Tough PDE on Separation of Variables

Question

I want to know on how to solve this question :

Given the PDE is : $U_t -tU -txU_x+tU_x=0$

Use separation of variables to find ALL possible solutions

Could someone help me this question because the PDE is too long and very frustrating. It took me 2 hours to try to solve but end up I cant solve it. This my first lesson in PDE. If someone can show me till the final step I would be much much happy and appreciated. I need to get this done because my final exam is coming soon.

Please guys help me out

This question is 25 marks. ( I got this question from past years midterm paper)

Answer 1

The trick with separation of variables is to write the solution $U$ as a product $X(x)T(t)$, so that your equation becomes

$$ X T'-t X T-t x X' T + t X' T = 0$$

Now divide through by $t X T$ so that we completely separate the dependence on $x$ and $t$ from each other:

$$\frac{T'}{t T} - 1 = (x-1) \frac{X'}{X} $$

Note that the left-hand-side only depends on $t$, while the right-hand side only depends on $x$. Thus, they are both constant, so we can set each side equal to some constant, $-\lambda$:

$$\begin{align} \frac{T'}{t T} - 1 &= -\lambda \\ (x-1) \frac{X'}{X} &= -\lambda \\ \end{align} $$

You may now solve each equation separately. I imagine you have initial conditions that will determine each component $X$ and $T$, and will determine $\lambda$.

EDIT

You will now see why I chose $-\lambda$ rather than $\lambda$. The solution of the $t$ equation is straightforward:

$$T(t) = T_0 e^{-(\lambda - 1) t^2/2} $$

The solution to the $x$ equation can be seen by rearranging terms:

$$(x-1)X'(x) + \lambda X(x) = 0$$

which may be rewritten as

$$ [(x-1)^{\lambda} X(x)]' = 0 \implies X(x) = X_0 (x-1)^{-\lambda}$$

Combining these solutions, we get

$$U(x,t) = X(x)T(t) = X_0 T_0 (x-1)^{-\lambda} e^{-(\lambda - 1) t} = K (x-1)^{-\lambda} e^{-(\lambda - 1) t^2/2}$$

You now need to specify some sort of initial conditions to determine $K$ and $\lambda$.

Answer 2

Let $U(x,t)=X(x)T(t)$ ,

Then $X(x)T'(t)-tX(x)T(t)-txX'(x)T(t)+tX'(x)T(t)=0$

$X(x)T'(t)=txX'(x)T(t)-tX'(x)T(t)+tX(x)T(t)$

$X(x)T'(t)=((x-1)X'(x)+X(x))tT(t)$

$\dfrac{(x-1)X'(x)+X(x)}{X(x)}=\dfrac{T'(t)}{tT(t)}=s$

$\begin{cases}\dfrac{(x-1)X'(x)+X(x)}{X(x)}=s\\\dfrac{T'(t)}{tT(t)}=s\end{cases}$

$\begin{cases}\dfrac{X'(x)}{X(x)}=\dfrac{s-1}{x-1}\\\dfrac{T'(t)}{T(t)}=ts\end{cases}$

$\begin{cases}X(x)=c_1(s)(x-1)^{s-1}\\T(t)=c_2(s)e^{\frac{t^2s}{2}}\end{cases}$

$\therefore U(x,t)=\int_s C(s)(x-1)^{s-1}e^{\frac{t^2s}{2}}~ds$ or $\sum\limits_s C(s)(x-1)^{s-1}e^{\frac{t^2s}{2}}$

$=e^{\frac{t^2}{2}}\int_s C(s)(x-1)^{s-1}e^{\frac{t^2(s-1)}{2}}~ds$ or $e^{\frac{t^2}{2}}\sum\limits_s C(s)(x-1)^{s-1}e^{\frac{t^2(s-1)}{2}}$

$=e^{\frac{t^2}{2}}\int_s C(s)\left((x-1)e^{\frac{t^2}{2}}\right)^{s-1}~ds$ or $e^{\frac{t^2}{2}}\sum\limits_s C(s)\left((x-1)e^{\frac{t^2}{2}}\right)^{s-1}$

$=e^{\frac{t^2}{2}}f\left((x-1)e^{\frac{t^2}{2}}+1\right)$

Compare the result with solving this PDE by using the method of characteristics:

$U_t-tU-txU_x+tU_x=0$

$U_t-t(x-1)U_x=tU$

$\dfrac{dt}{ds}=1$ , letting $t(0)=0$ , we have $t=s$

$\dfrac{dx}{ds}=-t(x-1)=-s(x-1)$ , letting $x(0)=x_0$ , we have $x=(x_0-1)e^{-\frac{s^2}{2}}+1=(x_0-1)e^{-\frac{t^2}{2}}+1$

$\dfrac{dU}{ds}=tU=sU$ , letting $U(0)=f(x_0)$ , we have $U(x,t)=f(x_0)e^{\frac{s^2}{2}}=f\left((x-1)e^{\frac{t^2}{2}}+1\right)e^{\frac{t^2}{2}}$