U(x,0)=0 implies u_x (x,0)=0 ?

[kingwinner]

Homework Statement

Let u(x1,x2,...,xn,t) be a function of n+1 variables. Let D be an open, bounded, connected set in R^n(with respect to x1,...,xn) and all functions are smooth.
Claim 1:
If u=0 on the boundary of D, then u_t=0 on the boundary of D.


Let u(x,y) be a function of 2 variables.
Claim 2:
If u(x,0)=0, then u_x(x,0)=0.

I don't understand either of the claims above.

Homework Equations

N/A

The Attempt at a Solution

I really can't think of a reason why these are true. Also, I cannot check it directly by differentiating u with respect to t (or x) because I don't even know what the function u is. I was only given that u is 0 at some very specific points.

Can someone please explain or prove it?
Any help is appreciated! :)

 

[HallsofIvy]

Homework Statement

Let u(x1,x2,...,xn,t) be a function of n+1 variables. Let D be an open, bounded, connected set in R^n(with respect to x1,...,xn) and all functions are smooth.
Claim 1:
If u=0 on the boundary of D, then u_t=0 on the boundary of D.
Let u(x,y) be a function of 2 variables.

Claim 2:
If u(x,0)=0, then u_x(x,0)=0.

I don't understand either of the claims above.

Homework Equations

N/A

The Attempt at a Solution

I really can't think of a reason why these are true. Also, I cannot check it directly by differentiating u with respect to t (or x) because I don't even know what the function u is. I was only given that u is 0 at some very specific points.

Can someone please explain or prove it?
Any help is appreciated! :)

If you don't understand a statement in "n" dimensions, start by reducing it to 1 dimension.
"Let u(x,t) be a function of 2 variables. Let D be an open, bounded connected set in R (The open interval (0,1) is a good example).

claim 1: if u(0,t)= 0 and u(1, t)= 0, for all t, then u_t(0,t)= 0, and u_t(1,t)= 0."

Which is, in fact, just "claim 2"! Can you think of any function for which that is NOT true? If not, why not?

 

[kingwinner]

I cannot think of any function for which that is NOT true, but I also have no idea how to prove it.
Suppose u(x,0)=0 for all x.
Then why is it true that u_x(x,0)=0?

u_x(x,0) means take the function u(x,y), FIRST take the partial derivative w.r.t. x, THEN evaluate at y=0. But here the trouble for me is that it seems like there is no way to recover what the function u(x,y) is. (we only know u(x,0)=0 for all x)
Is this the same as FIRST evaluating at y=0, and THEN take the resulting function of x and differenaite w.r.t. x? Why are they the same? This is not obvious to me...

Thanks!

 

[Dick]

The definition of u_x(x,y)=lim h->0 (u(x+h,y)-u(x,y))/h. If you put y=0...