I was trying to solve this PDE, but I haven't been able to fully solve the exercise: $$\partial_t f(x,t) = 2t\partial_{xx}^2f(x,t) + \delta(t^2-6t+5)\frac{e^{-\frac{x^2}{2}} }{\sqrt{2\pi}}, f(x,0) = \frac{e^{-\frac{x^2}{2}} }{\sqrt{2\pi}} $$ I started by noticing that $\delta(t^2-6t+5) = \frac{1}{4}[\delta(t-5)+\delta(t-1)]$ So I solved the pde for $t < 1$, which reduces to the simpler form: $$\partial_t f(x,t) = 2t\partial_{xx}^2f(x,t) $$ Using the Fourier transform, namely: $$f(x,t) = 1 / \sqrt{2\pi}\int_{-\infty}^{+\infty}\hat{f}(k,t)e^{ikx}dk, \hat{f}(k,t) = 1 / \sqrt{2\pi} \int f(x',t)e^{-ikx'}dx'$$ one easily gets to the easy ODE: $$\frac{d}{dt}\hat{f}(k,t) = -2k^2t\hat{f}(k,t) \Rightarrow \hat{f}(k,t) = \hat{f}(k,0)e^{-k^2t^2}$$ Now I need to compute $$ \hat{f}(k,0) = \frac{1}{\sqrt{2\pi}}\int f(x',0)e^{-ikx'}dx' = \frac{1}{2\pi}\int e^{-x'^2/2}e^{-ikx'}dx'$$ Using the formula for the Gaussian integral I got: $$ \hat{f}(k,0) = \frac{1}{\sqrt{2\pi}} e^{-k^2/2}$$ Therefore I finally obtain, anti-transforming: $$f(x,t) = \frac{1}{2\pi} \int e^{ikx}e^{-k^2(1/2+t^2)} $$ Which is again easily solvable using the well known formula $\int e^{i\beta y}e^{-ay^2}dy =\sqrt{\frac{\pi}{a}}e^{-\frac{\beta^2}{4a}}$. The result is: $$ f(x,t) = \frac{1}{2\pi(1+2t^2)}e^{-\frac{x^2}{4(1/2 + t^2)}}$$ I checked and it was all ok. Now the challenging part begins: I tried to consider the equation for $t \in ]1,5[$. The PDE reduces to: $$ \partial_t f(x,t) = 2t\partial_{xx}^2f(x,t) + \frac{1}{4}\delta(t-1)\frac{e^{-\frac{x^2}{2}} }{\sqrt{2\pi}} $$ I used the Fourier transform again and got: $$ \frac{d}{dt}\hat{f}(k,t) = -2k^2t\hat{f}(k,t) + \frac{1}{4}\delta(t-1)\frac{1}{2\pi} \int e^{-x^2/2} e^{ikx}dx \tag{1}$$ which leads to: $$ \frac{d}{dt}\hat{f}(k,t) = -2k^2t\hat{f}(k,t) + \frac{1}{4 \sqrt{2\pi}}\delta(t-1)e^{-k^2/2}$$ I tried to solve the ODE like this (making use of the sifting property in the last equality): $$ \hat{f}(k,t) = \hat{f}(k,0)e^{-k^2t^2} + \frac{1}{4 \sqrt{2\pi}}\int_{0}^{t} e^{-k^2(t-t')^2}\delta(t'-1)dt' = \hat{f}(k,0)e^{-k^2t^2} + \frac{1}{4 \sqrt{2\pi}} e^{-k^2(t-1)^2} $$ This would lead me to: $$ \hat{f}(k,t) = \frac{1}{\sqrt{2\pi}}e^{-k^2/2}e^{-k^2t^2} + \frac{1}{4\sqrt{2\pi}} e^{-k^2(t-1)^2}$$ Now using the anti-transform would mean solving an impossible integral. I can assume my procedure was wrong from the start, or maybe I wrongly solved the ODE $(1)$ which again wouldn't surprise me. If someone could help me and clarify how to solve this PDE for $t > 1$, I'd be no far from grateful\ P.S. I realized after @Winther's comment that I blindly assumed the integral was wrt $x$. Since I am integrating wrt $k$, I would obtain the following solution, deriving from the sum of two Gaussian integrals, like the ones before: $$ f(x,t) = \frac{1}{\sqrt{2\pi(1+2t^2)}}e^{-\frac{x^2}{2+4t^2}} + \frac{1}{\sqrt{64\pi(t-1)^2}}e^{-\frac{x^2}{4(t-1)^2}}$$ Nevertheless, the solution for $t>1$ is, according to my professor (doesn't tell me how to get this result though): $$ f(x,t) = \frac{1}{\sqrt{2\pi(1+2t^2)}}e^{-\frac{x^2}{2+4t^2}} + f_1(x,t)$$, where $$ f_1(x,t) = \frac{1}{4\sqrt{(4(t^2-1)+2)\pi}}e^{-\frac{x^2}{4(t^2-1)+2}}$$ whereas for $t>5$ the solution is: $$f(x,t) = \frac{1}{\sqrt{2\pi(1+2t^2)}}e^{-\frac{x^2}{2+4t^2}} + f_1(x,t) + f_2(x,t) $$, where $$f_2(x,t) = \frac{1}{4}\frac{1}{\sqrt{(4(t^2-5)+2)\pi} }e^{-\frac{x^2}{4(t^2-5)+2}} $$
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- 1$\begingroup$ "Now using the anti-transform would mean solving an impossible integral" Assuming you did it correctly then this function is just a sum of two terms that both are on the form $e^{-ak^2}$. If you know how to take the Fourier transform of a general gaussian then you can use this to take the inverse transform. Note that the fact that $a$ ($=(1/2 + t^2$ for ther first term and $=(t-1)^2$ for the second term) depends on $t$ is irrelevant for taking the inverse transform (which is wrt $x$). $\endgroup$Winther– Winther2023-07-03 14:27:01 +00:00Commented Jul 3, 2023 at 14:27
- $\begingroup$ Did you mean $\partial_{xx}$ or $\partial_x^2$ rather than $\partial_{xx}^2$? $\endgroup$Semiclassical– Semiclassical2023-07-03 14:29:43 +00:00Commented Jul 3, 2023 at 14:29
- 1$\begingroup$ $$\frac{\partial^2 f(x,t)}{\partial x^2}$$. You made me realized the integral is wrt $k$ not $x$, so I would get two gaussian integrals. $\endgroup$Claudio– Claudio2023-07-03 14:35:20 +00:00Commented Jul 3, 2023 at 14:35
- $\begingroup$ @Gonçalo you’re right, the right formula is indeed: $$ y(0)e^{\int -2k^2tdt}+\frac{1}{4\sqrt{2\pi}}e^{-k^2/2}\int_{0}^{t}e^{k^2t’^2}\delta(t’-1)dt’$$ which gives the correct solution. Problem’s still there. It just changes the second Gaussian integral to something like this $$\alpha e^{-\frac{x^2}{4(t^2+1)}}$$, which is still far from the expected solution. $\endgroup$Claudio– Claudio2023-07-03 16:47:08 +00:00Commented Jul 3, 2023 at 16:47
- $\begingroup$ Better take fourier tranform of your advisor solution and post it in the question and compare it with the solution you are getting and reverse engineer to find your problem. $\endgroup$user213498– user2134982023-07-03 17:30:42 +00:00Commented Jul 3, 2023 at 17:30
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1 Answer
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I think comments helped you with your problem but i see you are having to do lot of integrals to get the answer. So here is a thought for simplification.
Solve $$\partial_u g(x,u) = \partial^2_x g(x,u)$$
and now form the equation $f(x,t) = g(x,t^2)$
$$\partial_t f(x,t) = \partial_t g(x,t^2) = 2t \partial_u g(x,u) |_{u=t^2}$$
$$2t \partial_x^2 f(x,t) = 2t \partial^2_x g(x,t^2) = 2t \times \partial^2_x g(x,u)|_{u=t^2}$$
Hence $f(x,t) = g(x,t^2)$ is your solution.
So i think its enough to solve for $$\partial_u g(x,u) = \partial^2_x g(x,u)$$
As an example: $g(x,u) = e^{x} e^{t}$ is a solution and hence $$f(x,t) = e^{x} e^{t^2}$$ is a solution but does not satisfy boundary conditions.
See if these help writing it in simple form.