How to continue to find a solution to the PDE?

Question

I want to find a solution $u(x,y)$ for $\begin{cases}1 \cdot \frac{\partial u}{\partial x}+ (-x) \cdot \frac{\partial u}{\partial y}- x \cdot u = 0 \\u(x,-x^2)=1 \end{cases}$

I wanted to try it like in http://www1.maths.leeds.ac.uk/~kersale/Teach/M3414/Notes/m3414_1.pdf on page $11$ the example.

We put $a=1 ,\quad b=-x ,\quad c=-x ,\quad f=0$ and get the characteristic euqation $$\frac{dy}{dx}=\frac{-x}{1}=-x$$

Solving the ODE gives us $$y=-\frac{1}{2}x^2+c \quad c \in \Bbb R$$

Let $\eta$ be a constant,so $$\eta=\eta(x,y)=y+\frac{1}{2}x^2$$ now choose $x=\xi$ and get $$\eta=y+\frac{1}{2}\xi^2 \Rightarrow y= \eta-\frac{1}{2}\xi^2$$

Put $w(\eta,\xi)=u(x,y)$ and get $$\frac{\partial u}{\partial x}=\frac{\partial w}{\partial \xi}+\frac{\partial w}{\partial \eta}\cdot x=\frac{\partial w}{\partial \xi}+\xi \cdot\frac{\partial w}{\partial \eta}$$

$$\frac{\partial u}{\partial y}=0+\frac{\partial w}{\partial \eta}\cdot 1=\frac{\partial w}{\partial \eta}$$

our PDE becomes $$(\frac{\partial w}{\partial \xi}+\xi \cdot\frac{\partial w}{\partial \eta})-\xi \frac{\partial w}{\partial \eta}-\xi \cdot w=0$$

which we can write as $$\frac{\partial w}{\partial \xi} -\xi \cdot w=0$$

Can I say that $w=e^{1/2\xi^2}$ ? or how do I get the soultion to the original PDE ?

Answer 1

$$\frac{\partial u}{\partial x}+ (-x) \frac{\partial u}{\partial y}= x u $$ $$\frac{dx}{1}=\frac{dy}{-x}=\frac{du}{xu}$$ First characteristics, from $\quad\frac{dx}{1}=\frac{dy}{-x}$ : $$ 2y+x^2=c_1$$ Second characteristics, from $\quad\frac{dy}{-x}=\frac{du}{xu}$ : $$u\,e^y=c_2$$ General solution of the PDE : $$u\,e^y=F(2y+x^2)$$ $F$ is an arbitrary function, to be determined according to the boundary condition. $$u=e^{-y}F(2y+x^2)$$

Condition : $\quad u(x,-x^2)=1=e^{-(-x^2)}F(2(-x^2)+x^2)$ $$F(-x^2)=e^{-x^2}$$ Let $\quad X=-x^2$ $$F(X)=e^{X}$$ Now, the function $F$ is determined. We put it into the above general solution where $X=2y+x^2$ , then $F(2y+x^2)=e^{2y+x^2}$ .

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

IN ADDITION, comment about your calculus :

$$\frac{\partial w}{\partial \xi} -\xi \cdot w=0$$ $$w(\eta,\xi)=f(\eta)e^{\xi^2/2}$$ $$u(x,y)=f(y+\frac{x^2}{2})e^{x^2/2}$$ Condition :$\quad u(x,-x^2)=1=f(-x^2+\frac{x^2}{2})e^{x^2/2}=f(-\frac{x^2}{2})e^{x^2/2}$

Let $X=-x^2/2\quad;\quad 1=f(X)e^{-X} \quad;\quad f(X)=e^{X} \quad$ that we put into the general solution where $X=y+\frac{x^2}{2}$ , then $f(y+\frac{x^2}{2})=e^{y+\frac{x^2}{2}}$ $$u(x,y)=e^{y+\frac{x^2}{2}}e^{x^2/2}=e^{y+x^2}$$

Answer 2

Two things

First dont forget the function of $\eta$ $$\frac{\partial w}{\partial \xi} -\xi \cdot w=0$$ $$\int \frac{d w}{w} =\int \xi d\xi$$ $$\ln|w|=\frac {\xi ^2}{2}+h(\eta)$$ $$....$$

2 You have $\eta$ in function of x,y $$\implies \eta= y+\frac {x^2}2$$ And also $\xi$ in function of x y $$\implies \xi =x$$ Use them to get back to your original variables..