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

Question

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?

Answer 1

$$xu_x+(x+y)u_y=1$$ 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 : $$u(x,y)=\frac{y}{x}$$

Answer 2

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

Answer 3

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: