Tough PDE on Separation of Variables

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.

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

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I see that this is your 3rd question being asked about PDEs, and I am glad you are using this forum to gain better understanding. What you may want to do, however, is to accept answers that you choose to use. Doing this benefits the community and will encourage folks to respond to your questions. –  Ron Gordon Jan 26 '13 at 13:04

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 $\lambda$.

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Separation of variable could also be applied to sum of functions, am i right? So to get "all solutions" via separation of variables, he needs to do sum as well. –  mez Jan 26 '13 at 12:45
@Mezhang: what sum of functions? You are assuming that there is some boundary conditions, as we would see in a 2nd order equation. This is of first order, and without initial conditions, I do not know what the specific solution will be. –  Ron Gordon Jan 26 '13 at 12:47
U(x,t) = X(x)+T(t) –  mez Jan 26 '13 at 14:00
@Mezhang: please explain to me how that substitution separates the variables. –  Ron Gordon Jan 26 '13 at 14:35
U(x,t) = X(x)+T(t), $U_t -tU -txU_x+tU_x=0\;\Longrightarrow\;T'-tX+tT-txX'+tX'=0$, then write in the form of $a\frac{T'}{T}=\frac{X'}{X}$, then both have to be constant. Please correct me where I am wrong. –  mez Jan 26 '13 at 15:20

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 $U(x,t)=f(x_0)e^{\frac{s^2}{2}}=f\left((x-1)e^{\frac{t^2}{2}}+1\right)e^{\frac{t^2}{2}}$

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$(-1)$: sorry, but 1) This is not useful to OP, 2) it is far from clear what you are doing with superposition that properly stated initial conditions won't do; we typically superpose, for instance, in an angular spectrum type solution to the wave equation. But there, we know what the parameter being superpower means. You don't even know whether to sum or integrate. Also, 3) although it was a neat idea to check with MOC, it was a poor idea to present to OP, who is a beginner. –  Ron Gordon Jan 26 '13 at 19:51
"Superposed", tough typing on an iPad. –  Ron Gordon Jan 26 '13 at 20:03