Vtyjkyuk

Duhamel's principle gives the solutions of linear inhomogeneous evolution PDEs in terms of solutions of the associated homogenous problem. That is, the solution of the PDE
$$u_t(mathbfx,t)+sum_i=1^n a_i(mathbfx) u_x_i(mathbfx,t)=f(mathbfx,t), tag1 $$
can be expressed (using integrations) in terms of solutions of
$$u_t(mathbfx,t)+sum_i=1^n a_i(mathbfx) u_x_i(mathbfx,t)=0,$$
with varying initial conditions.

This is the multivariable analogue of the method of variation of parameters from ODEs, in which the inhomogeneous solution is expressed as a sum of the general solution of the associated homogeneous problem, and any particular solution.

I have been trying to generalize this

general soln of inhomo. = general soln of associated homo. + particular soln of inhomo.

to PDEs which are not of evolution type. That is, to PDEs for which there is no distinguished independent variable with the associated coefficient $=1$ in the PDE (without resorting to division by one of the coefficients).

For example, the PDE
$$x u_x+yu_y=f(x,y), $$
can be solved explicitly using the method of characteristics:
$$u(x,t)=phi left( fracxsqrtx^2+y^2,fracysqrtx^2+y^2 right)+int_1^sqrtx^2+y^2 f left( fracxsqrtx^2+y^2 s, fracysqrtx^2+y^2 s right) fracmathrmd ss. $$
Here $phi$ is an arbitrary function, which plays the role of the general solution of the associated homogeneous problem, while the integral part represents a a particular solution (the one satisfying $u equiv 0$ on the unit circle) of the inhomogeneous problem.

My questions are:

1. In the case of a general first order PDE $(1)$, is knowledge of the form of the associated homogeneous solution sufficient for solving $(1)$ in general? If so, how?




2. If a particular solution cannot be computed from the general solution of the associated homogeneous equation, is there any other way of forming a particular solution? It appears to me that it should be integral of the RHS along the corresponding characteristic curve with respect to some measure ($mathrmds/s$ in our example above).

Thank you!