1
$\begingroup$

I am studying partial differential equations by myself. Would you please help me with solving a problem?

We have the equation:
$4\frac{\partial^2u}{\partial x^2}-8\frac{\partial^2u}{\partial x\partial y}+4\frac{\partial^2u}{\partial y^2}=1$
Let $r=x+ky \quad and \quad s=x$, choose k such that the above equation becomes an easier one then find the answer according to:
$u(x,0)=\frac{1}{8}x^2$
$\frac{\partial u}{\partial y}(x,0)=1$


I tried to change variables:
$$\frac{\partial^2 u}{\partial x^2}=\frac{\partial^2 u}{\partial s^2}.(\frac{\partial s}{\partial x})^2+2\frac{\partial^2 u}{\partial s \partial r}.\frac{\partial s}{\partial x}.\frac{\partial r}{\partial x} +\frac{\partial^2 u}{\partial r^2}.(\frac{\partial r}{\partial x})^2+\frac{\partial u}{\partial s}.\frac{\partial^2 s}{\partial x^2}+\frac{\partial u}{\partial r}.\frac{\partial^2 r}{\partial x^2}=\frac{\partial^2 u}{\partial s^2}+2\frac{\partial^2u}{\partial s\partial r}+\frac{\partial^2 u}{\partial r^2}+0+0 \quad \longrightarrow \qquad u_{xx}=u_{ss}+2u_{sr}+u_{rr}$$
$$u_{yy}=0+0+k^2u_{rr}+0+0 \quad \longrightarrow \qquad u_{yy}=k^2u_{rr}$$
$$\frac{\partial u}{\partial y}=\frac{\partial u}{\partial s}.\frac{\partial s}{\partial y}+\frac{\partial u}{\partial r}.\frac{\partial r}{\partial y}=0+k\frac{\partial u}{\partial r}$$
Now how can I get $\frac{\partial}{\partial x}(\frac{\partial u}{\partial y})$?
I guess $\frac{\partial}{\partial x}(\frac{\partial u}{\partial y})=k\frac{\partial}{\partial x}(\frac{\partial u}{\partial r})=k\frac{\partial u}{\partial r}.\frac{\partial r}{\partial x}$, is it correct?



(This question probably will be edited in the future depend on the answers...)

$\endgroup$
3
  • $\begingroup$ You're not doing the second derivatives correctly. See this question: math.stackexchange.com/questions/2215654/… $\endgroup$ Commented Aug 5, 2017 at 7:22
  • $\begingroup$ With $k=1$ I got $u_{rr}+u_{ss}=1/4$. But the answer given below is a much simpler one. If you are working from a textbook, make sure everything is copied correctly. Check your use of the chain rule very carefully. $\endgroup$
    – jdods
    Commented Aug 5, 2017 at 11:30
  • $\begingroup$ The problem is exactly what I wrote! I don't know how to do derivatives correctly yet! I think it needs more time to learn with self-study... $\endgroup$
    – f44
    Commented Aug 6, 2017 at 8:59

1 Answer 1

5
$\begingroup$

A different simplification - You can 'factorise' the derivative operator as $\let\del\partial 4(\del_x^2 + 2\del_x\del_y + \del_y^2)u = 1$ and complete the square to get $$(\del_x+\del_y)^2 u = 1/4$$

If we set $r = x+y $ and $s=x-y$ (an orthogonal choice of coordinates) so that $x =\frac{r+s}{2}$ and $y=\frac{r-s}{2}$, The chain rule says $\del_r = \frac1{2}\del_x + \frac1{2}\del_y$ so the PDE reduces to $\del_r^2 u = 1$.

Really, any new choice of coordinates so that

$$ \begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix} a & b\\a & c\end{pmatrix} \begin{pmatrix}r\\s\end{pmatrix}$$ (and $a≠0,b≠c$ just so that its a legit choice) will work, so that the new derivative $\del_r$ will "hit $x$ and $y$ equally", $\del_r(u(x,y)) = \del_r(u(ar+bs,ar+cs)) = a\del_xu + a\del_y u$.


If you want to use $s=x$ and $r=x+ky$, this means that $x=s$ and $y=(r-s)/k$, so $$ \begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix} 0 & 1\\1/k & -1/k\end{pmatrix} \begin{pmatrix}r\\s\end{pmatrix}$$ So $\del_r = \frac{1}{k}\del_y $ and $\del_s = \del_x - \frac{1}k\del_y$. On inspection we see $\del_s+(k+1)\del_r = \del_x+\del_y$, which turns the PDE into

$$1/4 = (\del_s+(k+1)\del_r)^2 u = \del_s^2u + 2(k+1)\del_s\del_ru + (k+1)^2\del_r^2u$$

So we should use $k=-1$.

You might notice that if you relabel $s\leftrightarrow r$, this corresponds to $b=0,c=-1,a=1$ of the previous solution.

$\endgroup$
1
  • $\begingroup$ @Coder44 I have added a proper answer that solves the question via the method suggested. $\endgroup$ Commented Aug 6, 2017 at 17:35

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .