Solving 2nd Order PDE: u_{xx} - u_{tt} - au_{t} - bu = 0

[Somefantastik]

the book gives

$$u_{xx} - u_{tt} - au_{t} - bu = 0; 0<x<L, t>0$$

says if you multiply it by

$$2u_{t}$$

you can get

$$\left( 2u_{t}u_{x}\right)_{x} - \left( u^{2}_{x} + u^{2}_{t} + bu^{2}\right)_{t} -2au^{2}_{t} = 0$$

or

$$\frac{\partial}{\partial x} \left( 2 \frac{\partial u}{\partial t}\frac{\partial u}{\partial x} \right) - \frac{\partial}{\partial t} \left[ \left( \frac{\partial u}{\partial x} \right) ^{2} + \left( \frac{\partial u}{\partial t} \right) ^{2} + bu^{2} \right] - 2a\left(\frac{\partial u}{\partial t} \right)^{2} =0;$$

So far I have:

$$2 \frac{\partial u}{\partial t} \left( \frac{\partial^{2} u}{\partial x^{2}} - \frac{\partial^{2}u}{\partial t^{2}} - a\frac{\partial u}{\partial t} - b\frac{\partial u}{\partial t} \right) = 0;$$

$$2\frac{\partial u}{\partial t} \ \frac{\partial^{2}u}{\partial x^{2}} \ - \ 2 \frac{\partial u}{\partial t} \ \frac{\partial^{2}u}{\partial t^{2}} \ - \ 2a\left( \frac{\partial u}{\partial t} \right)^{2} - 2b\left(\frac{\partial u}{\partial t} \right)^2 = 0;$$

I can pull a d/dx out of the first term, to get $$\frac{\partial}{\partial x}\left(2\frac{\partial u}{\partial t} \ \frac{\partial u }{\partial x}\right)$$, and the $$- \ 2a\left( \frac{\partial u}{\partial t} \right)^{2}$$ is already there. How can I get the rest of it?

 

[tiny-tim]

$$2 \frac{\partial u}{\partial t} \left( \frac{\partial^{2} u}{\partial x^{2}} - \frac{\partial^{2}u}{\partial t^{2}} - a\frac{\partial u}{\partial t} - b\frac{\partial u}{\partial t} \right) = 0;$$

Hi Somefantastik!

I'm not swearing, but …

b u! ​

 

[Somefantastik]

$$2 \frac{\partial u}{\partial t} \left( \frac{\partial^{2} u}{\partial x^{2}} - \frac{\partial^{2}u}{\partial t^{2}} - a\frac{\partial u}{\partial t} - bu\right) = 0;$$

so now I need to show somehow that

$$- \frac{\partial}{\partial t} \left[ \left( \frac{\partial u}{\partial x} \right) ^{2} + \left( \frac{\partial u}{\partial t} \right) ^{2} + bu^{2} \right] \ = \ -2\frac{\partial u}{\partial t} \frac{\partial^{2}u}{\partial t^{2}} \ - \ 2bu\frac{\partial u}{\partial t}$$

$$-2\frac{\partial u}{\partial t} \frac{\partial^{2}u}{\partial t^{2}} \ - \ 2bu\frac{\partial u}{\partial t} \ = \ -\frac{\partial}{\partial t} \left(2\frac{\partial u}{\partial t}\frac{\partial u}{\partial t} + 2bu^{2} \right) \ = \ -\frac{\partial}{\partial t} \left[2\left(\frac{\partial u}{\partial t}\right)^{2} + 2bu^{2} \right]$$

I'm still missing the (du/dx)^2 term? And that 2 hanging out in there is wrong too.

 

[Emreth]

You should use chain rule for composite differentials, you can't pull out that d/dx term like that in the first post.The best way is, write out all the terms of the end result. Simplify and try to collect to get the first expression.Its pretty straightforward.You know chain rule right?

$$(AB)_{x}=A(B)_{x}+B(A)_{x}$$

Write out the squares explicitly, like (A*A),and use the chain rule.For example first term

$$\left( 2u_{t}u_{x}\right)_{x}=2(u_{t}u_{xx}+u_{xt}u_{x})$$

Second term

$$(u_{x}^2)_{t}=2(u_{xt}u_{x})$$

etc, etc
As you see, some of the terms will cancel out.you can do the rest.

 

[Somefantastik]

DUH, of course. I should know better. Thank you.