Problem 11.26 Challenge

a) จาก \(y = sin(x)cos(vt)\)

พิจารณา \(\frac{\partial y}{\partial t}= -vsin(x)\sin (vt)\)

\(\frac{\partial ^{2}y}{\partial t^{2}}= -v^{2}\sin (x)\cos (vt)\)

\(\frac{1}{v^{2}}\frac{\partial ^{2}y}{\partial t^{2}}= -\sin (x)\cos (vt)\cdots \cdots (1)\)

พิจารณา \(\frac{\partial y}{\partial x}= \cos (x)\cos (vt)\)

\(\frac{\partial ^{2}y}{\partial x^{2}}= -\sin (x)\cos (vt)\cdots \cdots (2)\)

\(\therefore (2) = (1)\)

\(\frac{\partial ^{2}y}{\partial x^{2}}=\frac{1}{v^{2}}\frac{\partial ^{2}y}{\partial t^{2}}\;\blacksquare\)

b)พิจารณา\(\frac{1}{2}\sin (x+vt)=\frac{1}{2}\sin (x)\cos (vt)+\frac{1}{2}\sin (vt)\cos (x)\)

\(\sin (x-vt)=\frac{1}{2}\frac{1}{2}\sin (x)\cos (vt)-\frac{1}{2}\sin (vt)\cos (x)\)

\(\sin (x+vt)+\frac{1}{2}\frac{1}{2}\sin (x-vt)=\sin (x)\cos (vt)\)

\(\therefore f(x+vt)=\frac{1}{2}sin(x+vt)\) และ\(g(x-vt)=\frac{1}{2}\sin (x-vt)\;\blacksquare\)