Solutions of a differential equation

Question

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?

Answer 1

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

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

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

Answer 2

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.

Answer 3

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