I have the following PDE , which have one solution:
$$ (y+xz)z_x + (x+yz)z_y = z^2 - 1$$ $$z(t,1) = t, t > 0$$
first I tried solving with the Lagrange method, but I had difficulty in solving the ODE : $\dfrac {dy}{dx} = \dfrac{x+yz}{y+xz}$
after that tried to solve it using the method of characteristics, thought I could solve all the 3 ODEs needed , it still seem pretty complicated solving it that way. it seems to me that I'm missing something.
your help/hints are appriciated.
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dz}{ds}=z^2-1$ , letting $z(0)=0$ , we have $z=-\tanh s$
$\therefore\begin{cases}\dfrac{dx}{ds}=y-x\tanh s~......(1)\\\dfrac{dy}{ds}=x-y\tanh s~......(2)\end{cases}$
$(1)+(2)$ :
$\dfrac{dx}{ds}+\dfrac{dy}{ds}=x+y-(x+y)\tanh s$
$\dfrac{d(x+y)}{ds}=(x+y)(1-\tanh s)$
$\dfrac{d(x+y)}{x+y}=(1-\tanh s)~ds$
$\int\dfrac{d(x+y)}{x+y}=\int(1-\tanh s)~ds$
$\ln(x+y)=s-\ln\cosh s+c_1$
$x+y=C_1e^s~\text{sech}~s$
$(1)-(2)$ :
$\dfrac{dx}{ds}-\dfrac{dy}{ds}=y-x-(x-y)\tanh s$
$\dfrac{d(x-y)}{ds}=(x-y)(-1-\tanh s)$
$\dfrac{d(x-y)}{x-y}=(-1-\tanh s)~ds$
$\int\dfrac{d(x-y)}{x-y}=\int(-1-\tanh s)~ds$
$\ln(x-y)=-s-\ln\cosh s+c_2$
$x-y=C_2e^{-s}~\text{sech}~s$
$\therefore\begin{cases}x=\dfrac{C_1e^s~\text{sech}~s+C_2e^{-s}~\text{sech}~s}{2}\\y=\dfrac{C_1e^s~\text{sech}~s-C_2e^{-s}~\text{sech}~s}{2}\end{cases}$
$x(0)=x_0$ , $y(0)=f(x_0)$ :
$\begin{cases}\dfrac{C_1+C_2}{2}=x_0\\\dfrac{C_1-C_2}{2}=f(x_0)\end{cases}$
$\begin{cases}C_1=x_0+f(x_0)\\C_2=x_0-f(x_0)\end{cases}$
$\therefore\begin{cases}x=\dfrac{(x_0+f(x_0))e^s~\text{sech}~s+(x_0-f(x_0))e^{-s}~\text{sech}~s}{2}\\y=\dfrac{(x_0+f(x_0))e^s~\text{sech}~s-(x_0-f(x_0))e^{-s}~\text{sech}~s}{2}\end{cases}$
$\begin{cases}x=x_0+f(x_0)\tanh s\\y=x_0\tanh s+f(x_0)\end{cases}$
$\therefore\begin{cases}x_0=\dfrac{x-y\tanh s}{1-\tanh^2s}=\dfrac{x+yz}{1-z^2}\\f(x_0)=\dfrac{y-x\tanh s}{1-\tanh^2s}=\dfrac{xz+y}{1-z^2}\end{cases}$
Hence $\dfrac{xz+y}{1-z^2}=f\left(\dfrac{x+yz}{1-z^2}\right)$
$z(t,1)=t$ :
$f\left(\dfrac{2t}{1-t^2}\right)=\dfrac{t^2+1}{1-t^2}$
$f\left(\dfrac{2\tanh t}{1-\tanh^2t}\right)=\dfrac{\tanh^2t+1}{1-\tanh^2t}$
$f(\sinh2t)=\cosh2t$
$f(t)=\sqrt{t^2+1}$
$\therefore\dfrac{xz+y}{1-z^2}=\sqrt{\dfrac{(x+yz)^2}{(1-z^2)^2}+1}$
$(xz+y)^2=(x+yz)^2+(1-z^2)^2$
Just to expand d-e's comment. Some instructive tutorials can be found here (I do not know the author)
We will follow the method described there.
The characteristic equation is $$ \frac{dx}{y+xz} = \frac{dy}{x+yz} = \frac{dz}{z^2 - 1}. $$ Using the fact $$ \frac{a}{b} = \frac{c}{d} \Longleftrightarrow \frac{a+ c}{b+ d} \Longleftrightarrow \frac{a- c}{b- d}. \qquad (1) $$ we get $$ \frac{d(x+y)}{(x+y)(z+1)} = \frac{dz}{(z+1)(z-1)}. $$ which is equivalent to $$ \frac{d(x+y)}{x+y} = \frac{d(z-1)}{z - 1}. $$ Integrating this yields the first constant of integration $$ C_1 = \frac{x+y}{z-1}. \qquad (2) $$
Similarly if we use the minus sign of (1), then $$ \frac{d(x-y)}{(x-y)(z-1)} = \frac{dz}{(z+1)(z-1)}, $$ or $$ \frac{d(x-y)}{x-y} = \frac{d(z+1)}{z+1}, $$ which yields the second constant of integration $$ C_2 = \frac{x - y}{z+1}. \qquad (3) $$
Now by the initial condition, we can impose a functional relationship $C_2 = G(C_1)$. With the help of (2) and (3), we get $$ \frac{x-y}{z+1} = G\left(\frac{x+y}{z-1} \right). \qquad (4) $$ The function $G$ can be sought from the initial condition $z(t, 1) = t$, or $x = t, y = 1, z = t$: $$ \frac{t-1}{t+1} = G\left(\frac{t+1}{t-1} \right), $$ which means $G(z) = 1/z$. Using this result in (4), we get $$ \frac{x-y}{z+1} = \frac{z-1}{x+y}, $$ or equivalently $$z^2 = 1 + x^2 - y^2,$$ agreeing with d-e's result.
BTW, the final answer is $z = \sqrt{x^2-y^2+1}$
– d_e Oct 24 '15 at 20:43