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Hi guys I just wanted to check if my logic is correct. The book we are using in class was a bit veague about it and mostly relies on inspection, but I wanted to prove it to myself.

PDE

$$u_{tt}+uu_{xx}=u^2$$

I think this is quasilinear. To prove this we need to show that the linear operator defined by the highest order derivatives is linear ie

$$L[u,\alpha u_{xx},\alpha, u_{tt}]=(\alpha u)_{tt}+u (\alpha u)_{xx}-u^2= \alpha (u_{tt}+uu_{xx}-u^2)=\alpha L[u, u_{xx}, u_{tt}] $$

And the other principal $L[v+u]=L[v]+L[u]$ comes directly from definition of derivative. Does this seem as a good argument?

1 Answers 1

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You cannot extract common factor $\alpha$ but the derivatives part, so the argument runs: it's quasilinear. Everytime I've read about cassification of Pde's it relies on "inspection". You have indeed:

$$L[\alpha u_{xx},\alpha, u_{tt}]=f(u)$$