Solve PDE in 2D

Question

Problem How should I go about solving this PDE:

$$ \phi_x+\phi_y=x+y-3c $$

Where $\phi = \phi(x,y)$, $c$ is a constant, and $\phi$ is specified on the circle

$$ x^2+y^2=1 $$

My Attempt to solve it I would like to use the method of characteristics, but then I get stuck because of the given initial condition. In fact, so far I have the characteristics equations

$$ \dot{{z}}(s)=x+y-3c $$ $$ \dot{{x}}(s)= 1 $$ $$ \dot{{y}}(s)= 1 $$

The last two are easy to solve but then I am not sure how to use the initial condition. If you know of a different/easier method to solve this PDE, feel free to let me know, thanks!

Your "initial condition" doesn't say anything about $\phi$. Perhaps you mean that $\phi(x,y)$ is specified on the circle $x^2 + y^2 = 1$? — Robert Israel, Dec 03 '13 at 05:18
oh yeah, sorry! I meant that $\phi$ is specified on the circle $x^2+y^2=1$. — johnsteck, Dec 03 '13 at 05:23

score 1 · Answer 1 · answered Dec 03 '13 at 07:37

As you say, the characteristic equations are $\dot{z} = x + y - 3 c$, $\dot{x} = 1$, $\dot{y} = 1$. So the characteristic curves are $x = x_0 + s$, $y = y_0 + s$, i.e. $x - y = \text{constant}$. But there's a problem with specifying the initial conditions on the circle $x^2 + y^2 = 1$: the characteristic curves through most points either don't intersect the circle at all (so the initial condition doesn't determine $\phi$ there) or intersect it in two points (so the initial conditions might not be consistent).

score 1 · Answer 2 · answered Dec 03 '13 at 07:45

This problem can also be solved by a substitution: $$ x = s+t,\;\;\; y = s-t,\;\;\; \psi(s,t)=\phi(s+t,s-t). $$ Then $$ \psi_{s} = \phi_{x}+\phi_{y} = x+y-3c = 2s-3c $$ Then there is a function $d(t)$ such that $$ \psi = s^{2}-3cs+d(t). $$ The function $d$ is determined from $\psi(s,t)$, which is assumed to be known on $1=x^{2}+y^{2}=2s^{2}+2t^{2}$.

Solve PDE in 2D

2 Answers2