Parabolic equation with Cauchy boundary condition

Question

Consider the domain $[0,1] \times [0,T]$ and the uniformly parabolic operator $L -\partial_t$ with smooth coefficient. I would like to obtain the existence of the problem \begin{equation} \left\{\begin{aligned} &L u -\partial_t u= F(u)& \hspace{10pt} &\text{for $(x,t) \in (0,1) \times (0,T]$} ;\\ & u(1,t) =f(t) & \hspace{10pt} &\text{for $t \in \big[0,T\big]$;}\\ & \partial_x u(1,t) =g(t) & \hspace{10pt} &\text{for $t \in \big[0,T\big]$;}\\ & u(x,0) =h(x) & \hspace{10pt} &\text{for $x \in \big(0,1\big)$.}\\ \end{aligned}\right. \end{equation} I know that this kind of boundary condition is called Cauchy boundary condition. I have also found some references about it. But some of them are too old and non-English. For example, references [4, 60, 73-77, 87, 96, 97, 105, 117, 118, 129, 138, 139] in the article "A noncharacteristic cauchy problem for the heat equation". May I have some other references concerning the existence problem? Thank you so much!!

Answer 1

Say that $L=\partial_x^2$ (the heat equation). Your problem is ill-posed in all reasonnable context : it does not admit a solution for generic data taken in spaces $L^2, C^\infty$ or even in distributional spaces, although it does when the data are analytic (Cauchy-Kowalevska).

Let us consider the simplified situation where $F\equiv0$ and $(0,T)$ is replaced by ${\mathbb R}$. Then the solution must obey $$\frac{d^2}{dx^2}\hat u(x,\tau)=i\tau\hat u(x,\tau)$$ where $\hat u$ is the Fourier transform in the time variable. Then $\hat u$ is a linear combination $$a_+(x)\exp(x\sqrt{i\tau})+a_-(x)\exp(-x\sqrt{i\tau}).$$ Since the ODE (in $x$) is second-order, your really need both exponentials, but one of the exponential is not even a tempered distribution because it grows fast as $\tau\to\pm\infty$. Thus you cannot invert the Fourier transform and recover $u$.

There is a general rule in Boundary-value problems for PDEs. Say that the boundary has a tangent hyperplane at a point $\bar p$. Replace the domain by the corresponding half-space and linearize the PDE at $\bar p$, to obtain a constant coefficient linear PDE and Boundary condition. Then make the Fourier transform in the tangential variables. In order that the problem be well-posed, it is necessary that the resulting ODE by solvable in the space of bounded functions. This rules out exponentials that grow as the normal coordinate enters the half-space. They remains only a few admissible exponentials, whose number must equal the number of boundary conditions.

In your case, there is only one bounded exponential, thus there must be one and only one boundary condition at $x=1$.