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How can I solve the following partial differential equation analytically ?

$\dfrac{\partial T}{\partial t}+u\dfrac{\partial T}{\partial x}=\alpha\dfrac{\partial^2T}{\partial x^2}$

where $u$ and $\alpha$ both are constants.

Thanks in advance.

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  • $\begingroup$ How to start ? Am I supposed to integrate the both sides of the equation directly ? $\endgroup$
    – hellomecha
    Mar 25, 2013 at 19:05

2 Answers 2

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Case $1$: $u=0$ and $\alpha=0$

Then $\dfrac{\partial T}{\partial t}=0$

$T(x,t)=C(x)$

Case $2$: $u\neq0$ and $\alpha=0$

Then $\dfrac{\partial T}{\partial t}+u\dfrac{\partial T}{\partial x}=0$

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

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

$\dfrac{dx}{ds}=u$ , letting $x(0)=x_0$ , we have $x=us+x_0=ut+x_0$

$\dfrac{dT}{ds}=0$ , letting $T(0)=C(x_0)$ , we have $T(x,t)=C(x_0)=C(x-ut)$

Case $3$: $\alpha\neq0$

Then $\dfrac{\partial T}{\partial t}+u\dfrac{\partial T}{\partial x}=\alpha\dfrac{\partial^2T}{\partial x^2}$

Note that this PDE is separable.

Case $3$a: $\text{Re}(\alpha t)\geq0$

Let $T(x,t)=F(x)G(t)$ ,

Then $F(x)G'(t)+uF'(x)G(t)=\alpha F''(x)G(t)$

$F(x)G'(t)=\alpha F''(x)G(t)-uF'(x)G(t)$

$F(x)G'(t)=(\alpha F''(x)-uF'(x))G(t)$

$\dfrac{G'(t)}{G(t)}=\dfrac{\alpha F''(x)-uF'(x)}{F(x)}=-\dfrac{4\alpha^2(f(s))^2+u^2}{4\alpha}$

$\begin{cases}\dfrac{G'(t)}{G(t)}=-\dfrac{4\alpha^2(f(s))^2+u^2}{4\alpha}\\\alpha F''(x)-uF'(x)+\dfrac{4\alpha^2(f(s))^2+u^2}{4\alpha}F(x)=0\end{cases}$

$\begin{cases}G(t)=c_3(s)e^{-\frac{t(4\alpha^2(f(s))^2+u^2)}{4\alpha}}\\F(x)=\begin{cases}c_1(s)e^{\frac{ux}{2\alpha}}\sin(xf(s))+c_2(s)e^{\frac{ux}{2\alpha}}\cos(xf(s))&\text{when}~f(s)\neq0\\c_1xe^{\frac{ux}{2\alpha}}+c_2e^{\frac{ux}{2\alpha}}&\text{when}~f(s)=0\end{cases}\end{cases}$

$\therefore T(x,t)=C_1xe^{\frac{2ux-u^2t}{4\alpha}}+C_2e^{\frac{2ux-u^2t}{4\alpha}}+\int_sC_3(s)e^{\frac{2ux-t(4\alpha^2(f(s))^2+u^2)}{4\alpha}}\sin(xf(s))~ds+\int_sC_4(s)e^{\frac{2ux-t(4\alpha^2(f(s))^2+u^2)}{4\alpha}}\cos(xf(s))~ds~\text{or}~C_1xe^{\frac{2ux-u^2t}{4\alpha}}+C_2e^{\frac{2ux-u^2t}{4\alpha}}+\sum\limits_sC_3(s)e^{\frac{2ux-t(4\alpha^2(f(s))^2+u^2)}{4\alpha}}\sin(xf(s))+\sum\limits_sC_4(s)e^{\frac{2ux-t(4\alpha^2(f(s))^2+u^2)}{4\alpha}}\cos(xf(s))$

Case $3$b: $\text{Re}(\alpha t)\leq0$

Let $T(x,t)=F(x)G(t)$ ,

Then $F(x)G'(t)+uF'(x)G(t)=\alpha F''(x)G(t)$

$F(x)G'(t)=\alpha F''(x)G(t)-uF'(x)G(t)$

$F(x)G'(t)=(\alpha F''(x)-uF'(x))G(t)$

$\dfrac{G'(t)}{G(t)}=\dfrac{\alpha F''(x)-uF'(x)}{F(x)}=\dfrac{4\alpha^2(f(s))^2-u^2}{4\alpha}$

$\begin{cases}\dfrac{G'(t)}{G(t)}=\dfrac{4\alpha^2(f(s))^2-u^2}{4\alpha}\\\alpha F''(x)-uF'(x)-\dfrac{4\alpha^2(f(s))^2-u^2}{4\alpha}F(x)=0\end{cases}$

$\begin{cases}G(t)=c_3(s)e^{\frac{t(4\alpha^2(f(s))^2-u^2)}{4\alpha}}\\F(x)=\begin{cases}c_1(s)e^{\frac{ux}{2\alpha}}\sinh(xf(s))+c_2(s)e^{\frac{ux}{2\alpha}}\cosh(xf(s))&\text{when}~f(s)\neq0\\c_1xe^{\frac{ux}{2\alpha}}+c_2e^{\frac{ux}{2\alpha}}&\text{when}~f(s)=0\end{cases}\end{cases}$

$\therefore T(x,t)=C_1xe^{\frac{2ux-u^2t}{4\alpha}}+C_2e^{\frac{2ux-u^2t}{4\alpha}}+\int_sC_3(s)e^{\frac{2ux+t(4\alpha^2(f(s))^2-u^2)}{4\alpha}}\sinh(xf(s))~ds+\int_sC_4(s)e^{\frac{2ux+t(4\alpha^2(f(s))^2-u^2)}{4\alpha}}\cosh(xf(s))~ds~\text{or}~C_1xe^{\frac{2ux-u^2t}{4\alpha}}+C_2e^{\frac{2ux-u^2t}{4\alpha}}+\sum\limits_sC_3(s)e^{\frac{2ux+t(4\alpha^2(f(s))^2-u^2)}{4\alpha}}\sinh(xf(s))+\sum\limits_sC_4(s)e^{\frac{2ux+t(4\alpha^2(f(s))^2-u^2)}{4\alpha}}\cosh(xf(s))$

Hence $T(x,t)=\begin{cases}C_1xe^{\frac{2ux-u^2t}{4\alpha}}+C_2e^{\frac{2ux-u^2t}{4\alpha}}+\int_sC_3(s)e^{\frac{2ux-t(4\alpha^2(f(s))^2+u^2)}{4\alpha}}\sin(xf(s))~ds+\int_sC_4(s)e^{\frac{2ux-t(4\alpha^2(f(s))^2+u^2)}{4\alpha}}\cos(xf(s))~ds&\text{when}~\text{Re}(\alpha t)\geq0\\C_1xe^{\frac{2ux-u^2t}{4\alpha}}+C_2e^{\frac{2ux-u^2t}{4\alpha}}+\int_sC_3(s)e^{\frac{2ux+t(4\alpha^2(f(s))^2-u^2)}{4\alpha}}\sinh(xf(s))~ds+\int_sC_4(s)e^{\frac{2ux+t(4\alpha^2(f(s))^2-u^2)}{4\alpha}}\cosh(xf(s))~ds&\text{when}~\text{Re}(\alpha t)\leq0\end{cases}$ or $\begin{cases}C_1xe^{\frac{2ux-u^2t}{4\alpha}}+C_2e^{\frac{2ux-u^2t}{4\alpha}}+\sum\limits_sC_3(s)e^{\frac{2ux-t(4\alpha^2(f(s))^2+u^2)}{4\alpha}}\sin(xf(s))+\sum\limits_sC_4(s)e^{\frac{2ux-t(4\alpha^2(f(s))^2+u^2)}{4\alpha}}\cos(xf(s))&\text{when}~\text{Re}(\alpha t)\geq0\\C_1xe^{\frac{2ux-u^2t}{4\alpha}}+C_2e^{\frac{2ux-u^2t}{4\alpha}}+\sum\limits_sC_3(s)e^{\frac{2ux+t(4\alpha^2(f(s))^2-u^2)}{4\alpha}}\sinh(xf(s))+\sum\limits_sC_4(s)e^{\frac{2ux+t(4\alpha^2(f(s))^2-u^2)}{4\alpha}}\cosh(xf(s))&\text{when}~\text{Re}(\alpha t)\leq0\end{cases}$

This is already the general solution of $\dfrac{\partial T}{\partial t}+u\dfrac{\partial T}{\partial x}=\alpha\dfrac{\partial^2T}{\partial x^2}$ . Note that when without any I.C.s, the form of $f(s)$ can choose arbitrary, but when I.C.s are given, the form of $f(s)$ and the choice whether using the integration kernel or using the summation kernel should choose wisely in order to accommodate the I.C.s to get the most nice form of the solution, especially the number of I.C.s is more than two.

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  • $\begingroup$ (G'(t))/(G(t))=(αF"(x)-uF'(x))/(F(x))=-(4α^2 〖(f(s))〗^2+u^2)/4α How did you get this one ? $\endgroup$
    – hellomecha
    Apr 2, 2013 at 16:47
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Hint: after a change of variables, it's the heat equation.

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  • $\begingroup$ In the heat equation you don't have first derivative of T with respect to x. $\endgroup$
    – hellomecha
    Mar 25, 2013 at 19:57
  • $\begingroup$ Yes, I know. That's what the change of variables is for. Try letting $x = y + c t$ for suitable $c$, and using $y$ and $t$ as the independent variables. Thus $T(x,t) = U(y+ct, t)$. $\endgroup$ Mar 26, 2013 at 0:04
  • $\begingroup$ Sorry, I didn't get it. $\endgroup$
    – hellomecha
    Mar 26, 2013 at 12:34

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