2
$\begingroup$

I am trying to solve $$J''(\tau) = h (J - B) \tag{1} $$ where $h$ is a positive real constant and $B$ is a real-valued, smooth and otherwise well-behaved function for $\tau \in (0, \infty)$. I have the boundary conditions

$$J(0) = 0 \\ \lim_{\tau \to \infty} J = \lim_{\tau \to \infty} B \equiv B_{\infty} \tag{2}$$ I'm using this equation in a physical model, and I have reason to expect $B_\infty$ to be positive (nonzero).

Here's my attempt at a solution:

A solution basis for the homogeneous equation is $J_1 = e^{\sqrt{h} \tau}$ and $J_2 = e^{-\sqrt{h} \tau}$.

The Wronskian evaluates to $-2\sqrt{h}$, a constant.

Then I use variation of parameters to write the general solution as

$$J = e^{\sqrt{h} \tau}\left[\int_{\tau}^{\infty} \frac{h \,B(s)}{2\sqrt{h}} e^{-\sqrt{h} s}ds + C_1\right] +e^{-\sqrt{h} \tau}\left[\int_0^{\tau} \frac{h \,B(s)}{2\sqrt{h}} e^{\sqrt{h} s}ds + C_2\right] \tag{3}$$

I have tried my best to get the signs right above, although there's still a chance I got it wrong, so please let me know if you spot something that's not quite right. In the first integral, I put $\tau$ as the lower limit, which canceled a minus sign out front. And the minus signs in the Wronskians cancel with the minus signs in front of $h B$.

To prevent the solution from diverging as $\tau \to \infty$, $C_1$ must be zero.

I have been able to use the $\tau = 0$ boundary condition to derive an expression for $C_2$.

But here's the problem. Consider once again the $\tau \to \infty$ boundary condition. If I plug in $C_1 = 0$ into the general solution (equation 3 above), and drop all terms that clearly vanish, I am left with

$$\lim_{\tau \to \infty} J = \lim_{\tau \to \infty} e^{-\sqrt{h} \tau} \int_0^{\tau} \frac{h \,B(s)}{2\sqrt{h}} e^{\sqrt{h} s}ds \tag{4}$$

Then, using L'Hospital's rule,

$$\lim_{\tau \to \infty} J = \frac{\frac{h \,B(\tau)}{2\sqrt{h}} e^{\sqrt{h} \tau}}{\sqrt{h} e^{\sqrt{h} \tau}} \tag{5} $$

which simplifies to

$$ \lim_{\tau \to \infty} J = \frac{1}{2} \lim_{\tau \to \infty} B \tag{6} $$

It seems that the only way that the second boundary condition (second line of equation 2) can be satisfied is if $B_{\infty} = 0$. But as I noted above, on physical grounds I expect $B_{\infty} = 0$ to be able to take arbitrary positive values. Have I done something wrong, or is it really the case that $B_{\infty}$ must equal zero in this situation?

$\endgroup$
2
$\begingroup$

I found my problem, finally. When going between equation 3 and equation 4, there are actually two terms that don't vanish, not just one, as $\tau \to \infty$ (the two integral terms). L'Hospital is needed both times to handle the indeterminate forms, and they both evaluate to $1/2 B_{\infty}$. So they sum to $B_{\infty}$ and all is consistent.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.