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Try let $v=x+y$ , $w=x-y$ ,

Then $\dfrac{\partial u}{\partial x}=\dfrac{\partial u}{\partial w}\dfrac{\partial w}{\partial x}=\dfrac{\partial u}{\partial w}$

$\dfrac{\partial u}{\partial y}=\dfrac{\partial u}{\partial w}\dfrac{\partial w}{\partial y}=-\dfrac{\partial u}{\partial w}$

$\dfrac{\partial^2u}{\partial y^2}=\dfrac{\partial}{\partial y}\left(-\dfrac{\partial u}{\partial w}\right)=-\dfrac{\partial}{\partial v}\left(\dfrac{\partial u}{\partial w}\right)\dfrac{\partial v}{\partial y}=-\dfrac{\partial^2u}{\partial vw}$

$\therefore-\dfrac{\partial^2u}{\partial vw}=\dfrac{\partial u}{\partial w}$

Let $z=\dfrac{\partial u}{\partial w}$ ,

Then $\dfrac{\partial z}{\partial v}=\dfrac{\partial^2u}{\partial vw}$

$\therefore-\dfrac{\partial z}{\partial v}=z$

$\dfrac{dz}{z}=-~dv$

$\int\dfrac{dz}{z}=\int-~dv$

$\ln z=-v+c_1(w)$

$z=c_2(w)e^{-v}$

$\dfrac{\partial u}{\partial w}=c_2(w)e^{-v}$

$u=\int c_2(w)e^{-v}~dw$

$u=C_1(v)+C_2(w)e^{-v}$

$u=C_1(x+y)+C_2(x-y)e^{-x-y}$

I did it correctly in Finding an analytical solution to the wave eqn using method of characterisitics , and did it not known whether correct or not in Solving $yu_{xx}+(x+y)u_{xy}+xu_{yy}=0$ , but why in here is wrong?

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8  
The second line is wrong. The multivariate chain rule is more complicated than this. –  Qiaochu Yuan Jul 8 '12 at 23:12
    
so how is correctly transform $\dfrac{\partial u}{\partial x}=\dfrac{\partial^2u}{\partial y^2}$ to a 2nd order PDE by letting $v=x+y$ , $w=x-y$ ? –  doraemonpaul Jul 9 '12 at 1:17

2 Answers 2

up vote 5 down vote accepted

Fleshing out @QiaochuYuan's comment a little: $$\begin{eqnarray*} \frac{\partial}{\partial x} &=& \frac{\partial v}{\partial x} \frac{\partial}{\partial v} + \frac{\partial w}{\partial x} \frac{\partial}{\partial w} \\ &=& \frac{\partial}{\partial v} + \frac{\partial}{\partial w} \\ \frac{\partial}{\partial y} &=& \frac{\partial v}{\partial y} \frac{\partial}{\partial v} + \frac{\partial w}{\partial y} \frac{\partial}{\partial w} \\ &=& \frac{\partial}{\partial v} - \frac{\partial}{\partial w}. \\ \end{eqnarray*}$$ The reason this is a nice change of variables for the wave equation is because $$\frac{\partial^2}{\partial x^2} - \frac{\partial^2}{\partial y^2} = 4 \frac{\partial}{\partial v} \frac{\partial}{\partial w}.$$ No such nice thing happens with the heat equation.

Addendum: Note that $$\begin{eqnarray*} \frac{\partial^2}{\partial x^2} &=& \left(\frac{\partial}{\partial v} + \frac{\partial}{\partial w}\right)^2 \\ &=& \frac{\partial^2}{\partial v^2} + 2 \frac{\partial}{\partial v} \frac{\partial}{\partial w} + \frac{\partial^2}{\partial w^2} \\ \frac{\partial^2}{\partial y^2} &=& \left(\frac{\partial}{\partial v} - \frac{\partial}{\partial w}\right)^2 \\ &=& \frac{\partial^2}{\partial v^2} - 2 \frac{\partial}{\partial v} \frac{\partial}{\partial w} + \frac{\partial^2}{\partial w^2}. \\ \end{eqnarray*}$$ The transformed equation is  $$\left(\frac{\partial}{\partial v} + \frac{\partial}{\partial w}\right)u(v,w) = \left(\frac{\partial^2}{\partial v^2}     - 2 \frac{\partial}{\partial v} \frac{\partial}{\partial w}     + \frac{\partial^2}{\partial w^2}\right) u(v,w).$$ There are no nice cancelations. This transformation just makes the PDE harder to solve.

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Your answer does not complete. How about $\dfrac{\partial^2u}{\partial x^2}$ and $\dfrac{\partial^2u}{\partial y^2}$ ? –  doraemonpaul Jul 9 '12 at 1:20
    
@doraemonpaul: Added. –  user26872 Jul 9 '12 at 1:26
    
How about $\dfrac{\partial^2u}{\partial xy}$ ? –  doraemonpaul Jul 9 '12 at 1:37
    
@doraemonpaul: What is $(a+b)(a-b)$? –  user26872 Jul 9 '12 at 1:41
    
$a^2-b^2$ . Do you mean $\dfrac{\partial^2u}{\partial xy}=\dfrac{\partial^2u}{\partial v^2}-\dfrac{\partial^2u}{\partial w^2}$ ? I don't understand why the partial derivative can first remove their dependent variable and then multiplicate each other? –  doraemonpaul Jul 9 '12 at 2:16

This equation is a heat equation. Usually we use the methods of seperation of variables or Fourier transforms.

Seperation of variables: Let $$ u=X(x)*Y(y),$$ then we get $$X'(x)Y(y)=X(x)Y''(y).$$ That is to say, $$X'/X=Y''/Y$$ Notice that the left side of the equation above is a function of $x$, while the right side the function of $y$. So if the left side equals the right side, then both of them must be constant. Thus $$X'/X=c,Y''/Y=c,$$ these are ODEs which are easy to solve.

Fourier Transform: Notice that we use Fourier Transform with respect of $y.$

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