# Thread: 2nd order PDE - help needed

1. ## 2nd order PDE - help needed

H''(n) + (1/2)n*H'(n) = 0.

Boundary conditions: H(0) = 1, H(inf) = 0.

I really don't know where to start ... any help will be greatly appreciated.

2. Originally Posted by Baluba
H''(n) + (1/2)n*H'(n) = 0.

Boundary conditions: H(0) = 1, H(inf) = 0.

I really don't know where to start ... any help will be greatly appreciated.
First separate and integrate

$\displaystyle \frac{H''}{H'} = - \frac{1}{2} n\, dn$

so

$\displaystyle \ln H' = \ln c_1 - \frac{1}{4} n^2\;\; \Rightarrow \;\;H' = c_1 e^{-n^2/4}$

then integrate again (on the second integration we'll keep it in integral form)

$\displaystyle H = c_1 \int_0^n e^{-t^2/4}dt + c_2$

$\displaystyle H(0) = 1\; \text{ gives}\; c_2 = 1$ while $\displaystyle H(\infty) = 0\; \text{ gives}\; c_1 \pi + 1 = 0$

giving the solution as

$\displaystyle H(n) = 1 - \frac{1}{\pi} \int_0^n e^{t^2/4}dt$

or in terms of error functions

$\displaystyle H(n) = 1 - \text{erf} \left(\frac{n}{2}\right)$.

3. Thanks a lot!