Solving a PDE such as $\frac{\partial}{\partial x}f(a+1,x)=f(a,x)+x\frac{\partial}{\partial x}f(a,x)$

Question

I am tying to understand what methods are available to find the analytical form of a function given a system of PDEs which it satisfies. For example, the function $f(a,x)=x^a$ is such that $f(a+1,x)=x^{a+1}=x f(a,x)$. Therefore I would like to understand what methods are available to go from this property: $$f(a+1,x)=x f(a,x)$$ to this definition: $$f(a,x)=x^a$$

My understanding is that derivation can lead to differential equations which can sometimes be solved. For example, the above equation for the property leads to: $$\frac{\partial}{\partial x}f(a+1,x)=f(a,x)+x\frac{\partial}{\partial x}f(a,x)$$

One difficulty I have is figuring out how to use this to find the analytical form of f (and since I am not advanced in Calculus I use Mathematica to help when necessary). Another difficulty I am encountering is related to the number of equations which are needed in this case; I do not expect the above property to necessarily yield a unique solution, but how many such properties are required to get at least one solution such as the one I started from? Also, I could just as easily add conditions by taking the partial derivatives for $a$, but I am not sure if this helps... Finally, in my browsing for answers for this, I am wondering if this needs to be considered a "Delay Differential Equation" rather than an ODE (or system of ODEs)...

The problem I am hoping to solve once I understand how to do this simpler version described aboe is actually a bit more general since it relates to the function $f(a,b,x)=x^a(1-x)^b$, which has the similar properties $x f(a,b,x)=f(a+1,b,x)$ and $(1-x) f(a,b,x)=f(a,b+1,x)$. Therefore I am hoping that the answer to this question will enable me to generalize this method to such a situation.

I have asked a similar question on the Mathematica forum since I am trying to use some of the functions provided by Mathematica to help solve this, and I hope that disclosing this will not be considered cross-posting but rather complementary posting as I am missing two sets of tools to solve this: some mathematical understanding related to how to elicit functions from PDEs, and some Mathematica knowledge as to how it can help me solve the problem once I understand the methodology.

I have also found the following related posts, but solutions seem to be more involved and (perhaps) more problem-specific:

• Constructing analytic solutions to the delay differential equation $f'(x) = x f(x-1) - f(x)$
• Is there a rational function $f(x,a)$ such that $\int_0^1 f(x,a) dx = \frac{\ln(a)}{a}$

Finally, this post relating to the number of conditions will likely help me determine the number of equations needed, though I have trouble understanding which equations actually "add value" to the solve (for example, I doubt that extending to a+2, a+3, ... will help): Is there any theorem that tells us how many ICs or BCs are needed for getting the determine solution of a PDE or a set of PDEs?

*** [EDIT 1] ***

One approach I have tried since posting requires making an assumption regarding the relation between $\frac{\partial}{\partial x}f(a,x)$ and $\frac{\partial}{\partial x}f(a+1,x)$. I have tried the following, which is obviously very much informed by the fact that I know the solution: $$ \frac{\partial}{\partial x}f(a+1,x)=(u+vx)\frac{\partial}{\partial x}f(a,x) $$

This reduces the problem to: $$ f(a,x)=(u+vx-x)\frac{\partial}{\partial x}f(a,x) $$

I write the following in Mathematica:

DSolve[(u + v x) D[f[a, x], x] == f[a, x], f[a, x], {a, x}]

And the result is $f[a, x] = (u + v x)^\frac1v C[1][a]$, which is a result I do not understand.

Although doing this exercise was not super helpful, it did get me to notice that I should perhaps bring the following more informative property into the mix: $$ x^n f(a,x)=f(a+n,x) $$ (and if it helps for my problem, I am ready to assume that n>0 as well as a>0).

Answer 1

Fourier transform will be your best ally to tackle the translation in $a$ without making the differential equation in $x$ more difficult.

Indeed, let's recall that $\mathfrak{F}[f(a+1,x)](t) = f(t,x)e^{it}$. In consequence, your differential equation becomes $$ \frac{\partial}{\partial x}f(t,x)e^{it} = f(t,x) + x\frac{\partial}{\partial x}f(t,x), $$ which can be rewritten as $$ \frac{\partial}{\partial x}\left((e^{it}-x)f(t,a)\right) = 0, $$ hence $$ f(t,x) = \frac{C}{e^{it}-x}, $$ where $C$ is constant. Then, the inverse Fourier transform is given by $$ f(a,x) = \int_\mathbb{R}\frac{\mathrm{d}t}{2\pi}\frac{Ce^{ita}}{e^{it}-x} $$ The (simple) poles of this integrand are located at $t_k = 2\pi ik - i\ln x, k\in\mathbb{Z}$, and their residues are $$ \mathrm{Res}_{z=t_k}\frac{e^{iza}}{e^{iz}-x} = \lim_{z\rightarrow t_k}\frac{(z-t_k)e^{iza}}{e^{iz}-x} = \lim_{z\rightarrow t_k}\frac{ia(z-t_k)e^{iza} + e^{iza}}{e^{iz}} = e^{it_k(a-1)} = e^{-2\pi k(a-1)}x^{a-1}, $$ thanks to L'Hospital's rule. However, only the residues of singularities with non-negative imaginary part (which depends on $x$ unfortunately) are to be taken into account in the final result, hence
$$ f(a,x) = iCx^{a-1}\sum_{k\in\mathbb{Z}}e^{-2\pi k(a-1)}H\left(k-\frac{\ln x}{2\pi}\right), $$ where $H$ is the Heaviside function. I'm not sure that there is a "more closed" form nonetheless.

Answer 2

On your first question about how to go from $$ f(a+1,x) = xf(a,x) $$ to $f(a,x) = x^a$, it is important to realize that is not implied--I mean there are many solutions other than $x^a$, as in the answer by Abezhiko.

Take the case $a\ge 0$. Take any function $f_0(a,x)$ defined for $0\le a\le 1$ and whatever $x$ you like, the only requirement (to make $f$ continuous in $a$) being $$ f_0(1,x) = xf_0(0,x). $$ Examples are $$ f_0(a,x) = \sin^2(\pi a), \qquad (a-1)g(x)+axg(x) {\rm \ \ any\ function\ } g. $$ Set $$ f(a,x) = f_0(a,x) {\rm\ when\ } 0\le a\le 1. $$ Then your first equation defines $f$ recursively for $a>1$: $$ f(a,x) = xf(a-1,x) = xf_0(a-1,x) {\rm \ when\ } 1 \le a \le 2, $$ $$ f(a,x) = xf(a-1,x) = x^2f_0(a-2,x) {\rm \ when\ } 2 \le a \le 3, $$ etc. In a nutshell, $$ f(a,x) = x^{[a]}f_0(a-[a],x). $$ That works because when you add 1 to $a$ the $f_0$ factor doesn't change but the exponential gives another factor of $x$.