Suppose $u(x,y) = \frac{y^{2}}{2} - g(ye^{-x})$ is the solution to a second-order linear PDE, where $g$ is some smooth function. If we have the additional boundary value condition that $u(0,y) = f(y)$, how would I write out the form of the solution to the boundary value problem?
We know that $u(0,y) = f(y) = \frac{y^{2}}{2} - g(y)$, but how could I write the solution $u(x,y)$ in terms of this $f$?
Note that we get the following relation: $$ g(y) = \frac{y^2}{2} - f(y) $$ so we can simplify $$ u(x,y) = \frac{y^2}{2} - g(ye^{-x}) = \frac{y^2}{2} - \frac{y^2e^{-2x}}{2} + f(ye^{-x}) $$