Answer to Question #248915 in Differential Equations for Athira

Question

How to find uniqueness of solution using characteristics curve and initial condition

Answer 1

Solution:

Let's take the problem from the system of characteristic ODEs that :

$\frac{d x}{p}=\frac{d y}{1}=\frac{d u}{u}$

A first characteristic equation comes from $\frac{d x}{p}=\frac{d y}{1}$ :

$x-p y=c_{1}$

A second characteristic equation comes from $\quad \frac{d y}{1}=\frac{d u}{u}$ :

$u e^{-y}=c_{2}$

The general solution of the PDE expressed on the form of an implicit equation is:

$\Phi\left(x-p y, u e^{-y}\right)=0$

$\Phi$ is an arbitrary function of two variables.

Equivalently, on explicit form :

$u e^{-y}=F(x-p y)$

F is an arbitrary function.

So, the general solution is :

$u(x, y)=e^{y} F(x-p y)$

Boundary condition:

$u(X, 0)=g(X)=e^{0} F(X-0)=F(X) \quad$ any X, doesn't matter the notation of the variable. Now the function F(X) is known :

F(X)=g(X)

We put it into the above general solution where X=x-p y.

The particular solution which satisfies the boundary condition is :

$u(x, y)=e^{y} g(x-p y)$