Question

Consider the differential equation: \[y''+\lambda y = 0\]
For the given boundary conditions, find the eigenvalues and corresponding eigenfunctions. \[y(0)+y'(0)=0,\qquad y(1)=0\]

Answer 1

Assume \(\lambda>0\). The characteristic equation is \(r^2+\lambda=0\), so \(r=\pm i\sqrt\lambda\). Then the characteristic solution is
\[y=C_1\cos\sqrt\lambda~t+C_2\sin\sqrt\lambda~t\]and the boundary conditions give the system
\[\begin{cases}
C_1\cos\sqrt\lambda+C_2\sin\sqrt\lambda=0&\color{lightgray}{y(1)=0}\\[1ex]
C_1+C_2\sqrt\lambda=0&\color{lightgray}{y(0)+y'(0)=0}
\end{cases}\]Solving both equations for \(\dfrac{C_1}{C_2}\), you get
\[\sqrt\lambda=\tan\sqrt\lambda\]so any eigenvalue \(\lambda>0\) satisfies this equation (infinitely many of these).
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Now assume \(\lambda=0\). The solution would then be
\[y=C_1+C_2t\]and the boundary conditions require
\[\begin{cases}C_1+C_2=0\\[1ex]C_1+C_2=0\end{cases}\]so there's an infinite number of linear solutions with \(C_1+C_2=0\).
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Now assume \(\lambda<0\), so the characteristic equation has roots \(r=\pm\sqrt\lambda\) and the solution becomes
\[y=C_1\cosh\sqrt\lambda~t+C_2\sinh\sqrt\lambda~t\]with the boundary conditions requiring that
\[\begin{cases}
C_1\cosh\sqrt\lambda+C_2\sinh\sqrt\lambda=0\\[1ex]
C_1+\sqrt\lambda~C_2=0\end{cases}\]Similar to the case where \(\lambda>0\), negative eigenvalues will be the negative roots to \(x-\tanh x\). This only has one root at \(x=0\), so there are no negative eigenvalues satisfying these boundary conditions.
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I'd have to look into some notes on BVPs to figure out the part about finding the eigenfunctions. Hope this was at least somewhat helpful.

Answer 2

i reckon you're crossing into Sturm-Liouville.

@Kainui

Answer 3

\[\ \newcommand\ps[1]{\left(#1\right)}}\]
\begin{align*}
y''+\lambda y &= 0
&y(0) + y'(0) = 0,\qquad
y(1) = 0\\[2ex]
m^2+\lambda &= 0
\end{align*}
\(\textbf{Case I}\)
\(\lambda = 0\), \(m=0,0\)
\begin{align*}
y &= A+Bx
& y(1)&=A+B=0\\
&& &B=-A\\
&= A(1-x)
& y(0)&=A\\[1ex]
y' &= -A
& y'(0)&=-A\\[1ex]
&& y'(0)+y(0)&=A-A=0\\[2ex]
y_0 &= 1-x
& \lambda_0&=0 \qquad n=0
\end{align*}
\(\textbf{Case II}\)
\(\lambda = -\mu^2 < 0\), \(m=\pm\mu\)
\begin{align*}
y &= Ae^{\mu x}+Be^{-\mu x}
& y(0) &=A+B\\[1ex]
y' &= A\mu e^{\mu x}-B\mu e^{-\mu x}
& y'(0)&=(A-B)\mu\\[2ex]
&& y(0)+y'(0)&=A+B+(A-B)\mu = 0\\
&& &A+B = 0\qquad
A-B = 0\\
&& &A = 0\qquad
B = 0\\
y &= 0
\end{align*}
\(\textbf{Case III}\)
\(\lambda = \mu^2 > 0\), \(m=\pm i\mu\)
\begin{align*}
y &= A\cos\mu x+B\sin\mu x
& y(1) &=A\cos\mu+B\sin\mu=0 \\
&& &B= -A\cot\mu\\
&= A\ps{\cos\mu x-\cot\mu\sin\mu x}
&y(0) &= A\\[1ex]
y' &= -A\mu\ps{\sin\mu x+\cot\mu\cos\mu x}
& y'(0)&= -A\mu\cot\mu\\[1ex]
&& y(0)+y'(0)&= A-A\mu\cot\mu=0\\[2ex]
y &= A\ps{\cos\mu_n x-\tfrac1{\mu_n}\sin\mu_n x}
&&\mu_n = \tan\mu_n\qquad
n = 1,2,3\dots\\[2ex]
y_n &= \cos\mu_n x-\tfrac1{\mu_n}\sin\mu_n x
&\lambda_n &= \mu_n^2
\end{align*}

Answer 4

\[\mu_n= 4.493, 7.725,10.904, 14.066, 17.221, \dots \]