Show $v(x,\tau)=e^{\frac{1}{2}(K-1)x-\frac{1}{4}(k+1)^2\tau}u(x,\tau)$

Question

Given $v(x,\tau)=e^{\alpha x+\beta\tau}u(x,\tau)$ where $\alpha,\beta$ are some constants.

The aim is to prove

$v(x,\tau)=e^{\frac{1}{2}(K-1)x-\frac{1}{4}(k+1)^2\tau}u(x,\tau)$

where $\frac{\partial u}{\partial \tau}=\frac{\partial^2u}{\partial x^2}$ for $-\infty < x < \infty$, $\tau >0$.

with $u(x,0)=max(e^{\frac{1}{2}(k+1)x}-e^{\frac{1}{2}(k-1)x},0)$

and $\alpha=-\frac{1}{2}(k-1),\beta=-\frac{1}{4}(k+1)^2$.

Actually, this one is the next question of:

Use change of variables to reduce a PDE to $\frac{\partial v}{\partial r}=\frac{\partial^2v}{\partial x^2}+(k-1)\frac{\partial v}{\partial x}-kv$

So I just have trouble here and do not quite know if the question linked above is enlightening (someone point out if it does?)

Thank you.

Answer 1

In general, you probably should give your questions a little more background. What kind of PDE where you trying to solve initially?

Anyways, following my hint we get the following:

$$u_t = \exp{\left[-(\alpha+\beta t)\right]}(v_t - \beta v)$$ $$u_{xx} = \exp{\left[-(\alpha+\beta t)\right]}(\alpha^2 v - 2 \alpha v_x + v_xx)$$

All together this means that, since $u$ satisfies the heat equation,

$$v_t = v_{xx} - 2\alpha v_x + (\beta + \alpha^2)v = 0$$

Now, just use your previous question since $v$ satisfies that equation, this means that

$-2 \alpha = k - 1$ and $\beta + \alpha^2 = -k$, which leads to your conclusion.