# derivating v(x, t) := u(x-4t, t) on x and t (partial derivatives)

#### MathNugget

##### New member
context: $u_t(x, t) - 4u_x(x,t) - 4u_{xx}(x, t)=0 \:$and $v(x, t) := u(x-4t, t)$
I want to prove:
$v_t(x,t)-v_{xx}(x,t)=0$
I know I have to go back from v(x,t) to u(x, t). I think it should work with the multivariate extension of the chain rule. so, firstly, going for x:
$\frac{dv}{dx}=\frac{du}{d(x-4t)}\frac{d(x-4t)}{dx}+\frac{du}{dt}\frac{dt}{dx}$The 2nd product is 0, but how do I do the first?
$\frac{du}{d(x-4t)}\frac{d(x-4t)}{dx}=\frac{du}{d(x-4t)}$
Is it as simple as calculating $\frac{dv}{dx}(x, t)=\frac{du}{d(x-4t)}(x-4t)=\frac{du}{dx}(x)?$

by the way, on last line I mean:$\frac{du}{d(x-4t)}(x-4t,t)=\frac{du}{dx}(x,t)$ (since u has 2 variables, not just 1). I'll edit later (if it's possible)

I want to prove:
vt(x,t)−vxx(x,t)=0v_t(x,t)-v_{xx}(x,t)=0vt(x,t)−vxx(x,t)=0

Are you sure it is [imath]v_t - v_{xx}[/imath], not [imath]v_t - 4 v_{xx}[/imath]?

What do you get for [imath]v_x[/imath] and [imath]v_{xx}[/imath] ? And for [imath]v_t[/imath]?

Are you sure it is [imath]v_t - v_{xx}[/imath], not [imath]v_t - 4 v_{xx}[/imath]?

What do you get for [imath]v_x[/imath] and [imath]v_{xx}[/imath] ? And for [imath]v_t[/imath]?
yes, it is $v_t(x, t)-v_{xx}(x,t)=0$so far I tried calculating $v_x=\frac{dv}{dx}$, but I am not sure of the results (they're at the end of my post).
if v(x, t)=u(x-4t, t), is derivating v on x (I don't know what the proper word would be) equivalent to derivating u(x-4t,t) on x-4t?

I don't know how to calculate $v_x, v_{xx}, v_t$, I guess that is the problem.
How am I supposed to derivate $v_x=\frac{dv}{dx}(x, t)=\frac{du}{dx}(x-4t, t) ?$

I don't know how to calculate $v_x, v_{xx}, v_t$, I guess that is the problem.
How am I supposed to derivate $v_x=\frac{dv}{dx}(x, t)=\frac{du}{dx}(x-4t, t) ?$
Is this a class assignment? Which subject are you studying?

I don't know how to calculate $v_x, v_{xx}, v_t$, I guess that is the problem.
How am I supposed to derivate $v_x=\frac{dv}{dx}(x, t)=\frac{du}{dx}(x-4t, t) ?$

By the way, the correct word for the verb form of "derivative" is not "derivate" but "differentiate". It's one of the oddities of mathematical English that I'd love to eliminate, but the usage is well-established!

I think it will help if you define another variable or two, since the source I showed doesn't seem to have an example quite like yours. Naming the intermediate expression makes things easier to follow.

Rather than [imath]v(x,t)=u(x−4t,t)[/imath], try calling it $v(x,t)=u(r,s)\\r=x-4t\\s=t$Then the other given looks like
$u_s(r, s) - 4u_r(r,s) - 4u_{rr}(r, s)=0$
Now apply the chain rule to find [imath]v_x[/imath] and [imath]v_t[/imath].

But also, check whether you either have an extra 4 in the given, or a missing 4 in the goal.

Is this a class assignment? Which subject are you studying?
PDE's (partial derivative equations). You can call it a class assignment...or an exam I already failed once. Trying to get some superficial understanding of things, since I cannot really grasp them on a deeper level.

I think it will help if you define another variable or two, since the source I showed doesn't seem to have an example quite like yours. Naming the intermediate expression makes things easier to follow.

Rather than [imath]v(x,t)=u(x−4t,t)[/imath], try calling it $v(x,t)=u(r,s)\\r=x-4t\\s=t$Then the other given looks like
$u_s(r, s) - 4u_r(r,s) - 4u_{rr}(r, s)=0$
Now apply the chain rule to find [imath]v_x[/imath] and [imath]v_t[/imath].

But also, check whether you either have an extra 4 in the given, or a missing 4 in the goal.
I'll check out the resource. by the way, you were right. It was just 1 [imath]u_{xx}[/imath], not 4 of them. And I cannot edit the initial post...

PDE's (partial derivative equations). You can call it a class assignment...or an exam I already failed once. Trying to get some superficial understanding of things, since I cannot really grasp them on a deeper level.
Practicing the Chain Rule for partial derivatives sounds like a good start. I'd follow up on post #8 by @Dr.Peterson and post the details of your work.

Good luck, and feel free to ask more questions.

I tried that (had some exams so it took me some weeks to get back to this (I failed that exam on first try, so it's time to prep for 2nd try).
with the substitution v(x, t) = u (r, s), r = x-4t, s=t

$v_x=\frac{dv}{dx} = \frac{du}{dr} \frac{dr}{dx} + \frac{du}{ds}\frac{ds}{dx}= u_r * 1 + u_s * 0 = u_r$
and now for dt,
$v_t=\frac{dv}{dt} = \frac{du}{dr} \frac{dr}{dt} + \frac{du}{ds}\frac{ds}{dt}= u_r * (-4) + u_t *1 = -4u_r + u_t?$
for $v_{xx}=\frac{dv}{dx^2} = \frac{du}{dr^2} = u_{rr}?$
looks like, if I substitute in the initial system, this is correct (considering there was a single [imath]u_{xx}[/imath], not 4 of them. maybe we did it ?

I tried that (had some exams so it took me some weeks to get back to this (I failed that exam on first try, so it's time to prep for 2nd try).
with the substitution v(x, t) = u (r, s), r = x-4t, s=t

$v_x=\frac{dv}{dx} = \frac{du}{dr} \frac{dr}{dx} + \frac{du}{ds}\frac{ds}{dx}= u_r * 1 + u_s * 0 = u_r$
and now for dt,
$v_t=\frac{dv}{dt} = \frac{du}{dr} \frac{dr}{dt} + \frac{du}{ds}\frac{ds}{dt}= u_r * (-4) + u_t *1 = -4u_r + u_t?$
for $v_{xx}=\frac{dv}{dx^2} = \frac{du}{dr^2} = u_{rr}?$
looks like, if I substitute in the initial system, this is correct (considering there was a single [imath]u_{xx}[/imath], not 4 of them. maybe we did it ?
Looks good so far, but what is the resulting equation for [imath]v[/imath] ?

A minor gripe: the convention is to use [imath]dx[/imath] for "ordinary" derivatives and [imath]\partial x[/imath] for partial ones, and you need another "square" in the "numerator" too. I.e., your last line should look
$v_{xx} = \frac{\partial^2 v}{\partial x^2} =...$

Looks good so far, but what is the resulting equation for [imath]v[/imath] ?

A minor gripe: the convention is to use [imath]dx[/imath] for "ordinary" derivatives and [imath]\partial x[/imath] for partial ones, and you need another "square" in the "numerator" too. I.e., your last line should look
$v_{xx} = \frac{\partial^2 v}{\partial x^2} =...$
Thank you. Soon-to-be 3rd full year of uni (hopefully getting the paper this year), never really understood when it's d and when it's the other one.

I thought the problem is over, but seems there's some steps left:

[imath]v_t(x, t) - v_{xx}(x, t)=[/imath][imath]u_t (r, t) - 4u_r(r, t) - u_{rr}(r, t)[/imath] and r = x-4t.
Forgot to mention here, t>0 and [imath]x \in \mathbb{R}[/imath], so maybe I could argue [imath]\forall y \in \mathbb{R}, \exists x \in \mathbb{R}[/imath] such that [imath]\forall t > 0[/imath] , y = x-4t, to prove r can be switched with x and solve the problem?

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