@Kitonum 

Dear friend, 

Thank you very much for your explanations.

Best Wishes

@Carl Love 

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.

@MapleMathMatt 

Dear friend,

Thank you very much.

You made me happy.

I also used Laplace-approach but in another way. 

Your way is so interesting.

Thanks a lot.

Yours Sincerely,

J.

@Carl Love 

Dear friend,

Thanks a lot for your reply.

You made me happy.

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.

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

@Kitonum 

Dear Friend,

Thanks a lot. You made me happy. 

I really do not know how I can appreciate you.

Thanks again.

Best wishes.

@tomleslie 

Dear Friend,

Thanks a lot. You made me happy. 

I really do not know how I can appreciate you.

Thanks again.

Best wishes.

@vv 

Dear Friend,

Thanks a lot. Yes, you are right.

@Carl Love 

Dear friend,
Thank you so much. 
Your descriptions are really excellent.
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 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 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.

You made me happy.

@vv 

Dear friend,

Thanks a lot.

You made me happy.

@mmcdara 

Dear friend,

Yes, you are right. Thank you so much.

Best Regards,

J.

@ecterrab 

Dear friend,

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

@acer 

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.

You made me happy. 

Thanks a lot.