## Simplify trigonometric functions to take the Laplace transfrom

I have one trigonometric function which is simplified as

f[t_,RV_,H_]:=0.000308148 H Cos[    0.0523599 - 0.0862193 RV - 0.172439 RV Sin[4.71239 - 6.28319 t]] +   0.032 Sin[2 \[Pi] (-0.1 + t)] +   H (-0.129421 -      0.0153853 Sin[       0.0523599 - 0.0862193 RV -         0.172439 RV Sin[4.71239 - 6.28319 t]])

This function acts as input to one simple ODE which I want to solve with Laplace transform where RV and H are just two arbitrary parameters.

My question is how to find the Laplace transform of this function. One idea I am trying to do is using the following rules to transform the identities.

Rules = {Cos[ p1_*Sin[p2_*u_]] ->     Sum[((-1)^k (p1 Sin[p2 u])^(2 k))/(2 k)!, {k, 0, Num}],    Sin[ p1_*Sin[p2_*u_]] ->     Sum[((-1)^k (p1 Sin[p2 u])^(2 k + 1))/(2 k + 1)!, {k, 0, Num}],    Sin[ p1_*Cos[p2_*u_]] ->     Sum[((-1)^k (p1 Cos[p2 u])^(2 k + 1))/(2 k + 1)!, {k, 0, Num}],    Cos[ p1_*Cos[p2_*u_]] ->     Sum[((-1)^k (p1 Cos[p2 u])^(2 k))/(2 k)!, {k, 0, Num}]}

Then apply the Laplace transform. Is there any other way, since I have several of these functions and I do not know exactly how to apply these rules repeatedly to functions like sin(sin(sin…))) and write them such that I can take the Laplace transform.

## Solving the heat equation using Laplace Transforms

I am trying to solve the 1-D heat equation using Laplace Transform theory. The equation is as follows. I don’t have the capability to write the symbols so I will write it out.

partial u/partial t = 2(partial squared u/ partial x squared) -x    boundary conditions are partial u/partial x(0,t)=1, partial u/partial x(2,t)=beta.

The problem asks the following: (a). For what value of beta does there exist a steady-state solution? (b). if the initial temperature is uniform such that u(x,0)=5 and beta takes the value suggested by the answer to part (a), derive the equilibrium temperature distribution.

I was able to get an equation that looks like U(x,s)=c e^(s/2)^1/2 -(1/s)((x/s)-u(x,0)). But I am not sure how to go from here to solve for beta using the boundary conditions. I need some assistance from someone.

## Solving Laplace PDE with DSolve

I’m trying to get an analytical solution of Laplace PDE with Dirichlet boundary conditions (in polar coordinates). I managed to solve it numerically with NDSolveValue and I know there is an analytical solution and I know what it is, but I would like DSolve to return it. But DSolve returns the input.

sol = DSolve[{Laplacian[       u[\[Rho], \[CurlyPhi]], {\[Rho], \[CurlyPhi]}, "Polar"] == 0,     DirichletCondition[u[\[Rho], \[CurlyPhi]] == 0,       1 <= \[Rho] <= 2 && \[CurlyPhi] == 0],     DirichletCondition[u[\[Rho], \[CurlyPhi]] == 0,       1 <= \[Rho] <= 2 && \[CurlyPhi] == \[Pi]],      DirichletCondition[      u[\[Rho], \[CurlyPhi]] == Sin[\[CurlyPhi]], \[Rho] == 1 &&        0 <= \[CurlyPhi] <= \[Pi]],      DirichletCondition[      u[\[Rho], \[CurlyPhi]] == 0., \[Rho] == 2 &&        0 <= \[CurlyPhi] <= \[Pi]]},     u, {\[Rho], 1, 2}, {\[CurlyPhi], 0, \[Pi]}];

## Graph of Laplace equation

I have solved the following Laplace equation

a0 = (1/Pi) Integrate[Cos[φ]^2 + Sin[φ]^3, {φ, 0, 2 Pi}] an = (1/Pi) Integrate[(Cos[φ]^2 + Sin[φ]^3)*Cos[n*φ], {φ, 0, 2 Pi}] Plot[an, {n, 0, 10}] bn = (1/Pi) Integrate[(Cos[φ]^2 + Sin[φ]^3)*Sin[n*φ], {φ, 0, 2 Pi}] ann = an*Cos[n*φ] bnn = bn*Sin[n*φ] a = Sum[((r/4)^n)*(ann + bnn), {n, 1, Infinity}] f[r_, φ_] := a0/2 + a ParametricPlot3D[{r,φ, f[r, φ]}, {r, 0, 1}, {φ, 0, 2 Pi}] Plot3D[f[r, φ], {r, 0, 1}, {φ, 0, 2 Pi}]

My problem is that Plot3D and ParametricPlot3D doesn’t work. Any help?

## am closed pretty much to solve laplace eq. but I need help with some thing

I have written the following Mathematica codes to solve the LAPLACE eq. using the finite differences method.

In[1]:= Remove[a, b, Nx, Ny, h, xgrid, ygrid, u, i, j] a = 0; b = 0.5; n = 4; h = (b - a)/n; xgrid = Table[x[i] -> a + i h, {i, 1, n}]; ygrid = Table[y[j] -> a + j h, {j, 1, n}]; eqnstemplate = {-4 u[i, j] + u[i + 1, j] + u[i - 1, j] + u[i, j - 1] +       u[i, j + 1] == 0}; BC1 = Table[u[i, 0] == 0, {i, 1, n - 1}]; BC2 = Table[u[i, 4] == 200 x[i], {i, 1, n - 1}]; BC3 = Table[u[4, j] == 200 y[j], {j, 2, n - 1}]; BC4 = Table[u[0, j] == 0, {j, 2, n - 1}]; Eqns = Table[eqnstemplate, {i, 1, n - 1}, {j, 1, n - 1}] /. xgrid /.      ygrid // Flatten; systemEqns = Join[Eqns, BC1, BC2, BC3, BC4] /. xgrid /. ygrid   Out[12]= {u[0, 1] + u[1, 0] - 4 u[1, 1] + u[1, 2] + u[2, 1] == 0,   u[0, 2] + u[1, 1] - 4 u[1, 2] + u[1, 3] + u[2, 2] == 0,   u[0, 3] + u[1, 2] - 4 u[1, 3] + u[1, 4] + u[2, 3] == 0,   u[1, 1] + u[2, 0] - 4 u[2, 1] + u[2, 2] + u[3, 1] == 0,   u[1, 2] + u[2, 1] - 4 u[2, 2] + u[2, 3] + u[3, 2] == 0,   u[1, 3] + u[2, 2] - 4 u[2, 3] + u[2, 4] + u[3, 3] == 0,   u[2, 1] + u[3, 0] - 4 u[3, 1] + u[3, 2] + u[4, 1] == 0,   u[2, 2] + u[3, 1] - 4 u[3, 2] + u[3, 3] + u[4, 2] == 0,   u[2, 3] + u[3, 2] - 4 u[3, 3] + u[3, 4] + u[4, 3] == 0, u[1, 0] == 0,   u[2, 0] == 0, u[3, 0] == 0, u[1, 4] == 25., u[2, 4] == 50.,   u[3, 4] == 75., u[4, 2] == 50., u[4, 3] == 75., u[0, 2] == 0,   u[0, 3] == 0}

I need this out put in matrix form just for the unknown variable with substituting the known value to get the linear system that arises from solving the laplacses eq.

## Why does the numerical inverse laplace function FT for small times give erroeous results and what is the alternative

I am trying to do the numerical laplace inverse of a very complicated transfer function, subject to a trapezoidal pulse input. For the sake of understanding, I will use a simple transfer function to pose my question. I want to find out the output value during pulse on times, but numerical laplace inverse gives highly erroneous results for such time scales. Here is the pulse input that I am giving

The mathematica code is the following:

Gtransfer[s_] = 1/(1 + s); T1 = 10^-9; T2 = 10*10^-9; T3 = 10^-9; Eo = 1;  Eint[t_] = Eo*10^9*t; Enet[t_] =    Eint[t] - Eint[t - T1]*UnitStep[t - T1] -     Eint[t - T1 - T2]*UnitStep[t - T1 - T2] +     Eint[t - T1 - T2 - T3]*UnitStep[t - T1 - T2 - T3]; Eins[s_] = LaplaceTransform[Enet[t], t, s]; fun1[s_] = Eins[s] Gtransfer[s];  thisDir =    ToFileName[("FileName" /.        NotebookInformation[EvaluationNotebook[]])[[1]]]; SetDirectory[thisDir]; << FixedTalbotNumericalLaplaceInversion.m t0 = 25 10^-9; Vnumerical = FT[fun1, t0](*Numerical*) Vanalytical =   N[InverseLaplaceTransform[fun1[s], s, t] /. t -> 25 10^-9](*Analytical*)

You will find that for values of t0<12*10^(-9), numerical inverse laplace is extremely erroneous. Could anyone suggest me as to why this is so and how to fix this? My original transfer function is just too complicated to invert analytically. Thanks in advance!

## A uniform upper bound for Fredholm index of quasi Laplace operators on a compact parallelizable manifold

Assume that $$M$$ is a compact parallelizable manifold. Is there an upper bound for the absolute value of Fredholm index of all operators in the form $$D=\sum_{i=1}^n \partial^2/\partial{X_i^2}$$ where $$\{X_1,X_2,\ldots,X_n\}$$ is a global smooth frame?

## How to obtain inverse Laplace transformation for the following function?

$$\hat{f} (s) = \frac{1}{(b\sqrt{s}\sinh{(\sqrt{as})}+a\sqrt{s}\cosh{(\sqrt{as})})}, \quad s>0$$

## Show the inverse laplace transform in general

My problem is, show that:

$$L^{-1}(\dfrac{s}{(s+a)(s+b)})=\dfrac{ae^{-at}-be^{-bt}}{a-b}$$

MY ATTEMPT:

I think that i have to use the convulution theorem, so

$$F(s)=\dfrac{1}{s+a}$$ $$G(s)=\dfrac{s}{s+b}$$

then

$$f(s)=L^{-1}F=e^{-at}$$ $$g(s)=L^{-1}G=-be^{-bt}+1$$

but when i try to integrate, it doesn´t work, i dont get what the problem says. Can someone help me please? I´m kind of desperate. Thanks in advance

## Use of the Leibniz integral rule in Laplace transform proof

My Laplace transform textbook presents the following theorem:

If $$\mathcal{L}\{ F(t) \} = f(s)$$, then $$\mathcal{L}\{ t F(t) \} = – \dfrac{d}{ds}f(s)$$ and in general $$\mathcal{L}\{ t^n F(t) \} = (-1)^n \dfrac{d^n}{ds^n} f(s)$$.

The proof then begins as follows:

$$\mathcal{L}\{ F(t) \} = \int_0^\infty e^{-st} F(t) \ dt$$

and differentiate this with respect to $$s$$ to give

\begin{align} \dfrac{df}{ds} &= \dfrac{d}{ds} \int_0^\infty e^{-st} F(t) \ dt \ &= \int_0^\infty -te^{-st} F(t) \ dt \end{align}

My understanding is that the author went from

$$\dfrac{d}{ds} \int_0^\infty e^{-st} F(t) \ dt$$

to

$$\int_0^\infty -te^{-st} F(t) \ dt$$

by using the Leibniz integral rule.

However, as you can see from the Wikipedia page, the Leibniz integral rule is only valid for $$\int_{a(x)}^{b(x)}, b(x) < \infty$$, whereas the Laplace transform has $$b(x) = \infty$$. Doesn’t this mean that the Leibniz rule is invalid?

I would greatly appreciate it if people could please take the time to clarify this.