Partial derivaives
𝐹 ′′ (𝑥)−𝑤 2 𝐹(𝑥) =0
We make the ansatz 12 that 𝐹(𝑥)= 𝑒 𝑘𝑥 for some k, this gives:
𝑘 2 𝑒 𝑘𝑥 −𝑤 2 𝑒 𝑘𝑥 =0
𝑒 𝑘𝑥 (𝑘 2 −𝑤 2 )=0
Since 𝑒 𝑘𝑥 >0 for all kx therefore:
𝑘 2 −𝑤 2 =0
𝑘 =±𝑤
∴ 𝐹(𝑥) = 𝑒 𝑤𝑥 𝑎𝑛𝑑 𝐹(𝑥) = 𝑒 −𝑤𝑥
This can be generalized for 𝐹(𝑥) = 𝐴𝑒 𝑤𝑥 and 𝐹(𝑥) = 𝐵𝑒 −𝑤𝑥 for some constants A, B. Since this is a linear second order differential equation, by principle of superposition:
𝐹(𝑥) = 𝐴𝑒 𝑤𝑥 +𝐵𝑒 −𝑤𝑥
Now we rewrite 𝐹(𝑥) in terms of the hyperbolic functions, since:
𝑒 𝑤𝑥 −𝑒 −𝑤𝑥 2
𝑒 𝑤𝑥 +𝑒 −𝑤𝑥 2
sinh(𝑤𝑥) =
,
cosh(𝑤𝑥) =
∴ 2 sinh(𝑤𝑥) = 𝑒 𝑤𝑥 −𝑒 −𝑤𝑥 ,
2 cosh(𝑤𝑥) = 𝑒 𝑤𝑥 +𝑒 −𝑤𝑥
∴ sinh(𝑤𝑥)+cosh(𝑤𝑥)=𝑒 𝑤𝑥 ,
cosh(𝑤𝑥) − sinh(𝑤𝑥) = 𝑒 −𝑤𝑥
This gives:
𝐹(𝑥) = 𝐴(sinh(𝑤𝑥) + cosh(𝑤𝑥)) + 𝐵(cosh(𝑤𝑥) − sinh(𝑤𝑥))
𝐹(𝑥) = 𝐴sinh(𝑤𝑥) +𝐵cosh(𝑤𝑥) 𝑓𝑜𝑟 𝑠𝑜𝑚𝑒 𝑛𝑒𝑤 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡𝑠 𝐴,𝐵
However, the boundary conditions require that:
𝐴sinh(0) + 𝐵cosh(0) = 0 ,
𝐴 sinh(𝑤𝐿) + 𝐵 cosh(𝑤𝐿) = 0
Since sinh(0)=0 and cosh(0)=1 this gives 𝐵 =0 . Therefore:
𝐴 sinh(𝑤𝐿) = 0
However, sinh(𝑥) = 0 only when 𝑥 =0 , therefore 𝐴 =0 , and so again we get a trivial solution for the case 𝜆 =−𝑤 2 , thus we move onto the third case.
(𝑖𝑖𝑖) 𝜆 = 𝑤 2 (𝑤 > 0 𝑤𝑙𝑜𝑔)
𝐹 ′′ (𝑥)+𝑤 2 𝐹(𝑥) =0
Like in the first case, we make the ansatz that 𝐹(𝑥)= 𝑒 𝑘𝑥 for some k, this gives:
𝑘 2 𝑒 𝑘𝑥 +𝑤 2 𝑒 𝑘𝑥 =0
𝑘 2 +𝑤 2 =0
12 Bronson and Costa (2014), 83-84.
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