Some background: define a process $Y_t=\frac{1}{t}\int_0^tX_sds$, where $X_t$ is a standard Brownian motion. Then I define a function $u(t,X_t,Y_t)$ and require it to be a martingale. Thus, by applying Ito formula and setting drift to zero I end up with a following p.d.e.: $$u_t+u_{xx}+\frac{1}{t}(x-y)u_y=0$$$x,y\in \mathbb{R}$, with some terminal condition $u(T,x,y)=f(s,y)$. I approach to solve this problem numerically with finite differences but I have a hard time understanding what is going on at time $0$. At the moment I can't even apply any implicit scheme due to singularity at zero. One can observe though, that due to Lebesgue integration theorem $\lim_{t->0} Y_t=X_0$, however, I divide by $t$, which also approaches zero and thus it is unclear how the pde behaves at that time. Any suggestions how to deal with time zero? This is where I am looking for the solution of this pde. Thanks!


If you multiply it all through by t, you get rid of the singularity. You can then also merge terms using the chain-rule to get:


Which may or may not be an easier form than if you didn't merge them together. Note at $t = 0$, this further simplifies (since t is not a function of x):

$(tu)_t + (x-y)u_y = u$

To solve numerically, which form is better depends on what you are doing. Consider the form where $t$ is not inside the derivatives and superscript $n$ indicates the current time step and $n+1$ indicates the time step to be solved. If you use a simple first order in time, second order central in space, you would get:

$t^n\frac{u^{n+1}_{ij}-u^{n}_{ij}}{dt} + t^n\frac{u^n_{i+1,j}-2u^n_{ij}+u^n_{i-1,j}}{dx^2} + (x_{ij}-y_{ij})\frac{u^n_{i,j+1}-u^n_{i,j-1}}{2dy} = 0$

keeping in mind that $dt < 0$.

Note how we discretized $t$. It's $t^n$ which means the solution of $u(t=0,x,y)$ is found with all the terms included. Another option is to use $t^{n+1/2}$ which is still not zero and all your terms are included (this is the average of $t^n$ and $t^{n+1}$ which in this case would be $0.5t^n$).

Looking at the case where $t=0$ and $x\neq y$ gives the solution $u_y = 0$. This means that any distribution in $x$ is valid provided it is constant in $y$. On it's own, this is not unique. However, when you solve the PDE and march in time, you will get a unique distribution in $x$ approaching $t=0$ and your final answer should then be constant in $y$.

  • $\begingroup$ thanks, but need more clarification: following your advise, I multiply by $t$ without merging: $tu_t+tu_{xx}+(x-y)u_y=0$, I don't have a singularity anymore but now that I still take limit as t approaches 0. Don't I run into trouble by having first and second term vanish(I see you had only second vanish) and I am left only with $(x-y)u_y=0$? since I know that $x=y$ does that imply that $u_y$=0 on $x\ne y$? And how do I solve that numerically, I just solve the new equation or I can solve the original up to some small value of $t$ and then modify the equation? $\endgroup$ – Kamil Nov 25 '12 at 3:07
  • $\begingroup$ @Kamil To solve this (and really any) PDE, you need both initial and boundary conditions. So does it make sense to solve at t=0 to begin with? Shouldn't that be the initial condition? Or are you starting at some other time and condition and marching backwards in time to t=0? $\endgroup$ – tpg2114 Nov 25 '12 at 4:06
  • $\begingroup$ @Kamil More to the point, if the PDE is correct then at $t=0$ you do get $u_y = 0$ if $x \neq y$ and your problem reduces to a trivial ODE at that time instant, ie. $u$ is a constant in $y$. But you don't have any information on variation in $x$ so all solutions such that $u$ is constant in $y$ hold, meaning there is not a unique solution at $t = 0$. $\endgroup$ – tpg2114 Nov 25 '12 at 4:10
  • $\begingroup$ you might note that that pde is backward in time(otherwise it is not well posed), so in fact, I have a terminal condition specified at $t=T$, and boundary conditions as well in the region $[0,x_{max}]\times [y_{min}, y_{max}]$. Using the pde theory I can say that inside of the domain and $(0,T)$ I have existence, uniqueness and infinitely smoothness. However, I have to understand what is at this point $t=0$. I have a suspicion that this term just disappears due to $\frac{x-y}{t}=0$ in the limit, but I am not able to justify that. $\endgroup$ – Kamil Nov 25 '12 at 16:11
  • $\begingroup$ @Kamil I updated the answer with what it would look like to solve numerically. $\endgroup$ – tpg2114 Nov 25 '12 at 21:47

In order to understand what happens at $t=0$ you shouldn't look at this as a point value in time -- i.e., you shouldn't assume that the PDE holds pointwise in time in the same way as, say, the Laplace equation doesn't hold pointwise in space. Rather, the PDE will hold in the weak form with regard to all possible test functions from appropriate test spaces. The weak form contains integrals over both space and time.

If you discretize the equation in the strong form, using for example finite differences in time, you get into trouble because of the singularity at $t=0$ (in the original form) or because certain terms disappear (after multiplying through with $t$).

On the other hand, if you discretized the weak form using test functions that exist in both space and time, you would realize that you have integrals of the form $$ \int_0^T \int_\Omega (tu(x,t))_t \phi(x,t) \; dx \; dt $$ where it doesn't make a difference what exactly happens at $t=0$. The same would be true for the term with the factor $1/t$.

  • $\begingroup$ I agree that I can consider a weak formulation of the pde, then I choose function $\phi(t,x,y)$ s.t. it is bounded, measurable and probably $C^{1,2,1}$ and multiply both sides, integrate over space and time and end up with three integrals. How do I discretize the integral though? I am not familiar with what you are talking about and if you could please provide some thorems/citations would be great. I am considering finite differences only. $\endgroup$ – Kamil Nov 27 '12 at 3:16
  • $\begingroup$ Well, this may be one of those cases where the finite difference method does not work very well while the finite element method does. It's such a non-standard equation that I don't have any references but you could look at how people deal with equations in cylindrical coordinates since there you have exactly the same problem. $\endgroup$ – Wolfgang Bangerth Nov 27 '12 at 4:04
  • $\begingroup$ But I can solve it just fine for any $t>0$, I can put any small number $\epsilon$ to solve it up to and I don't observe any problems, at least when I plot the function. So I wonder if there is no any simply because the singularity gets damped by another one and it is valid to claim that the numerical solution is $O(\epsilon)$ from what it should be at time $0$. $\endgroup$ – Kamil Nov 27 '12 at 4:08
  • $\begingroup$ It's possible but who knows in the absence of solid analysis. But for the record, just because a factor in the equation goes to zero does (i) not imply that any singularity must exist there, or (ii) even if one does exist, there is no reason to believe that the solution must go to infinity -- it could be that simply the gradient or time derivative goes to infinity but that the solution itself remains bounded. $\endgroup$ – Wolfgang Bangerth Nov 28 '12 at 2:12
  • $\begingroup$ Now that I am thinking, as the first answer has suggested, if I multiply by any $t$, I have another valid pde with no singularity, now plug $t=0$ in there. I am left with $(x-y)u_y=0$. Thus, can imply that this part in the original pde has to be $0$ in any event, i.e. $\frac{1}{t}(x-y)u_y=0$ even for $t=0$? Then, I just have $u_t+u_{xx}$ at time zero, which I Can solve numerically at the very last iteration of the numerical method. $\endgroup$ – Kamil Nov 28 '12 at 3:19

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