ReplaceAll error on solving system of differential equations

here’s my problem, I’m trying to solve symbolically a system of differential equations (simplified example):

fun[F_, P_, x0_, y0_, tau_, A_, t_] =   x /. First[    DSolve[{ x'[t] == -F * (x[t] - y[t]) +  x[t] * A * Exp[-t / tau] ,       y'[t] == +F * (x[t] - y[t]) - P * (y[t] - y0), x[0] == x0,       y[0] == y0}, {x, y}, t]] 

which doesn’t work:

ReplaceAll: {(x^\[Prime])[t]==A E^(-(t/tau)) x[t]-F y[t],(y^\[Prime])[t]==F  (x[t]-y[t]),x[0]==x0,y[0]==y0}is neither a list of replacement rules nor a valid dispatch table,      and so cannot be used for replacing, 

it only works if I take out the inhomogeneous term:

 fun[F_, P_, x0_, y0_, tau_, A_, t_] =       x /. First[        DSolve[{ x'[t] == -F * (x[t] - y[t]) +  x[t] * A ,           y'[t] == +F * (x[t] - y[t]) - P * (y[t] - y0), x[0] == x0,           y[0] == y0}, {x, y}, t]] 

in which case works fine. My question is: is this a mistake on my part or is this just because this goes beyond the capabilities of DSolve?

How to compute a system of ordinary differential equations with initial condictions over a continuous range

I have some questions about Mathematica programming and would appreciate if you could help me.

I want to solve a system of ordinary differential equations \ [Mu] ‘[t] and \ [Lambda]’ [t] and each equation contains a large number of terms so it is impractical to write them explicitly. I express these terms as two functions F1 and F2 that depend on two parameters P1 and P2 and \ [Lambda] [t] and \ [Mu] [t].

I have been able to solve this system for a couple of initial conditions \ [Lambda] [0] = ic1 and \ [Mu] [0] = ic2, but I would like to solve my system of equations for a continuum of values ​​\ [Lambda] [0] = {0, …., Pi / 2} and \ [Mu] [0] = {0, …., Infinity} and then get \ [Lambda] [t] and \ [Mu] [t] and use them to perform an integral on \ [Lambda]= \ [Lambda] [0] ={0, …., Pi / 2} and \ [Mu] =\ [Mu] [0] ={0, …., Infinity} that are precisely our initial conditions.

I integrate the product of a function G in the time t (where \ [Lambda] [t] and \ [Mu] [t] are taken into account for a certain initial condition defined by the continuous ranges of the integral) with the same function, but in t = 0 (where the initial conditions are taken into account with the continuous ranges of the integral).

The structure of the program is:

ode = {\[Mu]'[t] ==  F1[p1, p2, \[Lambda][t], \[Mu][t]], \[Lambda]'[t] ==  F2[p1, p2, \[Lambda][t], \[Mu][t]], \[Mu][0] == {0, ....,   Pi/2}, \[Lambda][0] == {0, ...., Infinity}};   Sol = NDSolve[ode, {\[Mu], \[Lambda]}, {t, 0, 1},`Method -> "Some method to choose"]     \[Mu]1[t_] := Evaluate[\[Mu][t] /. Sol] // First \[Lambda]1[t_] := Evaluate[\[Lambda][t] /. Sol] // First   data = ParallelTable[{t,  NIntegrate[   G[p1, p2, \[Mu]1[      t] "for the initial condition \[Mu]=\[Mu][0]", \[Lambda]1[      t] "for the initial condition \[Lambda]=\[Lambda][0]"] G[p1,     p2, \[Mu] "=\[Mu][0]", \[Lambda] "=\[Lambda][0]"] , {\[Mu] "=`\[Mu][0](initial condition)", 0,    Pi/2}, {\[Lambda] "=\[Lambda][0](initial condition)", 0,    Infinity}, Method -> {"Some method to choose"}]}, {t, 0, 1}];` 

How to apply the utility Packages for Numerical Differential Equation Solving

I have solved two coupled equations with {Solf, Solg} = NDSolveValue[eqns, {f, g}, {x, 0, L}, {t, 0, tmax}] and want to extract the computational grid {x,t} with InterpolatingFunctionDomain and InterpolatingFunctionGrid to observe the spatial mesh and to plot the time steps. In addition, sometimes it cannot compute a solution over the full time-interval that I specified, so I want to plot the solution that was computed in order to understand what might have gone wrong.

By reading the document, I know how to do this when using NDSolve to solve a single equation for h[x,t]. For example,

hSol = First[h/.NDSolve[eqn, f, {x, 0, L}, {t, 0, tmax}]]  hGrid = InterpolatingFunctionGrid[hSol]  Dimensions[hGrid] (*determine mesh points in each dimension*)  {Tini, Tfinal} = InterpolatingFunctionDomain[hSol][[2]] (*extract the time interval*)  tList = hGrid[[mesh-position, All, 2]]; (*here the 'mesh-position' can be chosen according to the output of Dimensions[hGrid]*) 

Then we can plot the time steps at the mesh-position using

ListLogPlot[tList, PlotRange -> All, Frame -> True, FrameLabel -> {"step", "t"}] 

But I cannot extend these to the case in which I used NDSolveValue to solve two coupled equations for f[x,t] and g[x,t], as mentioned above. For example, using hGrid = InterpolatingFunctionGrid[Solf] and Dimensions[hGrid] I obtained {1}, which was clearly wrong and should be in the form of {number-of-space-mesh, number-of-time-step, 2}.

I think the problem results form the structure of the output of {Solf, Solg} = NDSolveValue[...]. In this case the output is in this form: {InterpolatingFunction[{{0.,L},{0.,tmax}},<>],InterpolatingFunction[{{0.,L},{0.,tmax}},<>]}. Please help.

Here is an example from the document of NDSolveValue:

pde={\!\( \*SubscriptBox[\(\[PartialD]\), \(t\)]\(u[t, x]\)\)==\!\( \*SubscriptBox[\(\[PartialD]\), \(x\)]\((\((v[t, x] - 1)\)\  \*SubscriptBox[\(\[PartialD]\), \(x\)]u[t, x])\)\)+(16 x t-2 t-16 (v[t,x]-1)) (u[t,x]-1)+10 x E^(-4 x),\!\( \*SubscriptBox[\(\[PartialD]\), \(t\)]\(v[t, x]\)\)==\!\( \*SubscriptBox[\(\[PartialD]\), \({x, 2}\)]\(v[t, x]\)\)+\!\( \*SubscriptBox[\(\[PartialD]\), \(x\)]\(u[t, x]\)\)+4 u[t,x]-4+x^2-2 t-10 t E^(-4 x)};  bc={u[0,x]==1,v[0,x]==1,u[t,0]==1,v[t,0]==1,3 u[t,1]+(u^(0,1))[t,1]==3,5 (v^(0,1))[t,1]==E^4 (u[t,1]-1)};  {usol, vsol} = NDSolveValue[{pde, bc}, {u, v}, {x, 0, 1}, {t, 0, 2}] 

schedule Full, Differential and Transnational Backup in sql Server 2016

I studied about scheduling backup with SQL Server Agent and Maintenance Plans (which leads to the SQL Server Agent).

My first concern is what is the best way to schedule my backups?

I want Full Backup everyday, Differential Backup every hour and Transnational Log Backup every 15 mins. My second concern, Is this a good practice?

I noticed an issue when I used SQL Server Agent. My full and Differential backup overwrites. I am ok with having the Full Backup overwrites but not for the differential since when I need to recover it, I would only have one Differential backup which is not the purpose of differential backups. How can I Not let this overwrite on previous backups?

The last concern and question is regarding the way to implementing the backup scheduling. I am going to use Maintenance Plans , and may schedule three different backups. One for full, one for differential and one for transnational. Is this the best practice for a database that is used everyday by users?(at least 2000 transactions per day)

How to solve this partial differential equation for $f$?

I want Mathematica to solve for the function $ f$ .

$ f$ satisfies the following constraints.

$ \frac{\partial}{\partial x} f(x,y) = y$

$ \frac{\partial}{\partial y} f(x,y) = x$

$ f(0,0)=0$

It would seem obvious that $ f(x,y) = x y$

Yet I cannot coax Mathematica into returning this result.

Here is my attempt. Mathematica just returns it. What am I missing?

DSolve[     {        x == D[f[x, y] , y]      , y == D[f[x, y] , x]      , f[0, 0] == 0     }  , {f[x, y], f[x, y]} , {x, y} ] 

Keeping track of limit cycles via certain second order differential operator

Inspired by the two posts which are linked bellow we ask the following question:

Question: For a vector field $ X$ on the plane we define the differential operator $ D$ on $ C^{\infty}(\mathbb{R}^2)$ with $ D(f)=(\Delta\circ L_X-L_X\circ \Delta)(f)$ where $ \Delta$ is the standard Laplacian.

Is there a vector field $ X$ on the plane with $ 2$ nested closed orbits $ \gamma_1 \subset \gamma_2$ such that there exist a smooth function $ f$ for which $ D(f)$ does not vanish on the closur of the amnular region surrounded by $ \gamma_1$ and $ \gamma_2$ ?

On equation $ \Delta \circ \partial/\partial X=\partial/\partial X \circ \Delta$ on a Riemannian manifold

Elliptic operators corresponds to non vanishing vector fields

Remark: The first words of the titles of this post is inspired by a paper by C.C. Pugh and J.P Francois ” Keeping track of Limit cycles”

Continuity of a differential of a Banach-valued holomorphic map

Originally posted on MSE.

Let $ U$ be an open set in $ \mathbb{C}^{n}$ let $ F$ be a Banach space (in my case even a dual Banach space), and let $ \varphi:U\to F$ be a holomorphic map. I seem to be able to prove that the differential map $ D\varphi:U\times\mathbb{C}^{n}\to F$ defined by $ $ D\varphi (z,v)= \lim\limits_{t\to 0}\frac{\varphi(z+tv)-\varphi(z)}{t}$ $ is holomorphic.

Is there a reference for this assertion? (Or at least for continuity)

I tried to look into some sources on infinite-dimensional holomorphicity and could not find such a statement, but some of those sources are rather complicated, and so it is likely I missed it.

Inclusion of the spectrum of two differential operators defined on $L^2[-a,a]$ and $L^2[0, \infty)$

Let $ T$ be the formal operator defined by $ $ Tu:= \sum_{j=0}^{2n} a_j\frac{d^ju}{dx^j}$ $ where $ a_j \in \mathbb{C}$ . Consider the differential operators $ T_a: D(T_a)\subseteq L^2[-a,a] \to L^2[-a,a]$ and $ T_\infty: D(T_\infty)\subseteq L^2[0,\infty) \to L^2[0,\infty)$ defined by $ $ T_af:=Tf, \ T_bg:=Tg, \ f \in D(T_a), g \in D(T_\infty),$ $ where $ $ D(T_a):=\{ f \in L^2[-a,a] : Tf \in L^2[-a,a], f^{(j)}(-a)=f^{(j)}(a)=0 \mbox{ for } 0 \leq j \leq n-1\}$ $ and $ $ D(T_\infty):=\{ f \in L^2[0,\infty) : Tf \in L^2[0,\infty) , f^{(j)}(0)=0 \mbox{ for } 0 \leq j \leq n-1\}.$ $

Can we say that $ \sigma(T_a) \subseteq \sigma(T_\infty)$ ?. I know that the inclusion is true if we take $ Tu:=u”$ or $ Tu:=-u”-2u’$ , for example.

Thanks in advance for any help you are able to provide.

Restriction of a differential operator on $L^2(\mathbb{R})$

I’m reading a proof of a paper and I don’t understand an argument of the proof. That argument is the following:

Let $ A: D(A) \subseteq L^2(\mathbb{R}) \to L^2(\mathbb{R}) $ be a constant coefficients differential operator of order $ 2n$ . The restriction of $ A$ to $ L^2(I)$ ($ I$ is an interval in $ \mathbb{R}$ ) in the sense of quadratic forms is precisely the restriction which satisfies the Dirichlet boundary conditions (that boundary conditions are $ u(c)=u'(c)= \cdots = u^{(n-1)}(c)=0$ at any finite boundary point $ c$ ).

Can you help me to show the last statement, please?

I don’t know what “restriction in the sense of quadratic forms” means. Can you give me a reference on that subject, please?

Thanks in advance for any help.

Differential equation slope field and cauchy solution

Given the differential equation

$ 2y’=\frac{yx}{x^2+4}+\frac{x}{y}$

  • I have to draw the slope field in a rectangle called $ P$ , containing the point (2, 1).
  • Then in appropriate interval I have to find the solution of the Cauchy problem for the given differential equation with initial equation $ y(x_{0})=y_{0}$ where $ (x_{0}, y_{0})$ is inputed by clicking in the rectangle $ P$
  • and in the same rectangle $ P$ the graph of the found approximation of the Cauchy problem with given initial equation given above .

Here is my solution:

 function Plotslope x=-5:0.6:5; y=-6:0.6:6; delta=0.2;  hold on axis([-5,5,-6,6]) daspect([1,1,1])  for k=1:length(x)         for m=1:length(y)           eps=delta/(sqrt(1+ff(x(k),y(m))^2));           plot([x(k)-eps, x(k)+eps],...          [y(m)-eps*ff(x(k),y(m)),...            y(m)+eps*ff(x(k),y(m))],'k');         plot(x(k),y(m),'k.','LineWidth',0.2)          end end  [x0,y0]=ginput(1); plot(x0,y0,'bo') [T,Y]=ode45(@ff,[x0,5],y0); [T1,Y1]=ode45(@ff,[x0,-6],y0); plot(T,Y,'r',T1,Y1,'r')  function z=ff(x,y)      z=(y*x)/(2*(x^2+4))+x/2; end end   

I am sure I am finding the first 2 bullets, but how can I solve the 3rd one?