Consider the following Laplace boundary value problem (BVP)

$$\matrix{ {{\nabla ^2}\Phi (x,y) = 0,} \hfill & { - a \le x \le a} \hfill & { - b \le y \le b} \hfill \cr {{{\partial \Phi } \over {\partial y}}(a,y) = f(y)} \hfill & {} \hfill & {} \hfill \cr {{{\partial \Phi } \over {\partial y}}( - a,y) = f(y)} \hfill & {} \hfill & {} \hfill \cr {{{\partial \Phi } \over {\partial x}}(x,b) = 0} \hfill & {} \hfill & {} \hfill \cr {{{\partial \Phi } \over {\partial x}}(x, - b) = 0} \hfill & {} \hfill & {} \hfill \cr } $$

where $f(-y)=-f(y)$. We have prescribed the tangential derivatives of $\Phi(x,y)$ on the boundary instead of the function value.

The more general form of the problem in 2D or 3D can be written as

$$\matrix{ {{\nabla ^2}\Phi ({\bf{x}}) = 0,} \hfill & {{\bf{x}} \in \Omega } \hfill \cr {{\nabla _S}\Phi ({\bf{x}}) = \left[ {\left( {{\bf{I}} - {\bf{n}} \otimes {\bf{n}}} \right) \cdot \nabla \Phi } \right]({\bf{x}}) = {\bf{f}}({\bf{x}})} \hfill & {{\bf{x}} \in \partial \Omega } \hfill \cr } $$

where $\Omega$ is the domain of interest, ${\partial \Omega }$ is its boundary and $\bf{n}$ is the unit vector normal to the boundary. Here, $\nabla_S\Phi$, called the surface gradient, denotes component of $\nabla\Phi$ tangential to $\partial\Omega$ and $\bf{f}$ is a vector field on $\partial\Omega$, tangent to the boundary at all points. $\bf{I}$ is the identity mapping, $\cdot$ is the scalar product and $\otimes$ is the tensor product.


1) Is the solution to this BVP unique? If NO, what is the degree of non-uniqueness?

2) Is there a relation between the solution to this BVP and the one with Dirichlet boundary conditions, i.e., when we determine the function value on the boundary?

My Thought

I don't think that the solution is unique so I was thinking to relate this in some manners to the solution of the Laplace BVP with Dirichlet boundary conditions where the function value is prescribed over the boundary since this BVP has a unique solution.

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    $\begingroup$ Note that specific problem isn't always more simple than the general case of the same problem. $\endgroup$ – Michael Medvinsky Nov 17 '15 at 8:27
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    $\begingroup$ To the matter of this question. BC with directional derivative should work as well as the normal derivatives which is a general case of directional derivative. I have doubts thought about the specific case of tangential derivatives, I suspect it is not well-posed problem (means no solution in some cases or no unique solution in some other cases). The first step you should try is to reduce this problem to the problem with homogeneous BC and no homogeneous source. You can try to start with specific $f(x)$, e.g. basis functions of some basis or so. $\endgroup$ – Michael Medvinsky Nov 17 '15 at 8:28
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    $\begingroup$ try to read this (I didn't yet) ann.jussieu.fr/frey/papers/Navier-Stokes/… $\endgroup$ – Michael Medvinsky Nov 17 '15 at 8:43
  • $\begingroup$ From the probability point of view I would not expect this problem to be well-posed. That's because the Laplace equation is connected back to Brownian motion. With a Dirichlet boundary, you run the process until it hits the boundary, then record the boundary value and stop it, then average the boundary values that you get. $\endgroup$ – Ian Nov 17 '15 at 15:32
  • $\begingroup$ (Cont.) With a Neumann or Robin boundary (where the normal derivative is specified), you run the process until it hits the boundary and then reflect it back into the domain, possibly picking up some additional "heat" in the process (if the condition is inhomogeneous or Robin). Then the solution at a given point is the limiting average "temperature" if you start the process at that point. (In PDE terms, the solution to the Neumann or Robin Laplace problem is actually approached by the corresponding heat equation. One can generalize this to the Dirichlet case through a limit of Robin problems.) $\endgroup$ – Ian Nov 17 '15 at 15:34

Let $g$ be the restriction of $\Phi$ to the boundary of the square. Your boundary data determines $g$ up to a constant. You can then take this $g$ and solve the corresponding Dirichlet problem; standard theory tells that solutions exist uniquely. Therefore you can conclude that solutions are unique up to shifting by constants. (The fact that shifting a solution by a constant gives another solution is immediate, since the BVP only sees derivatives of $\Phi$.)

I assume that you want your function $\Phi$ to be continuous up to the boundary. You did not specify regularity assumptions, so I assume you work "classically enough" for this to make sense without any Sobolev spaces or such.

In general, knowing the gradient determines a function up to a constant, and now you know the boundary values up to a gradient. (The boundary is a manifold, not a Euclidean domain, so this can get a bit technical, but the same idea is true.) Let us see how to find $g$ in your specific case. First, $\partial_x\Phi(x,b)=0$, so $\Phi(x,b)=C$ for all $x$ for some constant $C$. Then, $\partial_y\Phi(a,y)=f(y)$, so $$ \Phi(a,y) = \Phi(a,b) + \int_b^yf(t)dt. $$ Since $g$ is continuous at the corner point $(a,b)$, you have $\Phi(a,b)=C$.

You assumed $f$ to be antisymmetric, so $\int_{-b}^bf=0$, and so $\Phi(a,-b)=C$. Now $\partial_x\Phi(x,-b)=0$, so in fact $\Phi(x,-b)=C$ for all $x$. And then you can use the same idea to find $\Phi$ on the remaining edge at $x=-a$.

You will get $$ \Phi(x,\pm b)=C\quad\text{and}\\ \Phi(\pm a,y)=C+\int_b^yf(t)dt. $$ These are your boundary values for a standard Dirichlet problem.

Alternatively, you could solve the ODE at each boundary part separately, obtaining four constants of integration. You will then get four equations to tie these together by continuity at the corners. One of these equations will be redundant — unless the boundary conditions are inconsistent and there is no solution.

Remark about your general formulation: In 2D it is fine, but in 3D not. There is no preferred tangent unit vector field. You either need to specify which vector field you choose or look at the whole tangential gradient instead. On the unit sphere $S^2\subset\mathbb R^3$ there is no globally defined unit tangent vector!

Added remark about the general case: Consider the boundary value problem for a harmonic function $\Phi$ in a bounded smooth domain $\Omega\subset\mathbb R^n$, $n\geq2$, with prescribed tangential gradient on $\partial\Omega$. Assume that the boundary $\partial\Omega$ is connected; this is true for balls, for example.

A solution need not exist; the boundary condition can be self-contradictory. To see why, consider the problem of finding a function $f:\mathbb R^n\to\mathbb R$ whose gradient is a prescribed vector field. No such function $f$ exists if the vector field does not satisfy some compatibility conditions. (For $n=3$ a sufficient and necessary compatibility condition is that the curl of the vector field vanish. The situation is a bit more complicated on manifolds than on Euclidean spaces, but the need for compatibility conditions remains. There can be non-local conditions, and the keyword is cohomology.)

Suppose that the boundary conditions are indeed such that there is at least one solution. How unique is this solution, then?

Suppose you had two solutions $\Phi$ and $\Psi$ to your BVP. Then $u=\Phi-\Psi$ solves the same (linear!) equation and has zero tangential gradient at the boundary. Since $\partial\Omega$ is connected, you can join any two points on it by a smooth curve on $\partial\Omega$. You can express the difference between the values of $u$ at these endpoints in terms of an integral of the gradient along the curve. This leads to the observation that $u$ is constant on the boundary. Now we can solve $u$ from a Dirichlet problem with constant boundary values, and the unique solution is well known to be constant. Since $u$ is constant throughout $\Omega$, the solutions $\Phi$ and $\Psi$ differ by a constant. That is, solutions are unique up to additive constants also in the general case.

If $\partial\Omega$ is not connected, the situations is more complicated. Then the same argument shows that the difference of any two solutions is constant on each component of $\partial\Omega$. If there are $m$ components in $\partial\Omega$, then the (affine) space of solutions to your BVP has dimension $m$.

  • $\begingroup$ Thanks for your thorough and nice answer for the special case. I think it is much better than the other two! :) About the general case, I meant the tangential gradient. Please feel free to edit the question for this part, if you can explain it better. :) $\endgroup$ – H. R. Nov 23 '15 at 9:22
  • $\begingroup$ @H.R., you are welcome. I will make the edit later when I have more time. I don't want to edit others' posts in a hurry. $\endgroup$ – Joonas Ilmavirta Nov 23 '15 at 9:25
  • $\begingroup$ OK, take your time. :) I am waiting for your edit and also the answer for the general case. :) $\endgroup$ – H. R. Nov 23 '15 at 9:26
  • $\begingroup$ @H.R., I edited my answer and your question. Feel free to re-edit the question or roll back if it didn't match your intentions. $\endgroup$ – Joonas Ilmavirta Nov 23 '15 at 12:54
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    $\begingroup$ @H.R., the general compatibility condition is indeed complicated. I don't want to discuss that here, since that would take us too far from your question. You can Google "de Rham cohomology". (Ask new questions about the new things if needed, of course!) Much of my work has to do with PDEs on manifolds, so I need a strong background in PDEs, but my knowledge comes from scattered small sources (not entire books or series), so I can't so much with the MESE question. Just don't plan too tightly and study what you seem to need. $\endgroup$ – Joonas Ilmavirta Nov 23 '15 at 16:15

I will deal only with the simplest version (rectangle) of the question. I will not use the symetry property of $f$.

$\bf\text{I. Connection to the Dirichlet problem.}$

Suppose that $\Phi$ is a solution of the system, and let $\Psi=\frac{\partial^2\Phi}{\partial x^2} = -\frac{\partial^2\Phi}{\partial y^2}$. Since $\Phi$ is harmonic, its partial derivatives are harmonic, too; and in particular, $\Psi$ is harmonic. Along the horizontal sides of the rectangle we have $$ \Psi(x,\pm b) = \frac{\partial^2\Phi}{\partial x^2}(x,\pm b) = \frac{\partial}{\partial x} \left( \frac{\partial\Phi}{\partial x}(x,\pm b) \right) = \frac{\partial}{\partial x}0 = 0.$$ Along the vertical sides we have $$ \Psi(\pm a,y) = -\frac{\partial^2\Phi}{\partial y^2}(\pm a,y) = -\frac{\partial}{\partial y}\left( = -\frac{\partial}{\partial y}(\pm a,y) \right) = -f'(y). $$ Hence, $\Psi$ is known along the boundary. For computing $\Psi$ we have to solve the Dirichlet problem.

$\bf\text{II. Uniqueness.}$

Clearly the solution is not unique: if $\Phi$ is a solution then $\Phi+C$ also satisfies the conditions. But this is the only freedom.

We already know that $\Psi=\frac{\partial^2\Phi}{\partial x^2}$ is unique. The function $-\frac{\partial^2\Phi}{\partial x\partial y}$ is the harmonic conjugate of $\Psi$, so $\frac{\partial^2\Phi}{\partial x\partial y}$ is determined up to a constant. Due to $\int_{-a}^a \frac{\partial^2\Phi}{\partial x\partial y} dx = \frac{\partial\Phi}{\partial y}(a,y)-\frac{\partial\Phi}{\partial y}(-a,y)=0$, that constant is determined; we have a unique $-\frac{\partial^2\Phi}{\partial x\partial y}$.

Then $$ \frac{\partial\Phi}{\partial x}(u,v)=f(v)+\int_{-a}^u \frac{\partial^2\Phi}{\partial x\partial y}(t,v)dt \quad\text{and}\quad \frac{\partial\Phi}{\partial y}(u,v)=\int_{-b}^v \frac{\partial^2\Phi}{\partial x\partial y}(u,t)dt $$ are uniquely determined. Hence, the partial derivatives of $\Phi$ are determined, so $\Phi$ is uniquely determined up to a constant term.

$\bf\text{III. Existence.}$

If $f$ is nice enough (say $f$ is continously differentiable and $f'(\pm b)=0$) then the Dirichet problem has a solution for $\Psi$. From that, we want to re-construct the function $\Phi$.

The function $\Psi$ is the real part of some holomorphic function $g$, $\Psi(x,y)=\Re g(x+yi)$. Then $g$ has a holomorphic antiderivative $g_1$, and $g_1$ also has a holomorphic antiderivative $g_2$ such that $g_1(-a-bi)=g_2(-a-bi)=0$.

Now take two real constants $A,B$ and let $$ \Phi(x,y) = \Re g_2(x+yi) +A(y+a)+B(x+a)(y+b) .$$ We show that the numbers $A,B$ can be chosen in such a way that $\Phi$ is a solution of the system.

Clearly, $\Phi$ is harmonic and $\Psi = \frac{\partial^2\Phi}{\partial x^2} = -\frac{\partial^2\Phi}{\partial y^2}$.

Along each horizontal side of the rectangle, $\frac{\partial\Phi}{\partial x}$ is constant, because $$ \frac{\partial^2\Phi}{\partial x^2}(x,\pm b) = \Psi(x,\pm b) = 0. $$ Along the bottom side this value is $$ \frac{\partial\Phi}{\partial x}= \frac{\partial\Phi}{\partial x}(-a,-b) = \Re g_1(-a-bi)=0. $$ Along the top side we have $$ \frac{\partial\Phi}{\partial x}= \frac{\partial\Phi}{\partial x}(a,b) = \Re g_1(a+bi)+2bB. $$ Now set $B=\frac{-\Re g_1(a+bi)}{2b}$; then we establish $\frac{\partial\Phi}{\partial x}=0$ along the top side as well. Notice that $\Phi$ is constant both along the top and the bottom sides of the rectangle.

Along each vertical side of the rectangle, $\frac{\partial\Phi}{\partial y}-f(y)$ is constant, because $$ \frac\partial{\partial y}\left(\frac{\partial\Phi}{\partial y}-f(y)\right) = \frac{\partial^2\Phi}{\partial y^2}-f'(y) = -\Psi-f'(y) = 0. $$ Along the left-hand side this constant is $$ \frac{\partial\Phi}{\partial y}-f= \frac{\partial\Phi}{\partial y}(-a,-b)-f(-b)=A-f(-b). $$ Chosing $A=f(-b)$ we get $\frac{\partial\Phi}{\partial y}=f$ along the left-hand side.

Along the right-hand side we have $\frac{\partial\Phi}{\partial y}-f=K$ with another constant $K$. By integrating along the vertical sides we can find that $K$ must be $0$: $$ 0 = \int_{-b}^b 0 dy = \int_{-b}^b \left(\frac{\partial\Phi}{\partial y}(-a,y)-f(y)\right) dy = \Phi(-a,b)-\Phi(-a,-b) - \int_{-b}^b f = \\ = \Phi(a,b)-\Phi(a,-b) - \int_{-b}^b f = \int_{-b}^b \left(\frac{\partial\Phi}{\partial y}(a,y)-f(y)\right) dy = \int_{-b}^b K dy = 2bK $$ $$ K=0. $$ Therefore, $\Phi$ is harmonic, $\frac{\partial\Phi}{\partial x}(x,\pm b)=0$ and $\frac{\partial\Phi}{\partial y}(\pm a,y)=f(y)$, so $\Phi$ satisfies the conditions.

  • $\begingroup$ +1 for the contribution. :) But I think you have made it a little complicated specially in the existence part! Is there a way to skip complex analysis. Also, in the uniqueness part, I don't understand your equations! Derivatives are with respect to $x$ and $y$ while the we have functions of $u$ and $v$! Also, we have $f(y)$ in the equations! $\endgroup$ – H. R. Nov 18 '15 at 19:32
  • $\begingroup$ Ok, I see. Tomorrow I will fix uniqueness. (Instead of $\Psi$ we need its harmonic conjugate.) $\endgroup$ – user141614 Nov 18 '15 at 20:50

Using the fundamental theorem of calculus and the Green's function $G$ for $\nabla$ that is $G(r) = r/[4\pi|r|^3]$, we can see in 3d that the following must be true:

$$\begin{align*}\nabla \Phi(r) &= -\int_V (G \circ s)(r') \times [\nabla \times (\nabla \Phi)|_{r'} ]\, dV' + \int_V (G \circ s)(r') (\nabla \cdot \nabla \Phi)|_{r'} \, dV\\ &\quad - \int_{\partial V} (G \circ s)(r') \times [\hat n(r') \times\nabla \Phi|_{r'}] \, dS' + \int_{\partial V} (G \circ s)(r') [\hat n(r')\cdot \nabla \Phi|_{r'}] \, dS'\end{align*}$$

(n.b. may be slightly wrong on the signs of these integrals, but the basic forms are correct)

where $s(r') = r - r'$ and $\hat n$ is the unit normal to the boundary. According to the identity $\nabla \Phi = \left( {\nabla \Phi .\hat n} \right)\hat n - \hat n \times \left( {\hat n \times \nabla \Phi } \right)$, specifying the tangential derivative is to specify $\hat n \times (\hat n \times \nabla \Phi)$. We know that $\nabla \times \nabla \Phi = 0$ identically, and $\nabla \cdot \nabla \Phi = 0$ as specified by the problem.

That eliminates both the volume integrals, but the surface integral with the normal derivative--corresponding to Neumann boundary conditions--still remains. So at this point, we can already conclude that the freedom to choose the normal derivative corresponds to freedom in $\nabla \Phi$.

So with the freedom to choose the value of $\Phi$ at one particular point (which we still have, as a constant of integration), the freedom to choose the normal derivative still allows for several distinct solutions (imagine selecting the value of $\Phi$ at a particular point on the boundary, and then integrating inward, normal to the boundary, to find $\Phi$ there; different values of the normal derivative will yield different values of $\Phi$).

In short, if you allow different values of the normal derivative, then the solution necessarily is not unique (aside from the ability to add or subtract a constant function).

  • $\begingroup$ Hence, you just answered my first question about uniqueness, right? :) Anyway, I couldn't understand your formula yet! $\endgroup$ – H. R. Nov 18 '15 at 20:53
  • $\begingroup$ Yes, that is so. I cannot offer any insight into the relationship with the Dirichlet problem, unfortunately. $\endgroup$ – Muphrid Nov 18 '15 at 20:54
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    $\begingroup$ All right. Finally, you could say that this formula suggests any vector field (as the gradient is) is determined by its tangential and normal components on the boundary as well as its divergence and curl within the volume. $\endgroup$ – Muphrid Nov 18 '15 at 20:56
  • $\begingroup$ Also, it would be good if you could determine the order of non-uniqueness. As the other answer shows about the special case, it is just a constant! $\endgroup$ – H. R. Nov 18 '15 at 20:57
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    $\begingroup$ I based this mostly on the multivector calculus chapter of Geometric Algebra for Physicists. What I personally have done that GAfP did not do is expand out the identity in terms of 3d vector calculus operations. $\endgroup$ – Muphrid Nov 19 '15 at 17:45

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