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Can anybody please explain equation $1$ in this answer in a simpler form? Also i cannot understand how from equation $1$ we can see that $u$ is the solution to the heat equation.

Tobsn
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Denis
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1 Answers1

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Hint

\begin{align} \frac{\mathrm d }{\mathrm dt}P_tf(x)&=\lim_{h\to 0}\frac{P_{t+h}f(x)-P_tf(x)}{h}\\ &=\lim_{h\to 0}P_t\left(\frac{P_hf(x)-f(x)}{h}\right)\\ &=P_t\left(\lim_{h\to 0}\frac{P_hf(x)-f(x)}{h}\right)\\ &=P_tAf(x)\\ &=AP_tf(x). \end{align} I let you justify each equality as a homework. For your other question, one can prove that the infinitesimal generator of the Brownian motion if given by $$Af(x)=\frac{1}{2}\Delta f(x).$$ Do it as a homework if this is not clear for you.

Surb
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  • Thanks. Putting the solution to $\partial_t u(t,x) = \frac{1}{2}\partial_x^2 u(t,x) \qquad u(0,x)=f(x)$ gives an extra term $\frac{1}{2} \partial_x^2 (\frac{t}{2}f''(x))$ May i ask why this last term vanishes? – Denis Aug 16 '20 at 10:56
  • Sorry, but I don't understand your question. I don't see where your extra term comes from. @Denis – Surb Aug 16 '20 at 11:00
  • actually, i am checking if $\mathbb{E}^x(f(B_t)) \approx f(x)+ \frac{t}{2} f''(x)$ is the solution to the heat equation. But it does not satisfy the heat equation, and we have the mentioned extra term... – Denis Aug 16 '20 at 11:26
  • As you remarked, $\mathbb E^x[f(B_t)]\approx f(x)+\frac{t}{2}f''(x),$ but they are not equal. They even don't have the same regularity... and obviously, $f(x)+\frac{t}{2}f''(x)$ doesn't satisfy heat equation @Denis – Surb Aug 16 '20 at 11:28
  • But in the linked post, it is argued that : "From (1) we see that $u(t,x) := \mathbb{E}^x(f(B_t))$ is the (unique) solution of the heat equation.." and he already defined $\mathbb E^x[f(B_t)]\approx f(x)+\frac{t}{2}f''(x)$ – Denis Aug 16 '20 at 11:32
  • Honnestly, I don't see any connection between the fact that $\mathbb E^x[f(X_t)]$ solve the heat equation, and the fact that $f(x)+\frac{t}{2}f''(x)$ should solve it... obviously, It won't solve the heat equation, since a priori $x\mapsto f(x)+\frac{t}{2}f''(x)$ is not derivable. @Denis – Surb Aug 16 '20 at 11:47
  • Sorry, i am really confused here. So, why in the linked answer, the person did the effort to obtain the approximation? and more importantly, if that is not the correct way, what is the correct way? any hint is really appreciated... – Denis Aug 16 '20 at 11:54
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    @Denis: Look, take the Cauchy problem $y'(x)=y(x)$, $y(0)=1$. The (unique) solution is given by $y(x)=e^x$. You have that $y(x)\approx 1+x$ (if $x\ll 1$), but $x\mapsto 1+x$ is not a solution of $y'(x)=y(x)$. – Surb Aug 16 '20 at 11:57
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    @Denis: For a Hint, you have that $$p(x,t)=\mathbb E^x[f(X_t)]=\frac{1}{\sqrt{2\pi t}}\int_{\mathbb R}f(y)e^{-\frac{(y-x)^2}{2t}},\mathrm d y.$$ Thus you can easily compute $\partial _tp$ and $\partial _{xx}p(x,t)$. – Surb Aug 16 '20 at 12:00
  • Thanks, just a last short question : Do you know why in the linked answer, the person tried to obtain $\mathbb E^x[f(B_t)]\approx f(x)+\frac{t}{2}f''(x)$? What did he want to show with obtaining this? – Denis Aug 16 '20 at 12:31
  • @Denis: $\mathbb E^x[f(X_t)]$ can be very diffcult to compute. Nevertheless, when $t$ is small, it should be close to $f(x)+\frac{t}{2}f''(x).$ – Surb Aug 16 '20 at 13:00