Solve $-x^2u_x+u_y=1$ with $u(e^y,y)=y$

Question

Solve this PDE with the method of characteristics: $$-x^2u_x+u_y=1$$ $$u(e^y,y)=y$$

---

Set up as usual: $$x_t=-x^2, y_t=1, u_t=1$$ $$x(0,s)=e^s, y(0,s)=s, u(0,s)=s$$ Solving this, we get: $y=t+s, u=t+s, x=1/(t+e^{-s})$

Thus $u=y$

---

Is this correct ?

Answer 1

By eliminating the parameterisation in $t$, we have

\begin{align} \frac{dx}{-x^{2}} = \frac{dy}{1} = \frac{du}{1} \end{align}

Solving the first equality yields

$$y - \frac{1}{x} = C_{1}$$

and solving the second equality gives

\begin{align} u &= y + f(C_{1}) \\ &= y + f \left(y - \frac{1}{x} \right) \end{align}

Applying the initial condition yields

$$y = y + f(y - e^{-y}) \implies f \equiv 0$$

and so

$$u = y$$

Note that instead of solving the second equality, we could have solved the first and third ratios. If we do this, we find

\begin{align} u &= \frac{1}{x} + f(C_{1}) \\ &= \frac{1}{x} + f \left( y - \frac{1}{x} \right) \end{align}

Applying the initial condition shows that $f \equiv \text{Id}$ and hence

$$u = y$$