# Finding solution of semi-linear PDE using Method of Characteristics

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

$$u_x+u_y+u=0$$

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?

• 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. May 25, 2019 at 18:00
• @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 $\frac{1}{4}e^{x+2y}$ May 25, 2019 at 18:40

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 $$u(x,y)=-\frac14e^{2(x-y)}e^{-x}+\frac14e^{x+2y}=\frac12e^x\sinh(2y).$$