reference request – Mean value formula for fractional heat equation

For the heat equation we have
$$u(z_0) = int_{Omega_r(z_0)}u(z) K_r(z_0-z),$$
where $$Omega_r(z_0) = left{z in mathbb{R}^{N+1}: Gamma(z_0-z) > 1/rright} quad K_r(z) = K_r(x,t) = frac{|x|^2}{4rt^2}$$
and $Gamma(z) = Gamma(x,t)$ is the fundamental solution at the origin of the heat equation.

Does something similar hold for the fractional heat equation $u_t + (-Delta)^su=0$?

differential equations – Heat transfer with functions defined on different domains

I’m attempting to model a situation in which a polymer initially at a higher temperature is sandwiched between two cooler metallic mold pieces with conductive boundary conditions in between the polymer and mold. On either outer side of the mold, there is convection with another fluid. Since the boundary conditions at either side of the sandwiched polymer where contact is made with the mold are the same, I’m focusing on only one of the mold pieces, with the mold spanning from $0 leq z leq d$ and the polymer spanning from $-w leq z leq 0$. The following equations model the setup:

$$frac{partial T_{mold}(z,t)}{partial t}=a_{mold}frac{partial^2T_{mold}(z,t)}{partial z^2}, 0 leq z leq d$$
$$frac{partial T_{poly}(z,t)}{partial t}=a_{poly}frac{partial^2T_{poly}(z,t)}{partial z^2}, -w leq z leq 0$$

Boundary conditions:
$$k_{mold}frac{partial T_{mold}(z,t)}{partial z}|_{z=d}=h_c(T_{mold}(z,t)-T_c)$$
$$-k_{mold}frac{partial T_{mold}(z,t)}{partial z}|_{z=0}=-k_{poly}frac{partial T_{poly}(z,t)}{partial z}|_{z=0}$$
$$-k_{mold}frac{partial T_{mold}(z,t)}{partial z}|_{z=-w}=-k_{poly}frac{partial T_{poly}(z,t)}{partial z}|_{z=-w}$$

Initial conditions:
$$T_{mold}(z,t=0)=T_m$$
$$T_{poly}(z,t=0)=T_p$$

I tried to approach solving this based on the wonderful iterative approach of @Alex Trounev from this post using the following code (the signs on NeumannValue might be incorrect due to my inexperience):

amold = 0.004;
Tm = 323;
apoly = 0.0001;
Tp = 433;
Tc = 303;
h = 5000;
d = 0.005;
w = 0.005;

poly(0)(z_, t_) := Tp;
Do(mold = 
  NDSolveValue({D(Tmold(z, t), t) - 
      amold*D(Tmold(z, t), {z, 2}) == 
     NeumannValue(-h (Tmold(z, t) - Tc)/
         kmold, z == d) + 
      NeumannValue(kpoly/kmold)*(D(poly(i - 1)(z, t), z) /. z -> 0), z == 0), 
    Tmold(z, 0) == Tm}, 
   Tmold, {z, 0, d}, {t, 0, 600}, 
   Method -> {"MethodOfLines", 
     "SpatialDiscretization" -> {"FiniteElement", 
       "MeshOptions" -> {"MaxCellMeasure" -> {"Length" -> 0.001}}}}); 
 poly(i) = 
  NDSolveValue({D(Tpoly(z, t), t) - 
      apoly*D(Tpoly(z, t), {z, 2}) == 
     NeumannValue(kmold/kpoly)*(D(mold(z, t), z) /. z -> 0), z == 0) + 
      NeumannValue(kmold/kpoly)*(D(mold(z, t), z) /. z -> 0), z == -w), 
    Tpoly(z, 0) == Tp}, 
   Tpoly, {z, -w, 0}, {t, 0, 600}, 
   Method -> {"MethodOfLines", 
     "SpatialDiscretization" -> {"FiniteElement", 
       "MeshOptions" -> {"MaxCellMeasure" -> {"Length" -> 
            0.001}}}});, {i, 1, 20})

However, I get a very unstable solution, and the kernel crashes after ~30 iterations. I also tried solving the system of PDEs without an iterative approach below, but I get the error that the boundary condition at $z=-w$ is not specified on a single edge of the boundary of the computational domain:

sys1 = {D(Tmold(z, t), t) - 
     amold*D(Tmold(z, t), {z, 2}) == 
    NeumannValue(-h (Tmold(z, t) - Tc)/
       kmold, z == d), 
   Tmold(z, 0) == Tm, 
   D(Tpoly(z, t), t) == 
    apoly*
     D(Tpoly(z, t), {z, 2}), -kmold*(D(Tmold(z, t), z) /. z -> 0) == -kpoly*(D(Tpoly(z, t), z) /. z -> 0), -kmold*(D(Tmold(z, t), z) /. z -> 0) == -kpoly*(D(Tpoly(z, t), z) /. z -> -w), 
   Tpoly(z, 0) == Tp};

heat = NDSolveValue(sys1, {Tmold,Tpoly}, {z, -w, d}, {t, 0, 600}, 
  Method -> {"MethodOfLines", 
    "SpatialDiscretization" -> {"FiniteElement", 
      "MeshOptions" -> {"MaxCellMeasure" -> {"Length" -> 0.001}}}})

I was wondering if anyone had insight into how I can improve either of these two approaches. Thank you again in advance for any help!

Including point heat sources in a 2D transient PDE heat equation

I want to be sure that I included the point heat sources correctly in my PDE.
The result should be:
enter image description here

I wrote:

q0 = 1;
Subscript(Q, 0) = 
  q0*(DiracDelta(x - x1)*DiracDelta(y - y1) + 
     DiracDelta(x - x1)*DiracDelta(y - y2)*DiracDelta(x - x2)*
      DiracDelta(y - y1) + DiracDelta(x - x2)*DiracDelta(y - y2));
heqn1 = Subscript(k, 
    CLS)*(D(Subscript(T, D)(x, y, t), x, x) + 
      D(Subscript(T, D)(x, y, t), y, y)) + Subscript(Q, 0) == 
  Subscript((Rho), CLS)*Subscript(Cp, CLS)*
   D(Subscript(T, D)(x, y, t), t)

I haven’t considered the time-dependency for the heat source just yet to simplify the notation.
Thank you in advance!

Graphing Heat Equation

I was able to solve the heat equation, but am unsure of how to graph the partial sums of the solution.

The problem:

a.) Solve the heat equation subject to

u(0, t) = 0, u(100, t)=0, t>0


u(x, 0) = 0.8x,  0 ≤ x ≤ 50
u(x, 0)=0.8(100-x), 50 < x ≤ 100

b.) Use the 3D plot application to graph the partial sums consisting of the first five nonzero terms of the solution in art a for 0 ≤ x ≤ 100, 0 ≤ t ≤ 200. Assume that k=1.6352. Experiment with various three-dimensional viewing perspectives of the surface.

I was able to get part a, but am unsure of how to code part b in Mathematica.

Prove an inequality involving the absolute value of the spatial derivative of the solution to the heat equation.

Given the initial value problem $$u_t=ku_{xx}$$ $x in mathbb{R}$, $t>0$ $$u(x,0)=f(x)$$ $x in mathbb{R}$ where $f$ is some square integrable function in $mathbb{R}$, prove there exist constant $c$ such that $$|u_x| leq c t^{-3/4} | f(x) |$$

My attempt, consider the solution to the heat equation $$u(x,t)=frac{1}{(4k pi t)^{1/2}} int_{- infty }^{infty } e^{-frac{(x-y)^2}{4kt}} f(y)dy$$
If I differentiate it I get $$u_x=frac{1}{(4k pi t)^{1/2}} int_{- infty }^{infty } e^{-frac{(x-y)^2}{4kt}} left( frac{-2(x-y)}{4kt} right) f(y)dy$$
Now $$|u_x|=left|frac{1}{(4k pi t)^{1/2}} int_{- infty }^{infty } e^{-frac{(x-y)^2}{4kt}} left( frac{-2(x-y)}{4kt} right) f(y)dy right| leq$$
$$left( frac{1}{(4k pi t)^{1/2}} int_{- infty }^{infty } e^{-frac{(x-y)^2}{2kt}} left( frac{-2(-1)(x-y)}{2kt} right) frac{1}{4kt} right)^{1/2} | f(x) |$$
I compute the integral using the error function by letting $z=frac{-(x-y)^2}{2kt}$ I get that $$|u_x| leq c t^{-3/2} | f(x) |$$ for $$c=frac{1}{4k (4k pi )^{1/2}} left( frac{sqrt{ pi } }{2} right)^{1/2}$$
But this is wrong since I got a $t^{-3/2}$ factor instead of the desired $t^{-3/4}$.
I spend a lot of time trying to figure out my mistake but couldn’t find it.
Can you help me spot my mistake?

Uniqueness of solution to heat equation when initial condition is a generalized function

Say $u(t,x)$ be a solution to the heat equation $$partial_t = partial_{xx} quad (t,x) in (0,T) times (-1,1)$$ subject to the initial/boundary conditions
$$u(0,x) = f(x), quad x in (-1,1), \
u(t,pm 1) = g^{pm}(t), quad t in (0,T),$$
with the usual compatibility conditions in corners: $f(pm 1) = g^{pm}(0)$. Suppose also that $f$ and $g$ are bounded and continuous. Then one can envoke the maximum principle or the energy method to prove that $u$ is the only solution.

What happens when $f$ or $g$ are unbounded? Say for instance, when $f(x) = delta_0(x)$ (point mass at zero) and $g^pm equiv 0$? This problem has a solution that can be easily represented as a series.

How does one go about proving uniqueness in such a situation?

In fact, come to think of it, how does one prove the uniqueness of the fundamental solution $v(t,x) = exp {-x^2 / (4t)}/ sqrt{4 pi t}$?

Is it some kind of weak uniqueness, where you show uniqueness of all classical solutions resulting from mollification of initial/boundary conditions? Is that the best one can do?

Any references would be deeply appreciated.