#### MathNugget

##### New member

- Joined
- Feb 1, 2024

- Messages
- 36

I want to prove:

[math]v_t(x,t)-v_{xx}(x,t)=0[/math]

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:

[math]\frac{dv}{dx}=\frac{du}{d(x-4t)}\frac{d(x-4t)}{dx}+\frac{du}{dt}\frac{dt}{dx}[/math]The 2nd product is 0, but how do I do the first?

[math]\frac{du}{d(x-4t)}\frac{d(x-4t)}{dx}=\frac{du}{d(x-4t)}[/math]

Is it as simple as calculating [math]\frac{dv}{dx}(x, t)=\frac{du}{d(x-4t)}(x-4t)=\frac{du}{dx}(x)?[/math]