Solving the PDE $x_{1}\dfrac{\partial f}{\partial x_{1}}+x_{2}\dfrac{\partial f}{\partial x_{2}}=e^{f(x_{1},x_{2})}-\alpha.$

Question

This post is closely related to Dirichlet to Neumann operator in the unit ball with Fourier Analysis.

I have transformed the exercise in the post above into a problem of finding solution of PDE:$$x_1\frac{\partial f}{\partial x_1} + x_2 \frac{\partial f}{\partial x_2} = e^{f(x_1,x_2)}-\alpha \text{ for } \alpha>0.$$

This is an exercise in Fourier analysis, so I am not prepared too many knowledge in differential equation.

There are two hints:

1. Use the Fourier Expansion; 2. Separate the argument into $\alpha\in\mathbb{N}$ and $\alpha\notin\mathbb{N}$.

However... I don't have any idea about how to solve this..

Any idea? Thank you!

Edit:

In the link above, it suggested that if $\alpha\notin\mathbb{N}$, then $f=\log\alpha$, and asked the reader to further discover what happens if $\alpha\in\mathbb{N}$.

So if $\alpha\notin\mathbb{N}$, we have $$x_1\frac{\partial}{\partial x_1}+x_2\frac{\partial f}{\partial x_2}=0 \text{??}$$

Edit 2:

Below is how I convert the exercise in the link above to this PDE:

Within the context of this exercise, we have the coinciding solution of the Dirichlet problem $$\Delta u=0 \text{ on } B_1$$ $$u=f \text{ on } \partial B_1 =\mathbb{S}^1$$ and of the Neumann problem $$\Delta u=0 \text{ on } B_1$$ $$\frac{\partial u}{\partial\nu}=e^f-\alpha \text{ on } \partial B_1=\mathbb{S}^1,$$ where $\dfrac{\partial u}{\partial\nu}= \nabla u\cdot \nu$ is the normal derivative of $u$ at the boundary with respect to the unit outer normal direction $\nu$.

Now, note that for a point $(x_1,x_2)\in\partial B_1 = \mathbb{S}^1$, we always have $\nu=(x_1,x_2)$. Also, by the solution of the Dirichlet problem, we know that $u=f$ on $\partial B_1=\mathbb{S}^1$, and thus on the boundary we have $$\frac{\partial u}{\partial\nu} = \nabla u\cdot \nu=x_1\frac{\partial u}{\partial x_1} + x_2 \frac{\partial u}{\partial x_2} = x_1 \frac{\partial f}{\partial x_1} + x_2\frac{\partial f}{\partial x_2},$$ but the boundary condition of Neumann problem is $$\dfrac{\partial u}{\partial\nu}=e^{f}-\alpha,$$ and thus we have $$x_1\frac{\partial f}{\partial x_1} + x_2\frac{\partial f}{\partial x_2} = e^{f(x_1,x_2)}-\alpha.$$

Edit 3: (initial value)

As "Ninad" pointed out, we need to an initial value to decide what is $C(t).$ And I believe perhaps the initial value is related to whether $\alpha$ is natural or not.

The exercise does not give what happens if $\theta=0$.

However, I missed one conditiona that $f\in C^\infty(\mathbb{S}^1)$, a infinity smooth $2\pi-$periodic function.

I don't know if this helps to provide us the initial value.

Answer 1

Use polar coordinates

$$x_1 = r\cos\theta$$

$$x_2 = r\sin\theta$$

to convert the PDE into an ODE since

$$x_1\frac{\partial}{\partial x_1} + x_2 \frac{\partial}{\partial x_2} = r\frac{\partial}{\partial r}$$

giving us the equation

$$r\frac{\partial f}{\partial r} = e^{f} - \alpha$$

which we can solve using separation of variables

$$ \frac{dr}{r} = \frac{df}{e^f-\alpha} = \frac{e^{-f}df}{1-\alpha e^{-f}} $$

$$\implies \log r + C(\theta) = \frac{1}{\alpha} \log| 1 - \alpha e^{-f} |$$

$$\implies f = \log\left(\frac{\alpha}{1-C(\theta)r^\alpha}\right)$$

If you have any initial conditions, you can apply them in clever ways in order to figure out what $C(\theta)$ has to be.

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$\textbf{EDIT}:$ Using the fact that $f$ has a harmonic extension, we can use the polar coordinates Laplacian:

$$\Delta u = \frac{1}{r}\frac{\partial}{\partial r}\left(r\frac{\partial u}{\partial r}\right) + \frac{1}{r^2}\frac{\partial^2 u}{\partial \theta^2}$$ to see what functions satisfy the harmonic condition.

$$\Delta f = \frac{\alpha^2 C(\theta) r^{\alpha -2}}{(1-C(\theta)r^\alpha)^2} + \frac{C''(\theta)r^{\alpha-2}}{1-C(\theta)r^\alpha} + \frac{(C'(\theta))^2r^{2\alpha - 2}}{(1-C(\theta)r^\alpha)^2} = 0$$

$$\implies \alpha^2 C(\theta) + C''(\theta) +\left[(C'(\theta))^2 - C(\theta)C''(\theta)\right] r^\alpha = 0$$

giving two ODEs that need to be simultaneously satisfied. Looking only at the first one, we have that

$$C''(\theta) + \alpha^2C(\theta) = 0 \implies C(\theta) = A\cos(\alpha\theta) + B\sin(\alpha\theta) $$

but notice that $C$ only has a nontrivial $2\pi$-periodic solution if $\alpha$ is an integer. Thus for noninteger $\alpha$ we can conclude that

$$C(\theta) = 0 \implies f = \log \alpha$$

For integer $\alpha$, plug and chug gives us the following equation:

$$(C'(\theta))^2 - C(\theta)C''(\theta) = 0 \implies \alpha^2 (A^2+B^2) = 0$$

which again gives us the trivial solution, which doesn't seem to be what your question implies happens.

Answer 2

$$x_1\frac{\partial f}{\partial x_1} + x_2 \frac{\partial f}{\partial x_2} = e^{f(x_1,x_2)}-\alpha \tag 1$$ Of course, solving in polar system simplifies the calculus. But it it is not necessary.

The Charpit-Lagrange characteristic ODEs are : $$\frac{dx_1}{x_1}=\frac{dx_2}{x_2}=\frac{df}{e^f-\alpha}$$ A first characteristic equation comes from solving $\frac{dx_1}{x_1}=\frac{dx_2}{x_2}$ : $$\frac{x_2}{x_1}=c_1$$ A second characteristic equation comes from solving $\frac{dx_1}{x_1}=\frac{df}{e^f-\alpha}$ : $$x_1^{-\alpha}\left(1-\alpha e^{-f} \right)=c_2$$ The general solution expressed on the form of implicite equation $c_2=F(c_1)$ is : $$x_1^{-\alpha}\left(1-\alpha e^{-f} \right)=F\left(\frac{x_2}{x_1}\right)$$ $F$ is an arbitrary function, to be determined according to some boundary condition.

Solving for $f$ leads to : $$\boxed{f(x_1,x_2)=\ln|\alpha|-\ln\left|1-x_1^{\alpha}F\left(\frac{x_2}{x_1}\right) \right|} \tag 2$$

It is easy to differentiate Eq.$(2)$ for $\frac{\partial f}{\partial x_1}$ and $\frac{\partial f}{\partial x_2}$ . Then putting them into Eq.$(1)$ and checking the equality proves that $(2)$ is solution of $(1)$.

Note that the particular case $F=0$ gives the trivial solution $f=\ln|\alpha|$

Note :

Since $F$ is an arbitrary function they are an infinity of equivalent forms of equations to express the solution. For example : $$f(x_1,x_2)=\ln|\alpha|-\ln\left|1-x_2^{\alpha}G\left(\frac{x_1}{x_2}\right) \right|$$ with arbitrary function $G$, related to arbitrary function $F$ through : $G(X)=X^{\alpha}F(1/X)$

Note :

In polar coordinates $\quad\begin{cases}x_1=\rho\cos{\theta}\\x_2=\rho\sin{\theta}\end{cases}$

$f(\rho,\theta)=\ln|\alpha|-\ln\left|1-\rho^{\alpha}\cos^{\alpha}(\theta)F\left(\cot(\theta)\right) \right|$ $$f(\rho,\theta)=\ln|\alpha|-\ln\left|1-\rho^{\alpha}H(\theta) \right|$$ where $H$ is an arbitrary function. If you want to have an example of smooth periodic solution on circle of radius $\rho$ , choose the function $H$ so that $H$ be periodic and $\big|\rho^{\alpha}H(\theta)\big|<1.$