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Given an equation in partial derivatives of the form $Af_x+Bf_y=\phi(x,y)$, for example $$f_x-f_y=(x+y)^2$$ How do I know which change of coordinates is appropiate to solve the equation? In this example, the change of coordinates is $u=x+y$, $v=x^2-y^2$, why?

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  • $\begingroup$ This is an inhomogeneous linear transport equation. You can solve it with the method of characteristics, I'm not sure where you got $v$ from though $\endgroup$ – Calvin Khor Jan 21 at 19:10
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lets write $\mathbf x = (x,y)^T$. The equation is the linear inhomogeneous transport equation, $$ \mathbf a \cdot \nabla f =b,$$ where $f=f(x,y)$ and specifically $\mathbf a = (a_1,a_2)^T=(1,-1)^T$ and $b = (x+y)^2$, but I will continue almost as if I don't know what $\mathbf a$ or $b$ are.

I'll sketch out a theory("The method of characteristics") assuming you are given initial data, that works in a large generality. In particular, with very minor modifications you can also allow the coefficient $\mathbf a$ of the derivative to be a function, or even depend on $f$ (but not $\nabla f$).

Suppose we are prescribed initial data $f(x,0) = f_0(x)$. (more general initial data along a hyper surface is possible) Consider the family of ODEs that define the "characteristic curves" $$\frac{d}{dt} \mathbf X(t;z) = \mathbf a, \quad \mathbf X(0;z) = \binom{z}{0} \\ \frac{d}{dt} F(t;z) = b(\mathbf X),\quad F(0;z) = f_0(z) $$ These can be solved, by Picard-Lindelof. In this case, $\mathbf X$ is in fact a globally invertible change of coordinates,x since we can explicitly write $$\mathbf X= \binom{z}{0} + t\mathbf a = \mathbf x \iff t= y/a_2,\quad z = x-ta_1 = x - ya_1/a_2$$ (the condition that $a_2 \neq 0$ is related to the fact that the initial data is prescribed on the $x$-axis: the important point is that $\mathbf a$ is not tangent to the hypersurface where the initial data is prescribed.) Let's write the inverse map of $\mathbf X$ as $\mathbf Z$ with the explicit formula $$\mathbf Z(\mathbf x):=\binom{t(\mathbf x)}{z(\mathbf x)} = \binom{y/a_2}{x - ya_1/a_2}$$

One can now check that if we define $$ f(\mathbf x) := F(\mathbf Z(\mathbf x))$$ then $f$ solves the original PDE. Indeed, by chain rule and inverse function theorem, $$ \nabla f(\mathbf x)^T = (\nabla F)^T(\mathbf Z(\mathbf x)) (\nabla \mathbf Z)^T(\mathbf x)=(\nabla F)^T(\mathbf Z(\mathbf x)) ((\nabla \mathbf X)^{-1}(\mathbf Z(\mathbf x)))^T$$ If we hide the point $\mathbf x$ where the functions are evaluated, this is written perhaps more legibly, $$ \nabla f^T = (\nabla F\circ \mathbf Z)^T ((\nabla \mathbf X)^{-1}\circ \mathbf Z)^T $$ Then the dot product $\mathbf a \cdot \nabla f$ is $$\mathbf a \cdot \nabla f= \nabla f^T \mathbf a = (\nabla F\circ \mathbf Z)^T ((\nabla \mathbf X)^{-1}\circ \mathbf Z)^T \mathbf a $$ Since $\mathbf a$ is precisely the first column of $\nabla \mathbf X\circ \mathbf Z$ (if we agree to write $t$ derivatives in the first column), $((\nabla \mathbf X)^{-1}\circ \mathbf Z)^T \mathbf a = \binom{1}{0}$. Thus

$$ \mathbf a \cdot \nabla f = (\nabla F\circ \mathbf Z)^T \binom{1}{0} = (\partial_t F)\circ \mathbf Z = b(\mathbf X\circ \mathbf Z) = b$$

as needed.

(See also Method of characteristics (quasilinear pde- nonlinear transport ), Explaining the method of characteristics )

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This is a linear PDE then

$$ f = f^h+f^p $$

with

$$ f^h_x-f^h_y = 0\\ f^p_x-f^p_y = (x+y)^2 $$

for the homogeneous solution we have $f^h(x,y) = \phi(x+y)$ by characteristics method. Now the particular is obtained proposing for $f^p$ a polynomial form as

$$ f^p = a x(x+y)^2+b y(x+y)^2 $$

and after substitution we have

$$ f^p_x-f^p_y - (x+y)^2=(a-b-1)(x+y)^2=0 $$

so the solution is

$$ f(x,y) = \phi(x+y) + a x(x+y)^2+b y(x+y)^2, \ \ \mbox{such that}\ \ a-b=1 $$

NOTE

Regarding the change of coordinates we know one coordinate for sure which is $u = x+y$ but we need two coordinates so we choose the other as $v = x-y$ because $u, v$ form a valid (independent) coordinate system. We could choose instead $x+y, x+ a y$ with $a \ne 1$ as well.

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