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Consider the function $u(x,y):{\mathbb{R}^2} \to {\mathbb{R}}$ and $u(x,y) \in {{C}^1}({\mathbb{R}^2})$. The function satisfies the following boundary value problem

$$c_1 u_x + c_2 u_y = f(x,y)$$

$$u|_{\partial \Omega} =0$$

Where $\Omega$ is an arbitrary simply connected domain in $\mathbb{R}^2$ with the boundary $\partial \Omega $ which can be assumed to be smooth as you like although weaker conditions may be needed. $c_1,c_2$ are some real constants.

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Find **a** function $f(x,y)$ such that $u(x,y) >0$ for all $(x,y) \in \Omega$.

In the next step, it can be aksed that how we can find **all** such $f(x,y)$? or What are the common properties of **all** such $f(x,y)$?

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Let $\Omega=[0,\pi]\times[0,\pi]$ and $u=\sin x \sin y$ .

We have, $u(0,y)=0$,$u(\pi,y)=0$, $u(x,0)=0$ and finally $u(x,\pi)=0$, so

$$u\big|_{\partial\Omega}=0$$

and hence

$$f(x,y)=c_1\cos x\sin y+c_2\sin x \cos y$$

Now you want to have the solution to

$$c_1u_x + c_2u_y=c_1\cos x\sin y+c_2\sin x \cos y$$

by characteristics.

Thus, $x_s=c_1$ so $x=c_1s+a$, similarly $y=c_2s+b$ where $a,b$ are arbitrary constant. Also, we can write

$$\eqalign{

& {u_s} = {c_1}\cos x\sin y + {c_2}\sin x\cos y \cr

& \,\,\,\,\,\, = {c_1}\cos ({c_1}s + a)\sin ({c_2}s + b) + {c_2}\sin ({c_1}s + a)\cos ({c_2}s + b) \cr} $$

so you got

$$u = \frac{1}{2} \left(\cos \left(a-b+c_1 s-c_2 s\right)-\cos \left(a+b+\left(c_1+c_2\right) s\right)\right)=

\frac{1}{2} \left(\cos \left(x-y\right)-\cos \left(x+y\right)\right)=\sin x \sin y$$

as we could expected.

Suppose the boundary is a closed smooth curve given by

$$\partial \Omega = \left\{ {\left( {x,y} \right)|x = p(t),y = q(t),{t_1} \leqslant t \leqslant {t_2}} \right\}\tag{1}$$

According to the notion of the method of characteristic, we make the following change of variables:

$$\left\{ \matrix{

x = x(s,t) \hfill \cr

y = y(s,t) \hfill \cr} \right.\tag{2}$$

and make the following definitions

$$\eqalign{

& z(s,t) = u(x(s,t),y(s,t)) \cr

& F(s,t) = f(x(s,t),y(s,t)) \cr}\tag{3}$$

Now we want our change of variables to come in handy, so we **require** that

$$\left\{ \matrix{

{{\partial x} \over {\partial s}} = {c_1},\,\,\,x({s_0},t) = p(t) \hfill \cr

{{\partial y} \over {\partial s}} = {c_2},\,\,\,y({s_0},t) = q(t) \hfill \cr} \right.\,\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,\left\{ \matrix{

x(s,t) = {c_1}(s – {s_0}) + p(t) \hfill \cr

y(s,t) = {c_2}(s – {s_0}) + q(t) \hfill \cr} \right.\,\tag{4}$$

where $s_0$ is some arbitrary real constant that you may choose whatever you want. Next, using **chain-rule**, $(3)$, $(4)$ and the PDE, we have

$${{\partial z} \over {\partial s}}\left( {s,t} \right) = F(s,t)\tag{5}$$

and hence

$$z(s,t) = \int\limits_{{s_0}}^s {F(\eta ,t)d\eta }\tag{6}$$

You can easily verify that $(6)$ satisfies both the PDE and BC. Now, we write $(6)$ in another form

$$u\left( {x(s,t),y(s,t)} \right) = \int\limits_{{s_0}}^s {f\left( {x(\eta ,t),y(\eta ,t)} \right)d\eta }\tag{7}$$

So, what I can say is that the $f$ you are looking for, must have the following property

$$\int\limits_{{s_0}}^s {f\left( {{c_1}(\eta – {s_0}) + p(t),{c_2}(\eta – {s_0}) + q(t)} \right)d\eta } > 0\tag{8}$$

Or equivalently, if you choose $s_0=0$

$$\int\limits_0^s {f\left( {{c_1}\eta + p(t),{c_2}\eta + q(t)} \right)d\eta } > 0\,\,\,\,\,\,for\,\,\,\,\forall \left( {s,t} \right):\left( {{c_1}s + p(t),{c_2}s + q(t)} \right) \in \Omega\tag{9}$$

**Some Suggestions**

1) I think you can get more out of $(9)$ but I didn’t have time to go further. Try some tricks like change of variables in the integral. Maybe you got something!

2) We have found $z(s,t)=u(x(s,t),y(s,t))$. Try to find $u(x,y)$ from $(7)$. You should think on how to write the RHS in $(7)$ in terms of $x(s,t)$ and $y(s,t)$. Alternatively, you can think of finding $s=s(x,y)$ and $t=t(x,y)$ from $(4)$.

3) One more suggestion that I can make is to investigate this problem when

$$\eqalign{

& {c_1} = {c_2} = 1 \cr

& \partial \Omega = \left\{ {\left( {x,y} \right)|x = \cos t,y = \sin t,0 \le t \le 2\pi } \right\} \cr} $$

and then try to find your desired $f(x,y)$. Examining special cases is always helpful. ðŸ™‚

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