# Solving a non linear differential equation

This is a first order differential equation: $$\frac{df_1}{dx} + \frac{(f_1)^2}{h^2} - \frac{2m}{h^2} \lambda \delta(x-pa)=-\frac{2mE_1}{h^2}$$

Where, h, $$\lambda$$ and $$E_1$$ are constants and and pa lies in [0,a] as 0<p<1 .

I have not been taught how to handle differential equations with a Dirac Delta function in it. Moreover, this is a non linear one. I came across this in a research paper and the answer is given but the method to solve it isn't. I have tried learning to use Laplace transform to solve this, but got stuck again because I didn't know how to do Laplace transform of the second term of the equation. Any help will be appreciated. Please, help me out.

The answer is: $$f_1=√2mE_1[cot (\frac{√2mE_1}{h} (x-b))]$$

Where b is constant of integration

P.s.: I know this might be rude but please don't vote this as a homework question because it isn't one. If you can't help just ignore.

• Would be better to show us the solution. Also plug the known solution into the equation, that should hint you how it can be solved.
– user65203
Commented Aug 12, 2020 at 14:37
• Would be better to hide the ugly constants and translate/rescale the variable. You can rewrite as $g'+g^2=a\delta(x)\pm1.$ ($g:=f/h$).
– user65203
Commented Aug 12, 2020 at 14:54

Using the notation in my comment, for $$x\ne0$$ the equation

$$g'+g^2=\pm1$$ is separable and the solution will be $$g$$ as the tangent or hyperbolic tangent of $$x$$, with constants.

The constants can differ on the left and on the right, and by introducing a discontinuity of height $$a$$, you will retrieve the Dirac Delta.

The solution can be like

$$x<0\to g(x)=-\tan(x+r),\\x>0\to g(x)= -\tan(x+s)$$

and there is a unit step at $$x=0$$ if $$\tan(r)-\tan(s)=1.$$

• The answer has a cotangent. Let me edit the question and add in. Commented Aug 12, 2020 at 14:55
• Done! @Yves Daoust Commented Aug 12, 2020 at 15:03
• then you how do we get a cot in the answer? Commented Aug 13, 2020 at 5:31
• @Korra: hem, relation between $\cot$ and $\tan$ ?
– user65203
Commented Aug 13, 2020 at 7:42
• cot is reciprocal of tan. Commented Aug 13, 2020 at 8:38