Solving a PDE with Method of Characteristics.

Question

I am currently trying to solve a PDE using Method of Characteristics, but am having a bit of trouble. The PDE is given by: $$xu_x+yu_y+zu_z=1$$ With Initial Conditions: $$u=0 \text{ on } x+y+z=1$$ Now I computed the Characteristics: $$x(s) = x_0 e^s, \>\>y(s) = y_0 e^s, \>\> z(s)=z_0 e^s$$ Where $x(0)=x_0$, and the analogous statement holds for $y_0, z_0$. Now, I don't get what the conclusion is. From these characteristics I get: $$e^s = \frac{x}{x_0}$$ And the same for the $y$ and $z$ cases, but what can I conclude about $u$ here?

Answer 1

Using the Lagrange-Charpit equations, we can write out the characteristic ODEs as

$$\frac{dx}{x} \stackrel{(1)}= \frac{dy}{y} \stackrel{(2)}= \frac{dz}{z} \stackrel{(3)}= \frac{du}{1}$$

Solving equality $(1)$ yields

\begin{align} \ln y &= \ln x + c_{0} \\ \implies y &= c_{0} x \\ \implies \frac{y}{x} &= c_{0} \quad (4) \end{align}

Using Componendo-Dividendo on equality $(1)$ and setting equal to $(2)$ yields

\begin{align} \frac{dx + dy}{x + y} &= \frac{d (x + y)}{x + y} \quad \text {(C.D)}\\ &= \frac{dz}{z} \\ \implies \ln(x + y) &= \ln z + c_{1} \\ \implies \frac{x + y}{z} &= c_{1} \quad (5) \end{align}

Using Componendo-Dividendo on equalities $(1)$ and $(2)$ and setting equal to $(3)$ yields

\begin{align} \frac{dx + dy + dz}{x + y + z} &= \frac{d (x + y + z)}{x + y + z} \quad \text {(C.D)}\\ &= \frac{du}{1} \\ \implies \ln(x + y + z) &= u + c_{2} \\ \implies \ln(x + y + z) - u &= c_{2} \quad (6) \end{align}

Hence, the solution is given implicitly by

\begin{align} \phi(c_{0}, c_{1}, c_{2}) &= 0 \\ \implies \phi \left( \frac{y}{x}, \frac{x + y}{z}, \ln(x + y + z) - u \right) &= 0 \\ \end{align}

where $\phi$ is an arbitrary differentiable function. We can also write this as

$$\ln(x + y + z) - u = f \left( \frac{y}{x}, \frac{x + y}{z} \right)$$

where $f$ is an arbitrary differentiable function. You can then apply your initial condition to determine $f$ and get the full solution (you should find $f \equiv 0$).