Semilinear equation with characteristics

Question

I have to solve the following equation by the method of characteristics $$\frac{\partial T}{\partial t}+v\frac{\partial T}{\partial z}=\frac{2 U}{RC_F}(T_M-T),$$ where $v,C_F,U,R,T_M$ are known constants. $T$ is a function $T(z,t)$ for $t>0$ and $0\leq z\leq L$. I have the initial conditions $T(z,0)=T_0(z)$ and $T(0,t)=T_I(t)$. My work so far: consider the characteristic system \begin{align} \frac{dt}{ds}=1\implies t(s)=s+a_1,\\ \frac{dz}{ds}&=v\implies z(s)=vs+a_2\\ \frac{dT}{ds}=\frac{2 U}{RC_F}\implies T(s)=T_M+a_3e^{-\frac{2 U}{RC_F}t}.\end{align}From here i either can obtain $T$ in function of $t$ or $z$,but not both. I don't really know how to proceed. Any help will be very appreciated!

Answer 1

$$\frac{\partial T}{\partial t}+v\frac{\partial T}{\partial z}=A(T_M-T)\quad;\quad A=\frac{2 U}{RC_F}=\text{constant.}$$ Characteristic ODEs : $$\frac{dt}{1}=\frac{dz}{v}=\frac{dT}{A(T_M-T)}$$ First characteristic equation from solving $\frac{dt}{1}=\frac{dz}{v}$ : $$z-vt=c_1$$ Second characteristic equation from solving $\frac{dt}{1}=\frac{dT}{A(T_M-T)}$ : $$(T_M-T)e^{At}=c_2$$ General solution of the PDE from implicit form $c_2=F(c_1)$ with arbitrary function $F$ : $$(T_M-T)e^{At}=F(z-vt)$$ $$\boxed{T(z,t)=T_M-e^{-At}F(z-vt)}$$ The function $F$ has to be determined according to the specified conditions $T(z,0)=T_0(z)$ and $T(0,t)=T_I(t)$.

One observe that $T_0(0)=T_I(0)$ is not true in general (the equality isn't specified in the wording of the problem). Thus the fonction $F$ must be a piecewise function.

Initial condition : $$T(z,0)=T_0(z)=T(z,t)=T_M-e^{0}F(z-0)$$ $$F(z)=T_M-T_0(z)$$ The function $F$ is determined. One put it into the above general solution where the argument is $(z-vt)$ : $$T(z,t)=T_M-e^{-At}\big(T_M-T_0(z-vt)\big)$$

Boundary condition : $$T(0,t)=T_I(t)=T_M-e^{-At}F(0-vt)\quad\implies\quad F(-vt)=\big(T_M-T_I(t)\big)e^{At}$$ Let $x=-vt\quad\implies\quad F(x)=\big(T_M-T_I(-\frac{x}{v})\big)e^{-\frac{A}{v}x}$.

The function $F$ is determined. One put it into the above general solution where the argument is $(z-vt)$ : $$T(z,t)=T_M-e^{-At}\big((T_M-T_I(t))e^{-\frac{A}{v}(z-vt)} \big)$$ $$T(z,t)=T_M-(T_M-T_I(-\frac{z-vt}{v}))e^{-\frac{A}{v}(z)}$$

Don't be surprised to have two different functions $F$ since each one is a part of a piecewise function each part valid on a domain with $x-vt=0$ as a frontier.

Solution satisfying the PDE and the two conditions : $$ T(z,t)=\begin{cases} T_M-e^{-At}\big(T_M-T_0(z-vt)\big)\quad\quad z>vt\\ T_M-e^{-\frac{A}{v}(z)}\big(T_M-T_I(-\frac{z-vt}{v})\big)\quad\quad\quad z<vt \end{cases}$$