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Hi

I have a nonlinear PDEs, solved using finite difference in the square

I get the following nonlinear system of equation. Is there any idea how correct the code and display the solution.

I will appreciate any help in this question.

 

restart;
n:=100;
h:=1/(n+1);

# Boundary condition

for j from 0 by 1  to n+1 do
u[0, j] = 0;
u[n+1, j] = 0;
u[j, 0] = 0;
u[j, n+1] = 0 ;
end do;
## Loop for interior point in the square
for i from 1 by 1 to  n do
for j from 1 by 1 to  n do
(u[i+1, j]-u[i, j])*(u[i+1, j]-2*u[i, j]+u[i-1, j])+h*(u[i, j+1]-2*u[i, j]+u[i, j-1]) = 0;
end do;
end do;
 

How can I solve this system of equations with unknown u[i,j], where i,j=1,..,n

 

Many thanks for any help

Hi,

Can somebody help me to find out why Maple can't completely solve this system of differential equations?

The answer to the previous command is

but I don't get the solution for u(x). This should be u(x)=-x+x^2/2.

Thanks for your help

 

Hello guys,

I have some system of differential equations,

How can i find  eigenvalues of this system?

If i have solution

res := evalf(dsolve(sys union ics, convert(x, list), type = numeric, method = rkf45))

and

sol := evalf(dsolve(sys union ics, convert(x, list), method = laplace))

AnSolution.mw

Thanks!

I want to get solutions of this system ,can anyone help me ?solutions.mw

Hi, I am new in Maple. If I have an electric network as in the figure, I want to get the Transfer Function V2(s)/Vi(s) from this equation system:

Vi=R1.I1+V1

Vi=R2.I2+V2

I1-I2=C1.dV1/dt

I2=C2.dV2/dt

Which are the commands that I may write to get this?? Before hand, Thanks by your answers!

Dear 

Hope you will be fine. My file takes to much time to solve the system of nonlinear algebraic equations for Iterations=8. please solve my problem I will be waiting for positive response.

Error_graph.mw

$$\textbf{x}' = \begin{bmatrix} -4 & -2 \\ 3 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}+\begin{bmatrix} -t \\ -2t-1 \end{bmatrix},\textbf{x}(0)=\begin{bmatrix} 3 \\ -5 \end{bmatrix}$$

As I know firstly, when the matrix is denoted by $A$, we must compute $e^{At}$ by diagonalizing $A$: if $A=PDP^{-1}$ for a diagonal $D$ then $e^{At} = P e^{Dt} P^{-1}$ where $e^{Dt}$ is a diagonal matrix with $(e^{Dt})_{ii} = e^{D_{ii} t}$...
 
How can I write The Maple code? maple.stackexchange)

restart: with(LinearAlgebra):

A := Matrix(2,2,[-4,-2,3,1]);

....

Hello

Hope everything going fine with you. I am facing problem to fine the exact (numerical) solution of the attached system of linear PDEs associated with BSc and ICs. I tried to solve it without BCs and ICs, with BCs and with ICs also all the times I failed. Please solve it either general, with ICs or BCs. You can try to solve it numrically. In attached file H(t) represent the unit step function. I am waiting your positive response.

PDEs_solve.mw 

With my best regards and sincerely.

Muhammad Usman

School of Mathematical Sciences 
Peking University, Beijing, China

I meet a interesting nonlinear system in the analysis of an mechanics problem. This system can be shown as following:

wherein, the X and Y is the solutions. A, B, S, and T is the symbolic parameters.

I want to express X and Y with A, B, S, T. Who can give me a help, thanks a lot!

PS:the mw file is given here.

A_symbolic_nonlinear_system.mw

Dears;

Hope everyone is fine. I am try to find the numerical solutions of system of nonlinear algabric equation via newton's raphson method in the attached file but failed. Please see the attachment and try to correct. You can solve it least square method if possible. I am waiting your positive response. 

Help_in_Newton.mw

With my best regards and sincerely.

Muhammad Usman

School of Mathematical Sciences 
Peking University, Beijing, China

Email: muhammadusman@pku.edu.cn

sol_L := dsolve({de_L, ic});

    {x(t) = (-y0 - x0) exp(-2 t) + (y0 + 2 x0) exp(-t),
      y(t) = -2 (-y0 - x0) exp(-2 t) - (y0 + 2 x0) exp(-t)}
How i can plot this?thanks

hy

i have to develop a code i which i have system of nonlinear equation 

i have to generate the matrix of that nonlinear equation then i want to do or apply any method say newton method and make a loop which help us to find a solution using some tolerance 

at the end i get a result in form of a table which give nth matrix then value of function matrix at nth value then error i-e xn-x(n-1) 

thanx in advance

Hello, 

I have a PDE system. When I use pdsolve it gets me the messege " pdsolve->Warning: System is inconsistent". Is there a way I can see which equations breaks the system down? 
For this system, it's difficult to see from ayeball where the problem is. 
Thank you! 

test.mw

 

 

 

I have tried to solve a matrix with the function "LinearSolve" as seen in the picture, but instead of solving it just gives me back the operation i wrote (3). My which is to solve an equation system a quick as possible - have a templet fill in the matrix and press enter. I thought this "LinearSolve" function was the easiest way of doing. I know that I can right click and choose the function, but I want it as a command.

 

Any solutions on how to use the "LinearSolve" command to solve an equation system?

 

i want to solve a system , A.b+B.X=0 , which A is 5*5 known matrix, B is 5*2 known matrix , and b is 5*1 and X is 2*1 unknown arbitary matrices ! i want to have solution for b and X . whatever they can be ! just equation to be solved !

restart:

A:=Matrix(5,[[-0.9800,0,0,-0.0160,0],[1.0000,0,0,0,0],[-2.1900,-9.7800,-0.0280,0.0740,0],[77.3600,-0.7700,-0.2200,-0.6700,0],[0,-79.9700,-0.0300,0.9900,0]]);

A := Matrix(5, 5, {(1, 1) = -.9800, (1, 2) = 0, (1, 3) = 0, (1, 4) = -0.160e-1, (1, 5) = 0, (2, 1) = 1.0000, (2, 2) = 0, (2, 3) = 0, (2, 4) = 0, (2, 5) = 0, (3, 1) = -2.1900, (3, 2) = -9.7800, (3, 3) = -0.280e-1, (3, 4) = 0.740e-1, (3, 5) = 0, (4, 1) = 77.3600, (4, 2) = -.7700, (4, 3) = -.2200, (4, 4) = -.6700, (4, 5) = 0, (5, 1) = 0, (5, 2) = -79.9700, (5, 3) = -0.300e-1, (5, 4) = .9900, (5, 5) = 0})

(1)

B:=Matrix((5,2),[[-2.44,0.58],[0,0],[0.18,19.62],[-6.48,0],[0,0]]);

B := Matrix(5, 2, {(1, 1) = -2.44, (1, 2) = .58, (2, 1) = 0, (2, 2) = 0, (3, 1) = .18, (3, 2) = 19.62, (4, 1) = -6.48, (4, 2) = 0, (5, 1) = 0, (5, 2) = 0})

(2)

b:=Matrix((5,1),[y[1],y[2],y[3],y[4],y[5]]);

b := Matrix(5, 1, {(1, 1) = y[1], (2, 1) = y[2], (3, 1) = y[3], (4, 1) = y[4], (5, 1) = y[5]})

(3)

X:=Matrix((2,1),[[x[1]],[x[2]]])

X := Matrix(2, 1, {(1, 1) = x[1], (2, 1) = x[2]})

(4)

M:=A.b+B.X

M := Matrix(5, 1, {(1, 1) = -.9800*y[1]-0.160e-1*y[4]-2.44*x[1]+.58*x[2], (2, 1) = 1.0000*y[1], (3, 1) = -2.1900*y[1]-9.7800*y[2]-0.280e-1*y[3]+0.740e-1*y[4]+.18*x[1]+19.62*x[2], (4, 1) = 77.3600*y[1]-.7700*y[2]-.2200*y[3]-.6700*y[4]-6.48*x[1], (5, 1) = -79.9700*y[2]-0.300e-1*y[3]+.9900*y[4]})

(5)

eval(M,y[1]=0)

Matrix([[-0.160e-1*y[4]-2.44*x[1]+.58*x[2]], [0.], [-9.7800*y[2]-0.280e-1*y[3]+0.740e-1*y[4]+.18*x[1]+19.62*x[2]], [-.7700*y[2]-.2200*y[3]-.6700*y[4]-6.48*x[1]], [-79.9700*y[2]-0.300e-1*y[3]+.9900*y[4]]])

(6)

 

 

 

Download solve.mw


actually the problem to be solved is M=0 ! which directly goes to y[1]=0;
after that , how can i find other unknowns so that M=0 is ok. tnx

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