I've been trying to work through some of the more difficult questions we've been given in class in regards to the method of characteristics for solving PDEs, but I've come a bit unstuck.
I've been asked to solve the following question;
$$x^2 u_x + e^{-y} u_y = 2y u^2$$ Subject to Cauchy data $u(s) = e^{-s}$ along $(x(s), y(s)) = (s,1)$
So, firstly, I set up the following ODEs; $$\frac{dx}{dt} = x^2, \frac{dy}{dt} = e^{-y}, \frac{du}{dt} = 2y u^2$$
Solving these three ODEs, I acquired the following expressions for $x$, $y$ and $u$... $$ x = \frac{1}{A - t}, y = \ln(t + B), u = \frac{1}{C - 2yt}$$ Where $A, B, C$ are all constants.
From there, I used my Cauchy data, and set $x = s$, $y =1$, and $u = e^{-s}$. Then, I set $t = 0$, in order to determine what my constants were.
I then got $A = \frac{1}{s}$, B = $e$, and $C = e^s$.
Now, rearranging my equation for $y$, I found that $t = e^y - e$, and using this, coupled with a rearrangement of my equation for $x$, I found an expression for $s$; $$s = \frac{x}{x(e^y-e)+1}$$.
Then, all that remained was to put everything into my expression for $u$; $$ u(x,y) = \frac{1}{e^{\frac{x}{x(e^y -e)+1}} - 2 y (e^y - e)}$$
However, this seems way too complicated,and I feel like I've made a mistake somewhere in my working.
I feel like my $\frac{du}{dt} = 2yu^2$ logic is kind of incorrect, and that I should be replacing the $y$ term here with $\ln(t + B)$, but even then, the integration becomes really horrible.
If anyone could provide some input, that would be fantastic. :)
EDIT: Yeah, I had made a mistake.
So, my equations and working for my $x$ and $y$ terms were fine, but my $u$ was bad.
To refresh, I've got that $y = \ln(t + e)$. Then,
$$\frac{du}{dt} = 2 y u^2 = 2 \ln(t+e) u^2$$ $$ \frac{du}{u^2} = 2 \ln(t+e) dt$$ $$ \frac{1}{u} = -2(t+e)\ln(t+e) + 2t + C$$
Now, using my Cauchy data, I have that $u = e^{-s}$. Setting $t=0$, we have... $$\frac{1}{e^{-s}} = -2 (0 + e) \ln(0+e) + 2 \times 0 + C$$ $$ C = e^s - 2e$$
Then, using some manipulation of the equation I got for $y$, I can write $1/u$ as follows; $$\frac{1}{u} = e^s - 4e + 2e^y (1 - y)$$
From here, I am assuming that I need to rearrange my equation for $x$ to make $s$ the subject, and then place that into the $e^s$ term, but still, this seems almost over the top. Is that really all I need to do??
Though it looks complicated, the solution is
$u(x,y) = \dfrac{1}{e^{\frac{x}{(e^y-e)x+1}} - 2 e^y(y -1)}$, which can be verified by substituting into the PDE.
The initial condition of the PDE is given by $u(s) = e^{-s}$ along $(x(s),y(s))=(s,1)$, which is $u_0(x,y) = u(x,1)=e^{-x}$.
The characteristic ODE are
$\dfrac{d x}{d t} = x^2$, $\dfrac{d y}{d t} = e^{-y}$, $\dfrac{d u}{d t} = 2 y u^2$,
with the initial conditions $x(0)= x_0$, $y(0)=1$ and $u(0) = u_0$ at $t=0$.
By solving the first two ODE, we obtain $x = \dfrac{x_0}{1-x_0 t} $ and $y = \ln (t + e )$.
By substituting $y$ into the third ODE, we obtain $\dfrac{d u}{d t} = 2 \ln (t + e ) u^2$ with initial $u(0)= u_0$.
Solve this ODE, then, $u = \dfrac{u_0}{1- 2 u_0 (t+e) (\ln (t+e) - 1) }$.
Obtain $x_0$ and $t$ from $x$ and $y$ respectively, and substitute $x_0$, $t$ and $u_0$ into $u$.
Thus, the solution is obtained, which is $u(x,y) = \dfrac{1}{e^{\frac{x}{(e^y-e)x+1}}-2 e^y(y -1)}$.