MathNugget
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- Joined
- Feb 1, 2024
- Messages
- 36
context: [math]u_t(x, t) - 4u_x(x,t) - 4u_{xx}(x, t)=0 \:[/math]and [math]v(x, t) := u(x-4t, t)[/math]
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]
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]