Solve the following PDE $u_x+yu_y=u$ where $u=u(x,y)$
with
- 1)$u(x,0)=e^x$, explain why several solutions are available.
- 2) $u(0,y)=y^2$
Here is what I did:
1) $$x_t=1, y_t=y, \ u_t=u$$ $$x(0,s)=s, \ y(0,s)=0, \ u(0,s)=e^s$$
Solving that, we get $x=t+s, \ y=e^t-1, \ u=e^te^s$ i.e. $t=x-s$ thus $u=e^{x-s}e^s=e^x$
2) $$x_t=1, y_t=y, \ u_t=u$$ $$x(0,s)=0, \ y(0,s)=s, \ u(0,s)=s^2$$
Solving that, we get $x=t, \ y=e^t+s, \ u=e^ts^2$ i.e. $t=x, s=y-e^x$ thus $u=e^{x}(y-e^x)^2$
My questions are:
1) Is my work correct ?
2) Why would several solutions be available for $u(x,0)$ ?
Thanks for your help !
Your question is almost identical to this question. Since it isn't an exact duplicate, I will repeat the same analysis made in the answer by JJacquelin.
$$u_x+yu_y=u$$ Set of characteristic ODEs: $\quad \frac{dx}{1}=\frac{dy}{y}=\frac{du}{u}$
First family of characteristic curves, from $\quad\frac{dx}{1}=\frac{dy}{y} \quad\to\quad ye^{-x}=c_1$
Second family of characteristic curves, from $\quad\frac{dx}{1}=\frac{du}{u} \quad\to\quad ue^{-x}=c_2$
General solution of the PDE : $\quad ue^{-x}=F\left(ye^{-x}\right)$ $$u(x,y)=e^xF\left(ye^{-x}\right)$$ $F$ is any function.
Condition 1: $\quad u(x,0)=e^x$.
$u(x,0)=e^x=e^xF\left(0e^{-x}\right)=e^xF(0) \quad\implies\quad F(0)=1$
Solution of the PDE with the condition : $$u(x,y)=e^xF\left(ye^{-x}\right)$$ where $F$ is no longer any function, but is any function which is equal to $1$ when $x=0$.
Since they are an infinity of functions $F(x)$ satisfying $F(0)=1$, the problem has an infinity of solutions.
For examples :
With $\quad F(X)=1+X\quad$ the particular solution is : $\quad u(x,y)=e^x\left(1+ye^{-x}\right)=e^x+y$
With $\quad F(X)=1+X^2\quad$ the particular solution is : $\quad u(x,y)=e^x\left(1+ye^{-x}\right)^2$
With $\quad F(X)=\cos(X)\quad$ the particular solution is : $\quad u(x,y)=e^x\cos\left(ye^{-x}\right)$
And so on : An infinity of such examples can be found.
Third family of characteristics curves, from $\frac{dy}{y}=\frac{du}{u} \quad\to\quad \frac{u}{y}=c_3$
General solution, with any differentiable function $F$: $$\frac{u}{y}=F(ye^{-x}) \quad\to\quad u=y\:F(ye^{-x}) $$
Condition 2: $\quad u(0,y)=y^2$.
$\quad u(0,y)=y^2=y\:F(ye^{0})=y\:F(y) \quad$ determines the function $F$ :
$$F(t)=t \quad \text{for any } t$$
Bringing this function $F$ into the above general solution gives
$$u(x,y)=y^2\:e^{-x}$$