Introduction 31
Applying the boundary conditions, we have
y(1) = 0 =c 1 ⇒c 1 =0
y(e)=0=c 2 sinkand the latter equation forcesc 2 =0,except for those values ofksatisfying
sink=0;thatis,c 2 is arbitrary when
k=π, 2 π, 3 π,...or
λ=−π^2 ,− 4 π^2 ,− 9 π^2 ,....
Therefore, the eigenvalues are
λn=−n^2 π^2 ,n=1, 2 , 3 ,...,and the associated eigenfunctions are
yn= sin(nπlnx),n=1, 2 , 3 ,....Example 4Do the same for
y′′+λy=0
y(0) =y(1) +y′(1) = 0.Case 1: λ< 0 ,λ=−k^2 ,k> 0
We have
y=c 1 coshkx+c 2 sinhkx.
Then,
y′=c 1 ksinhkx+c 2 kcoshkx
and, applying the boundary conditions, we have
y(0) = 0 =c 1
y(1) = 0 =c 2 (sinhk+kcoshk).So we must havec 2 = 0, except for those values ofk>0 satisfying
sinhk+kcoshk=0.Essentially, then, we wish to find all positive roots of the function
f(x)=sinhx+xcoshx.Now,f(0) = 0. Forx>0, let’s considerf′:
f′(x)=2coshx+xsinhx>0forx> 0.