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In the PDE $aU_x+bU_y = 0$.This is equivalent to $\frac a{\sqrt {a^2+b^2}}U_x+ \frac b{\sqrt {a^2+b^2}}U_y = 0$.
$\langle\frac a{\sqrt {a^2+b^2}}, \frac b{\sqrt {a^2+b^2}}\rangle$.$\nabla U=0$.
Then I can say that this is the directional derivative of U in the direction of $\widehat {\langle a,b \rangle}$.
This implies that the rate of change of $U$ in the direction of $\widehat{\langle a,b \rangle }$=0.That is we are on a level surface. Since the gradient is perpendicular to the level surface and the dot product is zero, tangent to the level curve at that point is orthogonal to the gradient.

Is there a relationship between the tangent to the level surface and the direction of the directional derivative?
Because in characteristic curves method I have seen that it says that "The directonal derivative in the direction of the vector $\langle a,b \rangle$ is zero. The curves on the xy plane with $\langle a,b \rangle$ as tangent vectors have slopes$\frac ba.$ Thus $\frac {dy}{dx}=\frac ba$ ".Please explain the relationship of directional derivative and its tangent vector.

In the PDE $aU_x+bU_y+cU = 0$ .This also has the directional derivative in the direction $\widehat {\langle a,b \rangle}$.
But here the rate of change of U in the direction of $\widehat{\langle a,b\rangle}$ is not zero.
That is we are not in a level surface. Hence can I say that the $\nabla U $ is perpendicular to the surface and the tangent at each point is in the direction $\langle a,b \rangle$.
I don't understand the relationship of characteristic curves have with the gradient and the tangent plane.

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  • $\begingroup$ From what you know about the dot product, $\vec{n}\cdot\nabla U=|\nabla U|\cos\alpha$ where $\alpha$ is the angle between the gradient and the direction in which you are taking the derivative. The above a special case. Of course, the fact that $\nabla U$ points perpendicular to the level surface, is one of the fundamental properties. $\endgroup$ – orion Mar 29 '14 at 9:20
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If you have a scalar function $U$, a point $x_0$ and a vector $v$, and $\gamma$ is a curve such that $\gamma(t_0) = x_0$, $\gamma'(t_0) = v$, then it holds $$ \frac{d}{dt}_{|t=t_0} U(\gamma(t)) = \frac{\partial U}{\partial v} (x_0) = U_x(x_0) v_x + U_y(x_0) v_y = \nabla U(x_0) \cdot v . $$

This contains all the information you need.In particular: the directional derivative along a direction $v$ is the scalar product of the gradient of $U$ with the vector $v$.

You first example with $v=(a,b)$ can be written as $\frac{\partial U}{\partial v} = 0$ which says that the function $U$ is constant in the direction $v$.

In the second example you have $\frac{\partial U}{\partial v} = c U$ or $\nabla U \cdot v = c U$ which says that the scalar product of the gradient of $U$ with the direction $v$ is equal to $U$. This cannot be simplified further... you cannot say which is the direction of the level sets, because you don't know the magnitude of the gradient.

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  • $\begingroup$ I don't understand what is done in the first expression $\endgroup$ – clarkson Mar 29 '14 at 13:06

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