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Question:

let $a,b,c$ be positive constants. Find $u=u(x,y)$ if is satisfies the partial differential equation

$$\dfrac{\partial^2 u}{\partial x^2}+\dfrac{\partial^2 u}{\partial y^2}=c$$ and the boundary condition $$u=0,\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1.$$

my try: I know this is screened Poisson equation $$\dfrac{\partial^2 u}{\partial x^2}+\dfrac{\partial^2 u}{\partial y^2}=c$$ I only find this poisson equation one of the solution $$u(x,y)=-\dfrac{c}{2}\cdot\dfrac{a^2b^2}{a^2+b^2}\left(1-\dfrac{x^2}{a^2}-\dfrac{y^2}{b}\right)$$

because $$\dfrac{\partial u}{\partial x}=-\dfrac{c}{2}\cdot\dfrac{a^2b^2}{a^2+b^2}\left(-\dfrac{2x}{a^2}\right)$$ $$\dfrac{\partial^2 u}{\partial x^2}=c\cdot\dfrac{b^2}{a^2+b^2}$$ and $$\dfrac{\partial^2 u}{\partial y^2}=c\cdot\dfrac{a^2}{a^2+b^2}$$ so $$\dfrac{\partial^2 u}{\partial x^2}+\dfrac{\partial^2 u}{\partial y^2}=c\cdot\left(\dfrac{a^2}{a^2+b^2}+\dfrac{b^2}{a^2+b^2}\right)=c$$ and when $$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\Longrightarrow u=0$$

and this solution is uniqueness? and How prove it? Thank you

Thank you very much!

jimjim
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math110
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4 Answers4

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You found one solution $u_0$ of the problem. Then for any solution $u$ of the problem the function $v:=u-u_0$ would satisfy $$\Delta v=0,\qquad v=0 \quad {\rm on}\quad \partial \Omega,$$ where $$\Omega:=\left\{(x,y)\Biggm| {x^2\over a^2}+{y^2\over b^2}\leq1\right\}$$ is an elliptical domain. Now a harmonic function which vanishes on the boundary of a compact domain has to vanish identically therein, as it cannot have a local extremum in the interior. So $v=0$, or $u=u_0$, which shows that your solution is unique.

Christian Blatter
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Hint. Try $$ u(x,y)=\frac{x^2}{a^2}+\frac{y^2}{b^2}-1. $$ Find a solution, and then show uniqueness.

Yiorgos S. Smyrlis
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The shape that your boundary conditions describe is in fact an ellipse. So if you switch to an elliptic coordinate system your pde should separate out nicely.

Kieran Cooney
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Let $$v(x,y)=u(x,y)+\dfrac{c}{2}\cdot\dfrac{a^2b^2}{a^2+b^2}\left(1-\dfrac{x^2}{a^2}-\dfrac{y^2}{b}\right)$$

Then $\Delta v=0$ in $\mathbb R^2$ and $v(x,y)=0$ on the boundary of the ellipse $U=\big\{\frac{x^2}{a^2}+\frac{y^2}{b^2}<1\big\}$. Thus, as the Dirichlet problem $$ \Delta v=0 \,\,\text{in $U$} \quad\text{and} \quad v=0 \,\,\text{on $\partial U$}, $$ has unique solution $v\equiv 0$, then $v$ vanishes in $U$ as well. But $v$ is a harmonic function in $\mathbb R^2$, and harmonic functions are real analytic, and thus it is expressible as a power series around $(0,0)$ $$ v(x,y)=\sum_{m,n=0}^\infty \frac{\partial^{m+n}v(0,0)}{\partial x^m\partial y^n}\frac{x^my^n}{m!n!}=0, $$ since $\frac{\partial^{m+n}v(0,0)}{\partial x^m\partial y^n}=0$, for all $m,n$.

Yiorgos S. Smyrlis
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