# Solving overdetermined, well posed, linear system of PDEs

Let $$f=f(u,v)$$ be a (given) solution of the following PDE,

$$\begin{equation} \frac{\partial^2 f}{\partial u\partial v}=f,\label{1}\tag{*} \end{equation}$$

and consider the overdetermined system(s) of PDEs

$$\begin{cases} \dfrac{\partial x}{\partial u}=f\cos\left(u-v\right)\\ \\ \dfrac{\partial x}{\partial v}=\dfrac{\partial f}{\partial v}\sin\left(u-v\right)\\ \\ \dfrac{\partial y}{\partial u}=f\sin\left(u-v\right)\\ \\ \dfrac{\partial y}{\partial v}=-\dfrac{\partial f}{\partial v}\cos\left(u-v\right). \end{cases}$$

Since $$f$$ solves \eqref{1} it is guaranteed that $$\frac{\partial^2 x}{\partial u\partial v}=\frac{\partial^2 x}{\partial v\partial u}$$ and similarly for $$y$$, so the system is well posed. The question is, is it possible to write down a general solution for $$x(u,v)$$ and $$y(u,v)$$ in terms of integrals of $$f$$ ?

If I try, for instance, to integrate the first equation to get $$x=\int \mathrm{d}u\;f\cos(u-v)+g(v),$$ with $$g(v)$$ some unknown function of $$v$$, and then plug in the second equation for $$x$$, I cannot solve for $$g(v)$$, so I need to try some other ansatz.

• Are these really the equations you want to consider? It would look more nice and symmetric if the derivatives of $x,y$ with respect to $u$ involved derivatives of $f$ with respect to $u$ just as the derivatives of $x,y$ with respect to $v$ involve derivatives of $f$ with respect to $v$. In that case it seems really promising to complexify the equation by defining $z \equiv x + i y, w \equiv u + i v$. Jul 17, 2021 at 22:06

You can do this as follows: Start with any values for $$(x,y)$$ at $$(0,0)$$, say $$(x,y)(0,0) = (x_0,y_0).$$ Integrate the first and third equations along the line $$v = 0$$: \begin{align*} x(u,0) &= x_0 + \int_{s=0}^{s=u} f(s,0)\cos(s)\,ds\\ y(u,0) &= y_0 + \int_{s=0}^{s=u} f(s,0)\sin(s)\,ds \end{align*} Next, for each $$u$$, integrate the second and fourth equations along the line $$(u,\cdot)$$ to get your solution: \begin{align*} x(u,v) &= x(u,0) + \int_{t=0}^{t=v} \frac{\partial f}{\partial v}(u,t)\sin(u-t)\,dt\\\ y(u,v) &= y(u,0) + \int_{t=0}^{t=v} \frac{\partial f}{\partial u}(u,t)\cos(u-t)\,dt. \end{align*}
• thanks for the analysis. I would like to observe the following. If we take $x(u,v)-x_{0}=\int^{u}dt\:f(t,v)\cos(t-v)$ then it satisfies the 1st eqtn. Now, $\partial x(u,v)/\partial v=\int^{u}dt\:\partial\left(f(t,v)\cos(t-v)\right)/\partial v=\int^{u}dt\:\partial\Big(\partial f(t,v)/\partial v\sin(t-v)\Big)/\partial t=\partial f(u,v)/\partial v\sin(u-v)$, then both equations are satisfied (similarly for $y(u,v)$) so by construction the solution should be just $x(u,v)-x_{0}=\int^{u}dt\:f(t,v)\cos(t-v)$, isn't it ? (By $\int^{u}$ I mean integration without evaluating a lower limit). Jul 20, 2021 at 16:30
• I mean $\int^t dt' \:\partial f(t',s)/\partial t'\equiv f(t,s)$. (This notation is, by the way, a common practice in physics) Jul 20, 2021 at 18:09
• This appears to be the indefinite integral, which means that the right side should be written $f(t,s) + C$, since adding any constant to $f$ also gives an antiderivative of $\partial f(t',s)/\partial t'$. The value of this constant is important in the formulas for the solution. Jul 20, 2021 at 19:16
• exactly, so I'm setting this value of $C$ to zero in my previous comment. My reasoning is this: the solution is unique up to an additive constante, so if I manage to find $one$ solution this is $the$ solution. Now, by the Leibniz rule the solution in my previous comment seems to satisfy the system, but I'm trying to check if this is the same that you kindly sketched. Jul 20, 2021 at 19:19