Find the value of the following PDE at $x=3$; $t=2$; $u(3,2)$
$$\frac{du}{dt} = (1-t)\frac{\partial^2{u}}{\partial^2{x}}$$
on the domain $-\infty< x < \infty$, $t\ge0$
For the given boundary conditions
$$u(x,0) = \frac{1}{1+x^2}$$
I have tried solving using fourier trasnform but apparently there is a method to get the value without having to solve for the complete solution ?
On
Starting from @DaveNine's answer, it is possible to compute exactly $$f(t)=\frac{1}{\sqrt{2 \pi}}\frac{1}{\sqrt{t(2-t)}}\int_{-\infty}^{\infty}\left( \frac{1}{1 + (3-y)^2} \right)\exp\left[{\frac{-y^2}{2t(2-t)}}\right]\, dy $$ A CAS gives $$\int_{-\infty}^{\infty} \frac{e^{-k^2 y^2}}{1+(3-y)^2}\,dy=\frac{\pi }{2} e^{-(8+6 i) k^2} \left(e^{12 i k^2} \text{erfc}((1+3 i) k)+i\, \text{erfi}((3+i) k)+1\right)$$ Then $$f(t)=\frac{\sqrt{\frac{\pi }{2}} e^{\frac{4-3 i}{(t-2) t}} \left(-\text{erf}\left((1+3 i) \sqrt{\frac{1}{4 t-2 t^2}}\right)+i e^{\frac{6 i}{(t-2) t}} \text{erfi}\left((3+i) \sqrt{\frac{1}{4 t-2 t^2}}\right)+e^{\frac{6 i}{(t-2) t}}+1\right)}{2 \sqrt{(2-t) t}}$$
Let $t=2-10^{-k}$ and get the results $$\left( \begin{array}{cc} k & f\left(2-10^{-k}\right) \\ 1 & 0.10530808447 \\ 2 & 0.10052118333 \\ 3 & 0.10005201191 \\ 4 & 0.10000520012 \\ 5 & 0.10000052000 \\ 6 & 0.10000005200 \\ 7 & 0.10000000520 \\ 8 & 0.10000000052 \end{array} \right)$$
Expanding $f(t)$ as series around $t=2$, $$f(t)=\frac{1}{10}-\frac{13 (t-2)}{250}+\frac{149 (t-2)^2}{12500}+\frac{132 (t-2)^3}{15625}-\frac{68121 (t-2)^4}{3125000}+O\left((t-2)^5\right)$$
It is possible to get a convolution integral that solves the problem via fourier transform, namely
$$u(x,t) = \frac{1}{\sqrt{2 \pi}}\frac{1}{\sqrt{t(2-t)}}\int_{-\infty}^{\infty}\left( \frac{1}{1 + (x-y)^2} \right)\exp\left[{\frac{-y^2}{2t(2-t)}}\right]\, dy$$
It is readily seen, however, that this solution only exists for $0<t<2$, so the evaluation of the solution at $u(3,2)$ does not exist. Numerical evidence, however, seems to suggest that
$$\lim_{t\rightarrow 2^-} u(3,t) = \lim_{t\rightarrow 2^-}\frac{1}{\sqrt{2 \pi}}\frac{1}{\sqrt{t(2-t)}}\int_{-\infty}^{\infty}\left( \frac{1}{1 + (3-y)^2} \right)\exp\left[{\frac{-y^2}{2t(2-t)}}\right]\, dy = 1/10$$
EDIT: Note that $u(x,t) = u(x,2-t)$ from the above solution, so that it is clear that
$$\lim_{t \rightarrow 0^+} u(x,t) = u(x,0) = u(x,2) = \dfrac{1}{1+x^2}$$
Hence we can compute the limit explicitly to be
$$\lim_{t->2^-}u(3,t) = \dfrac{1}{10}$$