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)?

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 $.

PDE by ParametricNDSolve, overdetermined?

I want to solve qr(z,t),qi(z,t),v(z,t) and T(z,t) with 4 PDEs. I tried method of parametricNDsolveValue with parameters of qr(z),qi(z),v(z) and T(z) with t as pars.
But, It does not work, even if I replace kr=1 and ki=1.

When I tried values such as sol(1)(0), several messages appear, such as overdetermined, cannot be transposed and et al. Could anyone help me out? Thanks!

       D(qi,z)==-K1(3 qi^2-qr^2)+K2(-qi^3*qr^(-1) + 3 qi*qr)-4K0*qi*qr,


differential equations – DSolve – incorrect solution to PDE

I’m trying to solve a simple Schrödinger equation in external field:

DSolve({-Derivative(2, 0)(psi)(x,t)/2 + F Sin((Pi) t)x  psi(x,t)==I Derivative(0, 1)(psi)(x,t),psi(x,0)==Exp(-x^2)}, psi, {x,t}, Assumptions-> t (Element) Reals && F (Element) Reals && x (Element) Reals)

And the solution that Mathematica spits out appears to be incorrect:

E^(I x (x/(-I + 2 t) - F t Sin((Pi) t)))/Sqrt(1 + 2 I t)

When I plug it to the original equation and FullSimplify I get:

(E^(I x (x/(-I+2 t)-F t Sin((Pi) t))) F t (2 (Pi) (I-2 t) x Cos((Pi) t)+Sin((Pi) t) (-4 x+F t (-I+2 t) Sin((Pi) t))))/(Sqrt(1+2 I t) (-I+2 t))==0

which obviously doesn’t appear to be zero for all F values (although it is for F=0).

Here’s a screenshot from my notebook:
enter image description here

I’m using MMA 12.3.0 on MacOS.

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
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).

Gto 0text{ as }rtoinfty,text{ where }r^2=(x-xi)^{2}+(y-eta)^{2}.

G=0text{ for }0<t<tau.

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
G=frac {1}{2pi}frac {H(t-tau-r)}{sqrt{(t-tau)^{2}-r^2}},text{ where }Htext{ is the Heaviside function.}

When I did the Fourier transform on $x$, I get
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}.

Then I did the Fourier transform on $y$ to get
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{$*$}

For $t<tau$, $hat{tilde{G}}=0$. 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
G=frac {1}{2pi}int_{0}^{infty}J_{0}(kr)sin {kt}, dk.

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
tilde {G}=frac {1}{2pi}K_{0}(sr),

where $K$ is modified Bessel function of the second kind. Any help are appreciated!

inequalities – Show `contraction-like’ inequality for solution of first-order PDE

I am interested in the following problem.

Let $D = mathrm{diag}(d_1, d_2, ldots, d_n) in mathbb{R}^n$ be positive definite, let $B, K in mathbb{R}^n$, and let $Gin L^infty((0, T)times (0, L); mathbb{R}^{n times n})$ be such that $|G|_{L^infty((0, T)times (0, L); mathbb{R}^{n times n})} leq R$ with $R>0$ as small as we want.
Show that there exists $M>0$ (independent of $y, overline{y}$) such that: if $y colon (0, T) times (0, ell) rightarrow mathbb{R}^n$ is solution to
partial_t y + D partial_x y + B y = G(t,x) overline{y}(x,t) & text{in }(0, T)times(0, ell)\
y(t, 0) = K overline{y}(t, ell) &text{for }t in (0, T)\
y(0, x) = 0 & text{for }x in (0, ell)

for some $overline{y}$ which is at least in $L^infty(0, T; H_x^1)$, then $y$ necessarily fulfills
|y|_{L^infty(0, T; L_x^2)} + M |y(cdot, ell)|_{L^2(0, T; mathbb{R}^n)} leq frac{1}{2} left( |overline{y}|_{L^infty(0, T; L_x^2)} + M |overline{y}(cdot, ell)|_{L^2(0, T; mathbb{R}^n)} right).

Here I used the shortened notation $L_x^2 = L^2(0, ell; mathbb{R}^n)$ and $H_x^1 = H^1(0, ell; mathbb{R}^n)$.

I tried the following (below I do not write the argument $t$ for $y$ and $overline{y}$ in the integrands in order to lighten the notation).
Using the governing equations and that $2leftlangle y, D partial_x y rightrangle = partial_x left( langle y, Dy rangle right)$ we get
frac{mathrm{d}}{mathrm{d}t} left( |y(t, cdot)|_{L_x^2}^2 right)
&= 2 int_0^ell left langle y, partial_t y right rangle dx \
&= – 2 int_0^ell left langle y, D partial_x y right rangle dx + 2 int_0^ell left langle y,- B y right rangle dx + 2 int_0^ell left langle y, G(t,x)overline{y} right rangle dx\
&= – left langle y(t, ell), D y(t, ell) right rangle + left langle y(t, 0), D y(t, 0) right rangle + 2 int_0^ell left langle y,- B y right rangle dx\
&quad + 2 int_0^ell left langle y, G(t,x)overline{y} right rangle dx,

where $langle cdot, cdot rangle$ denotes the inner product in $mathbb{R}^n$.
Then, using the boundary conditions we get
frac{mathrm{d}}{mathrm{d}t} left( |y(t, cdot)|_{L_x^2}^2 right)
&= – left langle y(t, ell), D y(t, ell) right rangle + left langle K overline{y}(t, ell), D K overline{y}(t, ell) right rangle + 2 int_0^ell left langle y,- B y right rangle dx\
&quad + 2 int_0^ell left langle y, G(t,x)overline{y} right rangle dx.

On the other hand,
frac{mathrm{d}}{mathrm{d}t}left( int_0^t left langle y(tau, ell), D y(tau, ell) right rangle dtau right)
&= left langle y(t, ell), D y(t, ell) right rangle.

Hence, denoting $eta(t) := |y(t, cdot)|_{L_x^2}^2 + int_0^t |D^frac{1}{2} y(tau, ell)|^2 dtau$ (and using the Cauchy-Schwarz and Young inequalities), we obtain that
frac{mathrm{d}}{mathrm{d}t} eta(t)
&=|D^frac{1}{2}Koverline{y}(t, ell)|^2 + 2 int_0^ell left langle y,- B y right rangle dx + 2 int_0^ell left langle y, G(t,x)overline{y} right rangle dx\
%&leq C_1 left(|overline{y}(t,ell)|^2 + |y(t, cdot)|_{L_x^2}^2 + |G|_{L^infty}|overline{y}(t)|_{L_x^2}^2right)\
&leq |D^frac{1}{2} K overline{y}(t,ell)|^2 + R |overline{y}(t)|_{L_x^2}^2 + (2|B| + R) eta(t).

Thus, by Gronwall’s inequality and the fact that $y(0, cdot)equiv 0$, we get
eta(t) &leq e^{(2|B|+R) t} int_0^t left(|D^frac{1}{2}Koverline{y}(tau,ell)|^2 + R|overline{y}(tau)|_{L_x^2}^2 right)dtau\
&leq e^{(2|B|+R) T} left(|D^frac{1}{2}Koverline{y}(cdot,ell)|_{L^2(0, T; mathbb{R}^n)}^2 + TR|overline{y}|_{L^infty(0, T; L_x^2)}^2 right)dtau.

|y|_{L^infty(0, T; L_x^2)}^2 + |D^frac{1}{2}y(cdot, ell)|_{L^2(0, T; mathbb{R}^n)}^2
leq e^{(2|B|+R) T} left(|D^frac{1}{2}Koverline{y}(cdot, ell)|_{L^2(0, T; mathbb{R}^n)}^2 + T R |overline{y}|_{L^infty(0, T; L_x^2)}^2 right).

So, we have obtained that
|y|_{L^infty(0, T; L_x^2)} + |D^frac{1}{2}y(cdot, ell)|_{L^2(0, T; mathbb{R}^n)} leq 2 e^{(|B|+tfrac{1}{2}R) T} left(|D^frac{1}{2}Koverline{y}(cdot, ell)|_{L^2(0, T; mathbb{R}^n)} + sqrt{TR} |overline{y}|_{L^infty(0, T; L_x^2)} right)

which also implies that
|y|_{L^infty(0, T; L_x^2)} + |y(cdot, ell)|_{L^2(0, T; mathbb{R}^n)} leq C e^{(|B|+tfrac{1}{2}R) T} left(|overline{y}(cdot, ell)|_{L^2(0, T; mathbb{R}^n)} + sqrt{TR} |overline{y}|_{L^infty(0, T; L_x^2)} right),

C= 2frac{max{1, |D^frac{1}{2}K|}}{min{1, min_{1leq ileq n}d_i}}.

However, I don’t see how to proceed further to obtain the desired inequality. It seems that the term $sqrt{TR} |overline{y}|_{L^infty(0, T; L_x^2)}$ can be made small via a certain choice of $T$ or $R$, but I do not know how to deal with the other term.

Is there another way to do the estimations for the term $|y(cdot, ell)|_{L^2(0, T; mathbb{R}^n)}$?

Any help, hint, solution, comment is appreciated. Thank you!

Also asked on MSE: https://math.stackexchange.com/questions/4144078/show-contraction-like-inequality-for-solution-of-first-order-pde

hyperbolic pde – Conservation law in the sense of measures

I am trying to understand the Lagrangian formulation of conservation laws in https://arxiv.org/abs/1608.02811 and I stumbled into the following problem.

Let $u$ be a $BV$ entropy-solution of the scalar conservation law
$$u_t(x,t) + f(u(x,t))_x = 0$$
It is claimed that then (since the derivatives of a $BV$ function are measures) it holds in the sense of measures that

$$D_tu + lambda D_xu = 0$$

$$lambda(t,x) = begin{cases}
f^{prime}(u(t,x)) & if quad (t,x)quad text{is continuity point of $u$} \
frac{f(u^+)-f(u^-)}{u^+-u^-}& if quad (t,x) quad text{is a point of jump for $u$}

Now, i tried to prove this using the structure theorem for the derivatives of $BV$ functions, but I can’t manage to recover the above formula. Does someone know how to do that? Thanks very much in advance.

Does this geometric PDE has a solution?

Let $s(theta), b(theta)$ be two smooth non-constant real-valued functions on $mathbb{S}^1$, and assume that $s$ never vanishes.

Do there exist a map $h:(0,1) times mathbb{S}^1 to mathbb{S}^1$, such that for every $r in (0,1)$, $h(r,cdot)$ is a diffeomorphism of $mathbb{S}^1$,

h_{theta}=s(theta)/s(h(r,theta)), tag{1}

h_{theta}^2+ 1/h_{theta}^2+big(rh_r+b circ h-bcdot h_{theta}big)^2/s^2 tag{2}

is independent of $r,theta$?

Are there some conditions on $s,b$ that imply such a solution $h$ exists?

Using equation $(1)$, the expression in $(2)$ can be written as
big(frac{s}{s circ h}big)^2+ big(frac{s circ h}{s}big)^2+ big(rh_r+bcirc h-b cdot frac{s}{s circ h}big)^2/s^2. tag{3}

This PDE arises when trying to build ‘concentric’ area-preserving diffeomorphisms of a given $2D$ shape, having constant singular values.

Help with this PDE

I’ve found this PDE in an exercise
f_x(x,y) + y^2 f_{yy}(x,y) + f(x,y) + g(x) = 0

and I would need some help to refresh PDE tools in order to being able to find a solution. Could anyone give me some hints to do it? Thanks!