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yungman
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I don't have the answer of these question. Can someone take a look at a) and tell me am I correct? I don't even know how to solve b)
a)
a) Show [tex]u(x,t)=F(x+ct) + G(x-ct)[/tex] is solution of [tex]\frac{\partial^2 u}{\partial t^2}= c^2\frac{\partial^2 u}{\partial x^2}[/tex]
let [tex]\alpha = x+ct,\beta = x-ct \Rightarrow \frac{\partial \alpha}{\partial x}= 1, \frac{\partial \alpha}{\partial t}= c,[/tex] and also [tex] \frac{\partial \beta}{\partial x}= 1, \frac{\partial \beta}{\partial t}= -c[/tex]
[tex]\frac{\partial u}{\partial x} = \frac{\partial F(\alpha)}{\partial \alpha} \frac{\partial \alpha}{\partial x} + \frac{\partial G(\beta)}{\partial \beta}\frac{\partial \beta}{\partial x} = \frac{\partial F(\alpha)}{\partial \alpha} + \frac{\partial G(\beta)}{\partial \beta}[/tex]
[tex]\frac{\partial^2 u}{\partial x^2} = \frac {\partial (\frac{\partial F(\alpha)} {\partial \alpha}) }{\partial \alpha} \frac{\partial \alpha}{\partial x} +
\frac {\partial (\frac{\partial G(\beta)}{\partial \beta })} {\partial \beta} \frac{\partial \beta}{\partial x}
= \frac{\partial^2 F(\alpha)}{\partial \alpha^2} + \frac{\partial^2 G(\beta)}{\partial \beta^2}[/tex]
[tex]\frac{\partial u}{\partial t} = \frac{\partial F(\alpha)}{\partial \alpha}\frac{\partial \alpha}{\partial t} + \frac{\partial G(\beta)}{\partial \beta}\frac{\partial \beta}{\partial t} = c[\frac{\partial F(\alpha)}{\partial \alpha} - \frac{\partial G(\beta)}{\partial \beta}][/tex]
[tex]\frac{\partial^2 u}{\partial t^2} = c\frac {\partial (\frac{\partial F(\alpha)} {\partial \alpha}) }{\partial \alpha} \frac{\partial \alpha}{\partial t} -
c\frac {\partial (\frac{\partial G(\beta)}{\partial \beta })} {\partial \beta} \frac{\partial \beta}{\partial t}
= c^2[\frac{\partial^2 F(\alpha)}{\partial \alpha^2} + \frac{\partial^2 G(\beta)}{\partial \beta^2}][/tex]
[tex]\Rightarrow \frac{\partial^2 u}{\partial t^2}= c^2\frac{\partial^2 u}{\partial x^2}[/tex]
Therefore:
[tex]u(x,t)=F(x+ct) + G(x-ct)[/tex] is solution of [tex]\frac{\partial^2 u}{\partial t^2}= c^2\frac{\partial^2 u}{\partial x^2}[/tex]
b) Transform [tex]\frac{\partial^2 u}{\partial t^2}= c^2\frac{\partial^2 u}{\partial x^2}[/tex] into [tex]\frac{\partial^2 u}{\partial \alpha \partial \beta}=0[/tex]
where [tex]u(x,t)=F(x+ct) + G(x-ct)[/tex]
and [tex]\alpha = x+ct,\beta = x-ct[/tex]
Can someone give me a hint how to go by this?
a)
Homework Statement
a) Show [tex]u(x,t)=F(x+ct) + G(x-ct)[/tex] is solution of [tex]\frac{\partial^2 u}{\partial t^2}= c^2\frac{\partial^2 u}{\partial x^2}[/tex]
The Attempt at a Solution
let [tex]\alpha = x+ct,\beta = x-ct \Rightarrow \frac{\partial \alpha}{\partial x}= 1, \frac{\partial \alpha}{\partial t}= c,[/tex] and also [tex] \frac{\partial \beta}{\partial x}= 1, \frac{\partial \beta}{\partial t}= -c[/tex]
[tex]\frac{\partial u}{\partial x} = \frac{\partial F(\alpha)}{\partial \alpha} \frac{\partial \alpha}{\partial x} + \frac{\partial G(\beta)}{\partial \beta}\frac{\partial \beta}{\partial x} = \frac{\partial F(\alpha)}{\partial \alpha} + \frac{\partial G(\beta)}{\partial \beta}[/tex]
[tex]\frac{\partial^2 u}{\partial x^2} = \frac {\partial (\frac{\partial F(\alpha)} {\partial \alpha}) }{\partial \alpha} \frac{\partial \alpha}{\partial x} +
\frac {\partial (\frac{\partial G(\beta)}{\partial \beta })} {\partial \beta} \frac{\partial \beta}{\partial x}
= \frac{\partial^2 F(\alpha)}{\partial \alpha^2} + \frac{\partial^2 G(\beta)}{\partial \beta^2}[/tex]
[tex]\frac{\partial u}{\partial t} = \frac{\partial F(\alpha)}{\partial \alpha}\frac{\partial \alpha}{\partial t} + \frac{\partial G(\beta)}{\partial \beta}\frac{\partial \beta}{\partial t} = c[\frac{\partial F(\alpha)}{\partial \alpha} - \frac{\partial G(\beta)}{\partial \beta}][/tex]
[tex]\frac{\partial^2 u}{\partial t^2} = c\frac {\partial (\frac{\partial F(\alpha)} {\partial \alpha}) }{\partial \alpha} \frac{\partial \alpha}{\partial t} -
c\frac {\partial (\frac{\partial G(\beta)}{\partial \beta })} {\partial \beta} \frac{\partial \beta}{\partial t}
= c^2[\frac{\partial^2 F(\alpha)}{\partial \alpha^2} + \frac{\partial^2 G(\beta)}{\partial \beta^2}][/tex]
[tex]\Rightarrow \frac{\partial^2 u}{\partial t^2}= c^2\frac{\partial^2 u}{\partial x^2}[/tex]
Therefore:
[tex]u(x,t)=F(x+ct) + G(x-ct)[/tex] is solution of [tex]\frac{\partial^2 u}{\partial t^2}= c^2\frac{\partial^2 u}{\partial x^2}[/tex]
b) Transform [tex]\frac{\partial^2 u}{\partial t^2}= c^2\frac{\partial^2 u}{\partial x^2}[/tex] into [tex]\frac{\partial^2 u}{\partial \alpha \partial \beta}=0[/tex]
where [tex]u(x,t)=F(x+ct) + G(x-ct)[/tex]
and [tex]\alpha = x+ct,\beta = x-ct[/tex]
Can someone give me a hint how to go by this?
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