Show that $\phi(x, t) = e^{-1(x - \alpha t)} + e^{b(x + b t)}$ is a solution of $u_t - u_{xx} = 0$, where $a$ and $b$ are positive numbers.

Question

I am trying to show that

$$\phi(x, t) = e^{-1(x - \alpha t)} + e^{b(x + b t)}$$

is a solution of

$$u_t - u_{xx} = 0$$

where $a$ and $b$ are positive numbers.

I'm thinking that this means that $u$ is a function of $\phi$, so that we have $u(\phi)$.

I got

$$u_t = u_{\phi}(ae^{-(x - at)} + b^2e^{b(x + bt)})$$

$$u_{xx} = \frac{\partial^2{u}}{\partial{\phi}^2} \left( \frac{\partial{u}}{\partial{x}} \right)^2 + \frac{\partial{u}}{\partial{\phi}} \frac{\partial^2{\phi}}{\partial{x^2}} = u_{\phi \phi}(-e^{-(x - at)} + be^{b(x + bt)})^2 + u_{\phi}(e^{-(x - at)} + b^2e^{b(x + bt)})$$

So

$$u_t - u_{xx} = u_{\phi}(ae^{-(x - at)} + b^2e^{b(x + bt)}) - u_{\phi \phi}(-e^{-(x - at)} + be^{b(x + bt)})^2 - u_{\phi}(e^{-(x - at)} + b^2e^{b(x + bt)})$$

Some of the terms cancel themselves out and leave us with

$$u_t - u_{xx} = - u_{\phi \phi}(-e^{-(x - at)} + be^{b(x + bt)})^2 $$

But we cannot have $-e^{-(x - at)} + be^{b(x + bt)} = 0$ for any positive $a, b$? We obviously require $b = 1$ for this case, but there is no $a$ for which this equals $0$? Have I made an error some where? I've looked over my work, but I cannot find any errors.

Answer 1

$\phi(x, t) = e^{-a(x - a t)} + e^{b(x + b t)}$

$\phi_t(x, t) = a^2e^{-a(x - a t)} + b^2e^{b(x + b t)}$

$\phi_x(x, t) = -ae^{-a(x - a t)} - be^{b(x + b t)}$

$\phi_{xx}(x, t) = a^2e^{-a(x - a t)} + b^2e^{b(x + b t)} =\phi_t(x, t) $