# Solving PDE with Laplace Transform method.

can someone assist me in solving the following PDE with the Laplace Transform method:

$$\qquad \qquad \qquad \qquad \qquad \quad u_t = u_{xx} - a^2 (u-T_0) \qquad 0 < x < 1, \qquad t>0$$

Boundary conditions: $$\qquad u_x (0,t) = 0 \quad \text{and} \quad u_x (1,t) = 0 \qquad \qquad t>0$$

Initial condition: $$\qquad \qquad u(x,0) = 0 \qquad \qquad \qquad \quad 0 < x < 1$$

The answer given in the notes is: $$\qquad u(x,t) = T_0(1-e^{-a^2t})$$

I attempted to solve the problem as follows:

The first thing I did was take the Laplace transform with respect to the variable t on both sides. (transformed variable os equal to s)

Taking Laplace transform on both sides gives me the following:

$$\qquad \qquad \qquad \qquad sU(x,s) - U(x,0) = U_{xx} - \frac 1s .[a^2.(u-T_0)]$$

Substituting the initial condition and rearranging the equation I get:

$$\qquad \qquad \qquad \qquad U_{xx} - sU(x,s) = \frac 1s .[a^2.(u-T_0)]$$

From here I attempted to solve the equation using the d-operator method which gives me the following complementary function:

$$\qquad \qquad \qquad \qquad U(x,s)$$ = $$A.\cosh(\sqrt {s} . x )$$ $$+ B.\sinh(\sqrt {s} . x )$$

This is the part where I get stucked as I don't know how to obtain the particular integral of $$\frac 1s .[a^2.(u-T_0)]$$

Can anyone please assist me in moving forward to solve this problem? Thanks a lot for your help!

• In the expression of particular integral, there is no $u(x,t)$ as it will be acted upon the laplace transform. The expression for P.I would be $\frac{-a^2 T_0}{s}$ Jun 15, 2021 at 14:52
• @Vedant Chourey Thank you! Jun 15, 2021 at 15:50

Taking the LT of your PDE results in \begin{align*} sU(x,s)-\lim_{t\to 0^-}u(x,t)&=U''(x,s)-a^2U(x,s)+\frac{a^2T_0}{s}\\ 0&=U''(x,s)-\left(s+a^2\right)U(x,s)+\frac{a^2T_0}{s}. \end{align*} You have an incorrect factor of $$1/s$$ multiplying $$U$$ on the RHS: the LT of $$u(x,t)$$ is so hard it's easy: $$U(x,s).$$ Now you have a linear, second-order ODE for $$U$$ in $$x.$$ You can crunch through that: the solution is $$U(x,s)=\frac{a^2T_0}{s(s+a^2)}.$$ Finally, you take the inverse LT. I would go with partial fractions here: $$U(x,s)=\frac{T_0}{s}-\frac{T_0}{s+a^2},$$ making your inverse LT $$u(x,t)=T_0\left(1-e^{-a^2t}\right),$$ as required.