PLAGUE said:
TL;DR Summary: I am new to partial differential equations and today, I was introduced to Lagrange's method of solving PDE's. Someone please help me understand this method.
I am new to partial differential equations and today, I was introduced to Lagrange's method of solving PDE's.
Here, is a proof that shows how Lagrange's method works. I understand the proof until it says,
I mean why should, "if u(x, y, z) = c1 and v(x, y, z) = c2 are two independent solutions of the system of differential equations (dx)/P = (dy)/Q = (dz)/R, then Ф(u, v) = 0 is a solution of Pp + Qq = R"?
I know we derived
View attachment 363293
But what's the link between this and the solution? I don't see any, unfortunately!
Great question — this part often feels like a leap when you're first seeing it. Let me try to break it down in a way that connects the dots:
You're dealing with the linear first-order PDE:
\[
P(x, y, z)\frac{\partial z}{\partial x} + Q(x, y, z)\frac{\partial z}{\partial y} = R(x, y, z),
\]
which we write symbolically as:
\[
Pp + Qq = R,
\]
where \( p = \frac{\partial z}{\partial x} \) and \( q = \frac{\partial z}{\partial y} \).
---
\textbf{Step 1: Solve the \emph{Characteristic System}}
The idea of Lagrange’s method is to reduce the PDE into ordinary differential equations (ODEs) along curves called \emph{characteristics}:
\[
\frac{dx}{P} = \frac{dy}{Q} = \frac{dz}{R}.
\]
This system describes the "flow" in 3D space \((x, y, z)\) along which the solution surface evolves.
---
\textbf{Step 2: Find Two Independent Integrals of This System}
Let’s say you solve this system and find two functionally independent combinations:
\[
u(x, y, z) = c_1, \quad v(x, y, z) = c_2,
\]
which means: along every characteristic curve, the values of \( u \) and \( v \) remain constant.
These define surfaces in \((x,y,z)\)-space. But any arbitrary function
\[
\Phi(u, v) = 0
\]
defines a level surface built from combinations of \( u \) and \( v \), and that surface will also be constant along the same characteristic curves.
---
\textbf{Step 3: Why is \( \Phi(u,v) = 0 \) a Solution?}
Because if \( u \) and \( v \) are constant along the characteristics, then so is any function of them — meaning \( \Phi(u,v) \) stays constant too.
Taking \( \Phi(u(x,y,z), v(x,y,z)) = 0 \) and differentiating implicitly via the chain rule shows that this function satisfies the original PDE automatically because its total derivatives align with the directions defined by the characteristic equations.
Thus:
\[
Pp + Qq = R
\]
is satisfied by any implicit solution of the form
\[
\Phi(u,v) = 0.
\]
---
\textbf{Analogy:}
Think of the characteristics like streamlines in a fluid. If two "labels" \( u \) and \( v \) are constant along those streamlines, then any function of \( u \) and \( v \) will also be constant — and that gives you the general solution to the PDE.
