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I have a PDE: $$ \frac{\partial^2 u}{\partial x^2} + a\frac{\partial^2 u}{\partial y^2} +bu = f(x,y) $$ where $a$ and $b$ are constants and $b>a>0$.

Also $\space u(x,0) = g(x)$, and $\space u(0,y) = h(y)$

How do i approach such a problem? I tried finding C.F. by separation of variables technique but got stuck on deciding the value of separation parameter. And how do i find the P.I.?

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    $\begingroup$ What does "C.F." mean$ $\endgroup$ – Robert Lewis Jan 15 at 19:52
  • $\begingroup$ CF: complementary function $\endgroup$ – titusarmah99 Jan 15 at 20:07
  • $\begingroup$ You haven't got enough boundary conditions. $\endgroup$ – Mattos Jan 15 at 22:31
  • $\begingroup$ @Mattos I am simply asking for hints, resources on solving specifically the above mentioned type of PDE. I went through lot of resources but couldn't find one $\endgroup$ – titusarmah99 Jan 16 at 2:32
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    $\begingroup$ The substitution $y= \sqrt{a}s$ reduces the equation to the inhomogeneous Hemholtz's equation $\nabla^2u + k^2u = f$. Then you solve it using Green's function $\endgroup$ – Dylan Jan 16 at 14:18
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First use the substitution $y = \sqrt{a} \cdot \bar{y}$ to obtain the inhomogeneous Helmholtz equation

\begin{cases} \nabla^2 u + k^2 u = f, & (x,\bar y) \in D \\ u = g, & y=0 \\ u = h, & x = 0 \end{cases}

where $D$ is the quarter plane $\{x > 0, \bar y > 0\}$ on $\Bbb R^2$

We can use the same method used for the Laplace equation here, with some modifications.


Step 1. Find the Green's function $G(x,x',\bar y,\bar y')$ on an open domain such that

$$ (\nabla^2 + k^2)G = \delta(x-x',\bar y-\bar y'), \quad (x',\bar y') \in \Bbb R^2 $$

There are several options, the most common one you'll find is

$$ G(\rho) = \frac{i}{4}[J_0(k\rho) + iY_0(k\rho)], \quad \rho = \sqrt{(x-x')^2+(\bar y-\bar y')^2} $$

where $J_0$ and $Y_0$ are Bessel functions. The real part of this function, with just $Y_0$, would also work here.


Step 2. Find the Green's function on the desired domain. We want $G_D(x,x',\bar y,\bar y')$ such that

\begin{cases} (\nabla^2 + k^2)G_D = \delta(\rho), & (x',\bar y') \in D \\ G = 0, & \bar y' = 0 \\ G = 0, & x' = 0 \end{cases}

Using the method of images, we find

$$ G_D(x,x',\bar y, \bar y') = \frac14 \big[G(x,x',\bar y,\bar y') - G(x,x',\bar y,-\bar y') - G(x,-x',\bar y, \bar y') + G(x,-x',\bar y,-\bar y')\big] $$


Step 3. Integrate the Green's function. Using Green's second identity

\begin{align} \iint_D \big[u(\mathbf{r}')\cdot(\nabla^2+k^2)G_D(\mathbf{r},\mathbf{r}') - G_D(\mathbf{r},\mathbf{r}')\cdot (\nabla^2+k^2)u(\mathbf{r}') \big]dA' &= \int_{\partial D} \big[u(\mathbf{r}')\nabla G_D(\mathbf{r},\mathbf{r}') - G_D(\mathbf{r},\mathbf{r'})\nabla u(\mathbf{r})\big]\cdot \hat{\mathbf{n}}ds' \\ u(\mathbf{r}) - \iint_D G_D(\mathbf{r},\mathbf{r}')f(\mathbf{r}') dA' &= \int_{\partial D} u(\mathbf{r}')\nabla G_D(\mathbf{r},\mathbf{r}')\cdot \mathbf{n} ds' \end{align}

$$ \implies u(x,\bar y) = \int_0^\infty\int_0^\infty G_Df(x',\bar y')dx' d\bar y' + \int_0^\infty g(x') \frac{\partial G_D}{\partial \bar y'}\Bigg\vert_{y'=0}dx' + \int_0^\infty h(y') \frac{\partial G_D}{\partial x'}\Bigg\vert_{x'=0} dy' $$

Note that this doesn't guarantee a unique solution, since there exists a family of homogeneous solutions

$$ u_h(x,y) = \sin(k_x x)\sin(k_y y), \quad k_x^2 + k_y^2 = k^2 $$

such that if $u(x,y)$ is a solution, then $u(x,y) + Cu_h(x,y)$ is also a solution.

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