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  1. Consider the first order partial differential equation $$ \frac{\partial u}{\partial t} + 3u^2 \frac{\partial u}{\partial x} = -\alpha u , \tag{1} $$ where $\alpha>0$ is a constant. This equation can be expressed in conservation form as $$ \frac{d}{dt} \int_{x_1}^{x_2} u\, dx + \int_{x_1}^{x_2} \frac{\partial}{\partial x} u^3\, dx = -\alpha \int_{x_1}^{x_2} u\, dx . \tag{2} $$ (a) Assume that $(1)$ has a solution with a shock which has trajectory $s(t)$ such that $x_1 < s(t) < x_2$. Use the conservation form $(2)$ to show that the jump condition across the shock is $$ \frac{ds}{dt} = \dot s = \frac{[u^3]}{[u]}, $$ where $[\, ]$ denotes the jump in the appropriate quantity across the shock.

I only know how to solve this problem when the right of $(1)$ to be $0$. Can someone help me to solve this problem? What should be changed for the right of equation $(2)$? Can I just turn it to $0$?

EditPiAf
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Serena
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  • Welcome to MSE! Remind that a good question has to be motivated, specific, typeset with MathJax, etc. But you are lucky: there is a related post which solves your problem. First try this method, and if you don't succeed, then edit your post to point out the problem. You may find also this post interesting. – EditPiAf Dec 07 '18 at 17:23

1 Answers1

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Recall from the proof of the Rankine-Hugoniat Condition, that $$\frac{d}{dt} \int_{x_1}^{x_2} u dx = \int_{x_1}^{s(t)} \frac{\partial u}{\partial t} dx + u(s - \epsilon) \frac{ds}{dt} + \int_{s(t)}^{x_2} \frac{\partial u}{\partial t} dx - u(s + \epsilon) \frac{ds}{dt}$$ for some $\epsilon$ which allows us to approximate $u$ with a continuously smooth function. Now, we also take note of several things: $$\int_{x_1}^{x_2} \frac{\partial }{\partial x} u^3 dx = \int_{x_1}^{s(t) - \epsilon} \frac{\partial }{\partial x} u^3 dx + \int_{s(t) + \epsilon}^{x_2} \frac{\partial }{\partial x} u^3 dx + [u^3]$$ Therefore, the complete equation becomes: $$\int_{x_1}^{s(t)} \frac{\partial u}{\partial t} dx + u(s - \epsilon) \frac{ds}{dt} + \int_{s(t)}^{x_2} \frac{\partial u}{\partial t} dx - u(s + \epsilon) \frac{ds}{dt} + \int_{x_1}^{s(t) - \epsilon} \frac{\partial }{\partial x} u^3 dx + \int_{s(t) + \epsilon}^{x_2} \frac{\partial }{\partial x} u^3 dx + [u^3] = -\alpha \int_{x_1}^{x_2} u dx$$ Solving for $\frac{ds}{dt}$:

$$\frac{ds}{dt} = \frac{\int_{x_1}^{s(t) - \epsilon} \frac{\partial }{\partial x} u^3 dx + \int_{s(t) + \epsilon}^{x_2} \frac{\partial }{\partial x} u^3 dx + [u^3] + \alpha \int_{x_1}^{x_2} u dx}{[u]}$$ Now, what should happen when you evaluate $\alpha \int_{x_1}^{x_2} u dx$?. Well, we notice that:

$$\alpha \int_{x_1}^{x_2} u dx = \alpha \int_{x_1}^{s(t)} u dx + \alpha \int_{s(t)}^{x_2} u dx$$ Therefore, can a conclusion be made about the quantity: $$\int_{x_1}^{s(t) - \epsilon} \frac{\partial }{\partial x} u^3 dx + \int_{s(t) + \epsilon}^{x_2} \frac{\partial }{\partial x} u^3 dx + \alpha \int_{x_1}^{s(t)- \epsilon} u dx + \alpha \int_{s(t) + \epsilon}^{x_2} u dx$$

Specifically, what happens when $x_1 \rightarrow s(t) - \epsilon$ and $x_2 \rightarrow s(t)+ \epsilon$? We can consider this limit because we can get arbitrarily closer to the shock.

In essence, shocks don't particularly depend on the inhomogenous component, unless that component was something like the Dirac Delta function, which adds some fun to the problem.

gdepaul
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