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How to find the limit limit(u(x,t),t=infinity), where u(x,t) is the solution of the Cauchy problem:
pde:=diff(u(x,t),t)=diff(u(x,t),x,x)+exp(-t) with the initial condition
ic:=u(x,0)=arctan(x) ?

discretization of pdes

April 16 2012 by ynht 0

Hi everyone,

I am trying to solve the following dimensionless system of pdes:

 

u1t=u1x+u2-u1

u2t=3.139uxx-2u2+u3+u1

u3t=-u3x+u2-u3

the boundary conditions are u1(5,t)=0, u2(0,t)=0, u2(5,t)=0, u3=1-heaviside(t-0.1)

initial conditions are u1(x,0)=0, u2(x,0)=0, u3(x,0)=0

 

pdsolve returns some results but they are not stable. Since I'm fairly new to maple, I don't know if it is possible to discretize...

how can i solve pdes of the following form in maple

[tex] tc(t,x)\frac{\partial a(t,x)}{\partial t} - ta(t,x)\frac{\partial c(t,x)}{\partial t} = a(t,x)c(t,x) + t(a(t,x))^2 and be able to find both functions a(t,x) and b(t,x)

i came into some questions when i wanted to solve a pde systerm as follows:

> pde := diff(u(x, y), y, y, y, y) = 0;
> sys := [pde, u(x, 0) = 0, u(x, a) = F0, (D[`$`(2, 2)](u))(x, 0) = 0, (D[2, 2](u))(x, a)-beta*(D[2](u))(x, a) = 0];
> pds := pdsolve(sys);
%;
Error, (in pdsolve/sys/info) found functions with same name but depending on different arguments in the given DE system: u(x, a), u(x, y). It is required an indication of the dependent variables

2nd argument error

March 16 2012 by nhty 0 Maple 15

Hi everyone,

I tried to solve the following system of equations with Maple 15:

 
You can copy paste the commands below to get the equations
pde1 := [diff(`#msub(mi("n"),mo("&uminus0;"))`(x, t), t) =
diff(`#msub(mi("n"),mo("&uminus0;"))`(x, t), x)+n[o](x, t)-`#msub(mi("n"),mo("&uminus0;"))`(x, t)]

 

pde2 := [diff(n[o](x, t), t) = 0.3139e-2*(diff(n[o](x, t), x, x))-2*n[o](x, t)+`#msub(mi("n"...

shooting method

March 14 2012 by kinesimario 0 Maple 15
why do i get this 
"Error, (in solve) cannot solve expressions with diff(diff(y(x), x), x) for x"
when i input these: 
> restart; with(DEtools); with(plots); with(linalg);
> ode2 := x^2*(diff(y(x), `$`(x, 2)))+x*(diff(y(x), x))+4*y(x) = -2*x+7; bc2 := y(1) = 7, y(4) = -1;
a := 1; b := 4
ic2 := bc2[1], (D(y))(a) = alpha
constraint := lhs(bc2[2])-rhs(bc2[2]); constraint = 0
 

Analytical solution to PDE given by...

February 23 2012 by Norge 0 Maple 15

Dear all,

Maple gives me the complete solution to the attached PDE. This solution is expressed in terms of KummerU and KummerM. What I want is to obtain the solution for real numbers only, i.e. I want to get rid of the possible imaginary numbers in the solution. This implies that the separation constant, _c1, must be less or equal to 2r. 

How can I do this?

Furthermore, the four arbitrary constants will need to be determined through boundary conditions....

i'm trying to solve a sysem of pde, but i get this error 

Error, (in PDEtools:-Library:-NormalizeBoundaryConditions) unable to isolate the functions {x(0, z, r), x(t, z, 0.5e-2), xeq(t, z, 0.5e-2), y(0, z, r), y(t, 0, r)} in the given boundary conditions {x(0, z, r) = 0, x(t, z, 0.5e-2) = xeq(t, z, 0.5e-2), y(0, z, r) = 0, y(t, 0, r) = 0}
>
> restart; with(PDEtools);
[CanonicalCoordinates, ChangeSymmetry, CharacteristicQ,

CharacteristicQInvariants,...

how can i overcome the error that i am being given for the problem below 3*t*d(t,x(t),diff(x(t),t))2 = 0

4200000*(diff(T(x, t), t))-0.1e-2*(diff(T(x, t), x))^2-0.1e-2*(diff(T(x, t), x, x))*T(x, t)-.445*(diff(T(x, t), x, x))-158780+4200*T(x, t)-50000/(exp(x))^200 = 0

with boundry conditions:

u[0](0.3e-1, t) = 37

Is there any command in Maple 15 to linearize an PDE non-linear? Or or there is a package or help file to do so?

 

I'm handling with Non-linear PDEs in my work and I would like to solve them by these methods.

 

 

Thank You

 

Washington Inacio

 

 

 

 

Hello,

I was wondering if anyone could help me with my problem.

I have an equation where I have an arbitrary function that I have been calling h(y^2-k^2*x^2) where h can be any function (e.g. sin, ln, ...).

I have to perform a few operations on the equation and I end up with a pde equal to 0. The operations I perform on my original equation leads maple to differentiate h at times with respect to x or y and so, for example, I end up with terms like D(h)(y^2-k^2*x^2)*2*y

Multilayer Heat Transfer - solved

December 31 2011 by wzelik 20 Maple 15

Hi.

Some times ago, I have asked about possibility to solve multilayer transient heat transfer problem.
The problem was solved originaly by Robert Israel. I enclosed his solution and explanation in file Xpde_oryginal.mw.
Thank's Robert!


There was a problem with correct results of numeric PDE in Maple 14 when I have used convection and radiation BC toogather. However Maplesoft have removed this bug in version 15, thus now I have obtained right results...

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