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Suppose we wish to solve the first order pde for the unknown function $f(x,y)$

$\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}=c(x,y)\Big(a(x,y)+b(x,y)\Big)$

We assume that the functions $a(x,y)$ and $b(x,y)$ are given, while $c(x,y)$ is arbitrary.

This is an inhomogeneous first order linear pde with constant coefficients, for which it is easy to compute a general solution in terms of the functions $a,b,c$ using the method of characteristics.

However, since $c(x,y)$ is arbitrary, I am interested in figuring out what conditions would need to be imposed on $c(x,y)$ so that the solution $f$ of the pde above also satisfies the condition

$\frac{\partial f}{\partial x}=c(x,y)a(x,y)$

I tried taking the derivative of the general solution to the pde with respect to $x$ and then equating that solution to the extra condition, but the expression seems quite messy. Maybe there's another approach?

Thanks!

Raul Gonzalez
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1 Answers1

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The system $$\begin{cases} \frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}=c(x,y)\Big(a(x,y)+b(x,y)\Big) \\ \frac{\partial f}{\partial x}=c(x,y)a(x,y) \end{cases}$$ can be viewed as an algebraic linear system of equations in the variables $\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} $.

Its solution is:

$$\frac{\partial f}{\partial x}=c(x,y) a(x,y),\frac{\partial f}{\partial y}=c(x,y) b(x,y), $$ and assuming $C^2$ regularity for $f(x,y)$ we must have $$\frac{\partial}{\partial y} \frac{\partial f}{\partial x}=\frac{\partial}{\partial x} \frac{\partial f}{\partial y}, $$ or in other words $$c_ya+ca_y=c_xb+cb_x. $$ This is a first order PDE for $c=c(x,y)$.

user1337
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