# Workaround for DSolve in V 12 when it gives undefined as solution to 1D heat PDE?

Bug introduced in 12.0 and fixed in 12.1.1

Reported to WRI CASE: 4278450

Comparing the following, all done from clean kernel

The strange thing is that V 12 can solve this same PDE without the assumptions

k    = 1/10; A = 60;
pde  = D[u[x, t], t] == k*D[u[x, t], {x, 2}];
bc   = u[0, t] == A;
ic   = u[x, 0] == 0;
sol  = DSolve[{pde, bc, ic}, u[x, t], {x, t}]


But it says in the above answer that it wants x>0,t>0, which is why I gave it the assumptions to help it, but then it returns undefined.

Something seems to have gone wrong in V 12 DSolve here, or may be in the Integrate? I do not know.

Do others see the same result on V 12?. Answer given by V 11.3 is the correct one.

Any workaround for V 12 to make it give same answer as V 11.3?

## Appendix

fyi, To verify that 11.3 solution is correct, I solved it by hand

Solving \begin{align} u_{t} & =ku_{xx\qquad}t>0,x>0\tag{1}\\ u\left( 0,t\right) & =A\nonumber\\ u\left( x,0\right) & =0\nonumber \end{align} And $$u\left( x,t\right) <\infty$$ as $$x\rightarrow\infty$$. This means $$u\left( x,t\right)$$ is bounded. This conditions is always needed to solve these problems.

Let $$U\left( x,s\right)$$ be the Laplace transform of $$u\left( x,t\right)$$. Defined as $$\mathcal{L} \left( u,t\right) =\int_{0}^{\infty}e^{-st}u\left( x,t\right) dt$$ Applying Laplace transform to the original PDE (1) gives $$sU\left( x,s\right) -u\left( x,0\right) =kU_{xx}\left( x,s\right)$$ But $$u\left( x,0\right) =0$$, therefore the above becomes $$U_{xx}-\frac{s}{k}U=0$$ The solution to this differential equation is $$U\left( x,s\right) =c_{1}e^{\sqrt{\frac{s}{k}}x}+c_{2}e^{-\sqrt{\frac{s}{k} }x}$$ Since $$u\left( x,t\right)$$ is bounded in the limit as $$x\rightarrow\infty$$ and $$k>0$$, therefore it must be that $$c_{1}=0$$ to keep the solution bounded. The above simplifies to $$U\left( x,s\right) =c_{2}e^{-\sqrt{\frac{s}{k}}x}\tag{2}$$ At $$x=0\,,u\left( 0,t\right) =A$$. Therefore $$U\left( 0,s\right) = \mathcal{L}\left( u\left( 0,t\right) \right) =\mathcal{L} \left( A\right) =\frac{1}{s}A$$. Hence at $$x=0$$ the above gives $$\frac{1}{s}A=c_{2}$$ Therefore (2) becomes $$$$U\left( x,s\right) =\frac{A}{s}e^{-\sqrt{\frac{s}{k}}x}\tag{3}$$$$ From tables, the inverse Laplace transform of the above is (since $$x>0,k>0$$) $$u\left( x,t\right) =A\operatorname{erfc}\left( \frac{x}{2\sqrt{kt}}\right)$$ With $$A=60,k=\frac{1}{10}$$ the above becomes

\begin{align*} u\left( x,t\right) & =60\operatorname{erfc}\left( \frac{x}{2\sqrt {\frac{1}{10}t}}\right) \\ & =60\operatorname{erfc}\left( \frac{\sqrt{\frac{5}{2}}x}{\sqrt{t}}\right) \end{align*}

• Yes, I have the same answer on version 12. Jul 9 '19 at 17:31

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