Find the Solution of $af_x + f_y = 0$, Where $a \not= -1$, and Where the Following Values Are Taken

Question

I have the following problem:

Find the solution of $af_x + f_y = 0$, where $a \not= -1$, and where the following values are taken

1. $f = e^{-x^2}$ on the line $y = -x$.
2. $f = 1$ on the line $y = -x$

For each of the above cases, what happens when $a = -1$?

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My work is as follows:

i)

$$\frac{dx}{dt} = a, \frac{dy}{dt} = 1, \frac{df}{dt} = 0$$

$$\therefore x(t) = at + C_1(s), y(t) = t + C_2(s), f = e^{-x^2}$$

$$x(0) = C_1(s) = -s, y(0) = C_2(s) = 0$$

$$\therefore x(s, t) = at - s, y(s, t) = t$$

Testing transversality condition:

$$|J| = \begin{vmatrix} a& 1 \\ -1 & 0 \end{vmatrix} = 0 - (-1) = 1$$

Therefore, the transversality condition holds.

ii)

$$\frac{dx}{dt} = a, \frac{dy}{dt} = 1, \frac{df}{dt} = 0$$

$$\therefore x(t) = at + C_1(s), y(t) = t + C_2(s), f = 1$$

$$x(0) = C_1(s) = -s, y(0) = C_2(s) = 0$$

$$\therefore x(s, t) = at - s, y(s, t) = t$$

Testing transversality condition:

$$|J| = \begin{vmatrix} a& 1 \\ -1 & 0 \end{vmatrix} = 0 - (-1) = 1$$

Therefore, the transversality condition holds.

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As can be seen, I'm confused as to how I should be solving such problems, as well as how the differences in $f$ affect these two problems and each of their solutions.

I would greatly appreciate it if people could please take the time to demonstrate how these problems are solved, with explanations/reasoning for the steps of the solution.

Answer 1

Let $u:=ax+y$ and $v:=x-ay$. Then, $u$ and $v$ are independent variables (by showing that $\dfrac{\partial u}{\partial v}=0$ and $\dfrac{\partial v}{\partial u}=0$) such that $$x=\frac{au+v}{a^2+1}\text{ and }y=\frac{u-av}{a^2+1}\,.$$ Note that $$\frac{\partial}{\partial u}=\left(\frac{\partial x}{\partial u}\right)\,\left(\frac{\partial}{\partial x}\right)+\left(\frac{\partial y}{\partial u}\right)\,\left(\frac{\partial}{\partial y}\right)=\frac{1}{a^2+1}\,\Biggl(a\,\left(\frac{\partial}{\partial x}\right)+\left(\frac{\partial }{\partial y}\right)\Biggr)\,.$$ Thus, if we write $F(u,v):=f\left(\frac{au+v}{a^2+1},\frac{u-av}{a^2+1}\right)=f(x,y)$, then the equation $$a\,\left(\dfrac{\partial}{\partial x}\,f(x,y)\right)+\left(\frac{\partial}{\partial y}\,f(x,y)\right)=0$$ is equivalent to $$\frac{\partial }{\partial u}\,F(u,v)=0\,.$$ That is, $F(u,v)=g(v)$ for some function $g$. Consequently, $$f(x,y)=g(x-ay)\text{ for all }x,y\,.$$

First, suppose that $a\neq -1$. Then, the solution $f(x,y)=g(x-ay)$ in Question (1) is given by $$g\big(x-a(-x)\big)=\exp\left(-\frac{x^2}{2}\right)\,,$$ or equivalently $$g(t)=\exp\left(-\frac{t^2}{2\,(a+1)^2}\right)\,.$$ For Question (2), the solution $f(x,y)=g(x-ay)$ satisfies $$g\big(x-a(-x)\big)=1\text{ or }g(t)=1\,.$$
Now, assume that $a=-1$. Then, Question (1) is impossible, as $g\big(x-a(-x)\big)=g(0)$ is constant but $\exp\left(-\dfrac{x^2}{2}\right)$ depends on $x$. For Question (2), a similar logic says that $$f(x,y)=g(x-ay)=g(x+y)\,,$$ where $g(0)=1$ (and there are no other conditions on $g$).

Answer 2

I have to write a separate answer for two reasons---(1) MathJax is misbehaving when there are too many equations and (2) there is only a small connection between this answer and my previous answer. The problem is here: suppose you have a partial differential equation of the form $$a\,\frac{\partial}{\partial x}\,f(x,y)+b\,\frac{\partial}{\partial y}\,f(x,y)=0\,,$$ where $a$ and $b$ are real constants which are not both zero. You are looking to find a new coordinate system $(u,v)$, where $u$ and $v$ are independent variables which are functions of $x$ and $y$, in such a way that $$\frac{\partial}{\partial u}\,G(u,v)=a\,\frac{\partial}{\partial x}\,g(x,y)+b\,\frac{\partial}{\partial y}\,g(x,y)\,,$$ if $G(u,v)=g(x,y)$. In other words, we want $$\frac{\partial}{\partial u}=a\,\frac{\partial}{\partial x}+b\,\frac{\partial}{\partial y}\,.\tag{*}$$

We are making a guess here. We suppose that $(u,v)=(px+qy,rx+sy)$ is a good coordinate system, where $p$, $q$, $r$, and $s$ are real constants with $(p,q)\neq (0,0)$ and $(r,s)\neq (0,0)$. That is, $$x=\frac{su-qv}{ps-qr}\text{ and }y=\frac{-ru+pv}{ps-qr}\,.$$ Because $$\frac{\partial x}{\partial u}=a\,\left(\frac{\partial x}{\partial x}\right)+b\,\left(\frac{\partial x}{\partial y}\right)=a\,,$$ we conclude that $$\frac{s}{ps-qr}=a\,.$$ Similarly, $$\frac{\partial y}{\partial u}=a\,\left(\frac{\partial y}{\partial x}\right)+b\,\left(\frac{\partial y}{\partial y}\right)=b$$ yields $$\frac{r}{ps-qr}=-b\,.$$
That is, $$ar+bs=(ps-qr)(-ab+ab)=0\,.$$

We need one last relation $$1=\frac{\partial u}{\partial u}=a\,\left(\frac{\partial u}{\partial x}\right)+b\,\left(\frac{\partial u}{\partial y}\right)=ap+bq\,.$$ It can be seen that the pair $(u,v)=(px+qy,rx+sy)$ with $(p,q)\neq (0,0)$ and $(r,s)\neq (0,0)$ has independent variables with the required condition (*) if and only if $ap+bq=1$ and $ar+bs=0$. One good choice is $(p,q):=\left(\frac{a}{a^2+b^2},\frac{b}{a^2+b^2}\right)$ and $(r,s)=\left(-\frac{b}{a^2+b^2},\frac{a}{a^2+b^2}\right)$. That is, $$(u,v)=\left(\frac{ax+by}{a^2+b^2},\frac{-bx+ay}{a^2+b^2}\right)$$ is a good new coordinate system.

You can also rescale or swap the signs of $(u,v)$ above, but when you rescale $u$ and $v$, the equation (*) becomes $$\frac{\partial}{\partial u}=k\,\left(a\,\frac{\partial}{\partial x}+b\,\frac{\partial}{\partial y}\right)$$ for some constant $k$. For example, $$(u_1,v_1)=(ax+by,-bx+ay)$$ is also a good coordinate system. Alternatively, $$(u_2,v_2)=(ax+by,bx-ay)$$ is another good coordinate system. There are infinitely many other ways to pick $(p,q,r,s)$.