$$ Greetings, $$ So, I have a heat equation to be solved for in the form of $$ \frac{\partial f(x,t)}{\partial t} = \frac{\partial^2 f(x,t)}{\partial x^2} $$ for t = [0,+inf) and x = (-inf,+inf)
and following I.C and B.C: $$ f(x,0)=1,x\geq0 $$ $$ f(x,0)=0,x<0 $$ $$ f(-\inf,t)=0 $$ $$ f(+\inf,t)=1 $$ I work out the solution roughly using the following procedure:
$f(x,t)=f(\eta)$, with $\eta = \frac{x}{\sqrt{t}}$
Obtain ODE as $$ \frac{d^2 f(\eta)}{d\eta^2}+\frac{1}{2}\eta\frac{df}{d\eta}=0 $$
Solve ODE using $u=\frac{df}{d\eta}$
Obtain $u=C_1e^{-\frac{1}{4}\eta^2}$
and here is where I started get confusing, I went ahead and solve for $f$ using $u$ obviously and end up with an integral that looks like $$ df = C_1e^{-\frac{1}{4}\eta^2} d\eta $$ The question also hitted that using error function to simplify the solution $$ erf(x) = \frac{2}{\sqrt{\pi}}\int_{0}^{x}e^{-y^2}dy $$ Thus, using $y=\frac{1}{2}\eta$, I can manage to get an integral that looks similar $$ f = C_1\int e^{-y^2} dy $$
So, my question is
1) What is the trick to convert the indefinite integral to a definite one
2) I can find the $C_2$ which is obtained after integrating the last integral as well as $C_1$ just by using the B.C. ONLY. What happens to the I.C. in this case? Does it do anything, and why?
First question post here,
Thank you all in advance
Cheers,