- #1

mathmari

Gold Member

MHB

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Hey!

Let $u(x,t), A(x)$ be functions, for which holds the following:

We have the pde $u_t+a(u)u_x=0$. Let $A'(u)=a(u)$ then the pde can be written as $u_t+A(u)_x=0$. We have the following integrals $$\int_{a-\epsilon}^au\cdot \left (\frac{x-a}{\epsilon}+1\right )\, dx+\int_a^budx+\int_b^{b+\epsilon}u\cdot \left (1-\frac{x-b}{\epsilon}\right )\, dx$$ and $$\frac{1}{\epsilon}\int_{a-\epsilon}^aA(u)\, dx-\frac{1}{\epsilon}\int_b^{b+\epsilon}A(u) \, dx$$ What do we get if we take the limit $\epsilon\rightarrow 0$ ? From the boundary th elimit goes to $0$ but the integrands go to $\infty$ or not? Or is the limit maybe something like the definition of the derivative? (Wondering)

Let $u(x,t), A(x)$ be functions, for which holds the following:

We have the pde $u_t+a(u)u_x=0$. Let $A'(u)=a(u)$ then the pde can be written as $u_t+A(u)_x=0$. We have the following integrals $$\int_{a-\epsilon}^au\cdot \left (\frac{x-a}{\epsilon}+1\right )\, dx+\int_a^budx+\int_b^{b+\epsilon}u\cdot \left (1-\frac{x-b}{\epsilon}\right )\, dx$$ and $$\frac{1}{\epsilon}\int_{a-\epsilon}^aA(u)\, dx-\frac{1}{\epsilon}\int_b^{b+\epsilon}A(u) \, dx$$ What do we get if we take the limit $\epsilon\rightarrow 0$ ? From the boundary th elimit goes to $0$ but the integrands go to $\infty$ or not? Or is the limit maybe something like the definition of the derivative? (Wondering)

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