Using Partial Derivatives To Prove Solution To Wave Equation

[Lancelot59]

I need to use partial derivatives to prove that
u(x,t)=f(x+at)+g(x-at)
is a solution to:
u_{tt}=a^{2}u_{xx}

I'm stuck on how I'm supposed to approach the problem. I'm lost as to what order I should do the derivations in. I tried making a tree diagram, and I came out like this. The arrow indicates what's under that variable.

u --> f--g
f--> x--t
g-->y--t

Earlier I asked my prof about the concept before I got to this problem and made up a random example using the following functions:

f(z)
z=x+y
x=t+\lambda
y=\lambda

So to find the partial derivative of f with respect to lambda the tree wound up looking like this:

f-->z
z-->x--y
x-->t--lambda
y-->lambda

With the final setup being this:
\frac{df}{dz}(\frac{\partial z}{\partial x}\frac{\partial x}{\partial \lambda}+\frac{\partial z}{\partial y}\frac{dy}{d\lambda})

Which I understand. You take a full derivative of f, which is the derivative of z. To get each part then you need to take partial derivatives of the x and y functions, and since x is dependent on t and lambda you need to partially derive it to lambda, and just take a full derivative of y.

Here however, I can't see how to apply this to this particular problem.

 

[Fredrik]

I need to use partial derivatives to prove that
u(x,t)=f(x+at)+g(x-at)
is a solution to:
u_{tt}=a^{2}u_{xx}

I'm stuck on how I'm supposed to approach the problem. I'm lost as to what order I should do the derivations in.

The partial derivative you write as u_{t}(x,t) is by definition the ordinary derivative of the function s\mapsto u(x,s) at t. In other words, if you define F by F(s)=u(x,s)=f(x+as)+g(x-as), you have u_t(x,t)=F'(t). Do you know how to take the (ordinary) derivative of F?

Reminder: the chain rule is (f\circ g)'(x)=f'(g(x))g'(x).Edit: By the way, I know that students of a course at this level often have a hard time understanding stuff like what I just said. I think that indicates that something is fundamentally wrong with how the concepts "function" and "derivative" are taught. Teachers seem to find a way to never say the things the students really need spelled out for them. You should never think of an expression like f(x) as a function. f is the function, and f(x) is a number in its range, the value of f at x. f' is another function, the derivative of f. f'(x) is a number, the value of f' at x. The functions f and f' can also be written as t\mapsto f(t) and t\mapsto f'(t) respectively. Here I've changed the variable name just to emphasize that the function is the same no matter what variable symbol is used. The expressions t\mapsto f(t) and x\mapsto f(x) are different expressions (different strings of text), but they represent the same thing, the function f. However, the expressions f(t) and f(x) represent different things, respectively the value of f at t and the value of f at x.

 

[Lancelot59]

The partial derivative you write as u_{t}(x,t) is by definition the ordinary derivative of the function s\mapsto u(x,s) at t. In other words, if you define F by F(s)=u(x,s)=f(x+as)+g(x-as), you have u_t(x,t)=F'(t). Do you know how to take the (ordinary) derivative of F?

Reminder: the chain rule is (f\circ g)'(x)=f'(g(x))g'(x).


Edit: By the way, I know that students of a course at this level often have a hard time understanding stuff like what I just said. I think that indicates that something is fundamentally wrong with how the concepts "function" and "derivative" are taught. Teachers seem to find a way to never say the things the students really need spelled out for them. You should never think of an expression like f(x) as a function. f is the function, and f(x) is a number in its range, the value of f at x. f' is another function, the derivative of f. f'(x) is a number, the value of f' at x. The functions f and f' can also be written as t\mapsto f(t) and t\mapsto f'(t) respectively. Here I've changed the variable name just to emphasize that the function is the same no matter what variable symbol is used. The expressions t\mapsto f(t) and x\mapsto f(x) are different expressions (different strings of text), but they represent the same thing, the function f. However, the expressions f(t) and f(x) represent different things, respectively the value of f at t and the value of f at x.

Yes, I didn't really follow that all too well. I don't see how u(x,t) (or u(x,s) in your example) is the total derivative of an arbitrary function F(t) (or F(s)).

 

[Fredrik]

Yes, I didn't really follow that all too well. I don't see how u(x,t) (or u(x,s) in your example) is the total derivative of an arbitrary function F(t) (or F(s)).

Apparently you also didn't see that I wrote that an expression like F(t) shouldn't be called a "function". ... F is a function and F(t) is a number.I didn't say that u is a total derivative of an arbitrary function. I did however say something that suggests that every partial derivative is an ordinary derivative. That was misleading, because the partial derivatives of a function from \mathbb R^2 into \mathbb R are functions from \mathbb R^2 into \mathbb R, while the ordinary derivative of a function from \mathbb R into \mathbb R is a function from \mathbb R into \mathbb R. So it isn't possible for a partial derivative to be equal to an ordinary derivative. I'll try to express myself more accurately this time.I think the notation u_t is causing confusion, so I'm going to change it. That expression represents the partial derivative with respect to the second variable of the function u, so it makes more sense to use a notation that doesn't mention a completely irrelevant variable name. I will therefore write D_2u instead of u_t. (D for "derivative" and 2 to remind us that it's the partial derivative with respect to the second variable).Let (x,t) be an arbitrary pair of numbers in the domain of D_2u. You need to find D_2u(x,t). This expression represents a number, not a function. It's a different number for each value of x and t, but that doesn't mean that the expression represents a function. D_2u is the function and D_2u(x,t) is its value at (x,t).D_2u is defined by

D_2u(a,b)=\lim_{h\rightarrow 0}\frac{u(a,b+h)-u(a,b)}{h}

for all real numbers a,b in the domain of u such that this limit exists. Now define F:\mathbb R\rightarrow\mathbb R by F(s)=u(x,s) for all s such that (x,s) is in the domain of u. These definitions tell us that

D_2u(x,t)=\lim_{h\rightarrow 0}\frac{u(x,t+h)-u(x,t)}{h}=\lim_{h\rightarrow 0}\frac{F(t+h)-F(t)}{h}=F'(t)

So to find D_2u(x,t), you just need to calculate F'(t).What you really want to find is what you wrote as u_{tt}(x,t), which in this notation is

D_2D_2u(x,t)=F''(t)

This gives you the left-hand side of the equation, and a very similar calculation gives you the right-hand side.

 

[jackmell]

May I suggest a more elementary approach since I feel he may be having some problems with this. Why not just let:

z=x+at

w=x-at

Then:

u(x,t)=f(z)+g(w)

Now:

\begin{aligned}<br /> \frac{\partial u}{\partial x}&amp;=\frac{\partial f}{\partial z}\frac{\partial z}{\partial x}+\frac{\partial g}{\partial w}\frac{\partial w}{\partial x}\\<br /> &amp;=\frac{\partial f}{\partial z}+\frac{\partial g}{\partial w}<br /> \end{aligned}<br />


\begin{aligned}<br /> \frac{\partial^2 u}{\partial x^2}&amp;=\frac{\partial}{\partial x}\left(\frac{\partial f}{\partial z}\right)+\frac{\partial}{\partial x}\left(\frac{\partial g}{\partial w}\right) \\<br /> &amp;=\frac{\partial^2 f}{\partial z^2}\frac{\partial z}{\partial x}+\frac{\partial^2 g}{\partial w^2}\frac{\partial w}{\partial x} \\<br /> &amp;=\frac{\partial^2 f}{\partial z^2}+\frac{\partial^2 g}{\partial w^2}<br /> \end{aligned}<br />

Alright, I did x. I bet you can do t now, get them in terms of w and z, back-substitute into the PDE and show that when you put everything on the left, it's zero.

 

[Fredrik]

\begin{aligned}<br /> \frac{\partial u}{\partial x}&amp;=\frac{\partial f}{\partial z}\frac{\partial z}{\partial x}+\frac{\partial g}{\partial w}\frac{\partial w}{\partial x}\\<br /> &amp;=\frac{\partial f}{\partial z}+\frac{\partial g}{\partial w}<br /> \end{aligned}<br />

You should probably use regular d's for the ordinary derivatives. (Edit: Or write f(z,w) instead of f(z)). Other than that, I don't have any objections.

\frac{\partial u}{\partial x}=\frac{df}{dz}\frac{\partial z}{\partial x}+\frac{dg}{dw}\frac{\partial w}{\partial x}=\frac{df}{dz}+\frac{dg}{dw}

You can also do this without defining w and z.

\frac{\partial u(x,t)}{\partial x}=\frac{d}{dx}u(x,t)=\frac{d}{dx}f(x+at)+\frac{d}{dx}g(x-at)

=f&#039;(x+at)\frac{d}{dx}(x+at)+g&#039;(x-at)\frac{d}{dx}(x-at)=f&#039;(x+at)+g&#039;(x-at)

Lancelot59: Here you have two easy-to-use recipes for how to do this, but it's still definitely worth the effort to try to understand the stuff in my previous post.

By the way, do you understand the first equality in my calculation in this post? (The first equality of the second calculation).

 

[Lancelot59]

You should probably use regular d's for the ordinary derivatives. (Edit: Or write f(z,w) instead of f(z)). Other than that, I don't have any objections.

\frac{\partial u}{\partial x}=\frac{df}{dz}\frac{\partial z}{\partial x}+\frac{dg}{dw}\frac{\partial w}{\partial x}=\frac{df}{dz}+\frac{dg}{dw}

You can also do this without defining w and z.

\frac{\partial u(x,t)}{\partial x}=\frac{d}{dx}u(x,t)=\frac{d}{dx}f(x+at)+\frac{d}{dx}g(x-at)

=f&#039;(x+at)\frac{d}{dx}(x+at)+g&#039;(x-at)\frac{d}{dx}(x-at)=f&#039;(x+at)+g&#039;(x-at)

Lancelot59: Here you have two easy-to-use recipes for how to do this, but it's still definitely worth the effort to try to understand the stuff in my previous post.

By the way, do you understand the first equality in my calculation in this post? (The first equality of the second calculation).

I sat down with my prof, and got the problem sorted out. I also got him to look over your post and I have a fairly good idea of what you were saying there. Thanks for your help.