## Solve Partial Differential Equation with Neumann Boundary Condition

I am trying to solve the heat equation with certain boundary conditions (one-dimensional in space).
I tried it this way:

``````heatequation= (D(T(x, t), t) - a (D(T(x, t), {x, 2}))) == 0; mysol =
NDSolve({heatequation, T(0, t) == 4, D(1, 0)(T(x, t))(0, t) == 100,
T(x, 0) == 293}, T, {x, 0, 1}, {t, 0, 600})
``````

I get this error: 0 is not a valid variable.

Which zero is this about? The one in the Neumann boundary? The stack trace does not make it clear.

## differential equations – How to implement limit boundary condition in solving PDE

I have to solve a partial differential equation for a function $$F(x,t)$$ where one of the boundary condition is formulated in terms of a limit:

$$lim_{xrightarrow +infty} e^x partial_xF(x,t)=0$$

Is it possible to implement this as a boundary condition in NDSolve (or possibly NDSolve`FiniteDifferenceDerivative)?

## at.algebraic topology – Surface separating the boundary of a cylinder

Let $$M^2$$ be a connected closed surface. Suppose there exists an smooth embedding from a connected closed surface $$N$$ into the interior of $$M times [0,1]$$ such that $$N$$ separates $$M times {0}$$ and $$M times {1}$$.

If $$N$$ is homeomorphic to $$M$$, can we prove that the region bounded by $$M times {0}$$ and $$N$$ is homeomorphic to $$M times [0,1]$$?

If the conclusion is true, can we generalize it to the higher dimensional case?

## Second order elliptic PDE problem with boundary conditions whose solutions depend continuously on the initial data

Consider the following problem
$$begin{cases} -Delta u+cu=f,&xinOmega\ u=g,&xinpartialOmega end{cases}$$
where $$Omegasubseteqmathbb R^n$$ is open with regular boundary, $$cgeq0$$ is a constant, $$fin L^2(Omega)$$ and $$g$$ is the trace of a function $$Gin H^1(Omega)$$. If we consider $$u$$ a weak solution to this problem, and define $$U=u-Gin H_0^1(Omega)$$, it is easy to see that $$U$$ is a weak solution to the following problem
$$begin{cases} -Delta U+cU=f+Delta G-cG,&xinOmega\ U=0,&xinpartialOmega end{cases}$$
It is also easy to see that we can apply Lax-Milgram theorem with the bilinear form
$$B(u,v)=int_Omegaleft(sum_{i=1}^nu_{x_i}v_{x_i}+cuvright)$$
and the bounded linear functional
$$L_f(v)=int_Omega(f-cG)v-int_Omegasum_{i=1}^n G_{x_i}v_{x_i}$$
to conclude there exists a unique weak solution $$U$$ to the auxiliary problem defined above. If we define $$u=U+Gin H^1(Omega)$$, it is clear then that this function will be a solution to the original problem.

Now to the question: I would like to prove that this solution $$u$$ depends continuously on the initial data, that is, that there exists a constant $$C>0$$ such that
$$lVert urVert_{H^1(Omega)}leq C(lVert frVert_{L^2(Omega)}+lVert GrVert_{H^1(Omega)})$$
I feel that the work I have done to prove that $$L_f$$ is bounded should be relevant for our purposes, because
$$lVert urVert_{H^1(Omega)}leqlVert UrVert_{H^1(Omega)}+lVert GrVert_{H^1(Omega)}$$
and
$$lVert UrVert_{H^1(Omega)}leq C B(U,U)^{1/2}= C|L_f(U)|^{1/2}$$
The problem is that I don’t know how to manipulate $$L_f(U)$$ to obtain the result. I have managed to prove a completely useless inequality, for it involves the norm of $$U$$.

P.S. The problem is that a priori $$Delta G$$ doesn’t have to be in $$L^2(Omega)$$, which makes it hard to use the $$H^2$$ regularity of $$U$$ (which would solve the problem instantly).

P.S.S. Also posted this question in SE.

## fa.functional analysis – Poincare Inequality for \$H^2\$ function satisfying homogeneous Robin boundary conditions

Let $$Omegasubsetmathbb{R}^3$$ be a bounded smooth domain. In general, for a Poincare inequality of the type
$$|u|_{L^2}le C |nabla u|_{L^2}$$
to hold for all $$uin Xsubset H^1(Omega)$$ and $$C$$ independent of $$u$$, then $$X$$ needs to be such that it doesn’t contain constant translates. That is, if we consider $$u+M$$ for large $$M>0$$, the left hand side of the inequality increases indefinitely while the right hand side is unchanged, so we need some extra constraint in the definition of $$X$$. So common choices are $$X=H^1_0(Omega)$$ or $$X={uin H^1(Omega)| int_Omega u,dx=0}$$.

Here’s my question. Suppose, we’d like to say that there exists $$C$$ such that for all $$uin X={uin H^2(Omega)|(partial_n u+u)_{|partialOmega}=0}subset H^2(Omega)$$ we have
$$|u|_{L^2}le C|nabla u|_{L^2}.$$
First, is this true? If so, how does one prove such a statement? Essentially the requirement that $$u$$ satisfies the homogeneous Robin condition $$(partial_n u+u)_{|partialOmega}=0$$ should at least formally rule out constant translates, since $$(partial_n u+u)_{|partialOmega}=0$$ is not invariant under translation of $$u$$ by constants.

My guess is that it IS true, however, the usual proof I know of such statements usually relies on some compactness argument. For example, if $$X$$ were simply $$H_0^1(Omega)$$, then for the sake of contradiction, if we assume that there exists a sequence $$u_nin H_0^1$$ such that
$$|u_n|_{L^2}ge n|nabla u_n|_{L^2}$$
then, defining $$v_n=u_n/|u_n|_{L^2}$$, we have
$$frac{1}{n}ge |nabla v_n|_{L^2}.$$
Thus we have a bounded sequence in $$H^1$$ and a subsequence that converges strongly in $$L^2$$ and weakly in $$H^1$$ to some $$vin H^1$$. Because $$|nabla v_n|_{L^2}to 0$$, $$v$$ is constant. And since the trace map is continuous (and weakly continuous) from $$H^1(Omega)$$ to $$H^frac{1}{2}(partialOmega)$$ we have that $$v$$ is in fact in $$H^1_0(Omega)$$ and therefore $$v=0$$. Then we have a contradiction because $$|v_n|_{L^2}=1$$ for each $$n$$ implies that $$|v|_{L^2}=1$$.

Now this argument doesn’t work for Robin boundary conditions because now the relevant (Robin) trace operator is continuous and weakly continuous from $$H^2(Omega)$$ to $$H^frac{1}{2}(partialOmega)$$. In particular, if $$v$$ is the weak $$H^1$$ limit of a sequence $$v_nin H^2$$, then $$v$$ could be in $$H^1$$ but not $$H^2$$ and thus the notion of the normal derivative $$(partial_n v)_{|partialOmega}$$ may not even make sense for $$v$$. And without being able to say $$(partial_n v+v)_{|partialOmega}=0$$, we can’t necessarily say that $$v=0$$ like we did in the previous paragraph. So this is where I’m stuck. Any help would be appreciated.

## complex analysis – Bounded analytic function that can not be extended to the boundary of the unit disc

I am trying to find a Luzin example as promised on the website encyclopediaofmath, but I don’t have access to Luzin collected works, which are referred to in the article. In particular, I don’t even read Russian, so I doubt I could make much use from them. Does anyone know the actual counterexample and could explain the construction process? I don’t need a proof that most can be extended, but would simply like to see an actual example of a function where it cannot be extended to the boundary on some points, potentially even on an infinite (uncountably?) subset of measure $$0$$ of the unit disc.

Cheers for any help on this.

## ordinary differential equations – Given boundary conditions and initial condition, solve the PDE

Question, solve the given PDE :

$$frac{partial C}{partial t} = a frac{partial^2 C}{partial x^2} -kC$$
where a, k are constants.

boundary conditions : $$C= C_0$$ at $$x=0$$, $$C=0$$ at $$x =$$ infinitum

initial condition : $$C=0$$ at $$t=0$$

My attempt :

$$C(x,t) = X(x)~T(t)$$
$$C = X~T$$
$$boxed{ frac{partial C}{partial t} = T^{prime} X,~ space frac{partial C}{partial x}= X^{prime}T, ~ frac{partial^2 C}{partial x^2} = X^{prime prime } T}$$
replacing the partial derivatives into the Original equation :
$$T^{prime}X = a ~X^{prime prime}T-k(X~T)$$
$$X (T^{prime} +KT)=a ~X^{prime prime}T$$
$$frac{T^{prime}}{T} = a ~frac{ X^{prime prime}}{X} -k$$
$$boxed{frac{T^{prime}}{T} = J, ~frac{ X^{prime prime}}{X} -k = J}$$
equating each side to a constant $$J = -lambda^2$$
$$boxed{frac{d T}{d t} = -lambda^2 T ,~ frac{d^2 X}{d x^2} = left (frac{-lambda^2 + k}{a} right) X}$$
solving each differential equation :
$$T(t) = Ae^{- lambda t} , ~ X(x) = B cos{left (x sqrt{frac{lambda^2 -k}{a}} right)} + C sin{left (-x sqrt{frac{lambda^2 -k}{a}} right) }$$
Note : D = AB and E = AC
$$boxed{C(x,t) = left(D cos{left (x sqrt{frac{+lambda^2 -k}{a}} right)} + ~E sin{left (x sqrt{frac{lambda^2 – k}{a}} right) } right) e^{- lambda t}}$$
first boundary condition $$C= C_0$$ at $$x=0$$ :
$$D e^{- lambda t} = C_0$$
second boundary condition $$C=0$$ at $$x =$$ infinitum : DNE
$$C(x,t) = C_0cos{left (x sqrt{frac{+lambda^2 -k}{a}} right)} + ~E sin{left (x sqrt{frac{lambda^2 – k}{a}} right) } e^{- lambda t}$$
using initial condition $$C=0$$ at $$t=0$$ :
$$C_0 cos{left (x sqrt{frac{lambda^2 -k}{a}} right)} = -E sin{left (x sqrt{frac{lambda^2 – k}{a}} right) }$$
final equation :
$$boxed{ C(x,t) = -E sin{left (x sqrt{frac{lambda^2 – k}{a}} right) } + E sin{left (x sqrt{frac{lambda^2 – k}{a}} right) }e^{- lambda t} }$$

Could you guys please verify the answer also is “second boundary condition : DNE” true?

## differential equations – How to solve the problem with a particular “Boundary value condition”?

When I try to solve the boudary condition problem with the following code

``````Clear("Global`*")
u = 0.1; v = 0.1; c1 = u (1 - 3 v)/(1 - 2 v); c2 =
u (1 - v)/(1 - 2 v); c3 = u (2 v)/(1 - 2 v); A = 0.6; B = 1.;
eqs = r''(R) + (c1/(c2*R) - R/(2 r(R)^2)) r'(R)^3 - 1/(2 R) r'(R) == 0;
bcs = {r'(A) == -A (c2/(c1*A^2 + c3*r(B)^2))^(1/2),
r'(B) == -B (c2/(c1*B^2 + c3*r(A)^2))^(1/2)};
NDSolve(Flatten@Join({eqs, bcs}), r, {R, A, B})
``````

it appears the warnings of “Power::infy”. Anyone can help explain this warning or solve this boudary condition problem?

## differential geometry – Calculate surface normals at the boundary of a Graphics3D object

First of all, we need 2 more options in `boundedOpenCone`. The option `BoundaryStyle -> Automatic` creates a `Line` on the boundary so we can easily locate the coordinates of point on the boundary. `PlotPoints -> 100` isn’t actually necessary, but will make the resulting boundary smoother.

``````boundedOpenCone(centre_, tip_, Rc_, vec1_, vec2_, sign_) :=
Module({v1, v2, v3, e1, e2,
e3},(*function to make 3d parametric plot of the section of a cone bounded between
two vectors:tvec1 and tvec2*){v1, v2, v3} = # & /@ HodgeDual(centre - tip);
e1 = Normalize(v1);
e3 = Normalize(centre - tip);
e2 = Cross(e1, e3);
ParametricPlot3D(
s*tip + (1 - s)*(centre + Rc*(Cos(t)*e1 + Sin(t)*e2)), {t, 0, 2 (Pi)}, {s, 0, 1},
Boxed -> False, Axes -> False, Mesh -> None, BoundaryStyle -> Automatic,
RegionFunction ->
Function({x, y, z},
RegionMember(HalfSpace(sign*Cross(vec1 - tip, vec2 - tip), tip), {x, y, z})),
PlotPoints -> 100, PlotStyle -> ColorData("Rainbow")(1)))

vec1 = {1, 0, 0}; vec2 = (1/Sqrt(2))*{1, 1, 0};
coneTip = {0, 0, 3};
cvec = {0, 0, 0};
Rc = Norm(vec1 - cvec);

pplot = boundedOpenCone(cvec, coneTip, Rc, vec1, vec2, -1);
``````

Then we modify `normalsShow` from the document of `VertexNormals` a little to preserve only the normals on the boundary:

``````boundarynormals(g_Graphics3D) :=
Module({pl, vl, boundaryindexlst = Flatten@Cases(g, Line(a_) :> a, Infinity)},
{pl, vl} = First@Cases(g,
GraphicsComplex(pl_, prims_, VertexNormals -> vl_,
opts___?OptionQ) :> {pl, vl}(Transpose)((boundaryindexlst))(Transpose),
Infinity);
Transpose@{pl, pl + vl/3});

vectors = boundarynormals@pplot;

``````

## complex analysis – Where do the boundary conditions go in 2D Laplace equation?

While reading Fourier Series and Boundary Value Problems by R.V. Churchhill, I came across a very strange solution strategy for solving the wave equation.

Consider the wave equation $$u_{tt} = a^2u_{xx}$$ subject to the boundary conditions $$u(x,0)=f(x)$$ and $$u_t(x,0) = 0$$.

By using the substitutions $$r = x+at$$ and $$s=x-at$$ the problem simplifies to $$u_{rs} = 0$$ which can be solved easily by successive integration getting $$u = g(r)+h(s)$$. If we now change coordinates and solve this subject to the two boundary conditions we have the solution:
$$u = frac{1}{2}(f(x+at)+f(x-at))$$
We only had 2 boundary conditions to start, but we should have had 4 free choices throughout the solution process. What happened to the other 2 choices?

I have a theory that it might be easier to hash this out if we make the strange substitution $$a=i$$ which will transform the wave equation into Laplace’s equation.
So we now have the equation $$u_{tt}+u_{xx}=0$$ subject to $$u(x,0)=f(x)$$ and $$u_t(x,0) = 0$$. With a solution given by:
$$u = frac{1}{2}(f(x+it)+f(x-it))$$

This shockingly will still give you actual solutions to the problem e.g. $$f(x)=x(2-x)$$ yields the soultion $$u = x(2-x) +t^2$$ which indeed is a solution to Laplace’s equation.

These are solutions you just cannot get using a Fourier method, because we do not have the periodic boundary conditions we are comfortable with. Still I am lost as to where these two free choices went? I wonder if their are certain conditions on $$f$$ that actually take the place of our boundary conditions, Does $$f$$ have to be a harmonic function? How does that limit the values of $$u$$?

I am confused and would appreciate any of your thoughts!