Geometrical Meaning of linear first order partial differential equations?

Question

Consider a linear first order differential equation:

$Pp + qQ = R$ where $p$ and $q$ respectively are derivative of function $f$ wrt $x$ and $y$ respectively and $f(x,y)= z$

Now we can say that at any point say $A$ on $f$ ie the point $A$ is on the graph of the function $f$ in 3D space, we can find the direction cosines of perpendicular to the function $f$ at $A$ as $(p,q,-1)$.

According to Sneddon book however, it also says that the first equation is basically an analytical way of saying that the direction ratios of line perpendicular to normal at $A$ ie tangential to the surface is $(P,Q,R)$ where They are functions of $x, y , z$ not involving any derivative term.

How is this possible to conclude from the given pde? Can someone please explain this?

Basically I am trying to prove Theorem 2 (See the picture)

Answer 1

To explain this, we need to understand where the geometric interpretation comes from.

For this it is crucial to know a bit about geometry on manifolds. Here, a vector $X$ is defined as the mapping $X : C(M) \to \mathbb{R}$ which maps functions on the manifold $M$ to the Reals. When we look at $X$ in a chart we see that it looks like a directional derivative acting on a function $X^i\frac{\partial}{\partial x^i}f = X^i\frac{\partial f}{\partial x^i} = X \cdot \nabla f$. Now with your $X = ({P,Q,R})^T$ and $ \nabla f=(p,q,-1)^T$ the equation $X \cdot \nabla f=0$ is your pde and graphically displays the orthogonality of $X$ and $\nabla f$

Answer 2

OK, so from what we've got so far (i.e., in the text of the question statement):

$p = \dfrac{\partial f(x, y)}{\partial x}, \tag 1$

$q = \dfrac{\partial f(x, y)}{\partial y}, \tag 2$

$z = f(x, y); \tag 3$

then the equation

$Pp + Qq = R \tag 4$

may be written

$P(x, y, z) \dfrac{\partial f(x, y)}{\partial x} + Q(x, y, z) \dfrac{\partial f(x, y)}{\partial y} = R(x, y, z); \tag 5$

if we define the function

$\phi(x, y, z) = f(x, y) - z, \tag 6$

then it is easy to see that the graph of $f(x, y)$, that is, the set of points $(x, y, z) \in \Bbb R^3$ such that (3) binds, is precisely the $0$-level surface of $\phi(x, y, z)$, since

$f(x, y) - z = \phi(x, y, z) = 0 \Longleftrightarrow z = f(x, y); \tag 7$

we also have the gradient of $\phi(x, y, z)$:

$\nabla \phi(x, y, z) = \left (\dfrac{\partial \phi(x, y, z)}{\partial x}, \dfrac{\partial \phi(x, y, z)}{\partial y}, \dfrac{\partial \phi(x, y, z)}{\partial z} \right ) = \left (\dfrac{\partial f(x, y)}{\partial x}, \dfrac{\partial f(x, y)}{\partial y}, -1 \right )$ $= (p(x, y), q(x, y), -1). \tag 8$

Now it is well-known that $\nabla \phi(x, y, z)$ is normal the the level surfaces of the function $\phi(x, y, z)$; since the graph of $z = f(x, y)$ is the $0$-level surface of $\phi(x, y, z)$, $\nabla \phi(x, y, z) = (p(x. y), q(x, y), -1)$ is normal to the graph of $z = f(x, y)$ at every point $A$ in this graph. This means in fact that for any vector $X$ tangent to said graph,

$\nabla \phi(x, y, z) \cdot X = (p(x. y), q(x, y), -1) \cdot X = 0, \tag 9$

and $X$ is tangent to the graph if (9) binds. It should be observed at this point that $\nabla \phi = (p, q -1)$ is not in general a unit vector; indeed we have

$\Vert \nabla \phi(x, y, z) \Vert = \Vert (p(x, y), q(x, y), -1) \Vert = \sqrt{(p^2(x, y) + q^2(x, y) + 1} \ge 1, \tag{10}$

with equality holding if and only if

$p^2(x, y) + q^2(x, y) = 0 \Longleftrightarrow p(x, y) = q(x, y) = 0; \tag{11}$

we thus see that $\nabla \phi(x, y, z)$ is a unit vector precisely when

$p(x, y) = \dfrac{\partial f(x, y)}{\partial x} = 0; q(x, y) = \dfrac{\partial f(x, y)}{\partial y}, \tag{12}$

that is, at those points $(x, y) \in \Bbb R^2$ which are critical points of $f(x, y)$; though this fact is perhaps not directly relevant to the present topic, it seems like a worthwhile result worth knowing in its own right.

We continue. We may find the direction cosines of $\nabla \phi = (p, q, -1)$ by normalizing it using (10); denoting them by $\alpha_x, \alpha_y, \alpha_z$ we have

$(\alpha_x, \alpha_y, \alpha_z) = \left ( \dfrac{p}{\sqrt{p^2 + q^2 + 1}}, \dfrac{q}{\sqrt{p^2 + q^2 + 1}}, \dfrac{-1}{\sqrt{p^2 + q^2 + 1}} \right ); \tag{13}$

$\alpha = (\alpha_x, \alpha_y, \alpha_z) \tag{14}$

is then of course a unit vector normal to the graph of $z = f(x, y)$.

We return to the equation (5), writing it as

$P(x, y, z) \dfrac{\partial f(x, y)}{\partial x} + Q(x, y, z) \dfrac{\partial f(x, y)}{\partial y} - R(x, y, z) = 0, \tag{15}$

or

$(P(x, y, z), Q(x, y, z), R(x ,y, z)) \cdot (\dfrac{\partial f(x, y)}{\partial x}, \dfrac{\partial f(x, y)}{\partial y}, -1)$ $= (P(x, y, z), Q(x, y, z), R(x ,y, z)) \cdot (p(x, y), q(x, y), -1) = 0; \tag{16}$

from what we have seen in (9), (16) shows that $(P(x, y, z), Q(x, y, z), R(x ,y, z))$ is tangent to the surface $\phi(x, y, z) = f(x, y) - z = 0$, which is the graph of the equation $z = f(x, y)$.

In the preceding discussion, we normalized the vector field $\nabla (x, y, z) = (p(x, y), q(x, y), -1)$; we may also normalize the vector field $(P, Q, R)$; we have

$\Vert (P, Q, R) \Vert = \sqrt{P^2 + Q^2 + R^2}, \tag{17}$

whence we may, as long as $\Vert (P, Q, R) \Vert \ne 0$, define the vector field $U(x, y, z)$:

$U = \dfrac{(P, Q, R)}{\Vert (P, Q, R) \Vert} = \left ( \dfrac{P}{\sqrt{P^2 + Q^2 + R^2}}, \dfrac{Q}{\sqrt{P^2 + Q^2 + R^2}}, \dfrac{R}{\sqrt{P^2 + Q^2 + R^2}} \right ); \tag{18}$

it is not at all clear, however, whether or not the equation (4)-(5) is any easier to solve when $(P, Q, R)$ is replaced by $U$, in which case it becomes

$ \dfrac{P}{\sqrt{P^2 + Q^2 + R^2}} \dfrac{\partial f(x, y)}{\partial x} + \dfrac{Q}{\sqrt{P^2 + Q^2 + R^2}} \dfrac{\partial f(x, y)}{\partial y} = \dfrac{R}{\sqrt{P^2 + Q^2 + R^2}}; \tag{19}$

(19) may, however, yield certain geometrical insights which are not so readily apparent from the un-nomralized form (5). But a discussion of this possibility would take us farther afield than time and space permit at this point.

Answer 3

Not sure if I understand the question correctly, but : from $Pp+Qq=R$ we rearrange to $$ (p,q,-1)\cdot (P,Q,R) = 0$$ so the vector $\mathbf P =(P,Q,R)$ is normal to $(p,q,-1)$. You noticed that the integral surface $S$ is the set of solutions to $F=0$, $$F(x,y,z) = f(x,y)-z , \quad S = \{ (x,y,z) : F(x,y,z) = 0 \}$$ with gradient $\nabla F = (p,q,-1)$. Therefore, if I take one point $\mathbf x_0$ in $S$ and evolve by the integral curve equation $$ \mathbf x' = \mathbf P(\mathbf x), \quad \mathbf x(0) = \mathbf x_0$$ then as $F(\mathbf x_0) = 0$ and $$ \frac{d}{dt}(F(\mathbf x)) = \nabla F \cdot \mathbf x' = (p,q,-1)\cdot (P,Q,R) = 0,$$ the resulting curve solves $F(\mathbf x) = 0$. Since $\mathbf x_0$ was arbitrary, any point in $S$ in contained in an integral curve.