# Definition of the eigenfunction?

1. Mar 11, 2009

### thepioneerm

I need another Definition of the eigenfunction

and the book the include this Definition

I found in SCHAUM’S DIFFERENTIAL EQUATIONS that:

Those values of Lamda for which nontrivial solutions do
exist are called eigenvalues; the corresponding
nontrivial solutions are called eigenfunctions.

2. Mar 11, 2009

### HallsofIvy

Well, that's pretty much the definition you will find in any book. I don't know exactly what your question is. Perhaps some examples would help.

The problem $d^2y/dx^2+ \lambda y= 0$ with boundary conditions y(0)= 0, y(1)= 0 has the "trivial solution" y(x)= 0 for all x. Does it have any other, "non-trivial", solutions?

The way we write the general solution depends on $\lambda$. If, for example, $\lambda= 0$, the equation is $d^2y/dx^2= 0$ and integrating twice,$y= C_1x+ C_2$. Then $y(0)= C_2= 0$ and $y(1)= C_1+ C_2= C_1= 0$ so both constants are 0 and y(x)= 0 for all x, the trivial solution. 0 is NOT an eigevalue.

If $\lambda$ is negative then we can simplify the problem by writing $\lambda= -\alpha^2$ where $\alpha$ is any non-zero number. The equation becomes $d^2y/dx^2- \alpha^2 y= 0$. That has characteristic equation $r^2- \alpha= 0$ which has roots $r= \alpha$ and $r= -\alpha$ so the general solution is $y(x)= C_1e^{\alpha x}+ C_2e^{-\alpha x}$ so $y(0)= C_1+ C_2= 0$ and $y(1)= C_1e^{\alpha}+ C_2e^{-alpha}= 0$. From the first equation, $C_2= -C_1$. Putting that into the second equation and factoring out $C_1$, we have $C_1(e^{\alpha}- e^{-\alpha}= 0$. Because $\alpha$ is not 0, $\alpha$ and $-\alpha$ are different and, since $e^x$ is a "one-to-one" function, $e^\alpha$ and $e^{-\alpha}$ are different. $(e^{\alpha}- e^{-\alpha}$ is not 0 so $C_1$ must be 0 which means $C_2= -C_1$ is also zero. That is $y(0)= 0(e^{\alpha x})+ 0(e^{-\alpha x})= 0$ for all x. Again, y(x)= 0 is the only solution so NO negative number is an eigenvalue.

If $\lambda$ is positive, we can write $\lambda= \alpha^2$ so the equation becomes $d^2y/dx^2+ \alpha^2y= 0$ which has characteristic equation $r^2- \alpha^2$ with solutions $r= i\alpha$ and $r= -i\alpha$ so the general solution is of the form $y(x)= C_1 cos(\alpha x)+ C_2 sin(\alpha x)$. Now, since cos(0)= 1 and sin(0)= 0, $y(0)= C_1= 0$. $y(1)= C_2 sin(\alpha)= 0$. If $sin(\alpha)$ is not 0, then $C_2$ is 0 and again we get y(x)= 0 for all x. But sine is 0 for any multiple of $\pi$, so if $\alpha$ is, say, $n\pi$, $C_2$ does NOT have to be 0: $y(x)= C sin(n\pi x)$ satisfies both the differential equation and the boundary conditions. $n\pi$ is an eigenvalue for for any integer n. The corresponding eigenfunctions are $y(x)= C sin(n\pi x)$ for C any non-zero number.

Notice by the way, that all of this depends not just on the equation but also on the boundary conditions. Also, as I said, y(x)= 0 for all x always satifies this equation so being an eigenvalue requires that the solution NOT be unique.

3. Mar 11, 2009

### thepioneerm

HallsofIvy

thank you very very much