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?


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?

  • $\begingroup$ Also, am I overly complicating this? $\endgroup$
    – Sidhha
    Dec 23, 2013 at 11:04
  • $\begingroup$ Find c2, and c3 when Y(0) = 0, 0 = c2*Ai(0) + c3*Bi(0), Ai(0) = known value, Bi(0) is a known value. You will have to find lambda form Y(y) and then use that lambda in X(x). Do not apply boundary conditions on u(x,y). I will look into it too. $\endgroup$ Dec 23, 2013 at 11:48
  • $\begingroup$ To have a bounded solution, $c_3 \rightarrow 0$ and then to have $Y(0)=0$, $c_2 \rightarrow 0$. I do not think the conclusion of $Y(0)=0$ is correct here. $\endgroup$
    – Sidhha
    Dec 23, 2013 at 12:10
  • $\begingroup$ If one of the Boundary Value is Y(0), you get an equation with C2 and C3 and I am wondering if there is someother BV for example Y'(0) = C2Ai'(0) +C3Bi'(0), Then you get a simultanious equation to find C2 and C3 $\endgroup$ Dec 23, 2013 at 12:18
  • $\begingroup$ We cannot have $Bi$ in the solution as I also mentioned in the edit since, $Bi(\infty) \rightarrow \infty$ and hence, $c_3 \rightarrow 0$. $\endgroup$
    – Sidhha
    Dec 23, 2013 at 12:23

2 Answers 2


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.

  • $\begingroup$ Could you please tell how you arrived at the solution? If you used some symbolic toolbox, can you share the code you have used. The solution seems to satisfy $u(x,0)=\cos(x)$. Also the coefficient $\dfrac{1}{8}$ should be $\dfrac{1}{2}$. $\endgroup$
    – Sidhha
    Dec 24, 2013 at 4:36
  • $\begingroup$ Also, if the solution contains terms with $Bi$, will it satisfy $\lim_{y \rightarrow \infty} u(x,y) = 0 .$ $\endgroup$
    – Sidhha
    Dec 24, 2013 at 4:50
  • $\begingroup$ Oops, yes, a slight problem there: just translate by $\pi/2$ in the $x$ direction. The point is that $\sin(x) = (e^{ix} - e^{-ix})/(2i)$. The constants were chosen to make $u \to 0$ as $y \to \infty$ if, as appears to be the case, $Ai(-(\sqrt{3}-i)t) + i Bi(-(\sqrt{3}-i)t)$ and its complex conjugate $Ai(-(\sqrt{3}+i)t) - i Bi(-(\sqrt{3}+i)t)$ go to $0$ as $t \to +\infty$. $\endgroup$ Dec 24, 2013 at 6:22
  • $\begingroup$ This was obtained with the help of Maple. I don't know why you want to change $1/8$ to $1/2$. $\endgroup$ Dec 24, 2013 at 6:47

Hint: I have not solved the whole problem but I will give the approach:

$f_{yy} - yf_{x} = 0$

Let $f = X_{x}.Y_{y}$

Then $X.Y^{''} - yY.X^{'} = 0$

$X.Y^{''} = yY.X^{'}$

$\frac{Y^{''}}{yY} = \frac{X^{'}}{X} = -\lambda$

${Y^{''}} + y\lambda Y = 0$

${X^{'}} + \lambda X = 0$

These are two ODEs in Y and X

Boundary Values are Y(0) = 0 and $\lim$ ( y tending to infinity) $Y(y) = 0$

X(x) = 0 is trivial so just leave it.

Boundary Values are used to find the general solution and the initial value is used to find the particular solution.

Initial Value f(x,0) = sin(x)

The method that I have used can be more generally summarized as follows

The Method of Separation of Variables:

  1. Separate the PDE into ODEs of one independent variable each. Rewrite the boundary conditions so they associate with only one of the variables.

  2. One of the ODEs is a part of a two-point boundary value problem. Solve this problem for its eigenvalues and eigenfunctions.

  3. Solve the other ODE.

  4. Multiply the results from steps (2) and (3)

Can you take it from here!!



  • $\begingroup$ The third equation which you wrote, should it not be: $X.Y'' - y.Y.X'=0$? I was trying to do separation of variables, but was stuck after this step. $\endgroup$
    – Sidhha
    Dec 23, 2013 at 7:54
  • $\begingroup$ You are right, I have changed in the ODE $\endgroup$ Dec 23, 2013 at 8:02
  • $\begingroup$ Ok, so I rechecked what I had done previously with separation of variables and I had done a stupid mistake there. Seems like I can solve it now. I will post the complete solution here. Thanks! $\endgroup$
    – Sidhha
    Dec 23, 2013 at 8:20
  • 1
    $\begingroup$ Why $Y(0)=0$? $Y(0)=0$ would make $u(x,0) = 0$, not $\sin(x)$. $\endgroup$ Dec 23, 2013 at 8:20
  • $\begingroup$ f(x,0) = X(x)Y(0), Thus it is either X(x) = 0 or Y(0) = 0 $\endgroup$ Dec 23, 2013 at 8:45

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