# Intuition behind boundary conditions for PDE

Suppose that I am trying to model the spread of heat through a $1$ dimensional rod of length $L$ in meters with $$\frac{\partial u}{\partial t} = D\frac{\partial^2 u}{\partial x^2}$$ where $u(x,t)$ is the temperature in Celsius and $D$ is a fixed coefficient. Set boundary conditions to $$\frac{\partial u(0,t)}{\partial x} = -1 \\ \frac{\partial u(L,t)}{\partial x} = 1$$ Do the boundary conditions last for the whole time that we are modeling the rod? In other words, if I want to model this for $t=2$ in minutes, will the rod constantly be losing $-1^\circ \textrm{C}$ at $x=0$ and gaining $1^\circ \textrm{C}$ at $x=L$ throughout the whole $2$ minutes?

EDIT: I understand the mistake in my hypothetical scenario, $\frac{\partial u(x,t)}{\partial x}$ is not the rate of change with respect to time but rather with length.

• The boundary conditions apply throughout, yes. That's why there's a $t$ in there. The boundary conditions could also depend on $t$, it just happens that they don't in this case. – Ian May 4 '16 at 1:31

The question is :

Given the boundary conditions : $$\frac{\partial u(0,t)}{\partial x} = -1 \quad\text{and}\quad \frac{\partial u(L,t)}{\partial x} = 1$$ Do the boundary conditions last for the whole time that we are modeling the rod? In other words, if I want to model this for $t=2$ in minutes, will the rod constantly be losing $-1^\circ \textrm{C}$ at $x=0$ and gaining $1^\circ \textrm{C}$ at $x=L$ throughout the whole $2$ minutes?

If I well understand the question, where is a mistake in it:

On one hand, of course the boundary conditions last for the whole time because that is what is stated.

On the other hand, it is false to say that the rod is constantly losing $-1^\circ \textrm{C}$ at $x=0$ and gaining $1^\circ \textrm{C}$ at $x=L$ throughout the whole $2$ minutes.

The partial derivative relatively to $x$ gives the gradient of $u(x,t)$ not the variation of $u(x,t)$ as the time $t$ changes.

Your assumption should be correct if the conditions where $\frac{\partial u(0,t)}{\partial t} = -1$ and $\frac{\partial u(L,t)}{\partial t} = 1$ : This isn't what is stated.

On a physical viewpoint, as they are stated the conditions mean that the flux of energy remains constant on the two boundaries. This doesn't imply that the temperatures are linear functions of time.