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I'm looking for a solution of the following PDE problem:

$$\begin{cases} -\frac{\partial^2u}{\partial x^2}-2\frac{\partial^2u}{\partial y^2}=f&\text{on}~U,\\ u=0&\text{on}~\partial U \end{cases}$$

for $U=\left\{(x,y)\in\mathbb{R}^2|\,x,y\geq0\right\}$ .

The solution should be represented as an integral of $f$ integrated against a kernel.

I tried so far to extend the Poisson kernel but it is somehow not working. I'm very glad for your help. Thank you so much.

Anastasia

doraemonpaul
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anastasia
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2 Answers2

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Hint: Make a transformation $(x, y) \mapsto (x, z)$ such that $2\frac{\partial^2}{\partial y^2} = \frac{\partial^2}{\partial z^2}$.

Tunococ
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$-\dfrac{\partial^2u}{\partial x^2}-2\dfrac{\partial^2u}{\partial y^2}=f(x,y)$

$\dfrac{\partial^2u}{\partial x^2}+2\dfrac{\partial^2u}{\partial y^2}=-f(x,y)$

Let $\begin{cases}p=y+\sqrt{2}ix\\q=y-\sqrt{2}ix\end{cases}$ ,

Then $\dfrac{\partial u}{\partial x}=\dfrac{\partial u}{\partial p}\dfrac{\partial p}{\partial x}+\dfrac{\partial u}{\partial q}\dfrac{\partial q}{\partial x}=\sqrt{2}i\dfrac{\partial u}{\partial p}-\sqrt{2}i\dfrac{\partial u}{\partial q}$

$\dfrac{\partial^2u}{\partial x^2}=\dfrac{\partial}{\partial x}\left(\sqrt{2}i\dfrac{\partial u}{\partial p}-\sqrt{2}i\dfrac{\partial u}{\partial q}\right)=\dfrac{\partial}{\partial p}\left(\sqrt{2}i\dfrac{\partial u}{\partial p}-\sqrt{2}i\dfrac{\partial u}{\partial q}\right)\dfrac{\partial p}{\partial x}+\dfrac{\partial}{\partial q}\left(\sqrt{2}i\dfrac{\partial u}{\partial p}-\sqrt{2}i\dfrac{\partial u}{\partial q}\right)\dfrac{\partial q}{\partial x}=-2\dfrac{\partial^2u}{\partial p^2}+2\dfrac{\partial^2u}{\partial pq}+2\dfrac{\partial^2u}{\partial pq}-2\dfrac{\partial^2u}{\partial q^2}=-2\dfrac{\partial^2u}{\partial p^2}+4\dfrac{\partial^2u}{\partial pq}-2\dfrac{\partial^2u}{\partial q^2}$

$\dfrac{\partial u}{\partial y}=\dfrac{\partial u}{\partial p}\dfrac{\partial p}{\partial y}+\dfrac{\partial u}{\partial q}\dfrac{\partial q}{\partial y}=\dfrac{\partial u}{\partial p}+\dfrac{\partial u}{\partial q}$

$\dfrac{\partial^2u}{\partial y^2}=\dfrac{\partial}{\partial y}\left(\dfrac{\partial u}{\partial p}+\dfrac{\partial u}{\partial q}\right)=\dfrac{\partial}{\partial p}\left(\dfrac{\partial u}{\partial p}+\dfrac{\partial u}{\partial q}\right)\dfrac{\partial p}{\partial y}+\dfrac{\partial}{\partial q}\left(\dfrac{\partial u}{\partial p}+\dfrac{\partial u}{\partial q}\right)\dfrac{\partial q}{\partial y}=\dfrac{\partial^2u}{\partial p^2}+\dfrac{\partial^2u}{\partial pq}+\dfrac{\partial^2u}{\partial pq}+\dfrac{\partial^2u}{\partial q^2}=\dfrac{\partial^2u}{\partial p^2}+2\dfrac{\partial^2u}{\partial pq}+\dfrac{\partial^2u}{\partial q^2}$

$\therefore-2\dfrac{\partial^2u}{\partial p^2}+4\dfrac{\partial^2u}{\partial pq}-2\dfrac{\partial^2u}{\partial q^2}+2\biggl(\dfrac{\partial^2u}{\partial p^2}+2\dfrac{\partial^2u}{\partial pq}+\dfrac{\partial^2u}{\partial q^2}\biggr)=-f\biggl(\dfrac{i(q-p)}{2\sqrt{2}},\dfrac{p+q}{2}\biggr)$

$8\dfrac{\partial^2u}{\partial pq}=-f\biggl(\dfrac{i(q-p)}{2\sqrt{2}},\dfrac{p+q}{2}\biggr)$

$\dfrac{\partial^2u}{\partial pq}=-\dfrac{1}{8}f\biggl(\dfrac{i(q-p)}{2\sqrt{2}},\dfrac{p+q}{2}\biggr)$

$u(p,q)=F(p)+G(q)-\dfrac{1}{8}\int_b^q\int_a^pf\biggl(\dfrac{i(t-s)}{2\sqrt{2}},\dfrac{s+t}{2}\biggr)ds~dt$

$u(x,y)=F(y+\sqrt{2}ix)+G(y-\sqrt{2}ix)-\dfrac{1}{8}\int_b^{y-\sqrt{2}ix}\int_a^{y+\sqrt{2}ix}f\biggl(\dfrac{i(t-s)}{2\sqrt{2}},\dfrac{s+t}{2}\biggr)ds~dt$

doraemonpaul
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