0
$\begingroup$

I have $r=Ax+Bt$ and $s=Cx+Dt$

I know $\frac{\partial u}{\partial x}=\frac{\partial u}{\partial r}A+\frac{\partial u}{\partial s}C$

But I don't understand how to differentiate this again with respect to $x$.

$\endgroup$
0
$\begingroup$

$$\frac{\partial }{\partial x}\left(\frac{\partial u}{\partial x}\right)=\frac{\partial }{\partial x}\left(\frac{\partial u}{\partial r}A\right)+\frac{\partial }{\partial x}\left(\frac{\partial u}{\partial s}C\right)$$

$$=\frac{\partial^2 u}{\partial x\partial r}A+\underbrace{\frac{\partial u}{\partial r}\frac{\partial A}{\partial x}}_{=0 }+\frac{\partial^2 u}{\partial x\partial s}C+\underbrace{\frac{\partial u}{\partial s}\frac{\partial C}{\partial x}}_{=0 }$$

$$=\frac{\partial^2 u}{\partial x\partial r}A+\frac{\partial^2 u}{\partial x\partial s}C$$

$\endgroup$
0
$\begingroup$

$\dfrac{\partial A}{\partial x}=\dfrac{\partial C}{\partial x}=0$ since $A$ and $C$ are constants so when you take the second partial derivative of $u$ with respect to $x$ you will get

$$\frac{\partial^2u}{\partial x^2}=\frac{\partial^2u}{\partial x\partial r}A+\frac{\partial^2u}{\partial x\partial s}C$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.