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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?

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  • $\begingroup$ The obvious solution is $f(x)=0 $ which is consistant with the boundary conditions $f(1)=0$ and $f'(0)=0$ $\endgroup$ – JJacquelin Dec 4 '14 at 7:31
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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.

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  • $\begingroup$ Are you saying that $f(x)=0$ is the only solution that satisfies the boundary conditions or the only non-negative solution that satisfies the boundary conditions? $\endgroup$ – Garimpeiro Dec 4 '14 at 14:41
  • $\begingroup$ If I am not wrong, only $f(x)=0$ can satisfy the boundary conditions. I gave the general solution and I did not find any $(c_1,c_2)$ other than $0$. $\endgroup$ – Claude Leibovici Dec 4 '14 at 14:45
  • $\begingroup$ Hmmm, my calculations show that for any real $k$, if $f_k(x)=e^x(k\cos(kx)-\sin(kx))$, then $f_k'(0)=0$ and $f_k''(x)-2f_k'(x)=-(1+k^2)f_k(x)$. Also, since $k\cos(k)-\sin(k)$ has infinitely many zeroes, there are infinitely many $k$ such that $f_k(1)=0$. Hence there should be infinitely many solutions that satisfy the boundary conditions. Unfortunately for my purposes, all of them (except $f(x)=0$) appear to change sign at least once. $\endgroup$ – Garimpeiro Dec 4 '14 at 15:02
  • $\begingroup$ I am just feeling stupid ! Look at Julian's answer. I apologize. $\endgroup$ – Claude Leibovici Dec 4 '14 at 15:05
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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 $$\lambda=1+\frac\pi4+k\,\pi,\quad k=0,1,2,\dots$$

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  • $\begingroup$ You are perfectly correct, indeed ! $\endgroup$ – Claude Leibovici Dec 4 '14 at 15:05

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