Incorrect results of DSolve

Question

Finding a fundamental solution of a certain linear partial differential operator, I try

DSolve[D[u[x, y], x] + I*D[u[x, y], y] ==DiracDelta[x, y] + I*DiracDelta[x, y], u[x, y], {x, y}]

{{u[x, y] -> C[1][-I x + y] + Inactive[Integrate][(1 + I) DiracDelta[-I x + y + I K[1], K[1]], {K[1], 1, x}]}}

and

DSolve[D[u[x, y], x] + I*D[u[x, y], y] == DiracDelta[x, y], u[x, y], {x, y}]

{{u[x, y] -> C[1][-I x + y] + Inactive[Integrate][ DiracDelta[-I x + y + I K[1], K[1]], {K[1], 1, x}]}}

It's clear both above results are meaningless in view of complex-valued arguments of DiracDelta.

My next attempts consist in the following u[x, y] -> r[x, y] + I*s[x, y], then

Expand[D[u[x, y], x] +  I*D[u[x, y], y] /. 
{D[u[x, y], x] ->D[r[x, y], x] + I*D[s[x, y], x],   D[u[x, y], y] -> D[r[x, y], y] + I*D[s[x, y], y]}]

I (r^(0,1))[x,y]-(s^(0,1))[x,y]+(r^(1,0))[x,y]+I (s^(1,0))[x,y]

and

DSolve[{-D[s[x, y], y] + D[ r[x, y], x] == DiracDelta[x, y],D[r[x, y], y] + D[s[x, y], x] == DiracDelta[x, y]}, {r, s}, {x, y}]

which returns the input. If we consider an approximation of DiracDelta in the weak topology (I don't go into deep.)

DSolve[D[u[x, y], x] + I*D[u[x, y], y] == (1 + I)/Pi*eps*1/(eps^2 + x^2)*1/Pi*eps*1/(eps^2 + y^2),u[x, y],{x, y}]

{{u[x, y] -> ( 1/(\[Pi]^2 (4 eps^4 + (x + I y)^4)))(1/2 + I/ 2) eps (2 (2 eps^2 - (x + I y)^2) ArcTan[x/eps] - 2 I (2 eps^2 + (x + I y)^2) ArcTan[y/eps] + eps (x + I y) (2 I ArcTan[(2 (x + I y) y)/( eps^2 + x^2 + 2 I x y - 2 y^2)] - Log[(eps^2 + x^2) (eps^2 + (x + 2 I y)^2)] + 2 Log[-eps^2 - y^2])) + C[1][-I x + y]}}

, then its weak limit is unclear to me. The operator under consideration is not pathological, e.g.

DSolve[D[u[x, y], x] + I*D[u[x, y], y] == I*x^2*y^3 + x + y^2,u[x, y], {x, y}]

works well. Could its fundamental solution be derived with Mathematica?

Answer 1

This is a partial linear first order differential equation with constant coefficients in the complex numbers with imaginery separated parts in the representation of $x,y$. Change to z may reduce the trouble. The $DiracDelta$ can be transformed easily.

Since the PDE is with complex coefficients this is closely related to solving path integrals and that is what the solution shows as the special not as the general solution a path integral.

The fundamental standpoint the $I$ is a symbol both joining imaginary and real part of the differential equation and separating it. You first have to solve both separately the real and the imaginary part and then the coupling. This is not an approximation for small $I$ it is hard coupling.

That is the path along which the given solution has to be understood.

It shows a general $u[x,y]$ in need of unification. It shows a general formula not a solution to the coupling.

Since this is a Dirac delta which might, might not contain the boundary conditions first order linear complex differential equation special attention has to be done to the boundary condition. In the case of Dirac delta, there are an only a limited amount of conditions that should be applied.

The use of $x,y$ guides Mathematica into this particular direction. It is like selecting the solution inner structure first.

In general wikipedia is not a good source for solving mathematical problems.

The given solution is a slight formal correct generalization of

DSolve[Derivative[1][u][z] == DiracDelta[z], u, z]

To which

{{u -> Function[{z}, C[1] + HeavisideTheta[z]]}}

This offer many hints in form of keywords to work up the theory necessary to understand the Mathematica solution to the problem. Perhaps research will give further hints: duckduckgo search for dirac delta first order partial linear differential equation general solution constant complex coefficients

Answer 2

The weak solution for eps->0 gives

Asymptotic[
DSolve[D[u[x, y], x] + I*D[u[x, y], y] == (1 + I)/Pi*eps*1/(eps^2 + x^2)*1/Pi*eps*1/(eps^2 + y^2), u[x, y], {x, y}][[1 ]][[1, 2]], eps -> 0]
(* C[1][-I x + y] *)

Result is in accordance with the DSolve-solution

DSolve[D[u[x, y], x] + I*D[u[x, y], y] ==DiracDelta[x] DiracDelta[y] (1 + I), u[x, y], {x, y}]
(* {{u[x, y] ->C[1][-I x + y] + 
Inactive[Integrate][(1 + I) DiracDelta[-I x + y + I K[1]] DiracDelta[K[1]], {K[1], 1, x}]}}*)

because the integralpart vanishs