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I have the following PDE:

$$\frac{\partial H(\lambda,r)}{\partial \lambda} - D \frac{1}{r^2} \frac{\partial}{\partial r} (r^2 \frac{\partial H(\lambda,r)}{\partial r}) = \delta(\lambda) \delta(\vec{r}) \label{eq:H},\quad \lambda \geq 0$$

The text I am reading suggests solving this equation with zero right-hand side and the boundary condition:

$$H(\lambda,r)\bigg|_{\lambda=0} = \delta(\vec{r})$$

I don't understand how exactly this boundary condition is derived and why the resulting problem is equivalent to the original PDE. I come from a physics perspective so I don't have much knowledge on the correct formal treatment of distributions.

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From physics perspective, at first, it is obvious, that right hand part is non-zero only at point (0,0), so physically it looks like boundary condition

And if you integrate equation through \lambda near zero, you get something like H(0+,r) - H(0-,r) = \delta(r) And we consider, that for negative \lambda H is equal to zero.

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