Compute $ \ \frac{dM}{dt} (t) \ $ in simplest form

Question

Diffusion Equation:

Let $ \ u(x,t) $ denote the concentration of a chemical which satisfies $$ u_t=2u_{xx}+\frac{2}{5} x , \ 0<x<4 , \ t>0 $$ $$ u_x(0,t)=0, \ u_x(4,t)=1, \ u(x,0)=2x $$ Let $ \ M(t)=\int_0^4 u(x,t) dx \ \ $ denote the total amount of chemical at time $ \ t \ $.

Then,

(a) What is the physical meaning of the boundary condition $ \ \frac{\partial u}{\partial x}(4,t)=1 \ ? $

(b) Compute $ \ \frac{dM}{dt} (t) \ $ in simplest form.

(c) Find $ \ M(t) \ $

Answer:

(a) I think $ \ \frac{\partial u}{\partial x}(4,t) =1 \ $ means that the flux across the boundary is constant $=1 $

That is there is no concentration difference .

(b)

$ M(t)=\int_0^4 u(x,t) dx \\ \Rightarrow \frac{dM}{dt}=\int_0^4 \frac{\partial u}{\partial t} (x,t) dx =\int_0^4 (2u_{xx}+\frac{2}{5} x) dx=\left[2u_x \right]_0^4+\frac{2}{5} \left[\frac{x^2}{2} \right]_0^4 =2u_x(4,t)-2u_x(0,t)+\frac{2}{5} \times 8=2+16/5=26/5 \\ \Rightarrow \frac{dM}{dt}=\frac{26}{5} $

Am I right so far?

If I am Right , then how to evaluate $ \ M(t) \ $ ?

Help me doing this.

Answer 1

Note that $$ M(0) = \int_0^4 u(x,0)\ dx = \int_0^4 2x\ dx = 16 $$

You now have the IVP $$ M'(t) = \frac{26}{5}, \quad M(0) = 16 $$

which gives $$ M(t) = \frac{26}{5}t + 16 $$