I've been trying to solve this one PDE using Laplace Transforms but I seem to get a conflicting answer with the back of the book and oddly enough the answer in the back doesn't even follow the initial conditions so I wanted to see if my answer had any noticeable errors:

The Question:

$\displaystyle{u_t = u_{xx},\\ u(x,0)=10e^{-x},\\ u(0,t)=10,\\ \lim_{x\to\infty}u(x,t)=0}$

My work:

$\mathcal{L}\{u_t\} = \mathcal{L}\{u_{xx}\}\\ sU(x,s)-u(x,0)=U_{xx}$

Plugging in the initial condition:


Solving for U (undetermined coefficients):

$\displaystyle{U = ae^{x\sqrt{s}}+be^{-x\sqrt{s}}+\frac{10}{s-1}e^{-x}}$

Because the limit of $x$ as it goes to $\infty$ must be $0$, then $a$ must be $0$. Leaving us with:

$\displaystyle{U = be^{-x\sqrt{s}}+\frac{10}{s-1}e^{-x}}$

Taking the Laplace transform of the other initial condition gives us $\mathcal{L}\{u(0,t)=10\}\to U(0,s)=\frac{10}{s}$:

$\displaystyle{U(0,s) = \frac{10}{s} = b + \frac{10}{s-1}\\ b = 10\left(\frac{1}{s}-\frac{1}{s-1}\right)}$

Thus giving us:

$\displaystyle{U = 10\left(\frac{1}{s}-\frac{1}{s-1}\right)e^{-x\sqrt{s}}+\frac{10}{s-1}e^{-x}\\ U = \frac{10}{s}e^{-x\sqrt{s}}-\frac{10}{s-1}e^{-x\sqrt{s}}+\frac{10}{s-1}e^{-x} }$

Applying the inverse Laplace transform:

$\displaystyle{\mathcal{L}^{-1}\{U\}=\mathcal{L}^{-1}\left\{\frac{10}{s}e^{-x\sqrt{s}}\right\} - \mathcal{L}^{-1}\left\{\frac{10}{s-1}e^{-x\sqrt{s}}\right\}+\mathcal{L}^{-1}\left\{\frac{10}{s-1}e^{-x}\right\}\\ u(x,t) = 10\text{erfc}(x\,/\,2\sqrt{t}) - 10\mathcal{L}^{-1}\left\{\frac{1}{s-1}\right\}\mathcal{L}^{-1}\left\{e^{-x\sqrt{s}}\right\} + 10e^{\,t-x}\\ u(x,t) = 10\text{erfc}(x\,/\,2\sqrt{t})+ 10e^{\,t-x}-10 \left(e^{t}\right)*\left(\frac{xe^{-x^2/4t}}{2\sqrt{\pi}t^{3/2}}\right)}$


$\displaystyle{u(x,t) = 10\text{erfc}(x\,/\,2\sqrt{t})+ 10e^{\,t-x}-10\int_0^t e^{t-\tau}\frac{xe^{-x^2/4\tau}}{2\sqrt{\pi}\tau^{3/2}}\,d\tau}$

This is what I got to. The answer in the back of the book is: $\displaystyle{u(x,t)=10e^{\,t-x}}+10\int_0^t(1+t-\tau+e^{\,t-x})\text{erfc}(x/2\sqrt{\tau})\,d\tau$ but theirs doesn't have the $u(0,t)=10$ condition satisfied graphically but mine does. I used Python to plug both of our expressions into the original PDE but both of ours didn't satisfy it but it might be due to that integral in the convolution causing weirdness perhaps. Note: I used a different inverse Laplace transform for the right term in the convolution as Wolfram alpha and another website showed that the one in book might be wrong (also the answer was wrong when trying that inverse Laplace transform as well.)

  • $\begingroup$ What's the name of the book? $\endgroup$
    – Franlezana
    Commented Feb 12, 2023 at 0:53
  • $\begingroup$ @Franlezana Introduction to Partial Differential Equations, Coleman, 2nd Edition $\endgroup$ Commented Feb 12, 2023 at 7:05

1 Answer 1


After looking back at the problem, I graphed $u_t$ & $u_{xx}$ and they both produced the same graph and looking at the graph of $u(x,t)$ the initial conditions were met so the answer I got was correct.


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