回答済み
Solve a system of Partial Differential Equations using function BVP4C
It looks to me like you have 9 dependent variables and only 7 equations and boundary conditions. Clearly, that can't be solved....

7年以上 前 | 1

回答済み
Solving 4th order PDE
You may be able to solve this in pdepe with two equations and two dependent variables h and P. The first equation would be your ...

8年弱 前 | 0

回答済み
Why bvp4c not works?
I think you want: dydt=@(x,y)[y(2); (x-2*y(2))/3];

約8年 前 | 0

回答済み
How to formulate boundary conditions for a PDE system?
pdepe is specifically designed for PDEs that are second-order in the spatial direction. That is why it requires you to specify b...

8年以上 前 | 0

回答済み
Using pdepe for 1d transcient heat conduction through a composite wall
You want to model this multi-region wall with a *single* pde and then write your pde function (pdefun) so that it returns a d...

8年以上 前 | 1

| 採用済み

回答済み
Spatial discretization has failed. Discretization supports only parabolic and elliptic equations, with flux term involving spatial derivative.
I see two problems. First, your boundary conditions are incorrectly defined. They should be: pl = [ul(1);ul(2);ul(3);ul(...

9年弱 前 | 0

| 採用済み

回答済み
How to solve a nonlinear Parabolic PDE IBVP
Since the problem is 1D, you can use the pdepe function.

9年以上 前 | 0

| 採用済み

回答済み
pdepe: why does spatial discretization fail?
I think you want the following in your pdebc function: function [pl,ql,pr,qr] = pdebc(xl,ul,xr,ur,t) pl = ul-0.3639; ...

9年以上 前 | 1

| 採用済み

回答済み
PDE Toolbox solution array along a line, to then find the mean value
The evaluate function is returning only one value because the length of your x and y vectors is one. Maybe you want something l...

10年弱 前 | 1

| 採用済み

回答済み
Exchange Problem with PDEPE solver
Torsten said: "Usually, if e.g. material properties change at the interface, the transmission conditions are continuity of te...

10年弱 前 | 0

回答済み
Spatial discretization has failed. (pdex1)
That error message is somewhat misleading. There are two problems in your code. The first is that your boundary condition fu...

10年弱 前 | 0

回答済み
How to solve second order partial differential equations ?
This problem is straightforward to solve using PDE Toolbox. I suggest you look at this example <http://www.mathworks.com/help/p...

10年弱 前 | 2

回答済み
elliptic PDE with variable coefficient
The input arguments, x and y are equal length row vectors of x and y coordinates where the a coefficient must be defined. If t...

約10年 前 | 0

| 採用済み

回答済み
How to increase the size of "s" using pdegeom?
If your concern is about producing a nice looking plot from pdegplot, you will need to edit that function by changing the line:...

約10年 前 | 0

| 採用済み

回答済み
Interpolation of data defined on triangular mesh
If you have access to the R2014b version of MATLAB, there is a new PDE Toolbox function, pdeInterpolant, that makes this operat...

約10年 前 | 1

| 採用済み

回答済み
Nonlinear Heat Transfer In a Thin Plate - bug in example?
What version of MATLAB are you running? The documentation page you are pointing to is for the R2014b version of MATLAB. The ...

約10年 前 | 0

回答済み
Applying Nonconstant Boundary Conditions for pde
The pdeGeometryFromEdges function is part of a new, simpler approach to defining boundary conditions in PDE Toolbox introduced ...

約10年 前 | 0

| 採用済み

回答済み
Solving an Hamilton Jacobi Bellman equation type /w nonlinear coefficients
Hi Matt, You *can*, in fact, express your equation in a form that the parabolic function in PDE Toolbox will accept. The ...

約10年 前 | 1

回答済み
1D diffusion-reaction model across two-layered slab using pdepe
I am doubtful that pdepe can be used to solve this system. I suggest using finite differences to discretize in the spatial di...

約10年 前 | 0

| 採用済み