# Generalized Fuchsian-type PDE

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.

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.

• FWIW setting $A(x,t) = (x^2t)^{-1}F(1-x,t)$ gives a much simpler-looking equation for $F(y,t)$: $y^3\partial_t \partial_y^3 F(y,t)=F(y,t)$. Commented Mar 6 at 15:40
• 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. Commented Mar 6 at 15:46
• 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.. Commented Mar 6 at 15:54
• To clarify, I added the additional constraints (used also in the simpler PDE) in the post, $i.e. a_i(0)=0$. Commented Mar 6 at 16:09
• $x^3t$ 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 $x^3t$. Commented Mar 6 at 22:33

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.

• 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 $A(x, 0)=1$. Commented Mar 6 at 12:54
• 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). Commented Mar 6 at 15:35

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.

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."

• 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. Commented Mar 7 at 15:17
• 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. Commented Mar 7 at 15:17
• @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. Commented Mar 8 at 15:42
• 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. Commented Mar 8 at 16:10