Help with Partial Differential Equation Arising from Generating Function

Question

I am considering a sequence of polynomials ${f_0}(x), {f_1}(x), {f_2}(x)\dots$ satisfying the recursive property ${f_k}(x)=(x^2+1){f_{k-1}}'(x)$, where ${f_0}(x)=x$.

Define $F(x,y)=\sum _{ n=0 }^{ \infty }{ \frac { 1 }{ n! } { f }_{ n }(x){ y }^{ n } } $, the exponential generating function for ${f_k}(x)$.

Using standard techniques, I found that $F(x,y)=x+(x^{ 2 }+1)\int { \frac { \partial F }{ \partial x } dy }$ (this might be wrong as I'm not terribly familiar with generating functions, but my question stands nonetheless).

My question is if there is there exists a (preferably nice) general solution to the above PDE?

Answer 1

$$F(x,y)=\sum _{ n=0 }^{ \infty }{ \frac { 1 }{ n! } { f }_{ n }(x){ y }^{ n } } $$ $$\frac{\partial F}{\partial y} = \sum _{ n=1 }^{ \infty }{ \frac { 1 }{ n! } { f }_{ n }(x)n{ y }^{ n-1 } } = \sum _{ n=1 }^{ \infty }{ \frac { 1 }{ (n-1)! } { f }_{ n }(x){ y }^{ n-1 } }$$ ${f_n}(x)=(x^2+1){f_{n-1}}'(x)$, where ${f_0}(x)=x$. $$\frac{\partial F}{\partial y} = \sum _{ n=1 }^{ \infty }{ \frac { 1 }{ (n-1)! }(x^2+1) { f _{ n-1 }}'(x){ y }^{ n-1 } } = (x^2+1)\sum _{ n=1 }^{ \infty }{ \frac { 1 }{ (n-1)! } { f _{ n-1 }}'(x){ y }^{ n-1 } }$$ $$\frac{\partial F}{\partial y} =(x^2+1)\sum _{ k=0 }^{ \infty }{ \frac { 1 }{ k! } { f _{ k }}'(x){ y }^{ k } }$$ $\frac{\partial F}{\partial x}=\sum _{ k=0 }^{ \infty }{ \frac { 1 }{ k! } { f _{ k }}'(x){ y }^{ k } } \quad\to\quad \frac{\partial F}{\partial y} =(x^2+1)\frac{\partial F}{\partial x}$ $$(x^2+1)\frac{\partial F}{\partial x}-\frac{\partial F}{\partial y}=0$$ Solving this PDE:

Set of characteristic ODEs : $\quad \frac{dx}{x^2+1}=\frac{dy}{1}=\frac{dF}{0}$.

From $dF=0 \to$ equation of a first family of characteristic curves : $\quad F=c_1$

From $\frac{dx}{x^2+1}=y \to$ equation of a second family of characteristic curves : $\quad \tan^{-1}(x)-y=c_2$

General solution of the PDE : $$F(x,y)=\Phi\left(\tan^{-1}(x)-y \right) \tag 1$$ where $\Phi$ is any differentiable function.

Condition : $F(x,0)=f_0(x)=x \quad\to\quad \Phi\left(\tan^{-1}(x)-0 \right)=x$

Let $t=\tan^{-1}(x)\quad\to\quad \Phi\left(t \right)=\tan(t)$

So, the function $\phi$ is known and can be put into (1) : $$F(x,y)=\tan\left(\tan^{-1}(x)-y \right)$$

Answer 2

After differentiating in $y$, you get the PDE $(x^2+1)\frac{\partial F}{\partial x}+\frac{\partial F}{\partial y}=0$. The ODE for the characteristic curves is $\frac{dx}{x^2+1}=dy$, or $d(y-\arctan(x))=0$, so $F(x,y)=\Phi(y-\arctan(x))$ is the general solution, which you could place in the equivalent form $$ F(x,y)=\Phi\left(\frac{\tan y-x}{1+x\tan y}\right). $$ Here, $\Phi$ is an arbitrary function. Not sure how nice it is, but it's explicit!