Wave Equation, stuck on a partial calculation


by Somefantastik
Tags: differentation, integral, partial
Somefantastik
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#1
Aug30-08, 04:21 PM
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Hey everybody,

My professor started our PDE I class in Chapter six, so I am having a hard time with the really basic stuff to get the theory down.

One of my questions to answer is to verify a solution by using direct substitution.

[tex]u(x,t) \ = \ \frac{1}{2}\left[\phi(x+t) \ + \ \phi(x-t) \right] \ + \ \frac{1}{2} \int^{x+t}_{x-t}\Psi(s)ds [/tex]

With initial conditions

[tex] u(x,t_{0}) = \phi(x) \ , \ \frac{\partial u}{\partial t} (x,t_{0}) = \Psi(x), \ and \ t_{0} = 0 [/tex]

satisfies [tex]\frac{\partial^{2}u}{\partial x^{2}} - \frac{\partial^{2}u}{\partial t^{2}} = 0 [/tex]

It was easy for me to plug and chug to show that [tex] u(x,t_{0}) = \phi(x) \ and \ \frac{\partial u}{\partial t} (x,t_{0}) = \Psi(x) [/tex]

Clearly my next step is to find [tex] \frac{\partial^{2}u}{\partial x^{2}} [/tex]

but that's the step on which I'm stuck. Can someone get me started? If someone can show me how to do the second partial w.r.t x, it would be a good exercise for me to figure out the second partial w.r.t t. I apply the chain rule and just get a bunch of garbage back, which means I'm messing it up somewhere.

Any input is appreciated.
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HallsofIvy
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#2
Aug30-08, 04:37 PM
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You need "Leibniz' formula":
[tex]\frac{d}{dx}\int_{\alpha(x)}^{\beta(x)} \phi(x,t)dt= \frac{d\beta}{dx}\phi(x,\beta(x))- \frac{d\alpha}{dx}\phi(x,\alpha(x))+ \int_{\alpha(x)}^{\beta(x)} \frac{\partial \phi}{\partial x}dt[/tex]

It's really just applying the chain rule correctly.
Somefantastik
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Aug31-08, 02:13 PM
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[tex]
\frac{d}{dx}\int_{\alpha(x)}^{\beta(x)} \phi(x,t)dt= \frac{d\beta}{dx}\phi(x,\beta(x))- \frac{d\alpha}{dx}\phi(x,\alpha(x))+ \int_{\alpha(x)}^{\beta(x)} \frac{\partial \phi}{\partial x}dt
[/tex]

I'm having trouble understanding this. It looks like the integrand you gave is a function of 2 variables, but the integrand I have is a function of one variable. I'm not sure what to do with that.

Also:
[tex] \frac{\partial}{\partial x} (\phi(x+t)) = ? [/tex]

is it [tex] = \phi '(x + t), \ or \ \phi_{x}(x+t) \ ? [/tex]

In the notation of a function, how do I write that? I guess I'm having a notational brain fart.

Somefantastik
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#4
Aug31-08, 03:04 PM
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Wave Equation, stuck on a partial calculation


I found this worked out in Walter Strauss's book, so I guess I don't need help anymore. Thanks for looking :)
HallsofIvy
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#5
Sep1-08, 03:29 AM
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Quote Quote by Somefantastik View Post
[tex]
\frac{d}{dx}\int_{\alpha(x)}^{\beta(x)} \phi(x,t)dt= \frac{d\beta}{dx}\phi(x,\beta(x))- \frac{d\alpha}{dx}\phi(x,\alpha(x))+ \int_{\alpha(x)}^{\beta(x)} \frac{\partial \phi}{\partial x}dt
[/tex]

I'm having trouble understanding this. It looks like the integrand you gave is a function of 2 variables, but the integrand I have is a function of one variable. I'm not sure what to do with that.
In that case,
[tex]\frac{\partial\phi}{\partial x}= 0[/tex]
and the formula becomes
[tex] \frac{d}{dx}\int_{\alpha(x)}^{\beta(x)} \phi(x,t)dt= \frac{d\beta}{dx}\phi(x,\beta(x))- \frac{d\alpha}{dx}\phi(x,\alpha(x)) [/tex]

Also:
[tex] \frac{\partial}{\partial x} (\phi(x+t)) = ? [/tex]

is it [tex] = \phi '(x + t), \ or \ \phi_{x}(x+t) \ ? [/tex]

In the notation of a function, how do I write that? I guess I'm having a notational brain fart.
Use the chain rule. If [itex]\phi'(u)[/itex] is the derivative of [itex]\phi[/itex] as a function of the single variable u, then
[tex]\frac{\partial\phi(x+t)}{\partial x}= \phi'(x+t)(1)= \phi'(x+t)[/tex]
Somefantastik
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#6
Sep3-08, 07:38 AM
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HallsofIvy, thanks for helping me.

If anyone finds this post and needs help with taking the derivative of an integral, I found this website that helped as well: http://mathmistakes.info/facts/Calcu...n/doi/doi.html


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