# Thread: Application of chain rule

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?

2. Yes
... if overline is the same as tilde, if we add a missing factor $k$ the third expression, and if we balance the parentheses.

Originally Posted by I like Serena
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!!!