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I am trying to solve the PDE $$x\frac{\partial u}{\partial y}+\frac{\partial u}{\partial x}=u,$$ subject to the conditions $u(x,0)=0$ and $u(0,y)=y$.

The characteristic equations are: $$\frac{dx}{dt}=1, \ \frac{dy}{dt}=x, \ \frac{du}{dt}=u.$$ Solving these ODEs gives \begin{align} x(t,s)&=t+A(s) \\ y(t,s)&=\frac{t^2}{2}+A(s)t+B(s) \\ u(t,s)&=C(s)e^t. \end{align}

Case $1$, $u(x,0)=0$: \begin{align} x(0,s)&=s\implies x(t,s)=t+s \\ y(0,s)&=0\implies y(t,s)=\frac{t^2}{2}+st \\ u(0,s)&=0\implies u(t,s)=0 \\ \therefore u(x,y)&=0 \end{align}

Case $2$, $u(0,y)=y$: \begin{align} x(0,s)&=s\implies x(t,s)=t \\ y(0,s)&=0\implies y(t,s)=\frac{t^2}{2}+s \\ u(0,s)&=0\implies u(t,s)=se^t \\ \therefore u(x,y)&=\left(y-\frac{x^2}{2}\right)e^x \end{align}

The solutions say that case $1$ is valid for $y\leq x^2/2$ and case $2$ is valid for $y>x^2/2$. Have absolutely no idea why this.

I solved this problem using a Laplace transform and I arrived at the same conditions. But how can this possibly be interpreted when using the method shown.

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$$x\frac{\partial u}{\partial y}+\frac{\partial u}{\partial x}=u$$ I agree with your calculus. A different presentation of the Charpit-Lagrange equations : $$\frac{dy}{x}=\frac{dx}{1}=\frac{du}{u}=dt$$ A first characteristic equation comes from $\frac{dy}{x}=\frac{dx}{1}$ : $$y-\frac{x^2}{2}=c_1$$ A second characteristic equation comes from $\frac{dx}{1}=\frac{du}{u}$ : $$e^{-x}u=c_2$$ General solution of the PDE on implicit form $c_2=F(c_1)$ : $$e^{-x}u=F\left(y-\frac{x^2}{2}\right)$$ $$u(x,y)=e^xF\left(y-\frac{x^2}{2}\right)$$ $F(X)$ is an arbitrary fonction to be determined according to the boundary conditions.

First condition alone :

$u(x,0)=0=e^xF\left(0-\frac{x^2}{2}\right)\quad\implies\quad F(X)=0\quad\text{on}\quad X<0$

Putting $F=0$ into the above general solution leads to $$u(x,y)=0$$ Of course this trivial solution was obvious by inspection of the PDE.

Second condition alone :

$u(0,y)=y=e^0F\left(y-\frac{0^2}{2}\right)=F(y)=y$

The function $F$ is determined : $F(X)=X$. We put it into the above general solution where $X=y-\frac{x^2}{2}$ .

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

Both conditions together :

In order to satisfy both conditions, necessarily the function $F(X)$ is a piecewise function now defined on two domains : $$F(X)=\begin{cases} 0 && X<0\\ X && X>0 \end{cases}$$ Since $X=y-\frac{x^2}{2}$ the boundary between the two domains is $y=\frac{x^2}{2}$. $$u(x,y)=\begin{cases} 0 && y<\frac{x^2}{2}\\ e^x (y-\frac{x^2}{2}) && y>\frac{x^2}{2} \end{cases}$$ This is in agreement with your own results.

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  • $\begingroup$ (+1) I like this method presented, thanks! My main question is though, if I were to use the method that I have shown, how am I too see the correct domains for the piecewise function of $u(x,y)$? For instance, for case $2$, how is it obvious that $y>\frac{x^2}{2}$ given that I have shown $u(x,y)=e^x\left(y-\frac{x^2}{2}\right)$? Why can't $y<\frac{x^2}{2}$? $\endgroup$ – Stuart-James Burney Apr 28 at 14:28

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