# PDEMOC17

Solve $\displaystyle u_x + u_y + u = 1\,$ with condition: $\displaystyle u=\sin(x)\,$ on $\displaystyle y=x^2+x\,$

The characteristic equations are

$\displaystyle \frac{dx}{1}=\frac{dy}{1}=\frac{du}{1-u}\,$

Using the first two equations,

$\displaystyle x=y+c_1 \implies c_1=x-y\,$

Using the second two equations,

$\displaystyle y=-\ln(1-u)+c_2 \implies c_2=y+\ln(1-u)\,$

So the general solution is an arbitrary function of these two constants.

$\displaystyle f(x-y,y+\ln(1-u))=0\,$

This can be equivalently written using another arbitrary function $\displaystyle g\,$ .

$\displaystyle y+\ln(1-u) = g(x-y)\,$

The general solution is

$\displaystyle u(x,y)=1-e^{g(x-y)-y}\,$

Now using the condition that $\displaystyle u=\sin(x)\,$ on $\displaystyle y=x^2+x\,$ ,

$\displaystyle \sin(x)=1-e^{g(-x^2)-x^2-x}\,$

$\displaystyle g(-x^2)=\ln(1-\sin(x))+x^2+x\,$

But we have an arbitrary function of $\displaystyle x-y\,$ in our solution, so set $\displaystyle -x^2=x-y\,$ .

$\displaystyle x=\sqrt{y-x}\,$

$\displaystyle g(-x^2)=g(x-y)=\ln(1-\sin\sqrt{y-x})+y-x+\sqrt{y-x}\,$

So now

$\displaystyle u(x,y)=1-e^{\ln(1-\sin\sqrt{y-x})+y-x+\sqrt{y-x}-y}\,$

The final solution is

$\displaystyle u(x,y)=1-(1-\sin\sqrt{y-x})e^{-x+\sqrt{y-x}}\,$