Problem on PDE existence and uniqueness properties

Question

Consider the partial differential equation $$u_x+2xu_y=0$$

Describe the existence and uniqueness properties for each auxiliary condition below . Include solutions possible or explain why none exists.

$(a). u(x,x^2)=2\\ (b). u(x,x^2)=e^{-x}\\ (c). u(0,y)=e^{-y}$

My attempt: Given PDE $u_x+2xu_y=0$

then $$\frac{dx}{1}=\frac{dy}{2x}=\frac{du}{0}$$

then $$2x-y-c=1, u=c_2$$

From option (a) $u(x,x^2)=2$ then $$x=t,y=t^2, u=2$$ also $$2t-t^2=c_1,2=c_2$$

Since we are not eliminate $c_1, c_2$ pde have no slotuion

is my approach is right?

Answer 1

$$\frac{dx}{1}=\frac{dy}{2x}=\frac{du}{0}$$ First characteristics from $\quad\frac{dx}{1}=\frac{dy}{2x}\quad\implies\quad y-x^2=c_1$.

Second characteristics from $\quad\frac{du}{0}=$finite $\quad\implies\quad u=c_2$.

General solution : $$u(x,y)=F(y-x^2)$$ with any differentiable function $F$.

Case $(a)$ :

$u(x,x^2)=2=F(x^2-x^2)=F(0)\quad$

All functions $F(X)$ with condition $F(0)=2$ are convenient. There is an infinity of solutions. For example :

$u(x,y)=2$

$u(x,y)=2+(y-x^2)$

$u(x,y)=2+\sin(y-x^2)$

and so on...

Case $(b)$ :

$u(x,x^2)=e^{-x}=F(x^2-x^2)=F(0)$

It is impossible that $e^{-x}=$constant. There is no solution.

Case $(c)$ :

$u(0,y)=e^{-y}=F(y-0)=F(y)$

The function is determined : $\quad F(X)=e^{-X}\quad$ Puting it into the general solution where $X=y-x^2$ leads to a unique solution :

$u(x,y)=e^{-(y-x^2)}$

Answer 2

Caveat: this is a longer answer, but I find it easier to understand just what's going on here.

Using the method of characteristics, you examine the function $u$ along a trajectory $(x(t),y(t))$ through some point $(x_0,y_0)$ (usually in a boundary value problem you choose a convenient pair for the initial conditions, i.e. where you have boundary data) converting the partial differential equation into an ordinary differential equation. Namely, you want $$ x'(t)=1\\ y'(t)=2x(t) $$ since then $\frac{d}{dt}u(x(t),y(t))=u_x+2x(t)u_y=0$ for $u$ which solve the pde. So along trajectories of the above sort, as $u(x(t),y(t))=u(x_0,y_0)$.

Solving the coupled system, we find $$ x(t)=t+x_0\\ y(t)=t^2+2x_0t+y_0 $$ and eliminating the parameter $t$, we have that $u$ is constant as long as $$ y=(x-x_0)^2+2x_0(x-x_0)+y_0=x^2-x_0^2+y_0 $$

We know that $$ u(x,y)=u(x,x^2+y_0-x_0^2)=u(x_0,y_0) $$

From here, it is obvious that the first condition is easy to satisfy. We know that the desired function is constant along the parabola $y=x^2$, this condition just fixes such a constant to be $2$.

For the second, it is clearly impossible, since we know that $u$ is constant on the parabola $y=x^2$, but $e^{-x}$ is not constant.

For the third, we have data on the line $x=0$, so it is convenient to set $x_0=0$. Then your condition yields $$ u(x,y)=u(0,y_0)=e^{-y_0} $$ but our choice of characteristic is valid for any $y_0$, so what is $y_0$ when $x_0$ is zero in terms of $x$ and $y$? It is $y-x^2$. Giving us the answer $$ u(x,y)=e^{-(y-x^2)} $$