## 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)?

## 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.

## differential equations – Unable to solve nonlinear PDE with NDSolve

Lately, I’ve been trying to solve the following PDE:
$$begin{equation} -v_0 |nabla F| + {bf f}cdot nabla F +Dnabla^2F = -1 end{equation}$$
inside a 2D region between two disks both centered in the origin with radii $$r=1$$ and $$l=5$$ respectively. The boundary conditions are $$F=0$$ on the inner circle and $$hat{n}cdotnabla F=0$$ on the external one. Here is the code I have written:

``````v0 = 1.; D1 = 0.01; f = 0.7;
r2 := 1; l2 := 5; cell2 := 0.001;
[CapitalOmega]2 =
RegionDifference[Disk[{0, 0}, l2], Disk[{0, 0}, r2]];
pde2 = D1 Laplacian[FF[x, y], {x, y}] + f D[FF[x, y], x] -
v0 Sqrt[D[FF[x, y], x]^2 + D[FF[x, y], y]^2] ;
dcond2 = DirichletCondition[FF[x, y] == 0, x^2 + y^2 == r2^2];

Fsol = NDSolveValue[{pde2 == -1 + NeumannValue[0., x^2 + y^2 == l2^2],
dcond2}, {FF[x, y]}, {x, y} [Element] [CapitalOmega]2,
Method -> {"PDEDiscretization" -> {"FiniteElement",
"MeshOptions" -> {"MaxCellMeasure" -> cell2}}}];
``````

However, as you can see I am getting some error messages but I don’t know why nor how to deal with them.
I actually think this is coming from the sqrt in the $$|nabla F|$$ term, but I can’t get rid of that. Hope it is clear enough and thanks in advance for your help!

## elliptic pde – Does unique continuation hold for the derivatives of solutions of PDE?

Let $$D subset mathbf{R}^n$$ be the unit ball, and $$u in C^2(D)$$ be a solution of the linear elliptic PDE
$$begin{equation} a^{ij} D_{ij} u + b^i D_i u + cu = 0 quad text{in D}, end{equation}$$
where the coefficients are regular, say of class $$C^d$$ for some integer $$d geq 1$$.

Assume that $$u(0) = 0$$, $$Du(0) = 0$$. Strong unique continuation means that $$u$$ has finite order of vanishing at the origin unless it vanishes identically: there exists $$N > 0$$ so that
$$r^{-n} int_{D_r} u^2 notin O(r^N)$$ as $$r to 0$$ if $$u not equiv 0$$.

Question. Does this also hold for its partial derivatives $$D_k u$$?

## 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?

## Non linear PDE solution

Could someone advise if it is possible to solve the following PDE with Mathematica? I am quite a beginner in Mathematica so any input would be highly appreciated.

$$displaystylefrac{1}{2} sigma^2frac{partial^2 u(x,y)}{partial x^2}+frac{1}{2} sigma^2frac{partial^2 u(x,y)}{partial y^2}+afrac{partial u(x,y)}{partial x}left(frac{frac{partial u(x,y)}{partial x}+frac{partial u(x,y)}{partial y}}{3frac{partial u(x,y)}{partial x}+frac{partial u(x,y)}{partial y}}right)^2-r u(x,y)=0.$$,
where $$sigma$$, a and r are come constants.

The domain for $$sigma=0.85$$, $$r=0.05$$ and $$a=50$$ is specified as follows
$$y leq 0.52 + 0.46 x; x leq 0.52 + 0.46 y; xgeq 0; y geq 0; x leq 0.97; y leq 0.97$$

witht the boundary conditions are

1. $$u(x,y)=0$$ for $$x=0,yleq0.46$$;
2. $$u(x,y)=249.5 mathrm{e}^{-34.5784 x} (-1 + mathrm{e}^{34.6 x})$$ for $$y=0, xleq0.46$$;
3. $$frac{partial u(x,y)}{partial x}=1$$ for $$0.46leq xleq 0.97$$, on $$x = 0.52 + 0.46 y$$;
4. $$frac{partial u(x,y)}{partial y}=0$$ for $$0.46leq yleq 0.97$$, on $$y = 0.52 + 0.46 x$$.

## hyperbolic pde – Solving two dimensional wave equation using Fourier/Laplace transform

The Green’s function for the wave equation in two-dimensions is defined by
begin{align*} frac {partial^{2}}{partial t^{2}}G(r,t)-left(frac {partial^{2}}{partial x^{2}}+frac {partial^{2}}{partial y^{2}}right)G(r,t)=delta(t-tau)delta(x-xi)delta(y-eta). end{align*}
begin{align*} Gto 0text{ as }rtoinfty,text{ where }r^2=(x-xi)^{2}+(y-eta)^{2}. end{align*}
begin{align*} G=0text{ for }0
I was asked to derive the solution using either Fourier transform in $$x$$ and $$y$$ or Laplace transform in $$t$$. The solution is given by
begin{align*} G=frac {1}{2pi}frac {H(t-tau-r)}{sqrt{(t-tau)^{2}-r^2}},text{ where }Htext{ is the Heaviside function.} end{align*}
When I did the Fourier transform on $$x$$, I get
begin{align*} frac {partial^{2}}{partial t^{2}}tilde{G}+omega^{2}_{1}tilde{G}-frac {partial^{2}}{partial y^{2}}tilde{G}=delta(t-tau)delta(y-eta)e^{iomega_{1}xi}. end{align*}
Then I did the Fourier transform on $$y$$ to get
$$begin{equation}label{eqn:c} frac {partial^{2}}{partial t^{2}}hat{tilde{G}}+left(omega^{2}_{1}+omega^{2}_{2}right)hat{tilde{G}}=delta(t-tau)e^{i(omega_{1}xi+omega_{2}eta)}.tag{*} end{equation}$$
For $$t, $$hat{tilde{G}}=0$$. For $$t>tau$$, $$hat{tilde{G}}=c_{1}cos{left(sqrt{w^{2}_{1}+w^{2}_{2}}tright)}+c_{2}sin{left(sqrt{w^{2}_{1}+w^{2}_{2}}tright)}$$. I wonder how to match with $$t=tau$$ to determine the values of $$c_{1}$$ and $$c_{2}$$. Once I get $$hat{tilde{G}}$$, I need to do an inverse Fourier transform and use polar coordinates to get
begin{align*} G=frac {1}{2pi}int_{0}^{infty}J_{0}(kr)sin {kt}, dk. end{align*}
Then I can get the solution by using integral tables. For the Laplace transform in $$t$$, I wonder how to show that the Laplace transform of G is
begin{align*} tilde {G}=frac {1}{2pi}K_{0}(sr), end{align*}
where $$K$$ is modified Bessel function of the second kind. Any help are appreciated!

## partial differential equations – Solve a first order PDE using the method of characteristics (\$ exp(u)u_x + frac{y}{x}u_y = 1\$)

I’m solving the PDE:

$$exp(u)u_x + frac{y}{x}u_y = 1$$

With characteristics given by

$$frac{dx}{exp(u)} = frac{dy}{y/x} = frac{du}{1}$$

Giving a corresponding solution of

$$c1 = x^2 -2y^2exp(u)\ c2 = u-frac{y^2}{2x}$$

Thus

$$u(x,t) = frac{y^2}{2x}+ F(x^2 – 2yexp(u))$$ .

Is this right?

## partial differential equations – What’s the name of this phenomenon occurring when numerically computing the explicit Euler method for this PDE IVP and why does it occur?

I’ve been asked to calculate numerically the solution of this IVP using Euler’s explicit method on Mathlab:

$$u_t-2u_xx=0$$
$$u(x,0)=sin (2pi x), quad xin(0,1)$$
$$u(0,t)=u(1,t)=0, quad tin(0,0.05)$$

However, I find the approximation is really bad for $$h = 0.04, k = 4 · 10^{-3}$$ (here the exact solution is shown in blue and the red dots correspond to the approximated values):

…then it gets considerably better for a slightly inferior value of $$k$$: here the same plot is shown for $$h = 0.04, k = 4 · 10^{-5}$$:

I’d like to know how is this phenomenon called and why does it occur for this particular IVP. Is there any conditions a numerical method has to meet in order not to present this kind of problems?

## elliptic pde – Reference request on Pucci extremal operators

While reading (1), I encountered with the concept “Pucci extremal operator” which is defined by:
$$M_Lambda^-(N):=left(sumtext{positive eigenvalues of }Nright)+Lambdaleft(sumtext{negative eigenvalues of }Nright),text{ and}$$
$$M_Lambda^+(N):=-M_Lambda^-(-N),$$
where $$Nintext{Sym}_{ntimes n}$$ and $$Lambdageq 1$$.

Then the author claims that the problem
$$M_Lambda^-(D^2u)leq 0leq M_Lambda^+(D^2u)$$
in viscosity sense includes all $$C^2$$ solutions to uniformly elliptic equations of the form $$tr(A(x)D^2u)=0$$ where $$Ileq A(x)leqLambda I$$. Since there is no citation on this, it seems this is well-known result in the field, but I am new to this and I want to know about its motivation, history, etc. I did some article search but I could not find a good reference book or paper that introduces the concept of Pucci extremal operator (of course this may be due to my lacking of searching skill). I will appreciate if anyone would explain the concept or give me some good reference on it. Thank you in advance.

(1) Mooney, Connor, A proof of the Krylov-Safonov theorem without localization, Commun. Partial Differ. Equations 44, No. 8, 681-690 (2019). ZBL1426.35124.