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Is $w_x(L,t)w_t(L,t) - w_x(0,t)w_t(0,t) $ equal to zero or nonpositive assuming any of the following?

$$w_{tt} (x,t) = w_{xx} (x,t), \ x \in (0,L), \ t \in (0,T) \ (1)$$

$$w_{tx} (x,t) = w_{xt} (x,t), \ x \in (0,L), \ t \in (0,T) \ (2)$$

$$w(x,0) = 0, x \in [0,L] \ (3)$$

$$w_x(0,t) = w(0,t), t \in [0,T] \ (4)$$

$$w_x(L,t) = -w(L,t), t \in [0,T] \ (5)$$


What I tried for nonpositive:

$$w_x(L,t)w_t(L,t) - w_x(0,t)w_t(0,t)$$

$$= -w(L,t)w_t(L,t) - w(0,t)w_t(0,t)$$

Can I say that is nonpositive though? Are w and its derivatives nonnegative?


What I tried for zero:

From (4) and (5), $$w_{xt}(0,t) - w_t(0,t) = w_{xt}(L,t) + w_t(L,t)$$

Note that $$\frac{\partial}{\partial x} w_x(x,t) w_t(x,t) - w_{tt}(x,t) w_t(x,t) = w_x(x,t) w_{xt}(x,t)$$

$$\to \frac{\partial}{\partial x} w_x(L,t) w_t(L,t) - w_{tt}(L,t)w_t(L,t) = w_x(L,t)w_{xt}(L,t)$$

$$\to - w_{tt}(L,t)w_t(L,t) = w_x(L,t)w_{xt}(L,t)$$

$$\to - w_{tt}(L,t)w_t(L,t) = w_x(L,t)w_{xt}(L,t)$$

$$\to - w_{xx}(L,t)w_t(L,t) = w_x(L,t)w_{xt}(L,t)$$

$$\to - w_{xx}(L,t)w_t(L,t) = w_x(L,t)w_t(L,t)$$

$$\to - w_{xx}(L,t) = w_x(L,t) (= -w(L,t)) \ or \ w_t(L,t) = 0$$

If $w_t(L,t) \ne 0$, can I guarantee that $w_x(L,t) = 0$?


Similarly, I get

$$\to - w_{xx}(0,t)w_t(0,t) = w_x(0,t)w_{xt}(0,t)$$

$$\to - w_{xx}(0,t)w_t(0,t) = w_x(0,t)w_t(0,t)$$

$$\to - w_{xx}(0,t) = w_x(0,t) (= w(0,t)) \ or \ w_t(0,t) = 0$$

If $w_t(0,t) \ne 0$, can I guarantee that $w_x(0,t) = 0$?

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