I am trying to solve the following linear pde where $u=f(x,y)$ in the domain $y \in (0,\infty)$:
$$y\dfrac{\partial{u}}{\partial x} = \dfrac{\partial^2 u}{\partial y^2}$$
with boundary conditions:
$$u(x,0)=\sin(x) $$
$$\lim_{y \rightarrow \infty} u(x,y) = 0 .$$
Can someone please suggest how do I proceed to get the analytical solution for this equation?
Solution:
Using method of separation of variables,
we assume solution to be of the form: $$u(x,y)=X(x).Y(y) $$
Inserting this to the pde, we get:
$$yX'Y=XY'' $$
$$\dfrac{X'}{X}=\dfrac{Y''}{yY}=-\lambda$$
Now we get the following ode's
$$X'+\lambda X=0 $$
$$Y'' + \lambda y Y = 0 $$
from which we get two general solutions of the following form:
$$X(x) = c_1\exp(-\lambda x) \qquad Y(y) = c_2Ai(\lambda^{1/3} y) + c_3Bi(\lambda^{1/3} y) $$ where $Ai$ and $Bi $ are Airy's functions and the general solution can be expressed as:
$$u(x,y) = c_1\exp(-\lambda x).(c_2Ai(\lambda^{1/3} y) + c_3Bi(\lambda^{1/3} y))$$
$c_3 \rightarrow 0 $ to satisfy second boundary condition since $Bi(\infty) \rightarrow \infty $ and hence, solution is of the form:
$$u(x,y) = A \exp(-\lambda x) Ai(\lambda^{1/3} y) $$
Putting $y=0$,
$$u(x,0) = A\exp(-\lambda x).\dfrac{1}{3^{2/3}}\Gamma({2/3}) = \sin(x) $$
How should I now proceed to find $A$ and $\lambda$, considering that $\lambda$ needs to be positive real value?
Best Answer
EDITED (to change cos to sin)
Solution seems to be $$\eqalign{\dfrac{\Gamma(2/3)\; 3^{2/3}}{8} & \left((-\sqrt{3}+i) e^{-ix} Ai(-(\sqrt{3}+i)y/2) \right.\cr & - (\sqrt{3} + i) e^{ix} Ai(-(\sqrt{3}-i)y/2) \cr & + (\sqrt{3} i + 1) e^{-ix} Bi(-(\sqrt{3}+i)y/2)\cr & + \left. (-\sqrt{3} i + 1) e^{ix} Bi(-(\sqrt{3}-i)y/2) \right)\cr} $$ where $Ai$ and $Bi$ are Airy functions.