finding positive solution to eigenvalue problem

Question

I'm trying to find the unique (up to a constant factor) positive solution to the eigenvalue problem: $$f''(x)-2f'(x)=-\lambda f(x) \ , \lambda>0 ,$$ with boundary conditions $f(1)=0$ and $f'(0)=0$.

I was able to solve the similar problem $f''(x)-f'(x)=-\lambda f(x)$ with the same boundary conditions.

In that case $f(x)=e^{\frac{x}{2}}(k\cos(kx)-\frac{1}{2}\sin(kx))$ where $k\approx 1.166$ is the smallest positive solution of $k\cos(k)-\frac{1}{2}\sin(k)=0$ and $\lambda=\frac{1}{4}(1+4k^2)$.

The analogous solution for the case with the $-2f'(x)$ term is $f(x)=e^x(k\cos(kx)-\sin(kx))$. However, there doesn't seem to be a suitable value of $k$ that satisfies the boundary conditions and makes $f$ a positive (or negative) function.

Any suggestions on how to proceed?

Answer 1

If you consider the differential equation $$f''(x)-2f'(x)+\lambda f(x) =0$$ and apply the standard methods, you should obtain $$f(x)=c_1 e^{\left(1-\sqrt{1-\lambda }\right) x}+c_2 e^{\left(1+\sqrt{1-\lambda }\right) x}$$ $$f'(x)=c_1 \left(1-\sqrt{1-\lambda }\right) e^{\left(1-\sqrt{1-\lambda }\right) x}+c_2 \left(1+\sqrt{1-\lambda }\right) e^{\left(1+\sqrt{1-\lambda }\right) x}$$ Now, apply the conditions $$f(1)=c_1 e^{1-\sqrt{1-\lambda }}+c_2 e^{1+\sqrt{1-\lambda }}=0$$ $$f'(0)=c_1 \left(1-\sqrt{1-\lambda }\right)+c_2 \left(1+\sqrt{1-\lambda }\right)=0$$ the only solution of which corresponding to $c_1=c_2=0$.

This just shows that the obvious solution $f(x)=0$, as already pointed out by JJacquelin, is the solution.

Answer 2

Claude's answer does not take into account the possiblity $\lambda>1$. Then the solutions are $$ f(x)=A\,e^x\cos((\lambda-1)x)+B\,e^x\sin((\lambda-1)x). $$ Imposing the boundary conditions we get $$ A\cos(\lambda-1)+B\sin(\lambda-1)=0,\quad A+B=0, $$ from where $$ \cos(\lambda-1)=\sin(\lambda-1). $$ Solutions are $$\lambda=1+\frac\pi4+k\,\pi,\quad k=0,1,2,\dots$$