## 220 Reputation

3 years, 171 days

## Thanks a lot....

Dear friend,

Thank you very much for your explanations.

Best Wishes

## @Carl Love  Yes, you are right. I...

Yes, you are right.

I think the reason is that in fracdiff, the form of the function must be specified when we use Laplace--option.

In my case, the form of first element, i.e. function, is unknown, hence we ecounter with error.

## Thanks a lot...

Dear friend,

Thank you very much.

I also used Laplace-approach but in another way.

Thanks a lot.

Yours Sincerely,

J.

## Thanks a lot...

Dear friend,

Yes you are right, but since this is a part of codes that I am writing and this one is an example, hence I should wirte it in general form.

Yours Sincerely,

J.

## A solution...

I found an alternative way to get a solution:

restart;
with(inttrans):
eq := fracdiff(u(x, t), t, 1/2) = t^(1/2)*sin(x)/GAMMA(3/2):
eval(invlaplace((solve(laplace(eq, t, s), laplace(u(x, t), t, s))), s, t),u(x,0)=0);

sin(x)*t

## Thanks a lot....

Dear Friend,

Thanks a lot. You made me happy.

I really do not know how I can appreciate you.

Thanks again.

Best wishes.

## Thanks a lot....

Dear Friend,

Thanks a lot. You made me happy.

I really do not know how I can appreciate you.

Thanks again.

Best wishes.

## Thanks a lot....

Dear Friend,

Thanks a lot. Yes, you are right.

## @Carl Love  Dear friend, Thank you...

Dear friend,
Thank you so much.
I had not paid any attention to these points that you mentioned.
Your knowledge in this field is very deep and you think professionally.
Many thanks for your explanations and clarifications.

## @Carl Love Dear friend,Thank you so...

Dear friend,

Thank you so much.

Yes, you are right. the previous did not work for the boundary condition.

But this one works. Also, the first issue (introducing the constant) works.

Thank you very much.

It is interesting to know that using convert((D[1\$3](y))(a, t), diff), both

D[1\$3](y)(a,t) = V

and

eval(diff(y(x, t), x\$2), [x = a]) = V

have the same result, but the former one works for boundary condition while the latter one does not work. However, by defining `a' as a constant, the latter one will work.

Thanks a lot.

## @Carl Love Dear friend,Many thanks ...

Dear friend,

Many thanks for your answer. I did not know these. You raised my knowledge.

Thanks a lot.

I want to use the aforementioned notation as a boundary condition for a differential equation.

Dear friend,

Thanks a lot.

## Thanks a lot....

Dear friend,

Yes, you are right. Thank you so much.

Best Regards,

J.

## Thanks a lot....

Dear friend,

Thank you so much. Thanks a lot for your nice and very clear explanations. You made me happy.

## Thanks a lot...

Dear friend,

Thank you so much.

I really do not know how I can appreciate you. Thank you very much for your time and efforts.