I am given the PDE:

$$u_x+u_y+u=e^{x+2y} \quad u(x,0)=0$$

I tried to do this using the method of the characteristics in the following way.

First I find the particular solution of this inhomogenous equation by substituting $u_p = Ae^{x+2y}$ giving me $A=\frac{1}{4}$ hence $$u_p=\frac{1}{4}e^{x+2y}$$

Now I try the homogenous equation


If I parameterize $x,y$ with $t$ and have now $u(x(t),y(t))$ I can write

$$\frac{du}{dt} = \frac{dx}{dt}\frac{\partial u}{\partial x} + \frac{dy}{dt}\frac{\partial u}{\partial y} $$

So equating like terms I have

$$\frac{dx}{dt}=1 \quad \frac{dy}{dt}= 1 \quad \frac{du}{dt}=-u$$

My initial conditions are $x(t=0)= x_0, y(t=0)=0$ and $u(t=0)=u(x(0),y(0))=u(x_0,0)=0$

So then I can solve these equations to get

$$x = t + x_0 \Rightarrow x_0=x-t \quad y = t \quad u = Ce^{-t}$$

Now I'm not 100% sure what $C$ should be but I assume it some function of the initial condition, i.e

$$u=F(x_0)e^{-t} \Rightarrow u=F(x-t)e^{-t} \Rightarrow u_0 = F(x_0)=0 ?$$

But the initial condition is $u_0 = 0$ so $u$ is $0$ at all $x$??

I'm getting stuck here but I followed similar method on other examples given and it works fine. Why is this initial condition $0$ giving me $u=0$, what am I missing?

  • $\begingroup$ Note that you modified $u$ to the homogeneous part by subtracting a particular solution. You need to remove the particular solution also from the initial condition to get the IC for the homogeneous part. $\endgroup$ May 25, 2019 at 18:00
  • $\begingroup$ @LutzL Why is that? I dont see that in my notes. I tried now with that and it works, do you have some reference or website i can see why this theory works? I mean by subtracting particular solution from IC? And I subtracted $e^{x+2y}$ which is not the particular solution, it is $4 \times \frac{1}{4}e^{x+2y}$, so a constant times the particular solution. So do I subtract $e^{x+2y}$ from the IC or $\frac{1}{4}e^{x+2y}$ $\endgroup$
    – hh4346
    May 25, 2019 at 18:40

1 Answer 1


You decompose your solution as $u=u_h+u_p$ where you determined $u_p$ per undetermined coefficients as $u_p(x,y)=\frac14e^{x+2y}$. Then indeed as constructed $$ (u_h)_x+(u_h)_y+u_h=0. $$ However, also in the initial condition you must reflect this decomposition to get $$ 0=u(x,0)=u_h(x,0)+u_p(x,0)\implies u_h(x,0)=-\frac14e^x. $$

Now as you found $u_h(x,y)=F(x-y)e^{-x}$, this leads in the initial conditions to $F(x)=-\frac14e^{2x}$ and thus $$ u(x,y)=-\frac14e^{2(x-y)}e^{-x}+\frac14e^{x+2y}=\frac12e^x\sinh(2y). $$


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