## 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)?

## How to use the Laplace transform of the option PDE to derive ordinary differential equations?

The Black-schole PDE
$$frac{partial V}{partial t}+frac{1}{2}sigma^2S^2frac{partial^2 V}{partial S^2}-rV+rSfrac{partial V}{partial S}=0$$

How to use the Laplace transform and the inversion method to solve it
The bound condition is $$V(0,t)=0$$

## 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!

`````` ClearAll("`Global`")
equ=With({v=v@z,qr=qr@z,qi=qi@z,T=T@t,B=10,n=100,Z=5/2,a0=0.01},
With({ve=1.72*Z*n*T^(-3/2)},
With({er=1+n*(B-1)/((B-1)^2+ve^2),ei=n*ve/((B-1)^2+ve^2)},
With({K1=Integrate(((1-er)Cos(qi*r^2)-ei*Sin(qi*r^2))Exp(-qr*r^2)*r,{r,0,100}),
K2=Integrate(((1-er)Sin(qi*r^2)+ei*Cos(qi*r^2))Exp(-qr*r^2)*r,{r,0,100}),
K0=Integrate(ei*Exp(-2*qr*r^2)*r,{r,0,100}),
kr=((er+(er^2+er^2)^(1/2))/2)^(1/2),
ki=((er+(er^2+er^2)^(1/2))/2)-er)^(1/2)},
{D(v,z)==-2*K1*v*qi-K2*v*qr^(-1)(qi^2-qr^2)-4K0*v*qr,
D(qr,z)==-4*K1*qr*qi-2*K2*(qi^2-qr^2)-4K0*qr^2,
D(qi,z)==-K1(3 qi^2-qr^2)+K2(-qi^3*qr^(-1) + 3 qi*qr)-4K0*qi*qr,
D(3/2*n*T/(511*10^3),t)+1/4*D((kr^2+er)a0^2*v^2*Exp(-2*ki*z),t)==-D(kr*a0^2*v^2*Exp(-2*ki*z),z)-n/2*ve/((B-1)^2+ve^2)*a0^2*v^2*Exp(-2*ki*z)
}
))));

ic={v(0)==1,qr(0)==1,qi(0)==0,T(0)==100};
sol=ParametricNDSolveValue({equ,ic},{v,qr,qi,T},{z,0,30},{t})
v1(t_,z_):=sol(t)(z)((1))
``````

## differential equations – DSolve – incorrect solution to PDE

I’m trying to solve a simple SchrÃ¶dinger 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:

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
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!

## 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
begin{align*} begin{cases} 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) end{cases} end{align*}
for some $$overline{y}$$ which is at least in $$L^infty(0, T; H_x^1)$$, then $$y$$ necessarily fulfills
begin{align*} |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). end{align*}

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
begin{align*} 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, end{align*}
where $$langle cdot, cdot rangle$$ denotes the inner product in $$mathbb{R}^n$$.
Then, using the boundary conditions we get
begin{align*} 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. end{align*}
On the other hand,
begin{align*} 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. end{align*}
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
begin{align*} 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). end{align*}
Thus, by Gronwall’s inequality and the fact that $$y(0, cdot)equiv 0$$, we get
begin{align*} 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. end{align*}
Consequently,
begin{align*} |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). end{align*}
So, we have obtained that
begin{align*} |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) end{align*}
which also implies that
begin{align*} |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), end{align*}
with
begin{align*} C= 2frac{max{1, |D^frac{1}{2}K|}}{min{1, min_{1leq ileq n}d_i}}. end{align*}

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

$$D_tu + lambda D_xu = 0$$

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

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 $$mathbb{S}^1$$,

$$h_{theta}=s(theta)/s(h(r,theta)), tag{1}$$
and
$$h_{theta}^2+ 1/h_{theta}^2+big(rh_r+b circ h-bcdot h_{theta}big)^2/s^2 tag{2}$$
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!