Prove this PDE has a unique (weak) real valued solution

Question

Let $\Omega = (0,1) \times (0,1)$ and let $k \in [0,\frac{3}{2}]$.

Prove the boundary value problem $$\frac{f_{xx}}{2y}+f_{yy}+2k^2\frac{f}{y} = 2k^2$$ subject to $$f(x,0)=0,f(x,1)=1/2,f(0,y)=y,f_{x}(1,y)=0$$ has a unique real valued solution on $\Omega $ for each $k \in [0,\frac{3}{2}]$ .

My approach: I tried the energy approach, suppose there are two solutions $f_1,f_2$ and let $g=f_1-f_2$, then $g$ satisfies a homogeneous PDE $$\frac{g_{xx}}{2y}+g_{yy}+2k^2\frac{g}{y} = 0$$ with homogeneous boundary conditions $$g(x,0)=g(x,1)=g(0,y)=g_{x}(1,y)=0.$$

Observe that the homogeneous PDE is equivalent to $$\bigtriangledown \cdot (\frac{1}{2y} \bigtriangledown g)+\frac{\partial}{\partial y}[(1-\frac{1}{2y})g_{y}]+\frac{2k^2g}{y} = 0$$

Multiply both sides PDE by $g$ and then integrate (need Divergence theorem and integration by parts formula) over $\Omega$ to have: $$\int\int_{\Omega}\frac{|\bigtriangledown g|^2}{2y}+(1-\frac{1}{2y})g_{y}^2-\frac{2k^2g^2}{y} \mathrm{d}y\mathrm{d}x=0$$

Next, if one can show the innequality $$\int\int_{\Omega}\frac{2k^2g^2}{y}\mathrm{d}y\mathrm{d}x \leq \int\int_{\Omega}(1-\frac{1}{2y})g_{y}^2 \mathrm{d}y\mathrm{d}x$$ holds then by non-negativity one can deduce $\bigtriangledown g =0$ and thus $g = 0$, therefore the solution is unique.

However it seems hard to prove $$\int\int_{\Omega}\frac{2k^2g^2}{y}\mathrm{d}y\mathrm{d}x \leq \int\int_{\Omega}(1-\frac{1}{2y})g_{y}^2 \mathrm{d}y\mathrm{d}x$$ as the Poincare (Sobolev) inequality seems not applicable to this case.

I know there is an proved inquality $\int\int_{\Omega} |\bigtriangledown g| \geq \frac{1}{2\pi^2}\int\int_{\Omega}|g|^2$, but it only applies to Dirichlet heat equation and it seems not useful in this particular problem

Answer 1

Let $f=g+y$ ,

Then $f_x=g_x$

$f_{xx}=g_{xx}$

$f_y=g_y+1$

$f_{yy}=g_{yy}$

$\therefore\dfrac{g_{xx}}{2y}+g_{yy}+2k^2\dfrac{g+y}{y}=2k^2$

$g_{xx}+2yg_{yy}+4k^2g+4k^2y=4k^2y$

$g_{xx}+2yg_{yy}+4k^2g=0$

with conditions $g(0,y)=0$ and $g_x(1,y)=0$

Let $g(x,y)=\sum\limits_{n=0}^\infty C(n,y)\sin\dfrac{(2n+1)\pi x}{2}$ so that it automatically satisfies $g(0,y)=0$ and $g_x(1,y)=0$ ,

Then $-\sum\limits_{n=0}^\infty\dfrac{(2n+1)^2\pi^2}{4}C(n,y)\sin\dfrac{(2n+1)\pi x}{2}+2y\sum\limits_{n=0}^\infty C_{yy}(n,y)\sin\dfrac{(2n+1)\pi x}{2}+4k^2\sum\limits_{n=0}^\infty C(n,y)\sin\dfrac{(2n+1)\pi x}{2}=0$

$\sum\limits_{n=0}^\infty\biggl(2yC_{yy}(n,y)+\biggl(4k^2-\dfrac{(2n+1)^2\pi^2}{4}\biggr)C(n,y)\biggr)\sin\dfrac{(2n+1)\pi x}{2}=0$

$\therefore2yC_{yy}(n,y)+\biggl(4k^2-\dfrac{(2n+1)^2\pi^2}{4}\biggr)C(n,y)=0$

According to http://eqworld.ipmnet.ru/en/solutions/ode/ode0206.pdf,

$C(n,y)=C_1(n)\sqrt{y}I_1\biggl(2\sqrt{\biggl(\dfrac{(2n+1)^2\pi^2}{8}-2k^2\biggr)y}\biggr)+C_2(n)\sqrt{y}K_1\biggl(2\sqrt{\biggl(\dfrac{(2n+1)^2\pi^2}{8}-2k^2\biggr)y}\biggr)$

$\therefore f(x,y)=y+\sum\limits_{n=0}^\infty C_1(n)\sqrt{y}I_1\biggl(2\sqrt{\biggl(\dfrac{(2n+1)^2\pi^2}{8}-2k^2\biggr)y}\biggr)\sin\dfrac{(2n+1)\pi x}{2}+\sum\limits_{n=0}^\infty C_2(n)\sqrt{y}K_1\biggl(2\sqrt{\biggl(\dfrac{(2n+1)^2\pi^2}{8}-2k^2\biggr)y}\biggr)\sin\dfrac{(2n+1)\pi x}{2}$

$f(x,0)=0$ :

$\sum\limits_{n=0}^\infty C_2(n)\lim\limits_{y\to0}\sqrt{y}K_1\biggl(2\sqrt{\biggl(\dfrac{(2n+1)^2\pi^2}{8}-2k^2\biggr)y}\biggr)\sin\dfrac{(2n+1)\pi x}{2}=0$

$\because\lim\limits_{y\to0}\sqrt{y}K_1\biggl(2\sqrt{\biggl(\dfrac{(2n+1)^2\pi^2}{8}-2k^2\biggr)y}\biggr)\neq0$ , $\therefore C_2(n)=0$

$\therefore f(x,y)=y+\sum\limits_{n=0}^\infty C_1(n)\sqrt{y}I_1\biggl(2\sqrt{\biggl(\dfrac{(2n+1)^2\pi^2}{8}-2k^2\biggr)y}\biggr)\sin\dfrac{(2n+1)\pi x}{2}$

$f(x,1)=\dfrac{1}{2}$ :

$1+\sum\limits_{n=0}^\infty C_1(n)I_1\biggl(2\sqrt{\dfrac{(2n+1)^2\pi^2}{8}-2k^2}\biggr)\sin\dfrac{(2n+1)\pi x}{2}=\dfrac{1}{2}$

$\sum\limits_{n=0}^\infty C_1(n)I_1\biggl(2\sqrt{\dfrac{(2n+1)^2\pi^2}{8}-2k^2}\biggr)\sin\dfrac{(2n+1)\pi x}{2}=-\dfrac{1}{2}$

As the solution of $C_1(n)$ is unique, thus proved the question.