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Definition of the eigenfunction?

  1. Mar 11, 2009 #1
    Please:

    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. jcsd
  3. Mar 11, 2009 #2

    HallsofIvy

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    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 [itex]d^2y/dx^2+ \lambda y= 0[/itex] 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 [itex]\lambda[/itex]. If, for example, [itex]\lambda= 0[/itex], the equation is [itex]d^2y/dx^2= 0[/itex] and integrating twice,[itex]y= C_1x+ C_2[/itex]. Then [itex]y(0)= C_2= 0[/itex] and [itex]y(1)= C_1+ C_2= C_1= 0[/itex] so both constants are 0 and y(x)= 0 for all x, the trivial solution. 0 is NOT an eigevalue.

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

    If [itex]\lambda[/itex] is positive, we can write [itex]\lambda= \alpha^2[/itex] so the equation becomes [itex]d^2y/dx^2+ \alpha^2y= 0[/itex] which has characteristic equation [itex]r^2- \alpha^2[/itex] with solutions [itex]r= i\alpha[/itex] and [itex]r= -i\alpha[/itex] so the general solution is of the form [itex]y(x)= C_1 cos(\alpha x)+ C_2 sin(\alpha x)[/itex]. Now, since cos(0)= 1 and sin(0)= 0, [itex]y(0)= C_1= 0[/itex]. [itex]y(1)= C_2 sin(\alpha)= 0[/itex]. If [itex]sin(\alpha)[/itex] is not 0, then [itex]C_2[/itex] is 0 and again we get y(x)= 0 for all x. But sine is 0 for any multiple of [itex]\pi[/itex], so if [itex]\alpha[/itex] is, say, [itex]n\pi[/itex], [itex]C_2[/itex] does NOT have to be 0: [itex]y(x)= C sin(n\pi x)[/itex] satisfies both the differential equation and the boundary conditions. [itex]n\pi[/itex] is an eigenvalue for for any integer n. The corresponding eigenfunctions are [itex]y(x)= C sin(n\pi x)[/itex] 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.
     
  4. Mar 11, 2009 #3
    HallsofIvy

    thank you very very much
     
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