MathNugget
New member
- Joined
- Feb 1, 2024
- Messages
- 36
Trying to solve this Cauchy problem:
[math]v_t (x, t) - v_{xx} (x, t) = 0 , x \in \mathbb{R}, t>0[/math][math]v (x, 0) = sin^2 (3x), x \in \mathbb{R}[/math].
I have seen a somewhat similar problem, solved by separating the variables. Meaning [imath]v(x, t) = A(x) B(t)[/imath].
Then I get [imath]v (x, 0) = A(x) B(0) = sin^2(3x)[/imath] so [imath]A(x) = \frac{sin^2(3x)}{B(0)}[/imath].
so then I calculate [imath]v_t, v_{xx}[/imath], plug them in first equation (and there probably is a problem here when I differentiate, or the method cannot be applied here):
[math]v_x(x, t)= \frac{6sin(3x)cos(3x)}{B(0)} = \frac{3sin(6x)}{B(0)}[/math][math]v_{xx} (x, t)= \frac{18cos(6x)}{B(0)}[/math]
And [imath]\frac{sin^2(3x)}{B(0)}B'(t) - \frac{18cos(6x)}{B(0)} B(t) = 0[/imath]
I suppose it's time to switch [imath]sin^2(3x)[/imath] or [imath]18cos(6x)[/imath] so that they're in the same thing. Let's take the first...
[math]sin^2(3x) = \frac{1-cos(6x)}{2}.[/math][math](1-cos(6x))B'(t) - 36cos(6x)B(t)= 0[/math]
From what I understood, I have to get rid of x for this to make sense. But I don't think I am getting there...
[math]v_t (x, t) - v_{xx} (x, t) = 0 , x \in \mathbb{R}, t>0[/math][math]v (x, 0) = sin^2 (3x), x \in \mathbb{R}[/math].
I have seen a somewhat similar problem, solved by separating the variables. Meaning [imath]v(x, t) = A(x) B(t)[/imath].
Then I get [imath]v (x, 0) = A(x) B(0) = sin^2(3x)[/imath] so [imath]A(x) = \frac{sin^2(3x)}{B(0)}[/imath].
so then I calculate [imath]v_t, v_{xx}[/imath], plug them in first equation (and there probably is a problem here when I differentiate, or the method cannot be applied here):
[math]v_x(x, t)= \frac{6sin(3x)cos(3x)}{B(0)} = \frac{3sin(6x)}{B(0)}[/math][math]v_{xx} (x, t)= \frac{18cos(6x)}{B(0)}[/math]
And [imath]\frac{sin^2(3x)}{B(0)}B'(t) - \frac{18cos(6x)}{B(0)} B(t) = 0[/imath]
I suppose it's time to switch [imath]sin^2(3x)[/imath] or [imath]18cos(6x)[/imath] so that they're in the same thing. Let's take the first...
[math]sin^2(3x) = \frac{1-cos(6x)}{2}.[/math][math](1-cos(6x))B'(t) - 36cos(6x)B(t)= 0[/math]
From what I understood, I have to get rid of x for this to make sense. But I don't think I am getting there...