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!!

  • $\begingroup$ Is $F$ linear, bounded? $\endgroup$ Nov 16, 2021 at 11:20
  • $\begingroup$ I want a smooth and bounded $F$. But I would like to see the existence result for linear and bounded $F$. $\endgroup$
    – mnmn1993
    Nov 16, 2021 at 11:50
  • $\begingroup$ Is there some other hypothesis on $L$? Is $L$ self adjoint/normal, or what do you know about its spectrum? $\endgroup$ Nov 16, 2021 at 15:06
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    $\begingroup$ I'm not an expert by any means but a general (?) existence result for the abstract Cauchy problem can be found in Fattorini's book (The Cauchy Problem 1984) chapter 2, for example Thm 2.1.1. This requires you to have some knowledge of the resolvent operator for the operator $A= L- F$ partially defined on a subset of $L^2([0,1])$ of functions that satisfy also your first 2 constraints. I hope an expert can say something more. $\endgroup$ Nov 16, 2021 at 15:43
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    $\begingroup$ you can plug in the boundary values in the choice of the Banach space $E$, where the operator $A$ is (partially) defined. See the examples in chapter 1 of Fattorini like the heat equation on a square. $\endgroup$ Nov 16, 2021 at 16:54

1 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 $x=1$.

  • $\begingroup$ Can you provide a few references for the above-mentioned general rule in boundary-value problems? $\endgroup$ Nov 18, 2021 at 19:11
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    $\begingroup$ @WarlockofFiretopMountain The book by Benzoni and myself concerns the hyperbolic situtation, but the strategy is the same for other type of PDEs. Actually the key word "Lopatinskii condition" is named that way because of Lopatinskii's seminal work on elliptic BVPs. $\endgroup$ Nov 18, 2021 at 20:53

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