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I have following PDE:

\begin{align} v \cdot \frac{\partial T}{\partial x} + \frac{\partial T}{\partial t} &= k \cdot (T-T_0)\\ T(x,0) &= T_0\\ T(0,t) &= T_1 \end{align}

It's first order PDE, so I tried to solve it by the method of characteristics:

$$ \frac{dx}{v} = \frac{dt}{1} = \frac{dT}{k(T-T_0)} $$

From $\frac{dx}{v} = \frac{dt}{1}$ I deduced:

\begin{align} x &= vt + C\\ C_1 &= x - vt \end{align}

From $\frac{dt}{1}=\frac{dT}{k(T-T_0)}$ I deduced:

\begin{align} T-T_0 &= C_2 \cdot e^{k\cdot t}\\ C_2 &= \frac{T-T_0}{e^{kt}} \end{align}


The problem

Now I have $C_1$ and $C_2$ and I'm not sure how I'm supposed to apply both conditions.

My attempt looks like this:

I've tried to express $t$ with $C_1$:

$$ t = \frac{x-C_1}{v} $$

And substitute it into the second expression with $C_2$:

$$ C_2 = \frac{T-T_0}{e^{k \frac{x-c_1}{v}}} $$

But when I try to apply the first condition:

$$ C_2 = \frac{T_0-T_0}{e^{0}} = \frac{0}{1} = 0 $$

And then the second one:

$$ 0 = \frac{T_1-T_0}{e^{\frac{-kC_1}{v}}} $$


I know, that in the method of characteristics I'm supposed to parametrize the T(x,t) function somehow, but I don't know, how should I do this when the equation is not homogeneous? Can you, please, help me?

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Taking it from your results : $$C_1 = x - vt$$ $$C_2 = \frac{T-T_0}{e^{kt}}$$ The general solution on implicit form is : $$\Phi\left(x-vt\:,\:\frac{T-T_0}{e^{kt}}\right)=0$$ where $\Phi(X,Y)$ is any differentiable function of two variables.

Solving $\Phi(X,Y)=0$ for $Y$ leads to the general solution on the explicit form : $$\frac{T-T_0}{e^{kt}}=F(x-vt)$$ $F$ is any differentiable function. $$T(x,t)=T_0+e^{kt}F(x-vt)$$

Up to this point, this is consistent with your results.

The hitch comes from the conditions : $\begin{cases} T(x,0)=T_0 \\T(0,t)=T_1 \end{cases}$

We observe that $T(0,0)=T_0=T_1$ . This is contradictory if $T_0\neq T_1$

As a consequence they are two cases :

First case : $T_0=T_1\quad \to\quad F(x)=0 \quad\to\quad$ The solution is trivial : $T(x,t)=T_0$

Second case : $T_0\neq T_1 \quad\to\quad$The particular point $(x=0\:,\:y=0)$ must be excluded from the conditions. The conditions become consistent thanks to the Heaviside step function $H$ : $$T(0,t)=T_0+(T_1-T_0)H(t)$$ $T_0+(T_1-T_0)H(t)=T_0+e^{kt}F(0-vt) \quad\to\quad F(-vt)=(T_1-T_0)H(t)e^{-kt}$

Change of variable : $X=-vt\quad\to\quad t=-\frac{X}{v}$ $$F(X)=(T_1-T_0)H\left(-\frac{X}{v}\right)e^{\frac{k}{v}X}$$ $$F(x-vt)=(T_1-T_0)H\left(-\frac{x-vt}{v}\right)e^{\frac{k}{v}(x-vt)}$$

$$T(x,t)=T_0+e^{kt}(T_1-T_0)H\left(-\frac{x-vt}{v}\right)e^{\frac{k}{v}(x-vt)}$$ $$T(x,t)=T_0+(T_1-T_0)H\left(t-\frac{x}{v}\right)e^{\frac{k}{v}x}$$

Check of conditions :

When $t=0$ and $x>0$ then $H\left(-\frac{x}{v}\right)=0\quad\to\quad T(x,0)=T_0$

When $x=0$ and $t>0$ then $H\left(t\right)e^{\frac{k}{v}0}=1\quad\to\quad T(0,t)=T_0+(T_1-T_0)=T_1$

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  • $\begingroup$ Thank you very much for your answer! Could you please add the solution in the simple case, when the second condition is $T(0,t=T_1), t > 0$? And could you, please, explain the general solution a little more? $\endgroup$ – Eenoku May 11 '16 at 11:58
  • $\begingroup$ $T(0,t=T_1)$ is not a correct writting for a condition. A condition must be written as $T(0,t)=...$what ? $\endgroup$ – JJacquelin May 11 '16 at 12:06
  • $\begingroup$ Sorry, I meant $T(0,t)=T_1, t>0$ of course... $\endgroup$ – Eenoku May 11 '16 at 12:35
  • $\begingroup$ You put $x=0$ into the general solution and you obtain $T(0,t)=T_1$. This is detailed on the last line of my main answer. What do you don't understand in the solution : $$T(x,t)=T_0+(T_1-T_0)H\left(t-\frac{x}{v}\right)e^{\frac{k}{v}x}$$ The function $H$ is the Heaviside step function. $\endgroup$ – JJacquelin May 11 '16 at 13:24

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