1

I form a PDE of the following type:

$$0=\frac{\partial V}{\partial \tau}+\frac{\beta -z}{v-\tau}\frac{\partial V}{\partial z}+\frac{1}{2}\frac{\partial^{2} V}{\partial z^{2}}$$

such that $\lim_{\tau\rightarrow v}V(\beta,\tau)=1$. I know that it is difficult to solve, but I would like to know if there exists any literature talking about PDE in which $\tau$ is undefined at one point here. I try to change it to $u=v-\tau$ and solve it, but it can be seen that the new PDE is not defined at $u=0$. Any literature talk about it? What is this type PDE called?

will_cheuk
  • 539
  • 5
  • 12

2 Answers2

0

Hint:

$\dfrac{\partial V}{\partial\tau}+\dfrac{\beta-z}{v-\tau}\dfrac{\partial V}{\partial z}+\dfrac{1}{2}\dfrac{\partial^2V}{\partial z^2}=0$

$\dfrac{\partial V}{\partial\tau}+\dfrac{z-\beta}{\tau-v}\dfrac{\partial V}{\partial z}+\dfrac{1}{2}\dfrac{\partial^2V}{\partial z^2}=0$

Let $\begin{cases}\tau_1=\tau-v\\z_1=z-\beta\end{cases}$ ,

Then $\dfrac{\partial V}{\partial\tau_1}+\dfrac{z_1}{\tau_1}\dfrac{\partial V}{\partial z_1}+\dfrac{1}{2}\dfrac{\partial^2V}{\partial z_1^2}=0$ with $\lim\limits_{\tau_1\to0}V(0,\tau_1)=1$

With reference the substitution concept in Change variables into Fokker-Planck PDE,

Let $\begin{cases}z_2=\dfrac{z_1}{\tau_1}\\\tau_2=\tau_1\end{cases}$ ,

Then $\dfrac{\partial V}{\partial z_1}=\dfrac{\partial V}{\partial z_2}\dfrac{\partial z_2}{\partial z_1}+\dfrac{\partial V}{\partial\tau_2}\dfrac{\partial\tau_2}{\partial z_1}=\dfrac{1}{\tau_1}\dfrac{\partial V}{\partial z_2}$

$\dfrac{\partial^2V}{\partial z_1^2}=\dfrac{\partial}{\partial z_1}\left(\dfrac{1}{\tau_1}\dfrac{\partial V}{\partial z_2}\right)=\dfrac{\partial}{\partial z_2}\left(\dfrac{1}{\tau_1}\dfrac{\partial V}{\partial z_2}\right)\dfrac{\partial z_2}{\partial z_1}+\dfrac{\partial}{\partial\tau_2}\left(\dfrac{1}{\tau_1}\dfrac{\partial V}{\partial z_2}\right)\dfrac{\partial\tau_2}{\partial z_1}=\dfrac{1}{\tau_1^2}\dfrac{\partial^2V}{\partial z_2^2}$

$\dfrac{\partial V}{\partial\tau_1}=\dfrac{\partial V}{\partial z_2}\dfrac{\partial z_2}{\partial\tau_1}+\dfrac{\partial V}{\partial\tau_2}\dfrac{\partial\tau_2}{\partial\tau_1}=-\dfrac{z_1}{\tau_1^2}\dfrac{\partial V}{\partial z_2}+\dfrac{\partial V}{\partial\tau_2}$

$\therefore-\dfrac{z_1}{\tau_1^2}\dfrac{\partial V}{\partial z_2}+\dfrac{\partial V}{\partial\tau_2}+\dfrac{z_1}{\tau_1^2}\dfrac{\partial V}{\partial z_2}+\dfrac{1}{2\tau_1^2}\dfrac{\partial^2V}{\partial z_2^2}=0$ with $\lim\limits_{\tau_2\to0}V(0,\tau_2)=1$

$\dfrac{\partial V}{\partial\tau_2}+\dfrac{1}{2\tau_2^2}\dfrac{\partial^2V}{\partial z_2^2}=0$ with $\lim\limits_{\tau_2\to0}V(0,\tau_2)=1$

$2\tau_2^2\dfrac{\partial V}{\partial\tau_2}+\dfrac{\partial^2V}{\partial z_2^2}=0$ with $\lim\limits_{\tau_2\to0}V(0,\tau_2)=1$

Which converts to a separable PDE.

doraemonpaul
  • 16,178
  • 3
  • 31
  • 75
-1

Differential equations like that one are called singular equations, and the points where the coefficients are undefined are called singular points. If you want to figure out how to solve these in general, look at the Frobenius method. It basically tries to look at how the function would have to scale $V \sim x^{\alpha} t^{\beta}$ near the singular points, then looking at an equation for $U$ where $V = x^{\alpha} t^{\beta} U(x,t)$. Sometimes your boundary conditions or other physical requirements (if this comes from modeling some real system) can limit the number of solutions and determine the powers so that you don't have to do the full general analysis.

I tried applying something like the Frobenius method to your problem, using the expansion $$ V(\tau_1,z_1) = \sum_{k=0}^{\infty} A_k(z_1) {\tau_1}^{r+k} $$ to get a sequence of differential equations for $A_k(z_1)$. Solving those with general $r$, I keep getting $A_k \sim {z_1}^{-(r+k+2)}$ for everything but $A_0$. That is, the required powers of $z_1$ increase faster than for $\tau_1$, so the result would not be well-defined at $(\tau_1,z_1) = (0,0)$ unless the sequence terminates after the first term. All that does is give $r=1$ and the very simple solution $V(\tau_1,z_1) = z_1/\tau_1$. This doesn't have the limit you asked for, but it does have a limit where $z_1$ and $\tau_1$ both go to zero simultaneously. I don't think your problem has a well-defined solution. (This isn't a proof that it doesn't, but it's all I can point you at for now.)

Jareth Holt
  • 109