I did read the answer to here, but my approach is a bit different and would like some suggestions.

Solve $u_x + u_y + u = e^{x+2y}$

One checks that $u = \frac{e^{x+2y}}{4}$ is a special solution to $u_x+u_y+u = e^{x+2y}$ (May I know if this step is by trial and error or there is a way to find out?)

Now I solve for $u_x+u_y + u = 0$, which is equivalent to solve $$\frac{u_x}{u}+\frac{u_y}{u} = -1$$ Let $v = \ln u$, and we have $v_x+v_y + 1 = 0$, then $$(v+x)_x+(v+x)_y = 0$$ now let $w = v+x$, we have $$w_x+w_y = 0$$ by the method of characteristics, $w$ is in the form of $f(x-y)$ and now $v = h(x-y)-x$.

Since $u = e^v$, we have $u = e^{h(x-y)-x}$. Thus the general solution is $$u =e^{h(x-y)-x}+\frac{e^{x+2y}}{4} $$

May I know if my approach is correct.

  • $\begingroup$ I like the use of the variable substitution, I believe this solution is correct, and could probably be generalizes to solving first order PDEs without involving the method of characteristics. $\endgroup$ – Kernel_Dirichlet Feb 1 '18 at 23:31

At first sight, I didn't understood your notation $h(x-y)$. So, I delete my previous comment.

I agree with your result : $u =e^{h(x-y)-x}+\frac{e^{x+2y}}{4}$

This is the same result as my result below, with the relationship $\quad e^{h(x-y)}=e^{x-y}F(x-y)$

The specific result according to the boundary condition is added below.

$$u_x+u_y=e^{x+2y}-u$$ The characteristic ODEs are : $\quad\frac{dx}{1}=\frac{dy}{1}=\frac{du}{e^{x+2y}-u}$

A first set of characteristic curves comes from $\quad \frac{dx}{1}=\frac{dy}{1}\quad\implies\quad x-y=c_1$

A second set of characteristic curves comes from $\quad \frac{dy}{1}=\frac{du}{e^{x+2y}-u}$

With $x=c_1+y\quad\to\quad \frac{dy}{1}=\frac{du}{e^{c_1+3y}-u} \quad\to\quad \frac{du}{dy}+u=e^{c_1+3y}$

The solution of this first order linear ODE is : $\quad u=\frac14 e^{c_1+3y}+c_2e^{-y}$

$u=\frac14 e^{(x-y)+3y}+c_2e^{-y}=\frac14 e^{x+2y}+c_2e^{-y}$ which gives the second family of characteristics :

$ue^y -\frac14 e^{x+2y}e^y=c_2 \quad\to\quad ue^y -\frac14 e^{x+3y}=c_2$

The general solution of the PDE expressed on the form of implicit equation is: $$\Phi\left(x-y\:,\: ue^y -\frac14 e^{x+3y} \right)=0$$ where $\Phi$ is any differentiable function of two variables.

Or, on explicit form : $\quad ue^y -\frac14 e^{x+3y}=F(x-y)$ $$u(x,y)=\frac14 e^{x+2y}+e^{-y}F(x-y)$$ Where $F(X)$ is any differentiable function of one variable $X$ with $X=x-y$.


$u(X,0)=0=\frac14 e^{X+0}+e^0F(X-0) \quad\implies\quad F(X)= -\frac14 e^{X}$

So, the function $F(X)$ is determined. Putting it into the above general solution with $X=x-y$ leads to the particular solution which satisfies the specified condition :

$u(x,y)=\frac14 e^{x+2y}+e^{-y}\left(-\frac14 e^{x-y}\right)$ $$u(x,y)=\frac14 e^{x}\left(e^{2y}-e^{-2y}\right)$$ $$u(x,y)=\frac12 e^{x}\sinh(2y)$$


You lost/excluded all solutions that take non-positive values by setting $v=\ln u$. You can repair this partially by replacing that with $v=\ln|u|$. In the end, setting $H(z)=e^{h(z)}$ and then allowing negative and zero values for $H$ gives the general solution $$ u(x,y)=H(x-y)e^{-x}+\frac{e^{x+2y}}4 $$


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