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I'm trying to determine if I did this problem properly.

Give the following ODE $y''-2 y'+(\lambda-1)y=0$ with $y(0)=0=y(1)$.

Find the eigenvalues and eigen-functions.

To do this, I rewrote

$$r^2-2r+1=1-\lambda+1 \Rightarrow r=1 \pm \sqrt{2-\lambda}.$$

I reduced this to three cases: $\lambda <2,\lambda =2, \lambda>2$.

Case 1: $\lambda<2$.

In this case we have $$y=C_1e^{(1+\sqrt{2-\lambda})x}+C_2e^{(1-\sqrt{2-\lambda})x}$$.

Using $y(0)=0$ the initial conditions I have $C_1=-C_2$. Using the second initial condition of $y(1)=0$, we now have

$$0=-C_2e^{(1+\sqrt{2-\lambda})}+C_2e^{(1-\sqrt{2-\lambda})}$$.

Some algebra later, I get $\lambda =2$.

Case 2: $\lambda=2$

In this case we have $y=C_1e^x+C_2xe^x$. From this, we get the trivial solution.

Case 3: $\lambda > 2$

In this case I have

$$y=C_1e^x\cos((\sqrt{2-\lambda}x))+C_2e^x\sin((\sqrt{2-\lambda}x)).$$

With $0=y(0)=C_1.$ Also, $0=y(1)=C_2e\sin((\sqrt{2-\lambda})).$

Moving things around we have

$$\arcsin(0)=\sqrt{2-\lambda} \Rightarrow \pi k=\sqrt{2-\lambda}=(\pi k)^2=2-\lambda\Rightarrow 2-(\pi k)^2=\lambda$$.

This leaves us $\lambda_k=c_k\sin(\pi k)$.

So our eigenvalues are $\lambda=2-(\pi k)^2$ and our eigen-functions are:

$$y_k=c_k\sin(\pi k)$$

where $k=0,1,\dots$.

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  • $\begingroup$ hope ok for readability. $\endgroup$ – Narasimham Dec 2 '16 at 19:07
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One way to save a little effort by eliminating special cases is to solve for the unique $y$ such that $y(0)=0,y'(0)=1$, and then plug in at $x=1$ to find the eigenvalue equation. In this case, $$ y(x)=C_1e^{(1+\sqrt{2-\lambda})x}+C_2e^{(1-\sqrt{2-\lambda})x} $$ may be written as $$ y(x)=D_1 e^{x}\sin(\sqrt{\lambda-2}x)+D_2e^{x}\cos(\sqrt{\lambda-2}x), $$ which makes it easy to solve $y(0)=0$, $y'(0)=1$. Automatically $D_2=0$, and $y'(0)=1$ gives $$ y(x)=e^{x}\frac{\sin(\sqrt{\lambda-2}x)}{\sqrt{\lambda-2}}. $$ This eliminates special cases because taking the limit as $\lambda\rightarrow 2$ also gives the correct solution for $\lambda=2$ as $y(x)=xe^x$. Now the eigenvalue equation becomes $$ y(1)=0\implies \frac{\sin(\sqrt{\lambda-2})}{\sqrt{\lambda-2}}=0 \tag{$\dagger$} $$ When you do it this way, you always end up with a power or McClaurin series in $\lambda$ for the eigenvalue equation, which eliminates special cases such as $\lambda=2$. For example, in this case, $$ 0 = \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}(\lambda-2)^{n}. $$ But you don't have to write out the series. Just know that there is one, and the special cases are limiting cases. You can see from $(\dagger)$ that the limiting case of $\lambda=2$ does not work, so that one is eliminated. The special case at $\lambda=2$ is no longer a special case. The other solutions are $\sqrt{\lambda-2}=n\pi$ for $n\ne 0$.

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  • $\begingroup$ Can you explain why you chose the first rewriting, and how you eliminated the case $\lambda>2$? (Also, the OP's other boundary condition was $y(1)=0$, not $y'(0)=1$). $\endgroup$ – rogerl Dec 17 '16 at 13:10
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    $\begingroup$ @rogerl : $\sqrt{\lambda -2} = i\sqrt{2-\lambda}$. And $\sinh(iz)=i\sin(z)$. So you can always write in terms of sin and cos or in terms of sinh and cosh instead. It generally pays to use sin and cos because it is almost never that an exponential or sinh or cosh term can be an eigenfunction satisfying two homogeneous endpoint conditions. There are exceptions, but the conditions have to be mixed, and you won't miss this using the sin and cos representations anyway. You know that all zeros of sin and cos are real, and you know what they are. $\endgroup$ – DisintegratingByParts Dec 17 '16 at 16:46

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