1
$\begingroup$

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!

$\endgroup$
  • 1
    $\begingroup$ Your "initial condition" doesn't say anything about $\phi$. Perhaps you mean that $\phi(x,y)$ is specified on the circle $x^2 + y^2 = 1$? $\endgroup$ – Robert Israel Dec 3 '13 at 5:18
  • $\begingroup$ oh yeah, sorry! I meant that $\phi$ is specified on the circle $x^2+y^2=1$. $\endgroup$ – johnsteck Dec 3 '13 at 5:23
  • $\begingroup$ A related problem. $\endgroup$ – Mhenni Benghorbal Dec 3 '13 at 5:46
1
$\begingroup$

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).

$\endgroup$
1
$\begingroup$

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 $1=x^{2}+y^{2}=2s^{2}+2t^{2}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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