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I am trying to find all the eigenvalues and eigenfunctions for the following boundary value problem \begin{eqnarray} \phi''(z) + \phi'(z) + \lambda \phi(z) &=& 0\\ \phi (0)&=& 0 \\ |\phi (\infty)| &<& \infty. \\ \end{eqnarray} The characteristic polynomial equation is $r^2 + r + \lambda$. The roots are easily obtained as $r_{1,2}=\dfrac{-1 \pm \sqrt{1 - 4\lambda}}{2}$.

Case $\lambda < \frac{1}{4}$ ($r_1$ and $r_2$ are distinct real roots)

Then the general solution to the differential equation is then, $$\phi(z) = c_1\exp\left[\left(\frac{-1 + \sqrt{1 - 4\lambda}}{2}\right)z\right] + c_2\exp\left[\left(\frac{-1 - \sqrt{1 - 4\lambda}}{2}\right)z\right].$$ Applying the first boundary condition gives us, $$\phi \left(0\right) = c_1 + c_2.$$ Case $\lambda = 0$ ($r_1 = r_2$)

Then the general solution to the differential equation is then, $$\phi(z) = c_1\exp\left[\frac{-z}{2}\right] + c_2 z\exp\left[\frac{-z}{2}\right].$$ Applying the first boundary condition gives us, $$\phi \left(0\right) = c_1.$$ Case $\lambda > \frac{1}{4}$ ($r_1$ and $r_2$ are complex roots) $$\phi(z) = c_1\exp\left[\frac{-z}{2}\right] \cos (z\frac{\sqrt{4\lambda -1}}{2}) + c_2 \exp\left[\frac{-z}{2}\right]\sin (z\frac{\sqrt{4\lambda -1}}{2}). $$ Applying the first boundary condition gives us, $$\phi \left(0\right) = c_1.$$

As suggested by others, in the comments the cases of interest are

  1. $\lambda < -\frac{1}{4}$ (distinct real roots case).
  2. $\lambda = -\frac{1}{4}$ (distinct real roots case).
  3. $-\frac{1}{4} < \lambda < 0$ (distinct real roots case).
  4. $\lambda = 0$ (distinct real roots case).
  5. $0 < \lambda < \frac{1}{4}$ (distinct real roots case).
  6. $\lambda = \frac{1}{4}$ (same real roots case).
  7. $\lambda > \frac{1}{4}$ (complex roots case).

I am not sure how to use the second boundary condition.

Any help will be greatly appreciated. Thank you.

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    $\begingroup$ I would recommend looking at the cases $\lambda<\frac{1}{4}$, $\lambda=\frac{1}{4}$, $\lambda>\frac{1}{4}$. Theses are the locations where you will have changes in behavior. $\endgroup$ – Aweygan Jul 12 '16 at 23:34
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    $\begingroup$ Actually, assuming $\lambda$ is real, the cases of interest are $\lambda<0,\lambda=0,0<\lambda<1/4,\lambda=1/4,\lambda>1/4$. $0$ is where the real part of one of the eigenvalues changes sign; $1/4$ is where they collide with each other. The change of sign is quite important for the second boundary condition. $\endgroup$ – Ian Jul 13 '16 at 1:06
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Case 1: $\lambda<0$. Then one of the independent solutions blows up at infinity. Thus its coefficient must be zero in order for the right boundary condition to be satisfied. But then the other must also be zero by the left condition. So there are no eigenfunctions.

Case 2: $\lambda>0$. The forms of the (real) solutions change as $\lambda$ passes through $1/4$, but despite this, both of the independent solutions go to zero at infinity. So the right boundary condition is automatically satisfied, and only the left boundary condition constrains the possible solution. Noting that it doesn't happen that both independent solutions are zero (so the left boundary condition is never trivial), there is a 1D subspace of eigenfunctions.

Case 2a: $0<\lambda<1/4$. In this case we can choose a solution basis given by two both real exponentials $e^{r_1 x},e^{r_2 x}$, and their coefficients must sum to zero by the left BC. So the eigenfunctions are $C(e^{r_1 x}-e^{r_2 x})$

Case 2b: $\lambda=1/4$. In this case we can choose a solution basis consisting of an exponential $e^{rx}$ and $xe^{rx}$ (where $r=r_1=r_2$). The first one does not vanish at zero while the second does, so the eigenfunctions are multiples of the second one, i.e. $Cxe^{rx}$.

Case 2c: $\lambda>1/4$. In this case we can choose a solution basis where one solution has a cosine and the other has a sine (neither with a phase shift). The one with a sine in it vanishes at zero, and the one with a cosine doesn't, so the eigenfunctions are multiples of the sine solution, i.e. $Ce^{Re(r_1)x}\sin(Im(r_1)x)$ (or $r_2$, it doesn't matter).

I'll leave $\lambda=0$ to you.

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  • $\begingroup$ Firstly thank you for your answer. Just wondering if it is not too much trouble, do you mind actually adding the general form of the eigenvalues and eignenfunctions for the case $\lambda > 0$ ? $\endgroup$ – Comic Book Guy Jul 13 '16 at 4:26
  • $\begingroup$ @ComicBookGuy I described the eigenfunctions. The eigenvalues are just the corresponding values of $\lambda$. $\endgroup$ – Ian Jul 13 '16 at 4:33

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