PDE uniqueness

[renegade05]

Hello there!

I am having problems with the following question:

$\displaystyle x u_x+y u_y - 3u=0$
With $\displaystyle u(x,1)=\phi(x)$

I solved it using method of characteristics to find the solution to be:

$\displaystyle u(x,y) = \phi(\frac{x}{y}) y^3$

With the data curve being $\displaystyle y=1$ (red line)
And characteristic base curves being $\displaystyle y=\frac{1}{\xi}x$ (black lines)

So plotting this with different values of $\displaystyle \xi$ we get the following idea:



Now the question asks to:
"explain how the problem needs to be restricted in order to have a unique solution."

So this is where I am stuck. Everything looks good to me - so what am I missing? What are these restrictions?

THANKS!

 

[Ishuda]

Hi,
Lets back up a step or two. By the (present) method of characteristics we have
x = a e^s
y = b e^s
u = c e^3s
putting these together we have
(x/y) = d
which is your equation $\displaystyle y=\frac{1}{\xi}x$ with d = $\displaystyle \xi$ = a/b. From that, letting
$\displaystyle \alpha_{\lambda} (\frac{x}{y})^{\lambda} = 1$
we have a solution of
u(x,y) = $\displaystyle \alpha_{\lambda} (\frac{x}{y})^{\lambda} y^3$
where the a, b, and c have been absorbed into the $\displaystyle \alpha_{\lambda}$. Since this is a homogenous equation, the full solution is the 'sum' of the general solutions, so
u(x,y) = $\displaystyle \Sigma_{\lambda} \alpha_{\lambda} (\frac{x}{y})^{\lambda} y^3$
Since
u(x,1) = $\displaystyle \Sigma_{\lambda} \alpha_{\lambda} x^{\lambda} = \phi(x)$
a further restriction is that $\displaystyle \phi(x)$ be expressible as such.

For example $\displaystyle \phi(x)$ is infinitely differentiable in some neighborhood of x=0 so that
u(x,y) = $\displaystyle \Sigma_{0}^{\infty}\space \alpha_{n} x^{n} y^{3-n}$
and
$\displaystyle \alpha_{n} = \frac{\phi^{(n)}(0)}{n!}$.

BTW: The old method of characteristics (separation of variables) which I learned gives the same answer, i.e. let u = X(x) Y(y) and go from there.

 

[HallsofIvy]

Hi,
Lets back up a step or two. By the (present) method of characteristics we have
x = a e^s
y = b e^s
u = c e^3s
putting these together we have
(x/y) = d
which is your equation $\displaystyle y=\frac{1}{\xi}x$ with d = $\displaystyle \xi$ = a/b. From that, letting
$\displaystyle \alpha_{\lambda} (\frac{x}{y})^{\lambda} = 1$
we have a solution of
u(x,y) = $\displaystyle \alpha_{\lambda} (\frac{x}{y})^{\lambda} y^3$
where the a, b, and c have been absorbed into the $\displaystyle \alpha_{\lambda}$. Since this is a homogenous equation, the full solution is the 'sum' of the general solutions, so

Since this is a linear homogeneous equation.

u(x,y) = $\displaystyle \Sigma_{\lambda} \alpha_{\lambda} (\frac{x}{y})^{\lambda} y^3$
Since
u(x,1) = $\displaystyle \Sigma_{\lambda} \alpha_{\lambda} x^{\lambda} = \phi(x)$
a further restriction is that $\displaystyle \phi(x)$ be expressible as such.

For example $\displaystyle \phi(x)$ is infinitely differentiable in some neighborhood of x=0 so that
u(x,y) = $\displaystyle \Sigma_{0}^{\infty}\space \alpha_{n} x^{n} y^{3-n}$
and
$\displaystyle \alpha_{n} = \frac{\phi^{(n)}(0)}{n!}$.

BTW: The old method of characteristics (separation of variables) which I learned gives the same answer, i.e. let u = X(x) Y(y) and go from there.

 

[renegade05]

HMMMMMMMMMMMMM..... This is what I did:

$\displaystyle \frac{dx}{x}=\frac{dy}{y}=\frac{du}{3u}$
$\displaystyle \frac{dx}{ds}=x \qquad x(0)=\xi \qquad x(s)=\xi e^s$
$\displaystyle \frac{dy}{ds}=y \qquad y(0)=1 \qquad y(s)=e^s$
$\displaystyle \frac{du}{ds}=3u \qquad u(0)=\phi(\xi) \qquad u(s)=\phi(\xi)e^{3s}$

Eliminating $\displaystyle \xi \enspace and \enspace s$:

$\displaystyle \xi=\frac{x}{e^s}=\frac{x}{y}$
$\displaystyle s=ln(y)$

Thus:
$\displaystyle u(x,y)=\phi(\frac{x}{y})y^3$

Then my restriction was because the data curve is at y=1 and the base curves are $\displaystyle y=\frac{x}{\xi}$ then all the characteristic base curves cross at (0,0), but have different values of u(x,y)- therefore there is a singularity. So we need to restrict y>0.

Is this wrong? If yes, I am still not sure of the reasoning and to what restrictions I must make. THANKS!

 

[Ishuda]

Since this is a linear homogeneous equation.

Of course, silly of me to forget that little (so very important) word.

 

[Ishuda]

If we let
$\displaystyle \phi(x) = x$
then what you have is
u(x,y) = (x/y) y³ = x y²
x u_x(x,y) = u(x,y)
y u_y(x,y) = 2 u(x,y)
and
x u_x + y u_y - 3 U = 0
and u is infinitely (partial) differential is all closed neighborhoods in the real plane. So there is no need for the y>0 restriction in this case. But, putting the y>0 together with $\displaystyle \phi(\frac{x}{y})$, maybe the restriction is
"If $\displaystyle \phi$ has at pole of order more than 3 at x=0, y must be restricted to y>0."

Now that you have pointed it out, also note that the solution I gave can also be written as
$\displaystyle u(x,y) = \Sigma_0^{\infty}\space \alpha_n\space (\frac{x}{y})^n y^3 = \phi(\frac{x}{y}) y^3$

 

[renegade05]

If we let
$\displaystyle \phi(x) = x$
then what you have is
u(x,y) = (x/y) y³ = x y²
x u_x(x,y) = u(x,y)
y u_y(x,y) = 2 u(x,y)
and
x u_x + y u_y - 3 U = 0
and u is infinitely (partial) differential is all closed neighborhoods in the real plane. So there is no need for the y>0 restriction in this case. But, putting the y>0 together with $\displaystyle \phi(\frac{x}{y})$, maybe the restriction is
"If $\displaystyle \phi$ has at pole of order more than 3 at x=0, y must be restricted to y>0."

Now that you have pointed it out, also note that the solution I gave can also be written as
$\displaystyle u(x,y) = \Sigma_0^{\infty}\space \alpha_n\space (\frac{x}{y})^n y^3 = \phi(\frac{x}{y}) y^3$

So I am confused - what is the restriction?

 

[Ishuda]

So I am confused - what is the restriction?

I would say something like "If [FONT=MathJax_Math]ϕ[/FONT] has at pole of order more than 3 at x=0, y must be restricted to y$\displaystyle \ge \epsilon \gt 0$." but I'm not sure of that. I would appreciate it if someone else could chime in here or if you would check with your instructor and post the answer here.