Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have this first-order non-homogeneous partial differential equation with initial condition:
$u_x + u_y=2u,\ u(x,0)=h(x)$
The following was what I tried:
By the method of characteristic curves, we have:
$dx/dt=1;\ dy/dt=1;\ du/dt=2u$
Then, we have $dy/dt=1$, and if we integrate both sides, we have $x-y=C_1$, meaning that the characteristic lines are $y-x=C_1$. And, we also have $du/dx=2u$. We will have $u=C_2 e^{2x}$, which is $u/e^{2x}=C_2$. So the question is where should I go next? My teacher did not give examples like these non-homogeneous equations. Any hints, please.

share|cite|improve this question
Just a correction to your question, the mentioned equation is in fact Homogeneous. – user128180 Feb 12 '14 at 8:15
Sorry, what I meant the PDE itself is Homogeneous, however, the initial conditions is not. – user128180 Feb 12 '14 at 8:17

3 Answers 3

up vote 2 down vote accepted

Consider the change of variables $(x,y)$ to $(a,v)=(x,x-y)$.

Then $u_x=u_a+u_v$ and $u_y=-u_v$. Your PDE then becomes $u_a=2u$. The solution to the PDE is $u(a,v)=C(v)e^{2a}$, where $C(v)$ is some function independent of $a$. Transforming back to the original variables gives $$u(x,y)=C(x-y)e^{2x}.$$ The initial condition then gives $u(x,0)=h(x)=C(x)e^{2x}\Rightarrow C(x)=h(x)e^{-2x}$. The solution of the PDE is then $$u(x,y)=h(x-y)e^{2y}.$$

share|cite|improve this answer
(-1): The OP wants us to solve this question by the method of characteristics rather than by the change of variables. – doraemonpaul Feb 5 '13 at 7:30
@doraemonpaul Clearly the OP was happy with this solution method, as it was accepted. Furthermore, my understanding of a variable change, in this case at least, is that it acts like a characteristic solution. – Daryl Feb 5 '13 at 9:17

Follow the method in

$\dfrac{dy}{dt}=1$ , letting $y(0)=0$ , we have $y=t$

$\dfrac{dx}{dt}=1$ , letting $x(0)=x_0$ , we have $x=t+x_0=y+x_0$

$\dfrac{du}{dt}=2u$ , letting $u(0)=f(x_0)$ , we have $u(x,y)=f(x_0)e^{2t}=f(x-y)e^{2y}$

$u(x,0)=h(x)$ :


$\therefore u(x,y)=h(x-y)e^{2y}$

share|cite|improve this answer

By method of characteristics: $u_x +u_y=2u$, and we know $du=u_xdx+u_ydy$, so $\frac {du}{dx}=u_x+\frac{dy}{dx} u_y$ and $\frac {dy}{dx}=1$ as shown by your analysis. So $u_x+u_y=\frac {du}{dx}$, substitute this to original equation we get:$\frac {du}{dx}=2u$, the. You can do the integration to get $ln(u)=2x+C$, which $C$ is a constant based on your initial condition. So $u(x,y)=e^{2x+C}$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.