## Are the following Mathematica codes correct for solving wave equation PDE?

I wanna solve the following PDE of wave equation using Mathematica.

$$u_{tt}=u_{xx}$$

$$00$$

Initial Conditions:

$$\begin{cases}u(x,0)=sin(x) \u_{t}(x,0)=1\end{cases}$$

Boundary Conditions:

$$\begin{cases}u(0,t)+u_{x}(0,t)=1\u(\pi,t)+u_{x}(\pi,t)=-1\end{cases}$$

• I know the boundary and initial conditions are inconsistent.

Are the following codes correct?

weqn = D[u[x, t], {t, 2}] == D[u[x, t], {x, 2}]; ic = {u[x, 0] == Sin[x],Derivative[0, 1][u][x, 0] == 1 }; bc = {u[0, t] + Derivative[1, 0][u][0, t] == 1,     u[Pi, t] + Derivative[1, 0][u][Pi, t] == -1}; sol = NDSolve[{weqn, ic, bc}, u, {x, 0, Pi}, {t, 0, 10}]; 

## PC Problem Solving and repair Online for $15 I can assist you with most of you IT Related problems from issues with word email setup problem solving , intalling software and removal of viruses I will Strive to assist as quickly as possible and if the problem is not resolved there will only be a small fee for time spent . I have 16 years of experience in the IT services and am confident that i will be able to assist by: Misterwizard Created: — Category: Virtual Assistant Viewed: 89 ## Solving a difficult differential equation Well, I’ve to solve the following DE: $$y(t)=x(t)\cdot\text{a}+\text{b}\cdot\ln\left(1+\frac{x(t)}{\text{c}}\right)+\int_0^tx(\tau)\cdot p(t-\tau)\space\text{d}\tau\tag1$$ And I’ve no idea how to start. Some background information: this DE describes a current in an electric circuit. For the values of $$a,b$$ and $$c$$ I know that they are real and positive. For $$c$$ I know that $$10^{-4}\le c\le10^{-16}$$. The function $$y(t)=k\cdot\theta\left(t-m\right)+(n-k)\cdot\theta\left(t-v\right)$$ where all the values of the constants are ral and positive and $$\theta$$ is the Heaviside Theta function. The function $$p(t-\tau)=\mathcal{L}_\text{s}^{-1}\left\{\frac{1}{\frac{1}{R_2+sl}+\frac{s}{sR_3+\frac{1}{z}}}\right\}_{\left(t-\tau\right)}$$ where all the constants are again real and positive. ## Solving T(n) = 3T(n-1) + 2 I am trying to get better at recurrence relation solving, so I am making my own simple relations and try to solve them. I have made the below recurrence T(n) = 3T(n-1) + 2 and we say that T(1) = 1 T(n) = 3 [3T(n-2) + 2] + 2 T(n) = 3 [3[3T(n-3) + 2] + 2] + 2 T(n) = 3 [3[3[3T(n-4) + 2] + 2] + 2] + 2 = T(n) = 81 T(n-4) + 8 . . . We say that k = n-1 T(n) = 3^k T(n-k) + 2*k T(n) = 3^k T(1) + 2*k  I am stuck on how to finish this and how to find out what the time complexity is in Big O. ## Microphone recording has slight static and crackling despite all attempts at solving it Exactly what the title states. I have a microphone that works well on Windows but regardless of distro, there’s always some slight crackling when using the microphone. I have tried everything, adding load-module module-hal-detect tsched=0 to /etc/pulse/default.pa, fiddling with the voice levels on alsamixer, you name it, I tried it, and it didn’t work. It continues to crackle no matter what I throw at it. I’d appreciate any solutions you might offer. Thanks. ## Solving$xu_x+yu_y=y\$ through the method of characteristics

Consider the following first order PDE

$$\begin{cases} x\partial_x{u}+y\partial_y{u}=y, \ u\big|_\Gamma=x. \end{cases}$$

where $$\Gamma:=\{(x,1)\}\subset \mathbb{R^2}$$

1. State the condition which guarantees that the initial surface $$\Gamma$$ is not characteristic.
2. Use the method of characteristics to find a solution of the PDE and discuss for which $$(x,y)\in\mathbb R^2$$ the solution exists.

Here is my attempted solution.

Following the method of characteristics, we can first write the general form as $$au_x+bu_y=f$$. Therefore, we have that

$$\frac{dx}{a}=\frac{dy}{b}=\frac{du}{f}$$ or $$\frac{dx}{x}=\frac{dy}{y}=\frac{du}{y}$$

In order to solve these equations, we need to find the value of two constants $$C_1$$ and $$C_2$$. First, let

$$\frac{dx}{x}=\frac{dy}{y}$$

Then,

$$\frac{1}{x}dx=\frac{1}{y}dy~~\Rightarrow~~~ ln(y) = ln(x)+C ~~\Rightarrow~~~ C=\ln\Big(\frac{y}{x}\Big) ~~\Rightarrow~~~ C_1=\frac{y}{x}$$

Next, let

$$\frac{dy}{y}=\frac{du}{y}$$

Then

$$\frac{dy}{du}=\frac{y}{y} ~~\Rightarrow~~~ dy=du~~\Rightarrow~~~ y=u+C ~~\Rightarrow~~~ C_2=y-u$$

So, we have that $$C_1=\frac{y}{x}$$ and $$C_2=y-u$$. We can combine the two constants such that $$C_2=F(C_1)$$ where $$F$$ is an arbitrary differentiable function. Then,

$$y-u =F\Big(\frac{y}{x}\Big)$$

or

$$u=y-F\Big(\frac{y}{x}\Big)$$

We now need to apply our initial data. We are given that $$u(x,1)=x$$. Therefore,

$$x=1-F\Big(\frac{1}{x}\Big) ~~\Rightarrow~~~ F\Big(\frac{1}{x}\Big)=1-x$$

Letting $$w=\frac{1}{x}$$, we see that $$x=\frac{1}{w}$$.

So, $$F(w)=1-\frac{1}{w}$$. Therefore,

$$u=y-F\Big(\frac{y}{x}\Big) = y-\Big(1-\frac{x}{y}\Big) = y-1+\frac{x}{y}$$

1. State the condition which guarantees that the initial surface $$\Gamma$$ is not characteristic.

I’m not sure. $$\Gamma:=\{(x,1)\}\subset \mathbb{R^2}$$ appears to be defined everywhere.

1. Use the method of characteristics to find a solution of the PDE and discuss for which $$(x,y)\in\mathbb R^2$$ the solution exists.

Assuming we don’t need to worry about

$$u=y-F\Big(\frac{y}{x}\Big)$$

and can directly apply what was found for $$u$$,

$$u= y-1+\frac{x}{y}$$

The solution exists for all $$x\in\mathbb R$$. For $$y\in\mathbb R$$, we need to require that $$y\neq{0}$$. Therefore, the solution exists for all $$(x,y)\in\mathbb R^2$$ where $$y\neq{0}$$.

I’m not sure if I am analyzing the correct material for the questions.

## Solving recurrence relations with two variables

I am trying to solve this recurrence relation with two variables:

$$T(n, k) = T(n – 1, k – 1) + T(n – 1, k)$$

The base cases are:

• $$T(n, k) = 1$$ if $$k = 0$$
• $$T(n, k) = 0$$ if $$k > n$$

I was wondering if standard techniques like characteristic polynomial and generating function would work in this situation.

## Solving a linear program with few nonzero variables

Consider the linear program:

$$A x = b, ~~~~~~ x\geq 0$$ where $$A$$ is an $$m$$-by-$$n$$ matrix, $$x$$ is an $$n$$-by-1 vector, $$b$$ is an $$m$$-by-1 vector, and $$m.

It is known that, if this program has a solution, then it has a basic feasible solution, in which at most $$m$$ variables are non-zero and at least $$n-m$$ variables are zero.

QUESTION: what is the run-time complexity of checking whether the system has a solution in which at most $$m-1$$ variables are non-zero?

NOTE: the question is a special case of Min-RVLS – finding a solution with a smallest number of non-zero variables. Min-RVLS is known to be NP-hard and hard to approximate within a multiplicative factor. Finding an additive approximation is hard too.

But, here our goal is much more modest – all we want is to find a solution with one less than the maximum ($$m$$). Is this special case easier?

## Solving Steady Navier Stokes Equation

I am trying to solve the Steady Navier Stokes equation in a two-dimensional domain.

I found a code online in freefem++ which at some point reads:

// //  Define u1, u1o, u2, u2o, p, trial functions. // Xh u1, u1o, u2, u2o; Qh p; // //  Define v1, v2, q, test functions. // Xh v1, v2; Qh q; // //  Define the problem. // problem navierstokes ( [u1,u2,p], [v1,v2,q] ) = int2d ( Th, qforder = 3 )(                             visc*(dx(u1)*dx(v1) + dx(u2)*dx(v2) + dy(u1)*dy(v1) + dy(u2)*dy(v2))                             + u1*dx(u1o)*v1 + u2*dy(u1o)*v1 + u1o*dx(u1)*v1 + u2o*dy(u1)*v1                             + u1*dx(u2o)*v2 + u2*dy(u2o)*v2 + u1o*dx(u2)*v2 + u2o*dy(u2)*v2                             - p*(dx(v1) + dy(v2))                             + q*(dx(u1) + dy(u2))                             - eps*p*q) - int2d ( Th, qforder = 3 )(                             u1o*dx(u1o)*v1 + u2o*dy(u1o)*v1                             + u1o*dx(u2o)*v2 + u2o*dy(u2o)*v2) + on(Inner,u1=0,u2=0) + on(Outer,u1=omega*(-y),u2=omega*x); 

I am able to solve the Stokes problem but I am having trouble working around the convective term in the Navier-Stokes equation.

Then they use a fixed point iteration to find the solution:

// //  Seek a solution of the nonlinear equations via a fixed point iteration. // for ( int i = 0; i < 10; i++ ) {     u1o = u1;     u2o = u2;     navierstokes; } 
• I do not understand what is the trial function (u1o,u2o) that they are using in the code…?
• How do I write down the weak formulation associated to this problem?
• Are they using Oseen, Stokes or Newton method?

## minimization problem solving and its step limits?

I am trying to solve the following minimization problem:

• x : is my unknown value (input) with complex elements and known size, here (4×1)
• y : is the output vector (known)
• c : is a ‘skaling’ vector

I am very new to this so my approach may seem basic. I simply loop over all combination of c (non-redundant) and according to the computed minimization value and condition I update my results. My questions are the following:

• Is this correct and is there a better approach to this?
• This code fails with a small step due to the huge size of combinations, so how to solve that?

My code:

from itertools import combinations from random import randint import numpy as np   def deg2rad(phase):     return round(((phase*3.14)/180),3)  def excitation(amplitude, phase):     return complex(round(amplitude * (np.cos(deg2rad(phase))),3), round(amplitude*(np.sin(deg2rad(phase))),3))  def compute_subject_equation_result(x):     M            = 12     difference   = []     y            = [randint(10, 20) for i in range(M)]      for m in range(0, M):               c  = np.array([randint(0, 9), randint(10,20), randint(0, 9), randint(0,20)]).reshape(4,1)         ch = c.conjugate().T         eq  = (y[m] - (abs(np.dot(ch, x))[0])**2)**2         difference.append(eq**2)     return round(sum(difference)[0], 3)  def compute_main_equation_result(x, beta):     norm1  = np.linalg.norm(x,1)     norm2  = np.linalg.norm(x,2)     return round(norm1 + beta*(norm2**2), 3)  def optimize(x, min_x, min_phi_x):     min_result      = 10**25      # compute the optimization formals and check for the min value     main_equation_result    = compute_main_equation_result(c, beta)     subject_equation_result = compute_subject_equation_result(c)      # update min value if min detected'       if subject_equation_result < epsilon and main_equation_result < min_result:         min_result = main_equation_result         min_x      = x         min_phi_x  = phx             return min_x, min_phi_x  # initialization phases    = [alpha for alpha in range(0, 361, 90)] beta      = 1 epsilon   = 10**25 min_x     = np.array([]) min_phi_x = np.array([])  phases_combinations = [list(comb) for comb in combinations(phases, 4)]  # start checking all combinations for phx in  phases_combinations:     phi1, phi2, phi3, phi4 = phx[0], phx[1], phx[2], phx[3]            # build the hypothesis for the excitations vector c      c = np.array([ excitation(1, phi1), excitation(1, phi2), excitation(1, phi3), excitation(1, phi4) ]).T.reshape(4,1)     min_x, min_phi_x = optimize(c, min_x, min_phi_x)     print('    --------------------------------------------------')     print('-----> new_min_c     = ', list(min_x))     print('-----> new_min_phi_c = ', min_phi_x) 

Remark: When trying phases = [alpha for alpha in range(0, 361, 1)] I get a Memory error. This, I can avoid using a higher step, However I am not sure about my approach in general nor of the step change effect on the accuracy