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!
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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: $(tu)_t+(tu)_{xx}+(x-y)u_y=u$ 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 $y$. |
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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 $1/t$. |
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