I have a equation like this:
$\dfrac{\partial y}{\partial t} = -A\dfrac{\partial y}{\partial x}+ B \dfrac{\partial^2y}{\partial x^2}$
with the following I.C
$y(x,0)=0$
and boundary conditions $y(0,t)=1$ and $y(\infty , t)=0$
I tried to solve the problem as follows:
Taking Laplace transform on both sides,
$\mathcal{L}(\dfrac{\partial y}{\partial t}) = – A \mathcal{L}(\dfrac{\partial y}{\partial x})+B \mathcal{L}(\dfrac{\partial^2 y}{\partial x^2})$
Now, on the L.H.S we have,
$sY-y(x,0)=sY$
$\mathcal{L}(\dfrac{\partial^2 y}{\partial x^2}) =\displaystyle \int e^{-st} \dfrac{\partial^2 y}{\partial x^2} dt$
Exchanging the order of integration and differentiation
$\displaystyle\mathcal{L}(\frac{\partial^2 y}{\partial x^2}) =\frac{\partial^2}{\partial x^2} \int e^{-st} y(x,t) dt$
$\displaystyle\mathcal{L}(\frac{\partial^2 y}{\partial x^2}) =\frac{\partial^2}{\partial x^2}\mathcal{L}(y) = \frac{\partial^2Y}{\partial x^2} $
$\displaystyle\mathcal{L}\frac{\partial y}{\partial x} = \frac{\partial Y}{\partial x}$
Now, combing L.H.S and R.H.S, we have,
$\displaystyle sY = – A \frac{\partial Y}{\partial x} + B \frac{\partial^2Y}{\partial x^2}$
Above equation might have three solutions:
If $b^2 – 4ac > 0 $ let $r_1=\frac{-b-\sqrt{b^2-4ac}}{2a}$ and $r_2 = \frac{-b+\sqrt{b^2-4ac}}{2a}$
The general solution is $\displaystyle y(x) = C_1e^{r_1x}+C_2 e^{r_2x}$
if $b^2 – 4ac = 0 $, then the general solution is given by
$ y(x)=C_1e^{-\frac{bx}{2a}}+C_2xe^-{\frac{bx}{2a}}$
if $b^2 – 4ac <0$ , then the general solution is given by
$y(x) = C_1e^{\frac{-bx}{2a}}\cos(wx) + C_2 e^{\frac{-bx}{2a}}\sin(wx)$
Since, A, and B are always positive in my problem, the first solution seems to be appropriate.
Now, from this point I am stuck and couldn't properly use the boundary conditions.
If anyone could offer any help that would be great.
"Solution added"
The solution of the problem is
$y(x,t)= \dfrac {y_0}{2} [exp(\dfrac {Ax}{B}erfc(\dfrac{x+At}{2\sqrt{Bt}}) + erfc(\dfrac{x-At}{2\sqrt{Bt}})$
Best Answer
Using your notation, we have
$$B Y''(x,s) - A Y'(x,s) - s Y(x,s) = 0$$
$$Y(0,s)=\int_0^{\infty} dt\: 1 \cdot e^{-s t} = \frac{1}{s}$$
$$\lim_{x \rightarrow \infty} Y(x,s) = 0$$
The general solution to the equation is
$$Y(x,s) = M(s) e^{r_+ x} + N(s) e^{r_- x}$$
where
$$r_{\pm} = \frac{A \pm \sqrt{A^2+ 4 B s}}{2 B}$$
where
$$M(s) + N(s) = \frac{1}{s}$$
$$M(s)=0$$
The latter equation was determined by the limit at $\infty$. (I am of course assuming that $B > 0$; the other case may be considered as well.)
Therefore
$$Y(x,s) = \frac{1}{s} \exp{\left[-\frac{\sqrt{A^2+4 B s}-A}{2 B} x \right ]}$$
It is this quantity that must be inverse LT'ed to get the solution to your equation, viz.
$$y(x,t) = \frac{1}{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \: \frac{1}{s} \exp{\left[-\frac{\sqrt{A^2+4 B s}-A}{2 B} x + s \, t\right ]}$$