MHB Is the Solution Unique? Investigating the Uniqueness of a Boundary Value Problem

[evinda]

Hello! (Wave)

Suppose that we have the following boundary value problem:

$$a^2 u_{xx}=u_t, 0<x<L, t>0 \\ u(x,0)=f(x) , 0 \leq x \leq L\\ u(0,t)=0, u(L,t)=0, t>0$$

By supposing that $u(x,t)=X(x) T(t)$ we find that the solution is of the form $u(x,t)=\sum_{n=1}^{\infty} c_n e^{-\frac{n^2 \pi^2 a^2 t}{L^2}} \sin{\frac{n \pi x}{L}}$

where $c_n=\frac{2}{L} \int_0^L f(x) \sin{\frac{n \pi x}{L}}$.

But do we know that this solution is unique? Or could there also be an other solution that will not be of the form $X(x) T(t)$ ?

 

[HallsofIvy]

What you have isn't "of the form X(x)T(t)"! It is a sum of such, basically a Fourier series. Now, what do you mean by "unique"? You may be able to write the solution in a different form but it will be the same function. It is easy to show that there is only one solution to this problem: suppose there were two, u_1(x, t) and u_2(x, t). Then u(x, t)= u_1(x, t)- u_2(x, t) satisfies the same differential equation but with all boundary and initial conditions equal to 0. And it is easy to show that only the zero function, 0 for all x and t, satisfies that.

 

[evinda]

What you have isn't "of the form X(x)T(t)"! It is a sum of such, basically a Fourier series. Now, what do you mean by "unique"? You may be able to write the solution in a different form but it will be the same function. It is easy to show that there is only one solution to this problem: suppose there were two, u_1(x, t) and u_2(x, t). Then u(x, t)= u_1(x, t)- u_2(x, t) satisfies the same differential equation but with all boundary and initial conditions equal to 0. And it is easy to show that only the zero function, 0 for all x and t, satisfies that.

Ok, I see. Thank you! (Smile)