1
$\begingroup$

Consider the following partial differential equation with an initial condition $u(x,0)=f(x)$:

\begin{equation} \frac{\partial}{\partial t} u(x,t)=g_{1}(x)\frac{\partial u}{\partial x}+g_{2}(x)\frac{\partial^2 u}{\partial x^2}+g_{3}(x)\frac{\partial^3 u}{\partial x^3} \end{equation}

where $g_{2}(x)=\frac{dg_{1}(x)}{dx}$, $g_{3}(x)=\frac{d^2g_{1}(x)}{dx^2}$, $x \in [a,b]$ and $t \in [0,1]$. Given that $f(x)$ and $g_{1}(x)$ are known functions, it is possible to numerically solve this IVP using the Euler method (assuming zero spatial derivatives at the boundaries). I, however, was wondering whether it would be possible to analytically find a kernel $k(x,\tau)$ such that the solution of this IVP at, e.g., $t=1$ be the following integral transform:

\begin{equation} u(x,1) = \int_{-\infty}^{\infty} k(x,\tau)f(\tau)d\tau \end{equation}

Thank you for your help. Guiding towards a reference would be highly appreciated as well.

Improve this question
$\endgroup$
4
  • $\begingroup$ Initially posted in math.stackexchange.com/q/4529237/1027701 $\endgroup$
    – Mirar
    Commented Sep 12, 2022 at 20:29
  • 1
    $\begingroup$ You certainly need to impose appropriate conditions on $g_1$. If, say, $g_1$ is a constant, then your PDE boils down to a transport equation with constant coefficients and it is well-known that the solution is NOT given by an integral kernel. If, however, $g_1$ is an affine function and hence your PDE is a heat equation with well-behaved drift term, then the solution will be given by $$u(x,t)=\int_a^b k(t;x,y)f(y)dy$$ for a family of $L^\infty$-kernels $(k(t;\cdot,\cdot))_{t>0}$ ("heat kernel") and, if $g_1$ is nice enough, it should be possible to find the heat kernel explicitly (Fourier?). $\endgroup$
    – Delio Mugnolo
    Commented Sep 13, 2022 at 14:09
  • $\begingroup$ Thank you for the comment. $g_{1}$ is a sufficiently smooth function that vanishes at $a$ and $b$. For instance, assume that $g_{1}$ is a Gaussian function. $\endgroup$
    – Mirar
    Commented Sep 13, 2022 at 14:48
  • $\begingroup$ Again, assuming "sufficiently smooth" won't in general help, as the example of a constant (!) $g_1$ shows. If you take something like a Gaussian, you will end up in a 3rd order equation, so you have to adapt your boundary conditions (only imposing zero spatial derivatives at the boundary won't suffice). On the whole line you can perhaps successfully Fourier-transform and then, if you're lucky, find an analytic solution for the corresponding ODE in dependence on $g_1$. But for bounded domains I see no hope - and generally no hope to find an analytic expression for your kernel, either. $\endgroup$
    – Delio Mugnolo
    Commented Sep 14, 2022 at 7:07

1 Answer 1

Reset to default
1
$\begingroup$

(Too long to be a comment.) $u(x,0)$ is not enough, you need to set some condition(s) on the $x$ coordinate. For instance, let's say that $u(0,t)=u(L,t)=0$ for some given $L$. In that case a natural ansatz is $u(x,t)=\sum_n c_ne^{\mathbb{i}\omega_n t}u_n(x)$, being $c_n$ determined by $u(x,0)$, and the eigenfunctions $u_n(x)$ satisfying the ODE $\mathbb{i}\omega_n u_n(x)=g(x)u_n'(x)+g'(x)u_n''(x)+g''(x)u_n'''(x)$ with $u_n(0)=u_n(L)=0$; the problem in this last equation is that the 3rd derivative is not self-adjoint on the space of functions satisfying the BC (and in general an odd derivative is not self-adjoint for sensible choices of Hilbert space) so it's hard to say something about the eigenvalues $\omega_n$. Also, as this is a 3rd order equation you need to supply one more condition. Even for simple polynomial values of $g(x)$ Mathematica does not return analytical solutions for $u_n(x)$, so I don't think that specific solutions can be written down.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .