Why Are Boundary Conditions Zero for \(\phi\) in Partial Differential Equations?

[omoplata]

The attached image is from "Numerical Partial Differential Equations: Conservation Laws and Elliptic Equations" by J.W. Thomas.

In the beginning the set of test functions, $\phi$ is defined.

They arrive at equation (9.2.11) by using $\phi(x,T) = \phi(a,t) = \phi(b,t) = 0$.

Where does this condition come from?

Thanks.

 

[omoplata]

I guess, to find out what's going on, I should find out what this set $C_0^1$ is first. It's not defined anywhere else on the book.

 

[Mark44]

The symbol C¹ means functions whose first derivative is continuous. From your attached page, I infer that C₀¹ means functions whose first derivatives are continuous, and that are 0 outside some rectangle in R².

 

[omoplata]

Thanks. That helps.

I guess from that you could say that, if the first derivative of $\phi$ is continuous, then $\phi$ is also continuous, so $\phi = 0$ at the edges of the rectangle? Therefore, $\phi(x,T) = \phi(a,t) = \phi(b,t) = 0$.

But in that case, why is it not required that $\phi(x,0) = 0$ ?

 

[Mark44]

Thanks. That helps.

I guess from that you could say that, if the first derivative of $\phi$ is continuous, then $\phi$ is also continuous, so $\phi = 0$ at the edges of the rectangle?

I don't think you can conclude that, at least based on the document you attached. It looks like there is more to the problem than what you scanned, so perhaps the answer is there.

Therefore, $\phi(x,T) = \phi(a,t) = \phi(b,t) = 0$.

But in that case, why is it not required that $\phi(x,0) = 0$ ?