# Solving PDE with Cauchy - Kowalewski Theorem

Hallo,

I have the following PDE that I am trying to solve via the Cauchy-Kowalewski Theorem. But I have no idea how to do it or if its possible. Maybe one of you has an idea. Here is the problem: Let $U \subset \mathbb{C}^{n}$ be some open subset witch contains zero. Well you can shrink $U$ arbitralily if you wish. Let $z_{j} = x_{j} + i y_{j}$ be the coordinates on $U$. I am looking for a real-valued analytic function $\beta$ defined on $U$ such that the equations are satisfied: $\frac{\partial \beta}{\partial x_{j}} = F_{j}(x,y,\beta)$ and $\frac{\partial \beta}{\partial y_{j}} = G_{j}(x,y,\beta)$,with $j = 1, ..., n$, where $F,G$ are also real analytic functions and with initial condition $\beta(x,0) = 0$, $\forall x \in U \cap \mathbb{R}^{n}$. Is it possible to solve such an system of equations via Cauchy-Kowalewski ? If yes, how? If no, why and is there any other method that can give me a solution? I hope that for a lot of answers and please excuse me if the question is too trivial. Thanks in advance.

Greeting Andrei

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If you are specifying the partial derivatives, you need necessarily compatibility conditions. As stated the PDE can be overdetermined. Do you know that $\partial_{y_k} F_j = \partial_{x_j} G_k$ for example (they need to be since partial derivatives commute on $\beta$)? Also, typically Cauchy-Kowelewski is used to solve a "initial value problem" where the data is prescribed on a co-dimension 1 set. You need something closer to Cartan-Kahler. – Willie Wong Oct 3 '12 at 7:18
Deane Yang's paper "Local Solvability of Overdetermined Systems Defined by Commuting First-Order Differential Operators" may help (1986, CPAM), I'll see if I can get him to say a few words here. – Willie Wong Oct 3 '12 at 7:19
Thanks that would be very nice! – Andrei Oct 3 '12 at 7:57
@Willie: You have to be a bit careful with the integrability conditions. Since $F$ and $G$ involve the unknown $\beta$, you can't just check whether $\partial_{y^k}F_j=\partial_{x^j}G_k$, since expanding these out will involve the partials of the unknown $\beta$, and you won't know whether you have equality until you know the solution $\beta$ that you are hoping to find, since these equations only have to hold for the solution $\beta$ that also satisfies the given initial conditions (assuming that it actually exists). – Robert Bryant Oct 3 '12 at 14:22
@Robert: can one not plug in the equation there? $$\partial_{y^k}(F_j(x,y,\beta)) = (\partial_{y^k}F_j)(x,y,\beta) + (\partial_\beta F_j)(x,y,\beta)G_j(x,y,\beta)$$ Of course this still depends on the unknown $\beta$, but if the integrability condition is only satisfied for isolated values of $\beta$ one may expect trouble. – Willie Wong Oct 3 '12 at 15:12

You don't need the Cauchy-Kowalewski Theorem for your problem. In fact, real-analyticity is a red herring here. What you are asking for is a function $\beta(x,y)$ such that the graph $\bigl(x,y,\beta(x,y)\bigr)$ is an integral manifold of the $1$-form $$\theta = d\beta - F_j(x,y,\beta)\ dx^j - G_j(x,y,\beta)\ dy^j$$ (sum on $j$ in both terms) defined on $\mathbb{R}^{2n+1} = \mathbb{C}^n\times \mathbb{R}$ (or some open neighborhood of $0$ in this space. This makes sense for smooth functions of course, and whether there is a solution or not doesn't depend on real-analyticity.
In fact, you have added the requirement that the $2n$-dimensional graph contain the $n$-dimension submanifold defined by $(x,y,\beta) = (x, 0, 0)$, and you can see from the above that $\theta$ vanishes on that graph if and only if the $F_j$ satisfy $F_j(x,0,0)\equiv0$, so this is certainly a necessary condition.
A sufficient condition, after that, would be, for example, that $d\theta\wedge\theta=0$, for then the Frobenius Theorem would apply. However, this is not necessary if all you are asking is that there be a solution to the specific 'initial value problem' you have posed. To get sufficient conditions, what you should do is use ODE to, for example, construct $\beta_1(x,y^1)$ satisfying the equation $$\frac{\partial\beta_1}{\partial y^1} = G_1(x,y^1,0,\ldots,0,\beta_1)$$ with the initial condition $\beta_1(x,0) = 0$. Then you need to check that $\theta$ vanishes on the $(n{+}1)$-dimensional graph $\bigl(x,y^1,0,\ldots,0,\beta_1(x,y^1)\bigr)$. Next, you construct $\beta_2(x,y^1,y^2)$ by solving the equation $$\frac{\partial\beta_2}{\partial y^2} = G_2(x,y^1,y^2,0,\ldots,0,\beta_2)$$ with the initial condition $\beta_2(x,y^1,0) = \beta_1(x,y^1)$, and so on. At each stage, you'll get more conditions on the functions $G_j$ and $F_j$ in order for the constructed graph to be an integral of $\theta$. When you get to the end, these will be the necessary and sufficient conditions for this particular initial value problem.