An eigenvalue problem

Consider the eigenvalue problem
$$u''=-\lambda u, u(0) = 0, u'(1) = 0.$$

• Find all values of $\lambda$ such that the problem has a non-trivial solution.
• For each of eigenvalue $\lambda$, what is the corresponding solution?
• Sketch the graphs of the solutions corresponding to the three smallest, positive eigenvalues.

Comments:

• I asked and answered a similar question here. It might help to look at that, particularly the first part, though this one is quite a bit simpler.
• It's fine with me if one person does the first two, purely algebraic parts and someone else uses Mathematica to sketch the graphs.

I took a very rough stab at this problem. I'm gonna skip some steps right now since we know that the general solution is $$u(x) = a\sin(\sqrt{\lambda}\,x) + b\cos(\sqrt{\lambda}\,x)$$

Plugging in $$u(0) = 0$$
you're left with something along the lines of
$$u(0) = b = 0$$
since we know that $$\sin(0)=0$$ and the entire "b" term goes away meaning $$u(x) = a\sin(\sqrt{\lambda}\,x)$$
Now you can take the derivative of that equation and plug in the variables to solve for lambda.
$$u'(x) = a\sqrt{\lambda}\cos(\sqrt{\lambda}\,x)$$
$$u'(1) = a\sqrt{\lambda}\cos(\sqrt{\lambda}\,x) = 0$$
$$a\sqrt{\lambda}\cos(\sqrt{\lambda}) = 0$$
$$\cos(\sqrt{\lambda}) = 0$$
$$\sqrt{\lambda} = \cos^{-1}(0)$$
$$\sqrt{\lambda} = \frac{\pi(2n+1)}{2}$$
$${\lambda} = [\frac{\pi(2n+1)}{2}]^{2}$$
$$u(x) = a\sin(x*\sqrt{[\frac{\pi(2n+1)}{2}]})$$

Assuming I did this right, those would be the values for lambda where the problem has a non-trivial solution. I'm operating under the assumption that n = -1, 0, 1 etc. Anyways, I'm not sure what the second bullet is asking for with the eigenvalues so if somebody else wants to chime in with that or the respective graphs, they're more than welcome to!

Looks great! Remember that $\lambda$ was a parameter of your general solution, which you have narrowed down to
$$u(x)=a\sin(\sqrt{\lambda}x).$$
I guess you just plug the $\lambda$ value that you completed back into the formula for the general solution.

Make sense? If so, you can just edit your answer to include it.

Sorry about that, I went back and edited it. Thank you!

Chiming in on the progress Qkhan has made,

Qkhan gave the general solution with our found $\lambda=[\frac{\pi(2n+1)}{2}]^{2}$ to be,

$$u(x) = a\sin(\sqrt{[\frac{\pi(2n+1)}{2}]^{2}}\,x)$$

Now for the graphs, the three smallest, positive eigenvalues would simply be when $n=0,1,2$ as the eigenvalues just increase as $n$ increases, thus giving $\lambda=(\frac{\pi}{2})^2,(\frac{3\pi}{2})^2,(\frac{5\pi}{2})^2$ and graphs respectively,

Untitled-2.jpg272x514 13 KB

Further if anyone wants to dig deeper into greater values of $n$ here is a neat mathematica code
P.S. I am not responsible for the great amount of fun that may occur as a result

Animate[Plot[Sin[x $(Pi (2 n + 1$)/2)], {x, 0, 1}, PlotRange -> {-1, 1}], {n, 0, 10, 1}, ControlPlacement -> Top]

Qurat, I think you did a great job of explaining this problem step by step. I am however slightly confused on a few things and was wondering if you or anyone else can explain it further to me. I am a very visual learner so seeing all the steps involved really helps me out.

You start of with the statement:

u$x$=asin$λ√x$+bcos$λ√x$
Plugging in
u$0$=0

you're left with something along the lines of
u$0$=b=0

I am a bit confused with this. Is it because λ*0=0, cos$0$=1, and b*1=b? If so I understand that. Then did you relate b to the initial statement of u$0$=0?

since we know that
sin$0$=0

and the entire "b" term goes away meaning
u$x$=asin$λ√x$

The next thing I am confused on is how u$x$=asin$λ√x$. I agree that the sin$0$=0 and I agree that since b=0 that term goes away. But since sin$0$=0, and you are left with a*0, wouldn't that term go away too?

Lastly, I understand your work with taking the derivative, however you lose me when you get to this step:

λ√=arccos$0$
λ√=π$2n+1$/2

Isn't arccos$0$=π/2? I don't see where the $2n+1$ is coming from.

Hi Sam - Maybe I can answer some of your questions:

You've got the idea on this one. Plugging in 0 for x $$u(x) = a\sin(\sqrt{\lambda}\,x) + b\cos(\sqrt{\lambda}\,x)$$ the first term will disappear $because sin(0$=0), while the second term will reduce to b $because cos(0$=1). So all we're left with is $u(0)=b$. Next, because of the given condition $u(0)=0$, we can equate b to 0.

At this point we know that $b=0$, so we're clear on why the second term disappears. However, now we're back dealing with the generic equation. In other words, we aren't plugging in 0 for x anymore. Thus, the first term is left as is.

arccos$0$ does equal $\pi/2$, but there are actually many more solutions. For example, $cos(\pi/2)=0$, but so do $cos(3\pi/2)$ and $cos(5\pi/2)$. We have to account for all of these solutions. Only odd multiples of $\pi/2$ will result in cosine equaling 0, so the term $2n+1$ accounts for this multiple. $n$ represents any integer. No matter what number you plug in to $n$, $2n+1$ will be an odd integer, which when multiplied by $\pi/2$ will result in cosine equaling 0.

Thank you Jacob! Your explanation really helped me out!

This has been confusing me as well. It's been brought up several times but seems to be glossed over. If someone could explain how to find the general solution it would be very much appreciated.

Jacob, thank you so much for explaining it so wonderfully! I was going to but I didn't see Sam's message until now. You took the words right out of my mouth.

@msmather To get the general solution, you recognize that $u''(x) = -{\lambda}\,u$ is an ordinary differential equation. Treating it like an auxiliary equation, you can go through the following steps to show that the general solution is: $u(x) = a\sin(\sqrt{\lambda}\,x) + b\cos(\sqrt{\lambda}\,x)$
$$u'' = -{\lambda}\,u$$
$$u'' + {\lambda}\,u = 0$$
$$m^{2}+{\lambda} = 0$$
You can already start to see that when you do the quadratic formula, this is going to be a complex root.
$$m = \pm\frac{\sqrt{-4{\lambda}}}{2} = \pm\frac{2i\sqrt{\lambda}}{2} = \pm i\sqrt{\lambda}$$
In Diff Eq we learned that when we have the form $c \pm ki$ the general solution will be $u(x) = e^{cx}[a\sin(kx)+b\cos(kx)]$ and since we have no "c" term in our solution solution for m, that's how we arrive at $u(x) = a\sin(\sqrt{\lambda}\,x) + b\cos(\sqrt{\lambda}\,x)$ because $\sqrt{\lambda}$ here is our "k" term in the complex conjugate.

I hope that clears it up. :${}$)

Thanks for the explanation!

I'm still confused on one part - can you explain the step going from $u''+{\lambda}u=0$ to $m^2+\lambda=0$? I haven't had diff eq for about two years now so some of these things are lost on me.

The derivation of $u(x)$ is nice but, to be clear, I don't think that it is essential. Once you have $u(x)$ in your hands, it's easy to verify that it works, which is all you really need.

I believe the term used for this is the "auxiliary equation." We have essentially related the second derivative of u to a scalar m^2, the first derivative of u to m^1 and the function u to m^0.
$$u''+Lambda*u=0$$ becomes $$m^2+Lambda*m^0=0$$
I hope this helps! I still have my ODE notes from last semester if you need a copy.

Thank you for asking this question! I was wondering the exact same thing. @sfrye your answer was very easy to understand and is greatly appreciated.