# Solve $xu_x+(x+y)u_y=1$ when $u(1,y)=y$

Solve the following PDE

$$xu_x+(x+y)u_y=1$$ when $$u(1,y)=y$$ using method of characteristics and find the projections of the characteristics on the xy plane

$$\begin{cases} a=x\\ b=x+y\\ c=1 \end{cases}\Rightarrow \begin{cases} \frac{dx}{dt}=x\iff \frac{dx}{x}=dt\iff ln(x)=t+c_1\iff x=ke^t\\ \frac{dy}{dt}=x+y\iff \frac{dy}{dt}=ke^t+y\iff y=(kt+m)e^t\\ \frac{du}{dt}=1\iff du=dt\iff u=t+c_2 \end{cases}$$

How do I solve $$?_1$$ and what should be my next step?

$$xu_x+(x+y)u_y=1$$ You have got the correct system of equation which can be written on this form : $$\frac{dx}{x}=\frac{dy}{x+y}=\frac{du}{1}=dt$$ Solving $$\frac{dx}{x}=\frac{dy}{x+y}$$ gives a first characteristic $$\frac{y}{x}-\ln|x|=c_1$$ Solving $$\frac{dx}{x}=\frac{du}{1}$$ gives a second characteristic $$u-\ln|x|=c_2$$ The general solution of the PDE is $$u-\ln|x|=F\left(\frac{y}{x}-\ln|x|\right)$$ $$u(x,y)=\ln|x|+F\left(\frac{y}{x}-\ln|x|\right)$$ where $$F$$ is an arbitrary function.

This function is determined according to the boundary condition :

$$u(1,y)=y=\ln|1|+F\left(\frac{y}{1}-\ln|1|\right)=F(y)$$

Thus $$F(y)=y$$ and as a consequence $$F\left(\frac{y}{x}-\ln|x|\right)=\frac{y}{x}-\ln|x|$$ .

$$u(x,y)=\ln|x|+\left(\frac{y}{x}-\ln|x|\right)$$

The final solution is : $$u(x,y)=\frac{y}{x}$$

Let us follow the method of characteristics.

• $$\frac{\text d}{\text d t}x = x$$, letting $$x(0) = x_0$$ we know $$x = x_0e^t$$;
• $$\frac{\text d}{\text d t}y = x+y$$, letting $$t(0) = t_0$$ we know $$y = (x_0 t + y_0)e^t$$;
• $$\frac{\text d}{\text d t}u = 1$$, letting $$u(0) = u_0$$ we know $$u = t + u_0$$.

Since the prescribed data is $$u(1,y) = y$$, we set $$x_0 = 1$$ and $$u_0 = y_0$$ -- in other words, when $$t=0$$ we impose $$x=1$$ and $$u=y$$. Using the equation of characteristics, we have \begin{aligned} x &= e^t\\ y &= (t+y_0) e^t\\ u &= t+ y_0 \end{aligned} \qquad\text{so that}\qquad u(x,y) = \frac{y}{x} . The characteristics are the curves $$y = x\, (\ln x + y_0)$$ for $$(x,y)$$ in $$\Bbb {R_+^*}\times \Bbb{R}$$, which are represented below:

The second equation is a linear inhomogeneous ODE of first order. The standard solution methods apply. For instance, $$e^{-t}$$ is an integrating factor, $$\frac{d}{dt}(e^{-t}y(t))=e^{-t}(y'(t)-y(t))=k\implies y(t)=(kt+m)e^t$$