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Reading a book, I saw that this function (where $H$ is the standard Heaviside function)

$$ u(t,x)= H(x-0.5t) $$

could be the solution, in the sense of distributions, of this equation

$$ u_t + (u^{2})_{x} = 0 $$

i.e

$$ \int_{0}^{\infty} \left(\int^\ u f_{t} + u^{2}f_{x} dx \right) dt = -\int^\ u_0(x)f(0,x) dx $$ for any test function $f = f(t,x)$.

Can anyone help me do this verification by computing the above integral?

Regards

Matthew Cassell
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cauchy
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  • It seems that you are considering the inviscid Burgers' equation $u_t + (\frac{1}{2}u^2)_x = 0$ instead (otherwise Rankine-Hugoniot would not be satisfied). You may have a look at this post. – EditPiAf Apr 12 '19 at 13:38

1 Answers1

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Basically when you integrate something times a heaviside function, you just get the integral of that function with domain such that the argument of said heaviside function is greater than $0$. Also note, heaviside function squared is just heaviside function.

$\int_0^\infty (\int u f_t + u^2 f_x dx) dt = \int_0^\infty (\int_{0.5t}^\infty (f_t + f_x) dx dt$

Split up integral and look at $f_x$ first. When we integrate $f_x$ wrt $x$ we get $f$.

So this term is simply $\int_0^\infty (f(t,\infty)-f(t,0.5t))dt $. My thinking is that $f(t,\infty)$ will be zero as it is a test function thus must be integrable.

$\int_0^\infty \int_0^{2x}f_t dt dx = \int_0^\infty (f(2x,x)-f(0,x)) dx$. Here we rearrange the domain in order change the order of integration.

We will happily cancel the $f(t,0.5t)$ and $f(2x,x)$ terms in our integrals.

Our final answer is $-\int_0^\infty f(0,x)dx$

$u_0(x) = H(x)$, so this matches the answer you have supplied.

fGDu94
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  • There must be some reason why $0.5$ is used in the answer, and it will be linked to the $u^2$ term. – fGDu94 Apr 12 '19 at 08:44