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I'm trying to solve the following differential equation and I'm stuck at what it appears to be simple calculations. I'm terribly sorry if this turns out to be really simple.

$(1)$ $X(f)=2f$

where $X=x_1^2 \frac \partial {\partial x_1}-x_2^2 \frac \partial {\partial x_2}$ in $\Bbb R^2$ with the identity chart $Id_{\Bbb R^2}=(x_1,x_2)$

and $f:\Bbb R^2 \to \Bbb R$,

$(2)$ $f(cosθ,sinθ)=cosθ+sinθ$.

Let $φ^Χ_t(p)=(\frac {x}{1-tx},\frac {y}{1+ty})$, where $p=(x,y)$, be the flow of $Χ$ and by denoting $h(t)=f(φ^Χ_t(p))$ we can make $(1)$ look like $h'(t)=2h(t)$ which can be easily solved to:

$e^{2t}f(x,y)=f(\frac {x}{1-tx},\frac {y}{1+ty})$

Then by use of the initial condition $(2)$ we have

$e^{2t}(cosθ+sinθ)=f(\frac {cosθ}{1-tcosθ},\frac {sinθ}{1+tsinθ})$ (this is as far as I can go)

I tried setting $u = \frac {cosθ}{1-tcosθ}, v=\frac {sinθ}{1+tsinθ} $ but I haven't been able to isolate $u,v$ from $θ, t$

Can you give me any hints? Is there any trick I'm not thinking of?

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    $\begingroup$ So the PDE is as follows? $$\left( x_1^2 \frac \partial {\partial x_1}-x_2^2 \frac \partial {\partial x_2} \right) f(x_1, x_2) = 2 (x_1+x_2)$$ $\endgroup$
    – md2perpe
    Aug 18, 2018 at 8:16
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    $\begingroup$ I'm not sure if you are asking me or hinting me. if $p=(x,y)$ then the pde would be $x^2 \frac {\partial f}{\partial x}(x,y)- y^2 \frac {\partial f}{\partial y}(x,y)=2f(x,y)$ $\endgroup$ Aug 18, 2018 at 12:53
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    $\begingroup$ I'm trying to understand what the PDE looks like. Why do you write $f(\cos\theta, \sin\theta) = \cos\theta + \sin\theta$ and not $f(x_1, x_2) = x_1 + x_2$? $\endgroup$
    – md2perpe
    Aug 18, 2018 at 13:24
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    $\begingroup$ That's an initial condition. It has that formula only for points on the unit circle. It should give me a way to calculate $f$ $\endgroup$ Aug 18, 2018 at 13:28
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    $\begingroup$ So the differential equation is $$x^2 \frac {\partial f}{\partial x}(x,y)- y^2 \frac {\partial f}{\partial y}(x,y)=2f(x,y)$$ with the following boundary condition on the unit circle $x^2+y^2=1$: $$f(x, y) = x + y$$ $\endgroup$
    – md2perpe
    Aug 18, 2018 at 13:34

3 Answers 3

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The differential equation is:

$$x^2 \frac {\partial f}{\partial x}(x,y)- y^2 \frac {\partial f}{\partial y}(x,y)=2f(x,y).$$

Express $f(x, y)$ as

$$f(x,y) = X(x) Y(y),$$

so

$$x^2 Y \dfrac{dX}{dx} - y^2 X \dfrac{dY}{dy} = 2XY $$

$$x^2 \frac{1}{X} \dfrac{dX}{dx} - y^2 \frac{1}{Y} \dfrac{dY}{dy} = 2$$

$$x^2 \frac{1}{X} \dfrac{dX}{dx} = y^2 \frac{1}{Y} \dfrac{dY}{dy} + 2$$

You can define

$$ x^2 \frac{1}{X} \dfrac{dX}{dx} = -k,$$

which means

$$ X(x) = C_1 e^{k/x}.$$

For $Y(y)$ you have

$$ -k = y^2 \frac{1}{Y} \dfrac{dY}{dy} + 2,$$

which means that the solution is

$$Y(y) = C_2 e^{(2+k)/y}.$$

Yes, at some point, I have to apply boundary conditions, unfortunately, I don't understand the boundary conditions you are using, so I cannot keep solving it.

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  • $\begingroup$ I think it might be too hard solving it using pde techniques. I found this on a geometry lecture and tried to give it a go using flows, but I only went so far myself $\endgroup$ Aug 20, 2018 at 14:40
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$$x^2 \frac {\partial f}{\partial x}(x,y)- y^2 \frac {\partial f}{\partial y}(x,y)=2f(x,y).$$ Search for the general solution (without taking account of the boundary condition) with the method of characteristics :

The characteristic ODEs are : $$\frac{dx}{x^2}=\frac{dy}{-y^2}=\frac{df}{2f}$$ A first characteristic equation comes from $\frac{dx}{x^2}=\frac{dy}{-y^2}$ : $$\frac{1}{x}+\frac{1}{y}=c_1$$ A second characteristic equation comes from $\frac{dx}{x^2}=\frac{df}{2f}$ : $$e^{2/x}f=c_2$$ The general solution expressed on the form of implicite equation is : $$\Phi\left(\frac{1}{x}+\frac{1}{y}\:,\:e^{2/x}f \right)=0$$ where $\Phi$ is an arbitrary function of two variables. This function has to be determined later according to boundary conditions.

Or, equivalently on explicit form : $e^{2/x}f=F\left(\frac{1}{x}+\frac{1}{y} \right)$

$$f(x,y)=e^{-2/x}F\left(\frac{1}{x}+\frac{1}{y} \right)$$ where $F$ is an arbitrary function. This function has to be determined later according to boundary conditions.

BOUNDARY CONDIION :

In the original wording of the question, the boundary condition is not clearly defined. A discussion took place in the comments.

To the question : Is the boundary condition $f(x,y)=x+y$ on the unit circle $x^2+y^2=1$ ? the OP answered "that should be it", which is not really affirmative. So, this supposed boundary condition can be suspected to be mistaken.

Supposing that the boundary condition is $f(x,y)=x+y$ on the unit circle $x^2+y^2=1$, thus $y=\pm\sqrt{1-x^2}$ , my comment is :

The function $F$ has to be determined from : $$x\pm\sqrt{1-x^2}=e^{-2/x}F\left(\frac{1}{x}+\frac{1}{\pm\sqrt{1-x^2}} \right)$$

In fact, it is theoretically possible to find the function $F$ but the calculus is rather arduous and the function $F$ is very complicated. This draw to think that something might be wrong in the wording of the question. The OP should re-examine what is really the boundary condition. To help him, it should be necessary that the OP re-edit his question with a detailed explanation how he got the above boundary condition.

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  • $\begingroup$ Unfortunately I found this on a lecture in differential geometry and this is how it was stated. That is why I re-edited my post clarifying that I used flows instead of actual pde techniques. $\endgroup$ Aug 20, 2018 at 14:36
  • $\begingroup$ The boundary condition was given as $f(cosθ,sinθ) = cosθ + sinθ$ by a professor.It could have been a mistake on the professor's part. $\endgroup$ Aug 20, 2018 at 14:48
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    $\begingroup$ I don't doubt about the professor. In fact I doubt about the link between the PDE and the boundary condition. In the PDE there is no $\theta$ but $x,y$ while in the condition there is no $x,y$ but $\theta$. Are you sure that the condition corresponds to the PDE or to another PDE coming from the first one, but already transformed by a change of variables ? $\endgroup$
    – JJacquelin
    Aug 21, 2018 at 8:22
  • $\begingroup$ I tried changing the coordinates in the field $X$ to polar but that made things so much worse. The problem is that on the lecture there are similarly stated pdes that really worked for me. For example this one doesn't look that different $Χ(f)=f$, $f(cosθ,sinθ)=cos^2(θ)$, where $X=x_1 \frac \partial {\partial x_1}+x_2 \frac \partial {\partial x_2}$ and it can be solved to $f(x,y)= \frac {x^2}{\sqrt{x^2+y^2}}$ $\endgroup$ Aug 21, 2018 at 13:37
  • $\begingroup$ The manner to present the problem is very confusing. The PDE is with variables $(x,y)$, so Cartesian. The boundary conditions are not Cartesian, but polar with variables $(?,\theta)$. One polar coordinate doesn't appear explicitly, may be $(r,\theta)$. Are you sure that the boundary condition is $f(\cos(\theta),\sin(\theta))=\cos(\theta)+\sin(\theta)$ on the unit circle (equation $\rho=1$) , or on another curve, for example on a radius (equation $\theta=$constant) ? Or something else ? Sorry I cannot help you better in such a fishy context. $\endgroup$
    – JJacquelin
    Aug 21, 2018 at 18:09
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The differential equation

$X(f) = 2f, \tag 1$

where $X$ is the vector field

$X = x_1^2 \dfrac{\partial}{\partial x_1} - x_2^2 \dfrac{\partial}{\partial _2}, \tag 2$

may also be written in the form

$\dfrac{\partial f}{\partial t} = 2f, \tag 3$

where $t$ is the running parameter along the integral curves of $X$; here we make the usual identification

$X \equiv \dfrac{\partial}{\partial t}. \tag 4$

It will be noted that in fact (3) is simply an ordinary differential equation; thus, along any trajectory of $X$, we may in the usual manner write

$X(\ln f) = \dfrac{\partial (\ln f)}{\partial t} = \dfrac{1}{f} \dfrac{\partial f}{\partial t} = 2, \tag 5$

which may be integrated 'twixt $t_0$ and $t$ to yield

$\ln \left ( \dfrac{f(t)}{f(t_0)} \right ) = \ln(f(t) ) - \ln(f(0)) = 2(t - t_0), \tag 6$

or

$f(t) = f(t_0)e^{2(t - t_0)}, \tag 7$

which expresses the evolution of $f$ along any trajectory of $X$ in terms of the parameter $t$ such that (4) binds.  The reader will no doubt recognize that (7) presents

$f:\Bbb R \to \Bbb R \tag 8$

as a function of the single parameter $t$, whereas the question specifies that

$f:\Bbb R^2 \to \Bbb R \tag 9$

is indeed dependent upon the two variables $x_1, x_2$; in reconciling these dual points of view we will exploit the fact that, along the integral curves of $X$ the coordinates $x_1, x_2$ must satisfy the differential equations

$\dot x_1 = x_1^2, \tag{10}$

$\dot x_2 = -x_2^2; \tag{11}$

these equations are both of the general form

$y = ay^2, \tag{12}$

and the solution is derived below in an appendix to this answer.  We in fact have:

$x_1(t) = x_{10}(1 - x_{10}(t - t_0))^{-1}, \; x_1(t_0) = x_{10}, \tag{13}$

$x_2(t) = x_{20}(1 + x_{20}(t - t_0))^{-1}, \; x_2(t_0) = x_{20}; \tag{14}$

then what we have written as

$f(t) = f(x_1(t), x_2(t)), \tag{15}$

and we may find $f(x_1, x_2)$ for arbitrary $x_1, x_2$ by discovering an $x_{10}, x_{20}$, $t$ and $t_0$ (if indeed such concurrently exist) such that (13) and (14) bind, where $x_{10}$ and $x_{20}$ are the coordinates of a point at which $f(t_0)$ is specified; typically, $x_{10}$, $x_{20}$ will lie in some submanifold, in the present instance in fact a curve in $\Bbb R^2$ that is, apparently, the circle $(\cos \theta, \sin \theta)$ on which we have

$f(\theta) = \cos \theta + \sin \theta; \tag{16}$

we may, via (13) and (14), express $x_1$, $x_2$ by means of a coordinate transformation which gives them in terms of $t$ and $\theta$:

$x_1(t, \theta) = \cos \theta (1 - \cos \theta (t - t_0))^{-1}, \tag{17}$

$x_2(t, \theta) = \sin \theta (1 + \sin \theta (t - t_0))^{-1}; \tag{18}$

in these coordinates we have, by (7),

$f(t, \theta) = e^{2(t - t_0)} (\cos \theta + \sin \theta); \tag{19}$

in principle, the transformation (17)-(18) may be reversed; in so doubg, the identity

$\sin^2 \theta + \cos^2 \theta = 1 \tag{20}$

may prove useful, allowing as it does the expression $\sin \theta$ in terms of $\cos \theta$.

Appendix:

$\dot y = ay^2, \; y(t_0) = y_0; \tag 1$

$y^{-2}\dot y = a; \tag2$

$y_0^{-1} - y^{-1} = \displaystyle \int_{y_0}^y y^{-2}dy = \int_{t_0}^t a \; ds = a(t - t_0); \tag 3$

$y^{-1} = y_0^{-1} - a(t - t_0) = y_0^{-1} - y_0^{-1} y_0 a(t - t_0) = y_0^{-1}(1 - y_0 a(t - t_0));\tag 4$

$y = y_0(1 - y_0 a(t - t_0))^{-1} = \dfrac {y_0}{1 - y_0 a (t - t_0)}; \tag 5$

we apply these calculations to find (locally) the integral curves of the vector field

$X = x_1^2 \dfrac{\partial}{\partial x_1} - x_2^2 \dfrac{\partial}{\partial _2} \tag 6$

with initial condition

$x_1(t_0) = x_{10}, \; x_2(t_0) = x_{20}; \tag 7$

the formula (5) applies to this situation when we take

$a = 1, \; y = x_1; \; a = -1, y = x_2; \tag 8$

we have

$x_1(t) = x_{10}(1 - x_{10}(t - t_0))^{-1}, \tag 9$

$x_2(t) = x_{20}(1 + x_{20}(t - t_0))^{-1}, \tag{10}$

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    $\begingroup$ That looks really nice. Thank you I was beginning to lose hope $\endgroup$ Aug 28, 2018 at 3:19
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    $\begingroup$ @Amontillado: thank you. I acutally have more on this but I rushed to get it posted by . . . . well, you know! May add it in a while, once I chill out from my last minute rush! Good question, thanks again, Cheers! $\endgroup$ Aug 28, 2018 at 3:21
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    $\begingroup$ Please do. Thank you! $\endgroup$ Aug 28, 2018 at 3:22
  • $\begingroup$ Could you please elaborate a bit more on your $(20)$ argument. I feel like I'm missing something $\endgroup$ Aug 28, 2018 at 9:49
  • $\begingroup$ @Amontillado: Yes in a few hours when I get going. It's 5 AM and I'm still in bed. Going back to sleep no. Cheers! $\endgroup$ Aug 28, 2018 at 12:42

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