458 Answers to Odd-Numbered Exercises
In the integral forbn, break the interval of integration ata; in the second
integral, make the change of variabley= 2 a−x. The two integrals cancel
ifnis even, and the coefficient is the same as Eq. (18) ifnis odd.
b. In the solution of Eqs. (1)–(4), the eigenfunctionφ(x)=sin(( 2 n−
1 )πx/ 2 a)has the propertyφ( 2 a−x)=φ(x),sothesumoftheseries
has the same property. This implies 0 derivative atx=a.15.W(t)=C 0 LA[
1 −
∑∞
n= 12 e−(λ^2 nDt)
(n− 1 / 2 )^2 π^2]
.
Section 2.6
- The graph ofv(x)is a straight line fromT 0 atx=0toT∗atx=a,where
T∗=T 0 +
ha
k+ha(T^1 −T^0 ).
In all cases,T∗is betweenT 0 andT 1.- Negative solutions provide no new eigenfunctions.
7.bm=λ^2 (^1 −cos(λma))
m[a+(κ/h)cos^2 (λma)].
9.bm=−^2 (κ+ah)cos(λma)
λm(ah+κcos^2 (λma)).
Section 2.7
1.λn=nπ/ln 2,φn=sin(λnln(x)).- a. sin(λnx),λn=( 2 n− 1 )π/ 2 a;
b. cos(λnx),λn=( 2 n− 1 )π/ 2 a;
c. sin(λnx),λnasolutionoftan(λa)=−λ;
d.λncos(λnx)+sin(λnx),λna solution of cot(λa)=λ;
e.λncos(λnx)+sin(λnx),λnasolutionoftan(λa)= 2 λ/(λ^2 − 1 ). - The weight functions in the orthogonality relations and limits of integra-
tion are:
a. 1+x,0toa;b.ex,0toa;c.x^12 ,1to2; d.ex,0toa. - Becauseλappears in a boundary condition.
- The negative value ofμdoes not contradict Theorem 2 because the coef-
ficientα 2 is not positive.