Asked this on Math Stack Exchange awhile ago but it got ignored then deleted.
To solve a differential equation of one variable, you need constraints equal to the number of derivatives.
For a partial differential equations, they are typically solved with functional constraints equal to the number of derivatives (though I'm not sure on this point). My question is if I have a linear operator $L$, with boundary condition functions $f_1 ... f_n$$f_1,\dots, f_n$, is there some generalized way to construct the solution $\Psi$ in terms of the operator and boundary conditions?
The typical problems in partial differential equations involve Fourier transforms to invert the operator in a roundabout way, but it's not clear to me what is the sufficient or necessary information need to solve this more generally.
Typical problems such as:
$\nabla^2 \Psi(x,y,z)=0$
$\Psi(x, y, 0)=f(x, y)$
$\implies \Psi(x,y,z)=\frac{1}{(2\pi)^2}\int dx^{\prime}\int dy^{\prime}\int d k_x \int d k_y \exp\left(-z\sqrt{k_x^2+k_y^2}+ik_x(x-x^{\prime})+ik_y(y-y^{\prime})\right) f(x^{\prime},y^{\prime})$
Require $$ \begin{cases} \nabla^2\Psi(x,y,z)=0\\ \\ \Psi(x, y, 0)=f(x, y) \end{cases}\\ $$ whose solution has the form $$ \begin{split} \implies \Psi(x,y,z) &=\frac{1}{(2\pi)^2}\int dx^{\prime}\int dy^{\prime}\int d k_x \int d k_y\\ &\qquad\exp\left(-z\sqrt{k_x^2+k_y^2}+ik_x(x-x^{\prime})+ik_y(y-y^{\prime})\right) f(x^{\prime},y^{\prime}), \end{split}$$ require reasoning about the specific problems, and are hard for me to generalize in arbitrary coordinate constraints such as:
$\Psi(x,y,z)=f(x-y,z+y)$
Is $$ \Psi(x,y,z)=f(x-y,z+y). $$ Is there a way to makeconstruct an 'equivalent source' term that encodes all the constraints?
$\nabla^2\Psi(x,y,z)=J(x,y,z)[f]$
$\implies \Psi(x,y,z)={\nabla^2}^{-1}J(x,y,z)[f]$
Or such as $$ \begin{split} \nabla^2\Psi(x,y,z)&=J(x,y,z)[f]\\ \implies \Psi(x,y,z)&=\big({\nabla^2}\big)^{-1}J(x,y,z)[f] \end{split}\quad? $$ Or for a general operator:
$\Psi=L^{-1}J[f]$, $$ \Psi=L^{-1}J[f]\quad? $$
