I'm trying to solve the Cauchy problem \begin{align*} u_{xx} + u_{yy} =0 \\ u(x,0) = x^2, u_y(x,0) = e^x \end{align*} for a solution near (0,0).
So far, I reduce the problem to an equivalent first order Cauchy problem, but I feel that it is harder to solve. Is there a way I can work directly with the problem without reducing it?
On
$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{{\rm u}\pars{x,y} = \fermi\pars{x + \ic y} + {\rm g}\pars{x - \ic y}}$
\begin{align} {\rm u}\pars{x,0} = x^{2} &\imp \fermi\pars{x} + {\rm g}\pars{x} = x^{2} \\ {\rm u}_{y}\pars{x,0} = \expo{x} \imp \ic\fermi'\pars{x} - \ic{\rm g}'\pars{x} = \expo{x}\quad&\imp \fermi\pars{x} - {\rm g}\pars{x} = -\ic\expo{x} + C \end{align} $C$ is a constant.
Then, $$ \fermi\pars{x} = \half\pars{x^{2} - \ic\expo{x} + C}\,,\qquad {\rm g}\pars{x} = \half\pars{x^{2} + \ic\expo{x} - C} $$
\begin{align} {\rm u}\pars{x,y}&=\half\bracks{\pars{x + \ic y}^{2} - \ic\expo{x + \ic y} + C} +\half\bracks{\pars{x - \ic y}^{2} + \ic\expo{x - \ic y} - C} \\[3mm]&=\Re\pars{\pars{x + \ic y}^{2} - \ic\expo{x + \ic y}} =\Re\pars{x^{2} + 2xy\ic - y^{2} - \ic\expo{x}\cos\pars{y} + \expo{x}\sin\pars{y}} \end{align}
$$\color{#00f}{\large% {\rm u}\pars{x,y} = x^{2} - y^{2} + \expo{x}\sin\pars{y}} $$
The Laplace equation is solved by the real (or imaginary) component of an analytical function, as a consequence of the Cauchy-Riemann equations. The function $$f(z)=z^2$$ $$u(x,y)=\operatorname{Re} f(x+iy)=x^2-y^2$$ satisfies $u(x,0)=x^2$ and $u_y(x,0)=0$. A function $$g(z)=g(x+iy)=-ie^z=-ie^{x}(\cos y+i\sin y)$$ then gives you $$u(x,y)=\operatorname{Re} g(x+iy)=e^x \sin y$$ which satisfies $u(x,0)=0$ and $u_y(x,0)=e^x$.
Because the Laplace equation is linear, just compose the final solution as $$u(x,y)=x^2-y^2+e^x \sin y$$
This is quite a general trick of writing the solution as a sum of two terms, one term that satisfies the function's value and has zero derivative at the boundary, and another term with zero value at the boundary and correct derivative.
You don't have to guess the analytical function (although it helps if you can). A function $f(z)=u(x,y)+iv(x,y)$ that has $v(x,0)=0$, can be constructed as an analytic continuation of the boundary condition. It satisfies $u_y=-v_x=0$ at $y=0$. The function $g(z)=p(x,y)+iq(x,y)$ with $p(x,0)=0$ on the other hand satisfies $p_y=-q_x$, so you can always write the solution as $$u(x,y)=\operatorname{Re}\left(u(x+iy,0)-i\int_0^{x+iy} u_y(t,0)dt\right)$$
If you simply want a series expansion around $(0,0)$, just write
$$u(x,y)=u(0,0)+u_x(0,0) x+u_y(0,0)y+\frac12 u_{xx}(0,0)x^2+\frac12 u_{yy}(0,0)y^2+u_{xy}(0,0)xy+\cdots$$ and using $u_{xx}=-u_{yy}=2$ and $u_{xy}=\frac{\partial }{\partial x} e^x$, you get $$u(x,y)=y+x^2-y^2+xy+\cdots$$