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  1. MHB Master
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    evinda's Avatar
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    #11 Thread Author
    A ok. And so
    $$\frac{\partial^2{u_k}}{\partial{\tilde{x_1}}^2}=k \frac{\partial}{\partial{\tilde{x_1}}} (u_{x_1}(k^2 \tilde{t}, k \overline{x_1}, \dots, k \overline{x_n}))=k (u_{x_1 x_1})(k^2 \tilde{t}, k \overline{x_1}, \dots, k \overline{x_n}))=k^2 (u_{x_1 x_1})(k^2 \tilde{t}, k \overline{x_1}, \dots, k \overline{x_n}))$$

    And thus $\Delta u_k=\sum_{i=1}^n \frac{\partial^2 u_k}{\partial{\tilde{x_i}^2}}=k^2 [u_{x_1 x_1} (k^2 \tilde{t}, k \mathbf{x})+ \dots+ u_{x_n x_n}(k^2 \tilde{t}, k \tilde{x})]=k^2 \Delta u (k^2 \tilde{t}, k \tilde{x})$


    Right?

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    I like Serena's Avatar
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    #12
    Yes
    ... if overline is the same as tilde, if we add a missing factor $k$ the third expression, and if we balance the parentheses.

  3. MHB Master
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    #13 Thread Author
    Quote Originally Posted by I like Serena View Post
    Yes
    ... if overline is the same as tilde, if we add a missing factor $k$ the third expression, and if we balance the parentheses.
    Yes, I see... Thanks a lot!!!

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