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Consider $$ \big(1+ t\partial_t\big) \left(\partial^3_x+ {6\over x}\partial^2_x + {6\over x^2}\partial_x\right)A(x,t)+ {t\over (1-x)^3} A(x,t)=0 $$ with the initial condition $A(x,0)=1$. In a small $t$ expansion

$A(x,t)= 1+ a_1(x)t + a_2(x) t^2 + a_3(x) t^3+...$

additional conditions are $a_i(0)=0$, where $i=1,2,3,..$.

The equation is invariant under the transformations $x \to {x\over x-1},~ t\to -t$, which fix the ratio $(1,6,6)$ in the $x$-derivative part.

This PDE looks like a Fuchsian-type equation, but normally Fuchsian equations assume only one variable – Does anyone know if this type of "generalised Fuchsian-type equation" was studied in the math literature?

I am interested in the PDE's analytic solutions, especially solutions that are non-perturbative in $t$, and symmetry properties.

Note added:

For a slightly simpler PDE $$ \big(1+ t\partial_t\big) \left(\partial^3_x+ {6\over x}\partial^2_x + {6\over x^2}\partial_x\right)A(x,t)+ t A(x,t)=0 $$ with the same boundary conditions, I found that the exact solution is given by a hypergeometric ${_0F_3}$ function: $$ A(x,t)=\;{_0F_3}\left(~;{{4\over 3}, {5\over 3},2}; -{x^3 t\over 27}\right). $$ This I believe corresponds to the small $x$ limit of the exact solution to the first PDE.

I adopted an indirect computation to arrive at the above solution so if anyone can re-derive this solution in a few lines I'd be happy to know! (If one knows the solution only depends on $x^3 t \equiv w$ one can set $A=A(w)$ then the PDE becomes an OPE which can be solved by Mathematica directly.)

In any case, I wonder if there are some methods which help obtain the exact solution to the original PDE.

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    $\begingroup$ FWIW setting $A(x,t) = (x^2t)^{-1}F(1-x,t)$ gives a much simpler-looking equation for $F(y,t)$: $y^3\partial_t \partial_y^3 F(y,t)=F(y,t)$. $\endgroup$ Commented Mar 6 at 15:40
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    $\begingroup$ The simpler PDE is then $\partial_t \partial_y^3 F(y,t)=F(y,t)$. In either case, I think this form highlights that your equation may not have a unique solution since without additional constraints, $\partial_y^3$ is not invertible. $\endgroup$ Commented Mar 6 at 15:46
  • $\begingroup$ Thanks yes – that leads to the equation form I posted in mathoverflow.net/questions/465993/a-4th-order-linear-pde. I realised that writing the PDE in terms of $A(x,t)$ might be more useful to talk about BCs or make a potential connection to Fuchsian-type equations.. $\endgroup$
    – Math2024
    Commented Mar 6 at 15:54
  • $\begingroup$ To clarify, I added the additional constraints (used also in the simpler PDE) in the post, $i.e. a_i(0)=0$. $\endgroup$
    – Math2024
    Commented Mar 6 at 16:09
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    $\begingroup$ $x^3t$ can be seen to be special in the simplified equation by a scaling argument. $A(kx,lt)$ is also a solution when $k^3=l^{-1}$. So if it happens that all the functions $A(kx,k^-3t)$ are the same that suggests looking for a function of $x^3t$. $\endgroup$ Commented Mar 6 at 22:33

3 Answers 3

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In your simplified case, I don't see how $A(x,0) = 1$. In fact, the overall factor of $t$ should for the solution to vanish for all $x$ at $t=0$.

Actually, I think there is no solution to your boundary value problem, at least not as written. Suppose that $A(x,t)$ has a (possibly distributional) Fourier transform in $y = \log x$ (or equivalently a Mellin transform in $x$), so that $A(x,t) = \int dk\, \alpha(k,t) e^{ikx}$. To satisfy the boundary condition, $\alpha(k,0) = \delta(k)$. Rewriting the simplified equation in terms of $y$, it has coefficients independent of $y$. Taking the Fourier transform of the equation then gives $$ \partial_t (t \alpha(k,t)) + \frac{(t \alpha(k,t))}{P(k)} = 0 \iff \partial_t (e^{t/P(k)} t \alpha(k,t)) = 0 , $$ where $P(k)$ obtained by Fourier transforming the $\partial_y$-dependent part of the operator acting on first term. The solution and its behavior for small $t$ must be of the form $$ \alpha(k,t) = \beta(k) \frac{e^{-t/P(k)}}{t} \sim \frac{\beta(k)}{t} - \frac{\beta(k)}{P(k)} + O(t) . $$ To satisfy the boundary condition $\lim_{t\to 0} \alpha(k,t) = \delta(k)$, you need $\beta(k) = 0$ and $\beta(k)/P(k) = \delta(k)$ at the same time, which is impossible.

I suspect that a similar argument would work also in the original problem. Though you would need to use an eigen-function expansion for a more complicated operator instead of a Fourier transform.

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  • $\begingroup$ Sorry, there was a typo: I meant $$ A(x,t)=\;{_0F_3}\left(~;{{4\over 3}, {5\over 3},2}; -{x^3 t\over 27}\right). $$ (i.e. no overall factor x^2 t) I also fixed this typo in the post. One can verify that this is an exact solution with $A(x, 0)=1$. $\endgroup$
    – Math2024
    Commented Mar 6 at 12:54
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    $\begingroup$ The argument from this answer would imply that this exact $A(x=e^y,t)$ solution cannot have a distributional Fourier transform in $y$ (unless I made a mistake). I don't know the properties of hypergeometric functions well-enough to immediately see why, but perhaps you can identify a reason. This might give a clue as to what function space to expect a solution in (if any). $\endgroup$ Commented Mar 6 at 15:35
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Non really an answer but a long comment with some (hopefully useful) suggestions. The equation you are studying seems tractable by using the method of multidimensional Mellin transform described by Szmydt and Bogdan in reference [1] below. In particular a general solution to your slightly simpler example can possibly be expressed by a multiple inverse Mellin transform (even if the $tA(x,t)$ poses some problem to deal with). Nevertheless I doubt that, using the same techniques described in the monograph will allow you to produce exact solutions: at most you'll possibly get an asymptotic solution for large $x$, $t$ or a series solution. Well, my two cents.

Reference

[1] Zofia Szmydt, Bogdan Ziemian, The Mellin transformation and Fuchsian type partial differential equations (English), Mathematics and Its Applications, East European Series 56, Dordrecht-Boston-Lancaster: Kluwer Academic Publishers. xiv, 223 p. (1992), ISBN: 0-7923-1683-5, MR1196461, Zbl 0771.35002.

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Here's one way to get the hypergeometric function for the "simpler" equation:

Consider the operator $x^3 (1 + t\partial_t)(\partial^3_{xxx} + \frac{6}x \partial^2_{xx} + \frac{6}{x^2} \partial_x)$, we can rewrite it as $$ (1 + T)(X+2)(X+1)X $$ where $T = t\partial_t$ and $X = x\partial_x$. These two operators have $t^\beta$ and $x^\alpha$ as eigenfunctions.

Your simpler equation is $$ (1 + T)(X + 2)(X + 1) X A + x^3 t A = 0 $$ assuming there is a series expansion of the form $$ A = \sum c_{\alpha\beta} x^\alpha t^\beta $$ the equation reduces to the recurrence $$ \alpha(\alpha + 1)(\alpha + 2) (1+\beta) c_{\alpha\beta} + c_{(\alpha-3)(\beta-1)} = 0$$ Your boundary conditions (at $x = 0$ and at $t = 0$) would require $$ c_{00} = 1, \quad c_{\alpha 0 } = c_{0\beta} = 0 \text{ for } \alpha,\beta > 0 $$ This is compatible with the only non-vanishing terms being those for which $\alpha = 3 \beta$; writing $b_\beta = c_{3\beta,\beta}$ the recurrence is $$ 3\beta(3\beta + 1)(3\beta + 2)(\beta + 1) b_\beta = - b_{\beta - 1} $$ This procedure actually gets a somewhat simpler power series, $$ A(x,t) = A_{\mathrm{simp}}(x,t) = \sum_{k = 0}^\infty \frac{(-1)^k\cdot 2}{(3k+2)! (k+1)!} x^{3k}t^k $$ (which is of course equivalent to your hypergeometric series expansion)


A quick word on the boundary conditions. The requirement that $c_{\alpha 0} = 0$ for $\alpha > 0$ implies via the equation that $c_{\alpha \beta} = 0$ whenever $\alpha > 3\beta$.

However, below the "diagonal" we do not have uniqueness; to get uniqueness you need to prescribe values of those $c_{\alpha\beta}$ with $\alpha = \{1,2\}$ and $\beta > 0$. Above we've made a choice to set such $c_{\alpha\beta} = 0$; this in turn implies that $c_{\alpha\beta} = 0$ when $3\beta > \alpha$.

For each other choice of the $c_{\alpha\beta}$ values with $\alpha = \{1,2\}$ one gains another series solution. Essentially the issue is that your equation is third order in $x$ within your principal part, and from the Fuchsian perpsective $0$ is a regular singular point, and so you expect there to be 3 independent solutions to the homogeneous problem "at each order in $t$".


For the original problem, the equation is

$$ (1 + T)(X+2)(X+1)X A + \frac{x^3t}{(1-x)^3} A = 0 $$

Let's set $\xi := \frac{x}{1-x}$ and try to expand $A$ in series form as $$ A = \sum c_{\alpha\beta} \xi^\alpha t^\beta$$ Conveniently we find that $$ X(\xi) = \xi + \xi^2$$ which leads to the following recursion relation (where I use the Pochhammer symbol notation) $$ (1+\beta)(2+\alpha)_3 c_{\alpha\beta} + 3(1+\beta)(1+\alpha)_3 c_{(\alpha-1)\beta} + 3(1+\beta) (\alpha)_3 c_{(\alpha-2)\beta} + (1+\beta)(\alpha-1)_3 c_{(\alpha - 3)\beta} + c_{(\alpha - 3)(\beta-1)} = 0 $$

The coefficients can be uniquely solved if one prescribes the boundary conditions $c_{00} = 0$, $c_{\alpha 0} = c_{0\beta} = 0$ for all $\alpha, \beta> 1$ (as you did) augmented with the choice that $c_{1\beta} = c_{2\beta} = 0$ for all $\beta$.

Unfortunately, the decay property of the coefficients is somewhat worse (at least, I cannot prove that it is better). In the "simplified problem" you have that as a function of $x^3 t$ the corresponding series has an infinite radius of convergence; this is reflected in you having found a way to write the solution as a hypergeometric function.

For the recursion relation in the present problem, the best I can do is something like $|c_{\alpha\beta}| \leq \frac{M^{(\alpha - 3\beta - 1)_+}}{\beta! (3\beta)!}$ (maybe not quite right, just did it very quickly), where $M$ is a global constant. If true this will allow the series to converge for all $\xi \lesssim \frac{1}{M}$ (and all $t$).

Assuming what I wrote above is correct, this will also justify your expectation that "as $\xi \to 0$ the solution converges to $A_{\mathrm{simp}}$ of the simplified problem."

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  • $\begingroup$ Thanks, I agree $T$ and $X$ make the PDE more compact. The original way I obtained the solution (to the simpler PDE): I started with the original PDE and solved for the perturbative solutions $a_i (x)$ (say $i=1$ to $i=10$) using the BCs described in the post. In this process I did not need any additional conditions. Then, I focused on the leading small-$x$ contributions from $a_i(x)$ to find a pattern allowing me to resum to get the hypergeometric function. After that, I found the hypergeometric function satisfies the simpler PDE. $\endgroup$
    – Math2024
    Commented Mar 7 at 15:17
  • $\begingroup$ For the original PDE, I am not sure if $\xi^{\alpha}$ is the best form – in the small $t$ expansion, the solutions $a_i(x)$ have $ln (1-x)$ structure. $\endgroup$
    – Math2024
    Commented Mar 7 at 15:17
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    $\begingroup$ @Math2024 (two above) if you solve pertubatively, you have to deal with the fact that your ODE for $x$ has nontrivial kernel (e.g. the function $x$) that is not ruled out by the BC that you provided. I don't see how you managed to get a unique solution. (one above) for small $x$ you have $\ln(1-x) \approx -x \approx \frac{x}{x-1}$. For $x$ closer to $1$ I don't see your point as unless you can show convergence of the series expansion these kinds of differences are immaterial. $\endgroup$ Commented Mar 8 at 15:42
  • $\begingroup$ Thanks. If one focuses on solving the original PDE in a small $t$ expansion with the given boundary conditions (including $a_i(0)=0$ for all $i$), one can directly verify that the solutions are uniquely fixed. However, I do not have anything to say about non-perturbative solutions at the moment. I see your point about 𝜉 now, but I guess I am interested in the solution with a more general 𝑥. I am not sure about $M$ yet. $\endgroup$
    – Math2024
    Commented Mar 8 at 16:10

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