Toronto Math Forum
APM3462015F => APM346Home Assignments => HA5 => Topic started by: Rong Wei on October 17, 2015, 02:55:31 PM

previous post should be HA5  P1, sorry about the typo!

and for question b, it should be x  ct instead of x  vt; otherwise, we couldn't have the solution.

I'll add my typed solutions.
(a) The ordinary heat equation is: \begin{equation}
u_t = ku_{xx} \end{equation}
Now consider $U(x,t) = u(x+ct,t)$. The partial derivatives needed for the heat equation are given by: \begin{equation}
\begin{cases}
U_t = cu_x + u_t \\
U_x = u_x \\
U_{xx} = u_{xx} \end{cases} \end{equation}
Now let's plug these into the heat equation: \begin{equation}
U_t = kU_{xx} \longrightarrow (u_t + cu_x) = k(u_{xx}) \longrightarrow u_t + cu_x = ku_{xx} \end{equation}
Therefore the heat equation with a convection term is obtained from the ordinary heat equation with a change of variables.
(b) Now we use the change of variabes $u(x,t) = U(xct,t)$ (Note: Thank you Rong Wei for pointing out the fact that this should be c here, as I was unable to solve the problem otherwise!). Now the partial derivatives are given by: \begin{equation} \begin{cases}
u_t = cU_x + U_t \\
u_x = U_x \\
u_{xx} = U_{xx} \end{cases} \end{equation}
Let's plug this into the heat equation with a convective term: \begin{equation}
u_t + cu_x = ku_{xx} \longrightarrow (cU_x + U_t) + c(U_x) = k(U_{xx}) \longrightarrow cU_x + cU_x + U_t = kU_{xx} \longrightarrow U_t = kU_{xx}
\end{equation}
So with this change of variables, the equation reduces to the familiar heat equation.
We can then use the usual formula to arrive at the solution: \begin{equation}
u(x,t) = \int_0^{\infty}G(xct,y,t)g(y)dy = \frac{1}{2\sqrt{\pi{}kt}}\int_0^{\infty}e^{(xcty)^2/(4kt)}g(y)dy \end{equation}
(c) I don't think we can use the method of continuation directly to solve IVBP with Dirichlet or Neumann boundary conditions as $x>0$ for the heat equation with a convection term on ${x>0, t>0}$. This is because we used a change of variables to define the problem.
(d)We have \begin{equation}
u(x,t) = v(x,t)e^{\alpha{}x + \beta{}t} \end{equation}
The partial derivatives are given by: \begin{equation} \begin{cases}
u_t = v_te^{\alpha{}x + \beta{}t} + \beta{}ve^{\alpha{}x + \beta{}t} \\
u_x = v_xe^{\alpha{}x + \beta{}t} + \alpha{}ve^{\alpha{}x + \beta{}t} \\
u_{xx} = v_{xx}e^{\alpha{}x + \beta{}t} + 2\alpha{}v_xe^{\alpha{}x + \beta{}t} + \alpha{}^2ve^{\alpha{}x + \beta{}t} \end{cases} \end{equation}
Let's plug them in to the heat equation with a convection term. \begin{equation}
v_te^{\alpha{}x + \beta{}t} + \beta{}ve^{\alpha{}x + \beta{}t} + cv_xe^{\alpha{}x + \beta{}t} + c\alpha{}ve^{\alpha{}x + \beta{}t} = ke^{\alpha{}x + \beta{}t}(v_{xx} + 2\alpha{}v_x + \alpha{}^2v) \end{equation}
Gathering like terms, we arrive at: \begin{equation}
v_t + v(\beta{} + c\alpha{}  k\alpha{}^2) + v_x(c2k\alpha{}) = kv_{xx} \end{equation}
For this to reduce to the heat equation, we want the coefficients in front of $v$ and $v_x$ to be zero. So: \begin{equation} \begin{cases}
\beta{} + c\alpha{}  k\alpha{}^2 \\
c2k\alpha{} \end{cases} \end{equation}
Solving first for $\alpha{}$, we get that $\alpha{} = \frac{c}{2k}$. We can then plug this into the equation for $\beta{}$ and solve to get that $\beta{} = \frac{c^2}{4k}$.
(e) I will work on this part in a little while. So far I think I have gotten all of the same solutions as Rong Wei. Added my solution below.
For the case of the halfline and Dirichlet boundary condition, we will have the solution: \begin{equation}
u(x,t) = \frac{e^{\alpha{}x + \beta{}t}}{2\sqrt{\pi{}kt}}\int_0^{\infty}[e^{(xy)^2/4kt}  e^{(x+y)^2/4kt}]g(y)dy \end{equation}
In the case of Neumann boundary conditions, we cannot use a similar method.

I think the e) solution posted by Emily is right! thank you

(e) I will work on this part in a little while. So far I think I have gotten all of the same solutions as Rong Wei. Added my solution below.
For the case of the halfline and Dirichlet boundary condition, we will have the solution: \begin{equation}
u(x,t) = \frac{e^{\alpha{}x + \beta{}t}}{2\sqrt{\pi{}kt}}\int_0^{\infty}[e^{(xy)^2/4kt}  e^{(x+y)^2/4kt}]g(y)dy \end{equation}
In the case of Neumann boundary conditions, we cannot use a similar method.
I don't think the solution is right here, unless we assume $v(x,0) = g(x)$. But usually we use $u(x,0) = g(x)$, then in this case \begin{equation}u(x,0) = v(x,0)e^{\alpha x} = g(x) \rightarrow v(x,0) = g(x)e^{\alpha x} \end{equation}
Dirichlet condition transforms to:
\begin{equation} u(0,t) = v(0,t)e^{\beta t} = 0 \rightarrow v(0,t) = 0 \end{equation}
Thus we need to solve \begin{equation}v_t = kv_{xx} \end{equation}
\begin{equation} v(x,0) = g(x)e^{\alpha x} \end{equation}
\begin{equation} v(0,t) = 0 \end{equation}
The answer is then the general result:
\begin{equation}
v(x,t) = \frac{1}{2\sqrt{\pi{}kt}}\int_0^{\infty}[e^{(xy)^2/4kt}  e^{(x+y)^2/4kt}]g(y)e^{\alpha y}dy \end{equation}
Then
\begin{equation}
u(x,t) = \frac{e^{\frac{c}{2k}x \frac{c^2}{4k}t}}{2\sqrt{\pi{}kt}}\int_0^{\infty}[e^{(xy)^2/4kt}  e^{(x+y)^2/4kt}]g(y)e^{\frac{c}{2k} y}dy \end{equation}

Oh I see, this makes sense. I think I misunderstood what the question was asking.

Shouldn't the integral bounds in equation (6) be from $\infty$ to $\infty$ instead of $0$ to $\infty$? Here the domain is not restricted yet. This is what is in Rong Wei's written solution also.

In the original problem on the line indeed integral is from $\infty$ to $+\infty$.
We cannot apply method of continuation directly to IBVP equation with the convective term because equation contains $\partial_x$ in the odd degree and replacing $x\mapsto x$ we change the equation. We cannot apply this method after change of $xct= x_{\mathsf{new}}$ because domain now is not $\{x>0\}$ but $\{x+ct>0\}$.
However as we plug $u=e^{\alpha x+\beta t}$ we get the normal heat equation and the domain $\{x>0\}$ and the boundary condition
$u_{x=0}=0\implies v_{x=0}=0$ and we can use method of continuation
$u_x_{x=0}=0\implies (v_x+\alpha v)_{x=0}=0$ and we cannot use method of continuation (unless $\alpha=0\iff c=0$).