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

• kingwinner
In summary: So u_x(x,0)=lim h->0 (0-0)/h= 0. So the derivative is 0. In summary, If u(x,0)=0 for all x, then ux(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 ut=0 on the boundary of D.

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

I don't understand either of the claims above.

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! :)

Last edited:
kingwinner said:

## 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 ut=0 on the boundary of D.
Let u(x,y) be a function of 2 variables.

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

I don't understand either of the claims above.

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 ut(0,t)= 0, and ut(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?

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 ux(x,0)=0?

ux(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!

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

## 1. What does the notation "U(x,0)" refer to in this equation?

The notation "U(x,0)" refers to the function U at a specific point (x,0), where 0 represents the initial time or starting point.

## 2. What does "u_x (x,0)" represent in this equation?

The notation "u_x (x,0)" represents the partial derivative of the function u with respect to x at the point (x,0).

## 3. Why is it important for U(x,0)=0 to imply u_x (x,0)=0?

This condition is important because it ensures that the function U remains constant at the initial time, meaning there is no change in the function's value. This, in turn, implies that the function's slope (represented by u_x) is also 0 at this point.

## 4. How does this equation relate to the concept of boundary conditions?

This equation is an example of a boundary condition, specifically an initial condition. It provides information about the function at the starting point (x,0) and helps to determine the behavior of the function for all subsequent points.

## 5. Can this equation be applied to any type of function?

Yes, this equation can be applied to any type of function, as long as the function has a well-defined initial value and partial derivative with respect to x. It is commonly used in the field of mathematics and physics to model various phenomena.

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