# Change of coordinates to solve an equation in partial derivatives

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?

• 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 – Calvin Khor Jan 21 at 19:10

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.

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.