$ – астон вілла олоф мэллбэрг May 3 '18 at 6:02
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Hint. Note that $y(x)=\sin(\pi x)$ is a solution of $$y'' - 2ay'+by = 0$$ when $a=0$ and $b=\pi^2$.

P.S. For $r_1=i\pi$ and $r_2=-i\pi$ we have that $e^{r_1} = e^{r_2}$ which does not imply that $e^{tr_1} = e^{tr_2}$ for all real $t$.

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  • $\begingroup$ Excellent. So what is the issue in the calculation above, which seems to hint that there are no non-trivial solutions? $\endgroup$ – астон вілла олоф мэллбэрг May 3 '18 at 5:57
  • $\begingroup$ @астонвіллаолофмэллбэрг See my P.S. $\endgroup$ – Robert Z May 3 '18 at 6:00
  • $\begingroup$ Better. It seems that the rule which applies trivially for real exponentiation does not hold true. Thank you for your answer. $\endgroup$ – астон вілла олоф мэллбэрг May 3 '18 at 6:00
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You did not consider the case of complex eigenvalues. You get non-trivial solutions that way. $$ y= e^{\alpha t}(c_1 \cos \beta t +c_2 \sin \beta t)$$ subject to $y(0)=y(1)=0$

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  • $\begingroup$ True. The complex case also leads to the edit I have made in my question above, namely one that would apply if $r_1,r_2$ were real. But thank you or pointing this out. $\endgroup$ – астон вілла олоф мэллбэрг May 3 '18 at 6:21

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