Solution of the parabolic equation

Question

The problem is set as follows: $$\frac{\partial W(\tau,z)}{\partial \tau}=-\frac{1}{\varepsilon} \frac{\partial(z_1 W(\tau,z))}{\partial z_1}-\frac{\partial(z_2 W(\tau,z))}{\partial z_2}+h^2\frac{\partial^2 W(\tau,z)}{\partial z_1^2}$$ where $\varepsilon>0,\, h>0$ with boundary conditions on $\Omega=\{z\in \mathbb{R}^2\vert z_1 \leqslant a,\,z_2 \in \mathbb{R} \}$, where $a>0$: $$W(\tau,z)\vert_{\partial \Omega}=0. $$ and initial conditions $$W(0,z)=\delta(z).$$ Is it possible to find a classical solution for that problem? The Dirac delta function for the initial condition can be substituted with some smooth enough function in order to simplify analysis. Can the solution of the stationary problem be obtained?

$\textbf{Edited on November 28}$

There is a mistake in the initial system. The corrected one looks as follows: $$\frac{\partial W(\tau,z)}{\partial \tau}=\frac{1}{\varepsilon} \frac{\partial(z_1 W(\tau,z))}{\partial z_1}+\frac{\partial(z_2 W(\tau,z))}{\partial z_2}+h^2\frac{\partial^2 W(\tau,z)}{\partial z_1^2}.$$

Answer 1

Not a solution, just a few thoughts about a product ansatz, which the solution is most likely not of the form. If $$ W(t,z_1,z_2)=w_1(t)w_2(z_1)w_3(z_2) $$ then one obtains the three equations \begin{align} \left\{ \frac{{\rm d}}{{\rm d}t} - c_1\right\} w_1(t) &= 0 \\ \left\{ \frac{{\rm d}^2}{{\rm d}z_1^2} w_2(t) - \frac{z_1}{h^2 \epsilon} \frac{\rm d}{{\rm d}z_1} + \frac{c_2}{h^2\epsilon} \right\} w_2(z_1) &= 0 \\ \left \{ \frac{\rm d}{{\rm d}z_2} + \frac{\epsilon c_1 + \epsilon + c_2 + 1}{z_2 \epsilon} \right\} w_3(z_2) &= 0 \end{align} with the solutions \begin{align} w_1(t) &= C_1 \, {\rm e}^{c_1 t} \tag{1} \\ w_2(z_1) &= C_{21} \, z_1 \, M\left(\frac{1-c_2}{2} , \frac{3}{2} , \frac{z_1^2}{2h^2\epsilon} \right) + C_{22} \, z_1 \, U\left(\frac{1-c_2}{2} , \frac{3}{2} , \frac{z_1^2}{2h^2\epsilon} \right) \tag{2} \\ w_3(z_2) &= C_3 \, z_2^{-\left(c_1 + 1 + \frac{c_2+1}{\epsilon}\right)} \tag{3} \end{align} where $c_1,c_2,C_1,C_{21},C_{22},C_3$ are arbitrary constants and $M,U$ are the two Kummer solutions.

You say that the solutions must vanish at $z_1=a, z_1=-\infty$ and $z_2=\pm \infty$:

If the first argument in the two Kummer solutions of $w_2(z_1)$ was a negative integer, they were polynomials of some degree and thus would not go to zero in the limit $z_1\rightarrow -\infty$. Thus $M$ and $U$ are not polynomials in which case we must have $C_{21}=0$ since $$M(a,b,z)\sim \frac{\Gamma(b)}{\Gamma(a)} \, {\rm e}^{z} \, z^{a-b}$$ diverges exponentially. For the second term we have $$ z_1 \, U\left(\frac{1-c_2}{2} , \frac{3}{2} , \frac{z_1^2}{2h^2\epsilon} \right) \sim |z_1|^{c_2} \qquad {\rm as} \quad z_1 \rightarrow -\infty $$ and so $c_2<0$. From (1) we must as well require $c_1<0$ and from (3) it follows $c_1+1+\frac{c_2+1}{\epsilon} > 0$ because of the boundary condition at $z_2=\pm \infty$.

Turning back to $U$ however it seems we need to find zeros in order to fulfill the boundary condition at $z_1=a$. But the total number of zeros for $U(a,b,z)$ is given by $T(a,b)=T(a-b+1,2-b)$ if $b\geq 1$ which here becomes $T\left(-\frac{c_2}{2},\frac{1}{2}\right)=0$ since the first argument $-c_2 >0$ (cf. https://dlmf.nist.gov/13.9).

Thus is seems there is no solution $\neq 0$ with this product-ansatz, since the boundary condition at $z_1=a$ can not be fulfilled.

Answer 2

It's just a drift of the solution. Using the same substitution for the correct system (each member enters with a positive sign) $$ W(\tau,z)=w_1(\tau)w_2(z_1)w_3(z_2)$$ the next equations can be obtained: $$ \begin{align*} \frac{d w_1(\tau)}{d\tau}&-c_1 w_1(\tau)=0\\ \frac{d w_3(z_2)}{d z_2}&-\frac{c_1-1+(c_2-1)/\varepsilon}{z_2}w_3(z_2)=0\\ \frac{d^2 w_2(z_1)}{d z_1^2}&+\frac{z_1}{h^2 \varepsilon} \frac{d w_2(z_1)}{d z_1}+\frac{c_2}{h^2\varepsilon}w_2(z_1)=0 \end{align*}$$ Solution for the first two equations can be presented as follows: $$ \begin{align*} w_1(\tau)&=K_1 e^{c_1 \tau}\\ w_3(z_2)&=K_3 z_2^{c_1-1+(c_2-1)/\varepsilon} \end{align*} $$ In order to solve the third system let's consider new variables: $$ x=\frac{z_1}{h \sqrt{\varepsilon}},\quad a=c_2-1,\quad w_2(x)=u(x)e^{\frac{-x^2}{4}} $$ and after substitution one can get a Weber's equation: http://mathworld.wolfram.com/WeberDifferentialEquations.html $$u''+(a+\frac{1}{2}-\frac{x^2}{4})u=0.$$ It's known that parabolic cylinder functions $D_a(x)$ are solutions for this equation. They are orthogonal on the whole real space for $a \in N$ (Watson, A Treatise on the Theory of Bessel Functions, 2nd ed., p.308): $$ \int_{-\infty}^{\infty}D_m(s)D_n(s) ds =\delta_{mn}n! \sqrt{2\pi} $$ Let's leave out for a while the problem of orthogonality of $\{D_a(x)\}$ for $a\in R$ and $x\in (-\infty,\frac{a}{h \sqrt{\varepsilon}})$ and realize that the function $$ w_{2,a}(x)=e^{\frac{-x^2}{4}}D_a(x) $$ is a solution for the most troublesome equation of ours. One need to find the eigenvalues $\{a_k\}$ such that the solution would satisfy the border condition. Thus, $$ D_{a_k}\left(\frac{a}{h \sqrt{\varepsilon}}\right)=0. $$ The general solution for our initial equation can be expressed as a series: $$ W(\tau,z)=\sum_{k=1}^{\infty} b_k e^{c_{1,k} \tau} z_2^{c_{1,k}-1+\frac{c_{2,k}-1}{\varepsilon}}e^{\frac{z_1^2}{4 h^2\varepsilon}} D_{a_k}\left(\frac{z_1}{h \sqrt{\varepsilon}}\right) $$ From the requirement of series convergence and the border condition for $z_2$ the constants $c_1,c_2$ should satisfy the next conditions: $$\begin{align*} c_{1,k}&<0\\ c_{2,k}&=a_k+1\\ c_{1,k}&-1+\frac{c_{2,k}-1}{\varepsilon}<0 \end{align*}$$ Coefficients $b_k$ are usually can be found from initial conditions. Let's assume that a delta function can be expanded as follows: $$ W(0,z)=\delta(z)=\sum_{k=1}^{\infty}f_k \psi_k(z_1,z_2), $$ where $$\psi_k(z_1,z_2)=\frac{1}{\sqrt{2\pi}k!}He_k(z_2) \cdot d_k \cdot D_{a_k}\left(\frac{z_1}{h \sqrt{\varepsilon}}\right).$$ Here $\frac{1}{\sqrt{2\pi}k!}He_k(z_2)-$ an orthonormal system of Hermite functions defined for $z_2\in (-\infty,\infty)$ and $d_k -$ are normalizing coefficients for the functions $D_{a_k}(z_1)$. I wonder on what domain shall we normalize them. It depends of what domain these functions can be transformed into orthogonal functions. In case the origin domain $(-\infty,a)$ is not satisfactory can we use an infinite interval $(-\infty,\infty)$ while expanding our initial condition into the series?

On the other hand, $$ W(0,z)=\sum_{k=1}^{\infty} b_k z_2^{c_{1,k}-1+\frac{c_{2,k}-1}{\varepsilon}}e^{\frac{z_1^2}{4 h^2\varepsilon}} D_{a_k}\left(\frac{z_1}{h \sqrt{\varepsilon}}\right) $$ Can we use this idea in order to define coefficients $b_k$?