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Unformatted text preview: ﬁes the boundary condition is f = 0, the trivial
solution.
Case 2: λ > 0. In this case, the diﬀerential equation
f (x) = λf (x), or f (x) − λf (x) = 0 has the general solution
√ f (x) = c1 e λx √ + c2 e− λx . It is convenient for us to change basis in the linear space of solutions, using
√
√
1√
cosh( λx) = (e λx + e− λx ),
2 √
√
1√
sinh( λx) = (e λx − e− λx )
2 instead of the exponentials. Then we can write
√
√
f (x) = a cosh( λx) + b sinh( λx),
87 with new constants of integration a and b. We impose the boundary conditions:
ﬁrst
√
f (0) = 0 ⇒ a = 0 ⇒ f (x) = b sinh( λx),
and then
f (L) = 0 √
b sinh( λL) = 0 ⇒ ⇒ ⇒ b=0 f (x) = 0, so we obtain no nontrivial solutions in this case.
√
Case 3: λ < 0. In this case, we set ω = −λ, and rewrite the eigenvalue
problem as
f (x) + ω 2 f (x) = 0, f (0) = 0 = f (L).
We recognize here our old friend, the diﬀerential equation of simple harmonic
motion. We remember that the diﬀerential equation has the general solution
f (x) = a cos(ωx) + b sin(ωx).
Once again
f (0) = 0 ⇒ a=0 ⇒ f (x) = b sin(ωx). b sin(ωL)...
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This document was uploaded on 01/12/2014.
 Winter '14
 Equations

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