Second partial derivative of d'Alembert's solution

[Somefantastik]

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.

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

With initial conditions

$$u(x,t_{0}) = \phi(x) \ , \ \frac{\partial u}{\partial t} (x,t_{0}) = \Psi(x), \ and \ t_{0} = 0$$

satisfies $$\frac{\partial^{2}u}{\partial x^{2}} - \frac{\partial^{2}u}{\partial t^{2}} = 0$$

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

Clearly my next step is to find $$\frac{\partial^{2}u}{\partial x^{2}}$$

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.

 

[HallsofIvy]

You need "Leibniz' formula":
$$\frac{d}{dx}\int_{\alpha(x)}^{\beta(x)} \phi(x,t)dt= \frac{d\beta}{dx}\phi(x,\beta(x))- \frac{d\alpha}{dx}\phi(x,\alpha(x))+ \int_{\alpha(x)}^{\beta(x)} \frac{\partial \phi}{\partial x}dt$$

It's really just applying the chain rule correctly.

 

[Somefantastik]

$$\frac{d}{dx}\int_{\alpha(x)}^{\beta(x)} \phi(x,t)dt= \frac{d\beta}{dx}\phi(x,\beta(x))- \frac{d\alpha}{dx}\phi(x,\alpha(x))+ \int_{\alpha(x)}^{\beta(x)} \frac{\partial \phi}{\partial x}dt$$

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:
$$\frac{\partial}{\partial x} (\phi(x+t)) = ?$$

is it $$= \phi '(x + t), \ or \ \phi_{x}(x+t) \ ?$$

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

 

[Somefantastik]

I found this worked out in Walter Strauss's book, so I guess I don't need help anymore. Thanks for looking :)

 

[HallsofIvy]

$$\frac{d}{dx}\int_{\alpha(x)}^{\beta(x)} \phi(x,t)dt= \frac{d\beta}{dx}\phi(x,\beta(x))- \frac{d\alpha}{dx}\phi(x,\alpha(x))+ \int_{\alpha(x)}^{\beta(x)} \frac{\partial \phi}{\partial x}dt$$

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,
$$\frac{\partial\phi}{\partial x}= 0$$
and the formula becomes
$$\frac{d}{dx}\int_{\alpha(x)}^{\beta(x)} \phi(x,t)dt= \frac{d\beta}{dx}\phi(x,\beta(x))- \frac{d\alpha}{dx}\phi(x,\alpha(x))$$

Also:
$$\frac{\partial}{\partial x} (\phi(x+t)) = ?$$

is it $$= \phi '(x + t), \ or \ \phi_{x}(x+t) \ ?$$

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 $\phi'(u)$ is the derivative of $\phi$ as a function of the single variable u, then
$$\frac{\partial\phi(x+t)}{\partial x}= \phi'(x+t)(1)= \phi'(x+t)$$

 

[Somefantastik]

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/CalculusFacts/learn/doi/doi.html