# A solve this eigenvalue problem hint the solution

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Unformatted text preview: ﬁes the boundary condition is f = 0, the trivial solution. Case 2: λ &gt; 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 &lt; 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.

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