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For the boundary value problem

$$y''+\lambda y=0,\ \ y(0)-y'(0)=0,\ \ y(1)=0$$

I believe only the trivial solution applies when $\lambda =0$. When $\lambda <0$ I get solutions of the form $c_1e^{-\sqrt{-\lambda}t}+c_2e^{\sqrt{-\lambda}t}$ and applying the boundary conditions,

$$c_1(1+\sqrt{-\lambda})+c_2(1-\sqrt{-\lambda})=0$$

$$c_1e^{-\sqrt{-\lambda}}+c_2e^{\sqrt{-\lambda}}=0$$

The second implies

$$c_1=-c_2e^{2\sqrt{-\lambda}}$$

which we can substitute into the first equation to get

$$1-\sqrt{-\lambda}=(1+\sqrt{-\lambda})e^{2\sqrt{-\lambda}}$$

I see no clear path to a solution for $\lambda$ but I do remark that the right-hand side is positive hence we require

$$1-\sqrt{-\lambda} > 0$$

so that $0>\lambda > -1$. That all such $\lambda$ would allow for a solution seems unlikely to me, but I can see no further solution method, except perhaps by using technology to plot the functions of $\lambda$ and find intersections. Is there a more satisfying solution?

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  • $\begingroup$ You can write the solutions to $\lambda = -k^{2}$ and $\lambda = k^{2}$ as $$y = A \cosh kx + B \sinh kx$$ and $$y = A \cos kx + B \sin kx$$ respectively. Applying your boundary conditions in both cases leads to non-trivial solutions provided $$k + \tanh k = 0$$ and $$k + \tan k = 0$$ respectively. $\endgroup$
    – mattos
    Nov 17, 2017 at 6:42
  • $\begingroup$ Are you sure that the second condition is $y(1)=0$ and not something else ? $\endgroup$ Nov 17, 2017 at 7:54
  • $\begingroup$ Do you know Laplace transform? $\endgroup$ Nov 17, 2017 at 8:22
  • $\begingroup$ @Claude Leibovici Yep, the given problem is correct. $\endgroup$
    – Addem
    Nov 18, 2017 at 4:47
  • $\begingroup$ @Jan Eerland I do but I'm not able to use them here. $\endgroup$
    – Addem
    Nov 18, 2017 at 4:48

1 Answer 1

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Negative $\lambda$

Following the comment by Mattos, let $\lambda=-k^2$ where $k>0$. The general solution of the ODE $y''=k^2y$ is $A\cosh kx +B\sinh kx$; this form is equivalent to $C_1e^{kx}+C_2e^{-kx}$ but is usually easier to handle with regard to initial conditions. The initial condition $y'(0)=y(0)$ yields $kB = A$, hence $$y(x) = B(k\cosh kx + \sinh kx)$$ The boundary condition $y(1)=0$ requires $B=0$, since $k\cosh k + \sinh k>0$. So, the only solution is the trivial one $y\equiv 0$.

Positive $\lambda$

Let $\lambda=k^2$ where $k>0$. The general solution of the ODE $y''=-k^2y$ is $A\cos kx +B\sin kx$. The initial condition $y'(0)=y(0)$ yields $kB = A$, hence $$y(x) = B(k\cos kx + \sin kx)$$ The boundary condition $y(1)=0$ requires $B=0$ (trivial solution), unless $k\cos 1 + \sin k=0$. So, nontrivial solution exists only when $\tan k = -k$. There are infinitely many such numbers $k$ (the line $y=-x$ intersects the graph of tangent infinitely many times), but there is no analytic formula for them.

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