Time Invariance of f_n for Quasi-Linear PDE Boundary Conditions

[Wuberdall]

Homework Statement

let y(x, t) be a solution to the quasi-linear PDE
\frac{\partial y}{\partial t} + y\frac{\partial y}{\partial x} = 0
with the boundary condition
y(0, t) = y(1, t) = 0
show that
f_n(t) = \int_0^1 y^n\,\mathrm{d}x
is time invariant for all n = 1, 2, 3,...

Homework Equations

The Attempt at a Solution

By differentiation under the integral sign i obtain
\frac{\mathrm{d}}{\mathrm{d}t}f_n(t) = \int_0^1ny^{n-1}\frac{\partial y}{\partial t}\,\mathrm{d}x \\<br /> = -n\int_0^1y^n\frac{\partial y}{\partial x}\,\mathrm{d}x \\<br /> = -n\underbrace{[y^{n+1}]_0^1}_{=0} + n^2\int_0^1y^n\frac{\partial y}{\partial x}\,\mathrm{d}x
where the last equality in valid by integration by parts.

My goal was to obtain a differential equation for f_n.

 

[Ssnow]

Hi, I don't understand why at the end it is $n^2\int_{0}^{1}y^n\frac{\partial y}{\partial x} dx$ instead of $n^2\int_{0}^{1}y^n dx=n^2f_{n}$

 

[Wuberdall]

Hi, I don't understand why at the end it is $n^2\int_{0}^{1}y^n\frac{\partial y}{\partial x} dx$ instead of $n^2\int_{0}^{1}y^n dx=n^2f_{n}$

Because,
\int_0^1 y^n\frac{\partial y}{\partial x}\mathrm{d}x = [y^{n+1}]_0^1 - \int_0^1 ny^{n-1}\frac{\partial y}{\partial x}\,y\,\mathrm{d}x
as
\int fg^\prime = fg - \int f^\prime g

 

[Ssnow]

ok sorry I lost a term deriving... yes it is no easy to find the form $f_{n}$ for a possible equation. The fact is that $f_{n}$ depends only by the time and also the quantity $F=\int_{0}^{1}y^{n}(x,t)\frac{\partial y(x,t)}{\partial x}dx$ is a function only of the time, so you equation will be

$\frac{df_{n}}{dt}=n^2F(t)$

 

[Ssnow]

Hi, but if you can think $y^n\frac{\partial y}{\partial x}=\frac{1}{n+1}\frac{\partial y^{n+1}}{\partial x}$ you have that the integral will be:

$\frac{n^2}{n+1}[y(1,t)^{n+1}-y(0,t)^{n+1}]=0$

so your equation is $\frac{d f_{n}}{dt}=0$