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$u_{tt} -c^2u_{xx}=F(x,t) $ , $ x>0$

$u(x,0)=f(x)$ , $ x\ge 0$

$u_t(0,t)=g(x) $ , $ x\ge0$

$u_x(0,t)=0 $ , $ t\ge0$

I did an even expansion to solve it for $-\infty<x<\infty$ I get that the solution for my problem is even with the variable $x$.

Now my question is why **$u_x(0,t)=0$ is true given that the solution $u(x,t)$ is even with x ?**

Michael Hardy
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Sijaan Hallak
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1 Answers1

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If $u(x,t)$ is even and differentiable in $x$ we have: $$ u_x(0,t) = \lim_{h \downarrow 0} \frac{u(h,t)-u(0,t)}{h} = -\lim_{h \downarrow 0} \frac{u(-h,t)-u(0,t)}{-h} =$$ $$ - \lim_{h \uparrow 0} \frac{u(h,t)-u(0,t)}{h} = - u_x(0,t)$$ So $u_x(0,t) = -u_x(0,t)$, thus $u_x(0,t) = 0$.

abcdef
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