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Question Number 184307 by Mastermind last updated on 05/Jan/23

Show that the boundary−value problem y′′+λy=0 y(0)=0, y(L)=0 has only the trival solution y=0 for the cases λ=0 and λ<0. let L be a non−zero real number. ?

$$\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{boundary}−\mathrm{value} \\ $$ $$\mathrm{problem}\:\mathrm{y}''+\lambda\mathrm{y}=\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\mathrm{y}\left(\mathrm{0}\right)=\mathrm{0}, \\ $$ $$\mathrm{y}\left(\mathrm{L}\right)=\mathrm{0}\:\mathrm{has}\:\mathrm{only}\:\mathrm{the}\:\mathrm{trival}\:\mathrm{solution} \\ $$ $$\mathrm{y}=\mathrm{0}\:\mathrm{for}\:\mathrm{the}\:\mathrm{cases}\:\lambda=\mathrm{0}\:\mathrm{and}\:\lambda<\mathrm{0}. \\ $$ $$\mathrm{let}\:\mathrm{L}\:\mathrm{be}\:\mathrm{a}\:\mathrm{non}−\mathrm{zero}\:\mathrm{real}\:\mathrm{number}. \\ $$ $$ \\ $$ $$ \\ $$ $$? \\ $$

Answered by FelipeLz last updated on 05/Jan/23

y′′+λy = 0 y = e^r → r^2 e^r +λe^r = 0 r^2 = −λ λ = 0 → r = 0 ∴ y(x) = c y(0) = 0 = c y(x) = 0 λ < 0 → r = ±(√(−λ)) ∴ y(x) = c_1 e^((√(−λ))x) +c_2 e^(−(√(−λ))x) y(0) = 0 → c_1 +c_2 = 0 ∴ y(x) = c(e^((√(−λ))x) −e^(−(√(−λ))x) ) y(L) = 0 → c(e^((√(−λ))L) −e^(−(√(−λ))L) ) = 0 L ≠ 0 → e^((√(−λ))L) −e^(−(√(−λ))L) ≠ 0 ∴ c = 0 y(x) = 0

$${y}''+\lambda{y}\:=\:\mathrm{0} \\ $$ $${y}\:=\:{e}^{{r}} \:\rightarrow\:{r}^{\mathrm{2}} {e}^{{r}} +\lambda{e}^{{r}} \:=\:\mathrm{0} \\ $$ $${r}^{\mathrm{2}} \:=\:−\lambda \\ $$ $$\: \\ $$ $$\lambda\:=\:\mathrm{0}\:\rightarrow\:{r}\:=\:\mathrm{0}\:\therefore\:{y}\left({x}\right)\:=\:{c}\: \\ $$ $${y}\left(\mathrm{0}\right)\:=\:\mathrm{0}\:=\:{c} \\ $$ $${y}\left({x}\right)\:=\:\mathrm{0} \\ $$ $$\: \\ $$ $$\lambda\:<\:\mathrm{0}\:\rightarrow\:{r}\:=\:\pm\sqrt{−\lambda}\:\therefore\:{y}\left({x}\right)\:=\:{c}_{\mathrm{1}} {e}^{\sqrt{−\lambda}{x}} +{c}_{\mathrm{2}} {e}^{−\sqrt{−\lambda}{x}} \\ $$ $${y}\left(\mathrm{0}\right)\:=\:\mathrm{0}\:\rightarrow\:{c}_{\mathrm{1}} +{c}_{\mathrm{2}} \:=\:\mathrm{0}\:\therefore\:{y}\left({x}\right)\:=\:{c}\left({e}^{\sqrt{−\lambda}{x}} −{e}^{−\sqrt{−\lambda}{x}} \right) \\ $$ $${y}\left({L}\right)\:=\:\mathrm{0}\:\rightarrow\:{c}\left({e}^{\sqrt{−\lambda}{L}} −{e}^{−\sqrt{−\lambda}{L}} \right)\:=\:\mathrm{0}\: \\ $$ $${L}\:\neq\:\mathrm{0}\:\rightarrow\:{e}^{\sqrt{−\lambda}{L}} −{e}^{−\sqrt{−\lambda}{L}} \:\neq\:\mathrm{0}\:\therefore\:{c}\:=\:\mathrm{0} \\ $$ $${y}\left({x}\right)\:=\:\mathrm{0} \\ $$ $$ \\ $$

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