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I'm trying to solve the following differential equation

$$ y'' + \dfrac{1}{2}(y')^2 = A \delta(x) + B \delta(x-a) + C $$

I tried two times, the first one using Laplace transforms, but I don't really know how to deal with the $(y')^2$ term. I found some papers discussing it, but I couldn't really use them since they overcomplicate my original equation. The second attempt has been to find a particular solution to add to the homogeneous one give by $y=\dfrac{1}{c_1+x}$, but I have no clue for where to start.

Does anyone have an idea?

Thank you all!

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    $\begingroup$ This is a Riccati equation. Call the RHS $f(x)$ and let $y'=\frac{2w'}{w}$, then your equation is equivalent to the linear equation $w''(x)=\frac{f(x)}{2}w(x)$ which may be solved directly $\endgroup$
    – Sal
    Mar 1 at 22:14
  • $\begingroup$ @Sal I'm not sure this change of variable is legitimate in the present case, because the expressions $w'/w$ and $fw$ might contain products of distributions. $\endgroup$
    – Abezhiko
    Mar 2 at 20:02
  • $\begingroup$ @Abezhiko The product $fw$ is not a problem because the equation may be solved away from $x=0$ and $x=a$ then patched together with a derivative discontinuity at these points. Knowing $y'$ cannot have Dirac delta contributions (else it wouldn't satisfy the original equation) is consistent with $w'/w$ having at worst discontinuities or isolated poles, because $w$ can only have isolated zeros and is continuous. Also $w$ may not even have zeros- depending on the signs of the constants and the boundary conditions (which are unstated) $\endgroup$
    – Sal
    Mar 2 at 21:32
  • $\begingroup$ @ OP you don't mention the boundary conditions, domain of solution, or anything about the constants- in particular $C$ plays the role of an eigenvalue in the $w$ equation. The character of solutions is different for $C<0$ or $C>0$. I can write an answer if you can clarify these points and also tell us where the problem is from $\endgroup$
    – Sal
    Mar 2 at 22:04

1 Answer 1

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Assuming $C\gt 0$, the differential equation can be rescaled by substituting $y=2 u$ and $x=\sqrt{2/C}\; t$, leading to the simpler form $$\ddot{u}(t)+\dot{u}(t)^2=1+\alpha\, \delta(t)+\beta \,\delta(t-\tau) \tag{1} \label{1}$$ with $\alpha = A/\sqrt{2C}$, $\beta=B/\sqrt{2C}$ and $\tau = a\sqrt{C/2}$. Comparing the LHS and the RHS of \eqref{1}, it is obvious that $u(t)$ must be continuous at the critical points $t=0$ and $t=\tau$ and only $\dot{u}(t)$ can have jumps at these points (see also the comments by @Sal above). Therefore, setting $\dot{u}(t)= \dot{w}(t)/w(t)$, leading to the linear differential equation $$ \ddot{w}(t)-w(t)= \alpha\, w(0)\, \delta(t) +\beta \,w(\tau)\, \delta(t-\tau) \tag{2} \label{2}$$should pose no problems, unless $w(t)$ acquires zeros for some unfortunate choice of parameters and/or initial conditions.

Restricting ourselves to the special case $\beta=0$, eq. \eqref{2} is solved by employing the ansatz $$w(t)=(c_1 e^t+c_2e^{-t}) \,\Theta(-t)+(c_3e^t+c_4e^{-t})\, \Theta(t), \tag{3} \label{3}$$ where $\Theta(t)$ is the Heaviside step function with $\Theta^\prime(t)=\delta(t)$. The continuity of $w(t) $ at $t=0$ requires $$c_1+c_2 = c_3+c_4 \tag{4} \label{4}$$ and inserting $\ddot{w}(t)$ into \eqref{2} yields the further condition $$c_3-c_4-c_1+c_2=\alpha (c_1+c_2) \tag{5} \label{5}.$$ As a result, $\dot{u}(t)$ is given by $$\dot{u}(t)= \frac{c_1 e^t -c_2 e^{-t}}{c_1 e^t+c_2e^{-t}} \, \Theta(-t)+\frac{c_3 e^t -c_4 e^{-t}}{c_3e^t+c_4e^{-t}} \, \Theta(t) \tag{6} \label{6}$$ and $u(t)$ is found by integrating \eqref{6}.

In the case $\beta \ne 0$, the ansatz $$w(t) =(c_1e^t+c_2e^{-t}) \, \Theta(-t)+(c_3 e^t+c_4 e^{-t}) \, \Theta(t) \Theta(\tau-t)+ (c_5 e^t+c_6 e^{-t}) \, \Theta(t-\tau) \tag{7} \label{7}$$ is used. The further steps of the calculation (analogously to the previous special case) are left as an exercise.

Note that \eqref{1} can be interpreted as the equation of motion of a particle under the influence of a constant force (the term $1$), a friction term $- \dot{u}^2$ (proportional to the square of the velocity) and two "kicks" at the times $t=0$ and $t=\tau$.

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  • $\begingroup$ Hello Hyperon, thank you for your detailed answer! I have one last question regarding expression (2) you obtained: from that moment on, is Laplace method well defined (since we're treating $w(0)$ and $w(\tau)$ as c-numbers), or are there any restriction making the use of your ansatz the only simple way to solve it? $\endgroup$
    – Fredrigo6
    Mar 3 at 0:52
  • $\begingroup$ (+1) I think, even for an unfortunate choice of parameters, the poles of $w'/w$ are first order and $y$ can still be found by integrating and taking the principle value. @Fredrigo6 the danger with using integral transforms on the equation for $w$ is that $w$ may not vanish sufficiently quickly at $\pm \infty$ so that the transform may not exist... $\endgroup$
    – Sal
    Mar 3 at 1:43
  • $\begingroup$ I fully agree with @Sal that more information about the context and the origin of the problem could be rather helpful. Note that the homogeneous equation (i.e. eq. (1) with $\alpha=\beta=0$) admits the well behaved solution $v(t)=\tanh (t-t_0)$ (with $v(t):=\dot{u}(t)$), but also the singular solution $v(t)=\coth(t-t_0)$ with $v(t) \to -\infty$ for $t \to t_0 -\epsilon$, sending $v(t)$ with initial condition $v(t_1) \lt -1$ for $t_1 \lt t_0$ to $-\infty$ within a finite time interval. Maybe in view of the context of your problem there might be a natural restriction on the domain? $\endgroup$
    – Hyperon
    Mar 3 at 7:13
  • $\begingroup$ In the case of the physical problem mentioned in the last sentence of my answer, the differential equation should in fact be $\dot{v} + v^2 {\rm sgn}(v) =1 +f(t)$, ensuring that the friction term acts always opposite to the velocity $v$ (damping). In this case, the solutions for $f(t)=0$ are given by $v(t)=\tan(t-t_0) \Theta(t_0-t)+\tanh(t-t_0) \Theta(t-t_0)$ (with natural domain $(t_0-\pi/2,\infty)$) for initial conditions with $v(t_1) \lt 1$ and $v(t)= \coth(t-t_0)$ (with domain $(t_0,\infty)$) and initial conditions $v(t_1)\gt 1$. In both cases $v(t) \to 1$ for $t \to \infty$. $\endgroup$
    – Hyperon
    Mar 3 at 7:31
  • $\begingroup$ I did not think about other methods to solve the problem but I am rather sure that they exist. $\endgroup$
    – Hyperon
    Mar 3 at 7:39

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