Method of Characteristics: Initial Conditions and the Unique Solution in This Linear PDE Example

Question

Find the general solution of the PDE $pu_x + u_y = u$, where $p$ is constant, in the domain $\{ (x, y) : x \in \mathbb{R}, 0 \le y \le Y \}$, subject to the initial condition $u(x, 0) = g(x), x \in \mathbb{R}$.

The parametric characteristic equations are

$$\frac{dx}{dt} = p, \frac{dy}{dt} = 1, \frac{du}{dt} = u$$

We can rewrite these in non-parametric form:

$$\frac{dx}{p} = dt, \frac{dy}{1} = dt, \frac{du}{u} = dt$$

$$\rightarrow \frac{dx}{p} = \frac{dy}{1} = \frac{du}{u}$$

And this can be written as two new ODEs:

$$\frac{dx}{dy} = p, \frac{du}{dy} = u$$

Solving these ODEs using separation of variables, we get that $x = py + k$ and $u = Ae^y$ on any single characteristic, with $k$ and $A$ being two related constants.

Question 1: What is meant here by "related"? What does it mean for two constants to be "related"? Why are they "related"?

The values of $k$ and $A$ are constant along any one characteristic. As such, we can write the general solution as

$$x = py + k, u = A(k)e^y$$

$A$ is written here as a function of $k$. Is this because, again, $A$ is "related" to $k$?

Geometrically, the characters are straight lines, along which the solution varies exponentially.

The parameter $k$ can be eliminated:

$$u(x, y) = A(x - py)e^y$$

To use the initial condition, we set $y = 0$, and so $u(x, 0) = A(x) = g(x)$, and our unique solution is

$$u(x, y) = g(x - py)e^y$$

Question 2: Something about this doesn't seem right? Using the initial condition, we found that $u(x, 0) = A(x) = g(x)$. So this is saying that $A$ as a function of $x$ is equal to $g$ as a function of $x$. But in $u(x, y) = A(x - py)e^y$, we have that $u(x, y) = A(k) = A(x - py)e^y$, not $u(x, y) = A(x) = A(x - py)e^y$? So I don't see how it makes sense to replace $A$ with the function $g$ to get $u(x, y) = g(x - py)e^y$? Something doesn't seem right here.

Thank you for any guidance you could offer.

Answer 1

"So this is saying that $A$ as a function of $x$ is equal to $g$ as a function of $x$"

Yes, it is exactly this way, so we can say too that $A$ as a function of $k$ is equal to $g$ as a function of $k$. In $u(x, y) = A(x - py)e^y$ we have $u(x, y) = A(k)e^y=g(k)e^y=g(x-py)e^y$ because $k=x-py$, as stated the ODEs.

"Related" means that there must be a functional relation between these constants $k$ and $A$, so is, "each $k$ has its $A$" determining a curve (as intersection of two surfaces $x-py=k$ and $u=Ae^y$) into a particular solution (and another value of $k$ has another correct value of $A$ to fit the curve so determined into the surface solution): remember that the solutions are surfaces constructed has a family of curves (the characteristics).

Answer 2

Let take the problem from the system of characteristic ODEs that you correctly got : $$\rightarrow \frac{dx}{p} = \frac{dy}{1} = \frac{du}{u}$$ A first characteristic equation comes from $\quad\frac{dx}{p} = \frac{dy}{1}$ : $$x-py=c_1$$ A second characteristic equation comes from $\quad\frac{dy}{1} = \frac{du}{u}$ : $$ue^{-y}=c_2$$ The general solution of the PDE expressed on the form of an implicit equation is : $$\Phi\left(x-py\:,\:ue^{-y} \right)=0$$ $\Phi$ is an arbitrary function of two variables.

Equivalently, on explicit form : $$ue^{-y}=F(x-py)$$ $F$ is an arbitrary function.

So, the general solution is : $$u(x,y)=e^yF(x-py)$$

Boundary condition :

$u(X,0)=g(X)=e^0F(X-0)=F(X)\quad$ 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 $X=x-py$.

The particular solution which satisfies the boundary condition is : $$u(x,y)=e^y g(x-py)$$

So, your result was correct. The variant to get it is equivalent. May be less confusing.