How do I solve this equation $u_{xy}+u_x+u_y=2$ ? I know how to solve this $u_{xy}+u_y=0$ (with a change of variables and integrating factor $e^x$ ) but in this case another term was added $(u_x)$, should I follow the same idea or it requires a completely different method?
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1In my opinion you should try to solve it using similar ideas to the already solved case. Then, if you fail you can write what you tried. – Yanko Apr 06 '18 at 19:28
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Did you mean "integrating factor $e^x$"? Now enhance that by the additional factor $e^y$ for symmetry and see what happens. – Lutz Lehmann Apr 06 '18 at 19:47
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@LutzL Yes, I edited. At which point will I add the integrating factor? With $v=u_y$, we have $v_x+u_x=2-v$ and then do I integrate respect to $x$? – Apr 06 '18 at 19:58
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5That PDE has a trivial particular solution $u_0 = x + y$. So write $u$ as $v + x + y$, you only need to solve $v_{xy} + v_x + v_y = 0$. now equation become symmetric wrt $x,y$. So change variable to $p = x+y$ and $q = x -y$ and the PDE becomes a separable one. – achille hui Apr 06 '18 at 20:04
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No, the idea was to try to set $v=e^{x+y}u$. Then $v_x=e^{x+y}(u_x+u)$ and $v_{xy}=e^{x+y}(u_{xy}+u_x+u_y+u)=2e^{x+y}+v$ which in the end is not so much simplified. – Lutz Lehmann Apr 06 '18 at 20:04
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@achillehui could you post full solution? I got $v_{pp}-v_{qq}+2v_p=0$ and don't know how to continue – Holo Apr 06 '18 at 22:23
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@achillehui see Holo's comment :P (full answer please) – Apr 07 '18 at 00:18
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$$u_{xy}+u_x+u_y=2$$ Let $\quad u(x,y)=x+y+v(x,y)\quad$ which leads to : $$v_{xy}+v_x+v_y=0$$ Search for particular solutions of the form $v=X(x)Y(y)$
$X'Y'+X'Y+XY'=0\quad;\quad\frac{X'}{X}=-\frac{Y'}{Y+Y'}=\lambda \qquad\begin{cases} X=e^{\lambda x} \\ Y=e^{-\frac{\lambda}{1+\lambda}y}\end{cases}$
Particular solutions : $\quad v_\lambda(x,y)=e^{\lambda x-\frac{\lambda}{1+\lambda}y}$
The general solution is any linear combination of the particular solutions.
General solution expressed on the form of integral : $$v(x,y)=\int f(\lambda)e^{\lambda x-\frac{\lambda}{1+\lambda}y}d\lambda$$ $f(\lambda)$ is an arbitrary function, in so far the integral be convergent. $$u(x,y)=x+y+\int f(\lambda)e^{\lambda x-\frac{\lambda}{1+\lambda}y}d\lambda$$ Or, equivalently : $$u(x,y)=x+y+\int g(\mu)e^{-\frac{\mu}{1+\mu}x+\mu y}d\mu$$ The function $f(\lambda)$ , or $g(\mu)$ , as well as the bounds of the integral, have to be determined according to the boundary conditions. This generally draw to solve an integral equation. Without well defined boundary conditions no further calculus is possible, even not to say if it is possible to analytically solve it and if there is a closed form for the solution.
Arctic Char
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JJacquelin
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1You have a propagating sign error starting in your formula for $v_\lambda$. – J.G. Apr 07 '18 at 09:48
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@Holo : I never assumed $v=XY$. Of course the general solution is not $v(x,y)=X(x)Y(y)$ as it is obvious at the end of my answer. In fact, I search for particular solutions of the form $v_\lambda(x,y)=X(x)Y(y$). Do not confuse particular solution and general solution. If there was no particular solution of this form, the method would failed and we would have to try something else. In the present case, particular solutions on this form were found. So we can continue the calculus. – JJacquelin Apr 07 '18 at 10:14
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@JJacquelin I see, So why can we express the general solution using integral? if it is linear combination shouldn't it be: $$\sum_{\lambda} f(\lambda)e^{\lambda x-\frac{\lambda}{1+\lambda}y}$$? Sorry for asking so many questions – Holo Apr 07 '18 at 10:38
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You are on the right track. The sum you wrote is the first step to understand. Instead of a discret sum : $\sum_k c_ke^{kx-\frac{k}{1+k}y}$ consider a "continuous sum", that is un integral. $k$ becomes a continuous variable, say $\lambda$, the coefficient $c_k$ becomes a function $f(\lambda)$, on an infinitesimal $d\lambda$ to be integrated instead of to be summed. The infinity of particular functions are taken into account with more or less weight according to the arbitrary function $f(\lambda)$. The case of discret sum is less general since it correspond to discretized functions only. – JJacquelin Apr 07 '18 at 12:26
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@JJacquelin can we just multiply by the infinitesimal? Even if it is in respect to $\lambda$ and not $x,y$ wouldn't it have 'side effects'? – Holo Apr 07 '18 at 12:40
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It is the common way to extend infinite series to integral. Sorry, It's not possible to rewrite the whole theory in a few lines. On the forum, the answer is often limited to an hint and rough information, by necessity and in interest of space. By the way, if you are not convinced, bring back the integral solution into the initial PDE and see that it satisfies. – JJacquelin Apr 07 '18 at 14:59
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@JJacquelin I see, can you give me some references, or the name of the theory? I would really appreciate it – Holo Apr 07 '18 at 17:10
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For another example, see : https://math.stackexchange.com/questions/2727054/generalized-solution-of-a-pde-when-boundary-conditions-are-unknown/2727426#2727426 – JJacquelin Apr 08 '18 at 06:50
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@JJacquelin I see! So(unless I am wrong) without the boundary condition the eigenvalues are not discrete, so we can make it integral, because the integral is in respect to the eigenvalue the infinitesimal won't harm the solution. – Holo Apr 08 '18 at 08:23
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Hint:
Let $u=v+x+y$ ,
Then $u_x=v_x+1$
$u_y=v_y+1$
$u_{xy}=v_{xy}$
$\therefore v_{xy}+v_x+1+v_y+1=2$
$v_{xy}+v_x+v_y=0$
Let $\begin{cases}p=x+y\\q=x-y\end{cases}$ ,
Then $v_x=v_pp_x+v_qq_x=v_p+v_q$
$v_y=v_pp_y+v_qq_y=v_p-v_q$
$v_{xy}=(v_p+v_q)_y=(v_p+v_q)_pp_y+(v_p+v_q)_qq_y=v_{pp}+v_{pq}-v_{pq}-v_{qq}=v_{pp}-v_{qq}$
$\therefore v_{pp}-v_{qq}+v_p+v_q+v_p-v_q=0$
$v_{qq}=v_{pp}+2v_p$
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doraemonpaul
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Thank you for your answer :) .Could you explain the first 3 dots equation? I don't see it. – Apr 25 '18 at 05:30