On Fri, 2013-05-03 at 17:46 +0800, asha g wrote:

> N = 41, iter = 111000;

> vvvv(iter,:) = V;

Okay. This means that 'vvvv' has (at least) 'iter' (=111,000) rows.

You haven't said how many columns it has though. I'm forced to assume

that it has N columns and that 'V' (whatever that is) is a 1-by-N (i.e.,

a 1-by-41) row vector.

>

> tmax = 55;

> x= linspace(0,tmax,niter);

I'm confused. What is 'niter'? Is it the maximum value of 'iter' or

something else entirely? Whatever it is (presumably a positive

integer), at least we know that

all(size(x) == [ 1, niter ])

> z = linspace(0,l,N);

all(size(z) == [ 1, N ])

> y = vvvv';

Size unknown. In the following, I'm going to assume that

all(size(y) == [ N, niter ])

because that seems the most likely given your setup. If that is not the

case, then you *really* need to tell us the size of 'y'.

> On using the following

> i = 1 : 5 : numel(x);

> j = 1 : 5 : numel(z);

> mesh(z(j), x(i), y(i,j))

> I get an error message at line mesh(z(j),x(i),y(i,j))

> error: A(I): Index exceeds matrix dimension.

That's seems consistent with the fact that numel(x) (i.e., 'niter') is

(much) larger than numel(z) (i.e., 'N').

> Maybe it needs to be rewritten

By 'it' I assume you mean the 'mesh' call, rather than your own

calculations. I already said in a previous e-mail that in an invocation

like

mesh(U, V, W)

in which 'U' and 'V' are vectors, then the following conditions must

hold

rows(W) == numel(V)

columns(W) == numel(U)

In other words, your 'mesh' invocation must then (assuming, as stated

above, that all(size(y) == [ N, niter ])) be written as

mesh(z(j), x(i), y(j,i))

Sincerely,

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